Home Teachers' Resources CHEMISTRY FORM 3 Comprehensive tutorial notes

CHEMISTRY FORM 3 Comprehensive tutorial notes

(a)Gas laws

  1. Matter is made up of small particle in accordance to Kinetic Theory of matter:

Naturally, there are basically three states of matter: Solid, Liquid and gas:

(i)A solid is made up of particles which are very closely packed with a definite/fixed shape and fixed/definite volume /occupies definite space. It has a very high density.

 

(ii) A liquid is made up of particles which have some degree of freedom. It thus has no definite/fixed shape. It takes the shape of the container it is put. A liquid has fixed/definite volume/occupies definite space.

 

(iii)A gas is made up of particles free from each other. It thus has no definite /fixed shape. It takes the shape of the container it is put. It has no fixed/definite volume/occupies every space in a container.

 

2.Gases  are affected by physical conditions. There are two physical conditions:

(i)Temperature

(ii)Pressure

 

  1. The SI unit of temperature is Kelvin(K).

Degrees Celsius/Centigrade(oC) are also used.

The two units can be interconverted from the relationship:

oC + 273= K

 K -273 = oC

 

Practice examples

  1. Convert the following into Kelvin.

(i) O oC

            oC  + 273  =  K substituting : O oC  +  273  =  273 K

 

(ii) -273 oC

            oC  + 273  =  K substituting :  -273oC  +  273  = 0 K

 

(iii) 25 oC

            oC  + 273  =  K substituting : 25 oC  +  273  =  298 K

(iv) 100 oC

            oC  + 273  =  K substituting : 100 oC  +  273  =  373 K

 

  1. Convert the following into degrees Celsius/Centigrade(oC).

(i) 10 K

         K -273 = oC  substituting:     10   –   273   =  -263 oC

(ii) (i) 1 K

         K -273 = oC  substituting:     1   –   273   =  -272 oC

 

(iii) 110 K

         K -273 = oC  substituting:     110   –   273   =  -163 oC

 

(iv) -24 K

         K -273 = oC  substituting:     -24   –   273   =  -297 oC

 

The standard temperature is 273K = 0 oC.

The room temperature is assumed to be 298K  = 25oC

 

  1. The SI unit of pressure is Pascal(Pa) / Newton per metre squared (Nm-2) . Millimeters’ of mercury(mmHg) ,centimeters of mercury(cmHg) and atmospheres are also commonly used.

 

The units are not interconvertible but Pascals(Pa) are equal to Newton per metre squared(Nm-2).

The standard pressure is the atmospheric pressure.

Atmospheric pressure is equal to about:

(i)101325 Pa

(ii)101325 Nm-2

          (iii)760 mmHg

          (iv)76 cmHg

(v)one atmosphere.

 

  1. Molecules of gases are always in continuous random motion at high speed. This motion is affected by the physical conditions of temperature and pressure.

Physical conditions change the volume occupied by gases in a closed system.

The effect of physical conditions of temperature and pressure was investigated and expressed in both Boyles and Charles laws.

 

  1. Boyles law states that

the volume of a fixed mass of a gas is inversely proportional to the pressure at constant/fixed temperature

Mathematically:

Volume   α             1 (Fixed /constant Temperature)

                              Pressure

V   α            1     (Fixed /constant T) ie PV = Constant(k)

                              P

From Boyles law , an increase in pressure of a gas cause a decrease in volume. i.e doubling the pressure cause the volume to be halved.

Graphically therefore a plot of volume(V) against pressure (P) produces a curve.

 

 

 

 

 

 

 

V

 

 

 

 

 

P

Graphically a plot of volume(V) against inverse/reciprocal of pressure (1/p) produces a straight line

 

 

 

 

 

V

 

 

 

 

 

                                                  1/P

 

For two gases then  P1 V1 =  P2 V2

P1  =   Pressure of gas 1

V1  =  Volume of gas 1

P2   =   Pressure of gas 2

V2   =   Volume of gas 2

 

Practice examples:

  1. A fixed mass of gas at 102300Pa pressure has a volume of 25cm3.Calculate its volume if the pressure is doubled.

Working

P1 V1 =  P2 V2      Substituting :102300  x  25  = (102300 x 2) x  V2

V2 = 102300  x  25     =    12.5cm3

(102300 x 2)

 

  1. Calculate the pressure which must be applied to a fixed mass of 100cm3 of Oxygen for its volume to triple at 100000Nm-2.

 

P1 V1 =  P2 V2      Substituting :100000  x  100  = P2 x (100 x 3)

V2 = 100000  x  100     =   33333.3333 Nm-2

(100 x 3)

 

3.A 60cm3 weather ballon full of Hydrogen at atmospheric pressure of 101325Pa was released into the atmosphere. Will the ballon reach stratosphere where the pressure is 90000Pa?

 

P1 V1 =  P2 V2      Substituting :101325  x  60  =  90000  x V2

V2 = 101325  x  60      =   67.55 cm3

90000

The new volume at 67.55 cm3 exceed ballon capacity of 60.00 cm3.It will burst before reaching destination.

 

7.Charles law states that“the volume of a fixed mass of a gas is directly  proportional to the absolute temperature at constant/fixed pressure

Mathematically:

Volume   α         Pressure  (Fixed /constant pressure)

                             

V   α         T           (Fixed /constant P) ie    V     = Constant(k)

                                                                T

From Charles law , an increase in temperature of a gas cause an increase in volume. i.e doubling the temperature cause the volume to be doubled.

Gases expand/increase  by 1/273 by volume on heating.Gases contact/decrease by 1/273 by volume on cooling at constant/fixed pressure.

The volume of a gas continue decreasing with decrease in temperature until at          -273oC /0 K the volume is zero. i.e. there is no gas.

This temperature is called absolute zero. It is the lowest temperature at which a gas can exist.

 

Graphically therefore a plot of volume(V) against Temperature(T) in:

(i)oC produces a straight line that is extrapolated to the absolute zero of -273oC .

 

 

 

 

 

 

 

V

 

 

 

 

 

-273oC                                          0oC

T(oC)

 

 

 

 

 

(ii)Kelvin/K produces a straight line from absolute zero of O Kelvin

 

 

 

 

 

V

 

 

 

 

 

                            0                                           T(Kelvin)

 

 

 

For two gases then      V1       =     V2

T1             T2

T1  =   Temperature in Kelvin of gas 1

V1  =  Volume of gas 1

T2  =   Temperature in Kelvin of gas 2

V2   =   Volume of gas 2

 

Practice examples:

  1. 500cm3 of carbon(IV)oxide at 0oC was transfered into a cylinder at -4oC. If the capacity of the cylinder is 450 cm3,explain what happened.

 

V1       =     V2    substituting           500              =         V2

T1             T2                               (0 +273)                (-4 +273)

 

500 x (-4 x 273)           =   492.674cm3

(0 + 273)

The capacity of cylinder (500cm3) is less than new volume(492.674cm3).

7.326cm3(500-492.674cm3)of carbon(IV)oxide gas did not fit into the cylinder.

 

  1. A mechanic was filling a deflated tyre with air in his closed garage using a hand pump. The capacity of the tyre was 40,000cm3 at room temperature. He rolled the tyre into the car outside. The temperature outside was 30oC.Explain what happens.

 

V1       =     V2    substituting           40000                    =         V2

T1             T2                               (25 +273)                      (30 +273)

 

40000 x (30 x 273)      =   40671.1409cm3

(25 + 273)

The capacity of a tyre (40000cm3) is less than new volume(40671.1409cm3).

The tyre thus bursts.

 

  1. A hydrogen gas balloon with 80cm3 was released from a research station at room temperature. If the temperature of the highest point it rose is -30oC , explain what happened.

 

V1       =     V2    substituting             80              =         V2

T1             T2                               (25 +273)                      (-30 +273)

 

=   80 x (-30 x 273)          =   65.2349cm3

(25 + 273)

The capacity of balloon (80cm3) is more than new volume (65.2349cm3).

The balloon thus remained intact.

 

  1. The continuous random motion of gases differ from gas to the other.The movement of molecules (of a gas) from region of high concentration to a region of low concentration is called diffusion.

The rate of diffusion of a gas depends on its density. i.e. The higher the rate of diffusion, the less dense the gas.

The density of a gas depends on its molar mass/relative molecular mass. i.e. The higher the density the higher the molar mass/relative atomic mass and thus the lower the rate of diffusion.

Examples

1.Carbon (IV)oxide(CO2) has a molar mass of 44g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus lighter/less dense than Carbon (IV)oxide(CO2). N2 diffuses faster than CO2.

2.Ammonia(NH3) has a molar mass of 17g.Nitrogen(N2)has a molar mass of 28g. (N2)is thus about twice lighter/less dense than Ammonia(NH3). Ammonia(NH3) diffuses twice faster than N2.

  1. Ammonia(NH3) has a molar mass of 17g.Hydrogen chloride gas has a molar mass of 36.5g.Both gases on contact react to form white fumes of ammonium chloride .When a glass/cotton wool dipped in ammonia and another glass/cotton wool dipped in hydrochloric acid are placed at opposite ends of a glass tube, both gases diffuse towards each other. A white disk appears near to glass/cotton wool dipped in hydrochloric acid. This is because hydrogen chloride is heavier/denser than Ammonia and thus its rate of diffusion is lower .

 

The rate of diffusion of a gas is in accordance to Grahams law of diffusion. Grahams law states that:

“the rate of diffusion of a gas is inversely proportional to the square root of its density, at the same/constant/fixed temperature and pressure”

Mathematically

R   α       and  since density is proportional to mass then   R α      

         √  p                                                                   √  m

For two gases then:

R1     =      R2        where: R1 and R2 is the rate of diffusion of 1st and 2nd gas.

√M2          √M1                 M1 and M2 is the molar mass of 1st and 2nd gas.

 

Since rate is inverse of time. i.e. the higher the rate the less the time:

 

For two gases then:

 T1   T2  where: T1 and T2 is the time taken for 1st and 2nd gas to diffuse.

M1   M2           M1 and M2 is the molar mass of 1st and 2nd gas.

 

Practice examples:

  1. It takes 30 seconds for 100cm3 of carbon(IV)oxide to diffuse across a porous plate. How long will it take 150cm3 of nitrogen(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,N=14.0=16.0)

 

Molar mass CO2=44.0        Molar mass NO2=46.0

Method 1

100cm3 CO2   takes         30seconds

150cm3       takes             150 x30   =  45seconds

100

 

T CO2           =     √ molar mass CO2  =>  45seconds   =     √  44.0

T NO2                  √ molar mass NO2           T NO2                           √ 46.0

 

T NO2 =45seconds x √ 46.0                =  46.0114 seconds

                                                            √  44.0

 

Method 2

100cm3 CO2   takes         30seconds

1cm3       takes                 100 x1   =     3.3333cm3sec-1

30

R CO2           =     √ molar mass NO2  =>  3.3333cm3sec-1   =     √  46.0

R NO2                 √ molar mass CO2             R NO2                                     √ 44.0

 

R NO2 = 3.3333cm3sec-1 x √ 44.0                 =    3.2601cm3sec-1

                                                                √  46.0

 

3.2601cm3            takes     1seconds

150cm3                 take            150cm3         =   46.0109seconds

                                                                                               3.2601cm3

 

  1. How long would 200cm3 of Hydrogen chloride take to diffuse through a porous plug if carbon(IV)oxide takes 200seconds to diffuse through.

 

Molar mass CO2 = 44g    Molar mass HCl = 36.5g

T CO2           =     √ molar mass CO2  =>  200 seconds   =     √  44.0

T HCl              √ molar mass HCl            T HCl                                √ 36.5

 

T HCl = 200seconds x √ 36.5              =  182.1588 seconds

                                                            √  44.0

  1. Oxygen gas takes 250 seconds to diffuse through a porous diaphragm. Calculate the molar mass of gas Z which takes 227 second to diffuse.

 

Molar mass O2 = 32g       Molar mass Z = x g

T  O2           =     √ molar mass  O2          =>   250 seconds   =     √  32.0

T  Z                  √ molar mass  Z                            227seconds               √   x

 

√  x =  227seconds  x  √ 32                 =  26.3828 grams

                                                              250

  1. 25cm3 of carbon(II)oxide diffuses across a porous plate in 25seconds. How long will it take 75cm3 of Carbon(IV)oxide to diffuse across the same plate under the same conditions of temperature and pressure. (C=12.0,0=16.0)

 

Molar mass CO2 = 44.0        Molar mass CO = 28.0

Method 1

25cm3 CO   takes   25seconds

75cm3       takes               75 x25   =  75seconds

25

 

T CO2           =     √ molar mass CO2  =>  T CO2seconds   =     √  44.0

T CO                √ molar mass CO                  75                                    √ 28.0

 

T CO2 =75seconds x √ 44.0                =  94.0175 seconds

                                                            √  28.0

 

Method 2

25cm3 CO2   takes           25seconds

1cm3       takes                 25 x1   =     1.0cm3sec-1

25

R CO2           =     √ molar mass CO  =>   x cm3sec-1   =         √  28.0

R CO               √ molar mass CO2            1.0cm3sec-1                                  √ 44.0

 

R CO2 = 1.0cm3sec-1 x √ 28.0             =    0.7977cm3sec-1

                                                                √  44.0

 

0.7977cm3            takes     1 seconds

75cm3                 takes         75cm3           =   94.0203seconds

                                                                                              0.7977cm3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(b)Introduction to the mole, molar masses and Relative atomic masses

 

  1. The mole is the SI unit of the amount of substance.
  2. The number of particles e.g. atoms, ions, molecules, electrons, cows, cars are all measured in terms of moles.
  3. The number of particles in one mole is called the Avogadros Constant. It is denoted “L”.

The Avogadros Constant contain 6.023 x10 23 particles. i.e.

 

1mole = 6.023 x10 23 particles                                 =  6.023 x10 23

2 moles =  2 x 6.023 x10 23 particles                       =  1.205 x10 24

0.2 moles =  0.2 x 6.023 x10 23 particles                =  1.205 x10 22

0.0065 moles =  0.0065 x 6.023 x10 23 particles     =  3.914 x10 21

  1. The mass of one mole of a substance is called molar mass. The molar mass of:

(i)an element has mass equal to relative atomic mass /RAM(in grams)of the element e.g.

Molar mass of carbon(C)= relative atomic mass  = 12.0g

6.023 x10 23 particles  of carbon  = 1 mole  =12.0 g

 

Molar mass of sodium(Na) = relative atomic mass  = 23.0g

6.023 x10 23 particles  of sodium  = 1 mole  =23.0 g

 

Molar mass of Iron (Fe)  =  relative atomic mass  = 56.0g

6.023 x10 23 particles  of iron  = 1 mole  =56.0 g

 

(ii)a molecule has mass equal to relative molecular mass /RMM (in grams)of the molecule. Relative molecular mass  is the sum of the relative atomic masses of the elements making the molecule.

The number of atoms making a molecule is called atomicity. Most gaseous molecules are diatomic (e.g. O2, H2, N2, F2, Cl2, Br2, I2)noble gases are monoatomic(e.g. He, Ar, Ne, Xe),Ozone gas(O3) is triatomic e.g.

 

Molar mass Oxygen molecule(O2) =relative molecular mass =(16.0x 2)g =32.0g

6.023 x10 23 particles  of Oxygen molecule = 1 mole  = 32.0 g

 

Molar mass chlorine molecule(Cl2) =relative molecular mass =(35.5x 2)g =71.0g

6.023 x10 23 particles  of chlorine molecule = 1 mole  = 71.0 g

 

Molar mass Nitrogen molecule(N2) =relative molecular mass =(14.0x 2)g =28.0g

6.023 x10 23 particles  of Nitrogen molecule = 1 mole  = 28.0 g

 

(ii)a compound has mass equal to relative formular mass /RFM (in grams)of the molecule. Relative formular mass  is the sum of the relative atomic masses of the elements making the compound. e.g.

 

(i)Molar mass Water(H2O) = relative formular mass =[(1.0 x 2 ) + 16.0]g =18.0g

 

6.023 x10 23 particles  of Water molecule = 1 mole  = 18.0 g

6.023 x10 23 particles  of Water molecule has:

2 x 6.023 x10 23 particles  of Hydrogen atoms

          -1 x 6.023 x10 23 particles  of Oxygen atoms

 

(ii)Molar mass sulphuric(VI)acid(H2SO4) = relative formular mass

=[(1.0 x 2 ) + 32.0 + (16.0 x 4)]g =98.0g

 

6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) = 1 mole = 98.0g

6.023 x10 23 particles of sulphuric(VI)acid(H2SO4) has:

2 x 6.023 x10 23 particles  of Hydrogen atoms

1 x 6.023 x10 23 particles  of Sulphur atoms

           -4 x 6.023 x10 23 particles  of Oxygen atoms

 

(iii)Molar mass sodium carbonate(IV)(Na2CO3) = relative formular mass

=[(23.0 x 2 ) + 12.0 + (16.0 x 3)]g =106.0g

 

6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) = 1 mole = 106.0g

6.023 x10 23 particles of sodium carbonate(IV)(Na2CO3) has:

2 x 6.023 x10 23 particles  of Sodium atoms

1 x 6.023 x10 23 particles  of Carbon atoms

           -3 x 6.023 x10 23 particles  of Oxygen atoms

 

(iv)Molar mass Calcium carbonate(IV)(CaCO3) = relative formular mass

=[(40.0+ 12.0 + (16.0 x 3)]g =100.0g.

 

6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) = 1 mole = 100.0g

6.023 x10 23 particles of Calcium carbonate(IV)(CaCO3) has:

1 x 6.023 x10 23 particles  of Calcium atoms

1 x 6.023 x10 23 particles  of Carbon atoms

           -3 x 6.023 x10 23 particles  of Oxygen atoms

(v)Molar mass Water(H2O) = relative formular mass

=[(2 x 1.0 )+ 16.0 ]g =18.0g

 

6.023 x10 23 particles of Water(H2O) = 1 mole = 18.0g

6.023 x10 23 particles of Water(H2O) has:

2 x 6.023 x10 23 particles  of Hydrogen atoms

 -2 x 6.023 x10 23 particles  of Oxygen atoms

Practice

  1. Calculate the number of moles present in:

(i)0.23 g of Sodium atoms

Molar mass of Sodium atoms = 23g

Moles =      mass in grams        = >  0.23g   =   0.01moles

Molar mass                      23

(ii) 0.23 g of Chlorine atoms

Molar mass of Chlorine atoms  = 35.5 g

Moles =      mass in grams        = >  0.23g   = 0.0065moles /6.5 x 10-3 moles

Molar mass                   35.5

(iii) 0.23 g of Chlorine molecules

Molar mass of Chlorine molecules  =( 35.5 x 2) = 71.0 g

Moles =      mass in grams        = >  0.23g   = 0.0032moles /3.2 x 10-3 moles

Molar mass                    71

(iv) 0.23 g of dilute sulphuric(VI)acid

Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g

Moles = mass in grams   = >  0.23g       = 0.0023moles /2.3 x 10-3 moles

Molar mass               98

 

  1. Calculate the number of atoms present in:(Avogadros constant L = 6.0 x 10 23)

(i) 0.23 g of dilute sulphuric (VI)acid

Method I

Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g

Moles = mass in grams   = >  0.23g       = 0.0023moles /2.3 x 10-3 moles

Molar mass               98

 

1 mole has 6.0 x 10 23 atoms

2.3 x 10-3 moles has    (2.3 x 10-3 x 6.0 x 10 23)   =   1.38 x 10 21 atoms

1

Method II

Molar mass of H2SO4 = [(2 x 1) + 32 + (4 x14)] = 98.0g

98.0g = 1 mole has 6.0 x 10 23 atoms

0.23 g therefore has    (0.23 g x 6.0 x 10 23 )      =   1.38 x 10 21 atoms

98

(ii)0.23 g of sodium carbonate(IV)decahydrate

 

Molar mass of Na2CO3.10H2 O=

[(2 x 23) + 12 + (3 x16) + (10 x 1.0) + (10 x 16)] = 276.0g

Method I

Moles = mass in grams     = >  0.23g   =   0.00083moles /

Molar mass                  276          8.3 x 10-4 moles

 

1 mole has 6.0 x 10 23 atoms

8.3 x 10-4 moles has    (8.3 x 10-4 moles x 6.0 x 10 23)   =   4.98 x 10 20 atoms

1

Method II

276.0g = 1 mole has 6.0 x 10 23 atoms

0.23 g therefore has    (0.23 g x 6.0 x 10 23 )      =   4.98 x 10 20 atoms

276.0

(iii)0.23 g of  Oxygen gas

 

Molar mass of O2 =  (2 x16) = 32.0 g

 

Method I

Moles = mass in grams     = >  0.23g   =   0.00718moles /

Molar mass                  32          7.18 x 10-3 moles

 

1 mole has  2 x 6.0 x 10 23 atoms in O2

7.18 x 10-3moles has (7.18 x 10-3moles x 2 x 6.0 x 10 23) =8.616 x 10 21atoms

1

 

Method II

32.0g = 1 mole has 2 x 6.0 x 10 23 atoms in O2

0.23 g therefore has    (0.23 g  x  2  x  6.0  x  10 23 )      =   8.616 x 10 21atoms

32.0

 

(iv)0.23 g of  Carbon(IV)oxide gas

 

Molar mass of CO2 =  [12  + (2 x16)] = 44.0 g

Method I

Moles = mass in grams     = >  0.23g   =   0.00522moles /

Molar mass                  44          5.22 x 10-3 moles

 

1 mole has  3 x 6.0 x 10 23 atoms in CO2

7.18 x 10-3moles has (5.22 x 10-3moles x 3 x 6.0 x 10 23) =9.396 x 10 21atoms

1

Method II

44.0g = 1 mole has 3 x 6.0 x 10 23 atoms in CO2

0.23 g therefore has    (0.23 g  x  3  x  6.0  x  10 23 )      =   9.409 x 10 21atoms

44.0

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(c)Empirical and molecular formula

 

1.The empirical formula of a compound is its simplest formula. It is the simplest whole number ratios in which atoms of elements combine to form the compound. 2.It is mathematically the lowest common multiple (LCM) of the atoms of the elements in the  compound

3.Practically the empirical formula of a compound can be determined as in the following examples.

 

To determine the empirical formula of copper oxide

 

(a)Method 1:From copper to copper(II)oxide

Procedure.

Weigh a clean dry covered crucible(M1).Put two spatula full of copper powder into the crucible. Weigh again (M2).Heat the crucible on a strong Bunsen flame for five minutes. Lift the lid, and swirl the crucible carefully using a pair of tong. Cover the crucible and continue heating for another five minutes. Remove the lid and stop heating. Allow the crucible to cool. When cool replace the lid and weigh the contents again (M3).

 

Sample results

 

Mass of crucible(M1) 15.6g
Mass of crucible + copper before heating(M2) 18.4
Mass of crucible + copper after heating(M3) 19.1

 

Sample questions

 

  1. Calculate the mass of copper powder used.

Mass of crucible + copper before heating(M2)    =     18.4

Less        Mass of crucible(M1)                            =  – 15.6g

Mass of copper                                                           2.8 g

 

  1. Calculate the mass of Oxygen used to react with copper.

Method I

Mass of crucible + copper after heating(M3)       =    19.1g

Mass of crucible + copper before heating(M2)    =  – 18.4g

Mass of Oxygen                               =   0.7 g

Method II

Mass of crucible + copper after heating(M3)       =       19.1g

Mass of crucible                                                  =   –  15.6g

Mass of copper(II)Oxide                                               =       3.5 g

Mass of copper(II)Oxide                                               =        3.5 g

Mass of copper                                                              =      – 2.8 g

Mass of Oxygen                                                  =        0.7 g

  1. Calculate the number of moles of:

(i) copper used (Cu = 63.5)

number of moles of copper  = mass used     =>     2.8     =    0.0441moles

Molar mass            63.5

(ii) Oxygen used (O = 16.0)

number of moles of oxygen  = mass used     =>     0.7     =    0.0441moles

Molar mass           16.0

 

  1. Determine the mole ratio of the reactants

Moles of copper       =  0.0441moles   =    1       =>  Mole ratio Cu: O = 1:1

Moles of oxygen          0.0441moles         1

5.What is the empirical, formula of copper oxide formed.

CuO (copper(II)oxide

 

  1. State and explain the observations made during the experiment.

Observation

Colour change from brown to black

Explanation

               Copper powder is brown. On heating it reacts with oxygen from the air to form black copper(II)oxide

 

  1. Explain why magnesium ribbon/shavings would be unsuitable in a similar experiment as the one above.

Hot magnesium generates enough heat energy to react with both Oxygen and Nitrogen in the air forming a white solid mixture of Magnesium oxide and magnesium nitride. This causes experimental mass errors.

 

 (b)Method 2:From copper(II)oxide to copper

Procedure.

Weigh a clean dry porcelain boat (M1). Put two spatula full of copper(II)oxide powder into the crucible. Reweigh the porcelain boat (M2).Put the porcelain boat in a glass tube and set up the apparatus as below;

 

 

Pass slowly(to prevent copper(II)oxide from being blown away)a stream of either dry Hydrogen /ammonia/laboratory gas/ carbon(II)oxide gas for about two minutes from a suitable generator.

When all the in the apparatus set up is driven out ,heat the copper(II)oxide strongly for about five minutes until there is no further change. Stop heating.

Continue passing the gases until the glass tube is cool.

Turn off the gas generator.

Carefully remove the porcelain boat form the combustion tube.

Reweigh (M3).

 

Sample results

 

Mass of boat(M1) 15.6g
Mass of boat before heating(M2) 19.1
Mass of boat after heating(M3) 18.4

 

Sample questions

 

  1. Calculate the mass of copper(II)oxide used.

Mass of boat before heating(M2)                =   19.1

Mass of empty boat(M1)                            =  – 15.6g

Mass of copper(II)Oxide                                 3.5 g

  1. Calculate the mass of

 (i) Oxygen.

Mass of boat before heating(M2)                =    19.1

Mass of boat after heating (M3)                  =  – 18.4g

Mass of oxygen                                         =    0.7 g

(ii)Copper

Mass of copper(II)Oxide                    =   3.5 g

Mass of oxygen                                  =    0.7 g

Mass of oxygen                                  =    2.8 g

 

  1. Calculate the number of moles of:

(i) Copper used (Cu = 63.5)

number of moles of copper  = mass used     =>     2.8     =    0.0441moles

Molar mass            63.5

(ii) Oxygen used (O = 16.0)

number of moles of oxygen  = mass used     =>     0.7     =    0.0441moles

Molar mass           16.0

 

  1. Determine the mole ratio of the reactants

Moles of copper       =  0.0441moles   =    1       =>  Mole ratio Cu: O = 1:1

Moles of oxygen          0.0441moles         1

5.What is the empirical, formula of copper oxide formed.

CuO (copper(II)oxide

 

  1. State and explain the observations made during the experiment.

Observation

Colour change from black to brown

Explanation

               Copper(II)oxide powder is black. On heating it is reduced by a suitable reducing agent to brown copper metal.

 

  1. Explain why magnesium oxide would be unsuitable in a similar experiment as the one above.

Magnesium is high in the reactivity series. None of the above reducing agents is strong enough to reduce the oxide to the metal.

 

  1. Write the equation for the reaction that would take place when the reducing agent is:

 

(i) Hydrogen

CuO(s)         +        H2(g)           ->       Cu(s)           + H2O(l)

(Black)                                    (brown)                    (colourless liquid form

on cooler parts )

(ii)Carbon(II)oxide

CuO(s)        +        CO (g)         ->       Cu(s)           + CO2(g)

(Black)                                              (brown)                    (colourless gas, form

white ppt with lime water )

(iii)Ammonia

3CuO(s)       +        2NH3(g)      ->       3Cu(s)    +   N2 (g)  +      3H2O(l)

(Black)                                               (brown)                 (colourless liquid form

on cooler parts )

 

  1. Explain why the following is necessary during the above experiment;

 

(i)A stream of dry hydrogen gas should be passed before heating copper (II) Oxide.

Air combine with hydrogen in presence of heat causing an explosion

 

(ii)A stream of dry hydrogen gas should be passed after heating copper (II) Oxide has been stopped.

Hot metallic copper can be re-oxidized back to copper(II)oxide

 

(iii) A stream of excess carbon (II)oxide gas should be ignited to burn

 

Carbon (II)oxide is highly poisonous/toxic. On ignition it burns to form less toxic carbon (IV)oxide gas.

  1. State two sources of error in this experiment.

 

(i)All copper(II)oxide may not be reduced to copper.

(ii)Some copper(II)oxide may be blown out  the boat by the reducing agent.

 

4.Theoreticaly the empirical formula of a compound can be determined as in the following examples.

 

(a)A oxide of copper contain 80% by mass of copper. Determine its empirical formula. (Cu = 63.5, 16.0)

 

% of Oxygen  =  100% – % of Copper  =>  100- 80   =  20% of Oxygen

 

Element Copper Oxygen
Symbol Cu O
Moles present =    % composition

Molar mass

 80

63.5

20

16

Divide by the smallest value 1.25

1.25

1.25

1.25

Mole ratios 1 1

 

Empirical formula   is CuO

 

(b)1.60g of an oxide of Magnesium contain 0.84g by mass of Magnesium. Determine its empirical formula(Mg = 24.0, 16.0)

 

Mass of Oxygen  =  1.60  –  0.84  =>  0.56 g of Oxygen

 

Element Magnesium Oxygen
Symbol Mg O
Moles present =    % composition

Molar mass

 0.84

24

0.56

16

Divide by the smallest value 0.35

0.35

0.35

0.35

Mole ratios 1 1

Empirical formula   is MgO

 

(c)An oxide of Silicon contain 47% by mass of Silicon. What is its empirical formula(Si = 28.0, 16.0)

 

Mass of Oxygen  =  100  –  47  =>  53% of Oxygen

 

 

Element Silicon Oxygen
Symbol Si O
Moles present =    % composition

Molar mass

 47

28

53

16

Divide by the smallest value 1.68

1.68

3.31

1.68

Mole ratios 1 1.94   =  2

Empirical formula   is SiO2

 

 

(d)A compound  contain 70% by mass of Iron and 30% Oxygen. What is its empirical formula(Fe = 56.0, 16.0)

 

Mass of Oxygen  =  100  –  47  =>  53% of Oxygen

 

Element Silicon Oxygen
Symbol Si O
Moles present =    % composition

Molar mass

 47

28

53

16

Divide by the smallest value 1.68

1.68

3.31

1.68

Mole ratios 1 1.94   =  2

 

Empirical formula   is SiO2

2.During heating of a hydrated copper (II)sulphate(VI) crystals, the following readings were obtained:

Mass of evaporating dish =300.0g

Mass of evaporating dish + hydrated salt  = 305.0g

Mass of evaporating dish + anhydrous salt  = 303.2g

Calculate the number of water of crystallization molecules in hydrated copper (II)sulphate(VI)

(Cu =64.5, S = 32.0,O=16.0, H = 1.0)

Working

Mass of Hydrated salt = 305.0g -300.0g  = 5.0g

Mass of anhydrous salt = 303.2 g -300.0g  = 3.2 g

Mass of water in hydrated salt = 5.0g -3.2 g  = 1.8g

Molar mass of water(H2O) = 18.0g

Molar mass of anhydrous copper (II)sulphate(VI) (CuSO4) = 160.5g

Element/compound anhydrous copper (II)

sulphate(VI)

Oxygen
Symbol CuSO4 O
Moles present =    composition by mass

Molar mass

 3,2

160.5

1.8

18

Divide by the smallest value 0.0199

0.0199

0.1

18

Mole ratios 1  5

 

The empirical formula of hydrated salt = CuSO4.5H2O

Hydrated salt has five/5 molecules of water of crystallizations

 

 

  1. The molecular formula is the actual number of each kind of atoms present in a molecule of a compound.

 

The empirical formula of an ionic compound is the same as the chemical formula but for simple molecular structured compounds, the empirical formula may not be the same as the chemical formula.

 

The molecular formula is a multiple of empirical formula .It is determined from the relationship:

 

(i)            n  =    Relative formular mass

                              Relative empirical formula

where  n  is a whole number.

 

(ii) Relative empirical formula x  n  = Relative formular mass

where  n is a whole number.

 

Practice sample examples

 

  1. A hydrocarbon was found to contain 92.3% carbon and the remaining Hydrogen.

If the molecular mass of the compound is 78, determine the molecular formula(C=12.0, H =1.0)

 

Mass of Hydrogen  =  100  –  92.3  =>  7.7% of Oxygen

 

 

Element Carbon Hydrogen
Symbol C H
Moles present =    % composition

Molar mass

 92.3

12

7.7

1

Divide by the smallest value 7.7

7.7

7.7

7.7

Mole ratios 1 1

 

Empirical formula   is CH

The molecular formular is thus determined :

n  =    Relative formular mass                                  =   78        =   6

                    Relative empirical formula                                  13

 

The molecular formula is   (C H )  x  6   =  C6H6

 

  1. A compound of carbon, hydrogen and oxygen contain 54.55% carbon, 9.09% and remaining 36.36% oxygen.

If its relative molecular mass is 88, determine its molecular formula(C=12.0, H =1.0, O= 16.0)

 

Element Carbon Hydrogen Oxygen
Symbol C H O
Moles present =    % composition

Molar mass

 54.55

12

9.09

1

36.36

16

Divide by the smallest value 4.5458

2.2725

9.09

2.2725

2.2725

2.2725

Mole ratios 2 4 1

 

Empirical formula   is C2H4O

The molecular formula is thus determined :

n  =    Relative formular mass                                  =   88        =   2

                    Relative empirical formula                                  44

 

The molecular formula is   (C2H4O )  x  2   =  C4H8O2.

 

4.A hydrocarbon burns completely in excess air to form 5.28 g of carbon (IV) oxide and 2,16g of water.

 If the molecular mass of the hydrocarbon is 84, draw and name its  molecular structure.

 

Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then:

Mass of carbon in CO2 =                     Mass of C in CO2     x   mass of CO2       =>

                                                  Molar mass of CO2                                                    

12    x   5.28       =   1.44g√

44

Mass of Hydrogen in H2O =      Mass of C in H2O     x   mass of H2O      =>

                                                  Molar mass of H2O                                                    

  2    x   2.16      =   0.24g√

18

 

Element Carbon Hydrogen
Symbol C H
Moles present =    mass

Molar mass

 1.44g

12

0.24g

1

Divide by the smallest value 0.12

0.12

0.24

0.12

 

Mole ratios 1 2

Empirical formula   is CH2

The molecular formular is thus determined :

n  =    Relative formular mass           =   84    =   6√

                    Relative empirical formula         14

 

The molecular formula is   (CH2 )  x  6   =  C6H12.

molecular name   Hexene/Hex-1-ene    (or any position isomer of Hexene)

Molecular structure

H       H       H       H       H       H

 

H       C       C       C       C       C       C       H

 

 

H       H       H       H

  1. Compound A contain 5.2% by mass of Nitrogen .The other elements present are Carbon, hydrogen and Oxygen. On combustion of 0.085g of A in excess Oxygen,0.224g of carbon(IV)oxide and 0.0372g of water was formed. Determine the empirical formula of A (N=14.0, O=16.0 , C=12.0 , H=1.0)

 

Mass of N in A      =   5.2%   x   0.085   =   0.00442 g

 

Mass of C in A      =     12          x   0.224       = 0.0611g

                                      44

Mass of H in A      =      2          x   0.0372       = 0.0041g

                                      18

Mass of O in A      =  0.085g –  0.004442g   =   0.0806g  (Mass of C,H,O)

=> 0.0611g  + 0.0041g    =   0.0652g (Mass of C,H)

0.0806g  (Mass of C,H,O)- 0.0652g (Mass of C,H) =  0.0154 g

Element Nitrogen Carbon Hydrogen Oxygen
Symbol N C H O
Moles present =   mass

Molar mass

0.00442 g

14

 0.0611g

12

0.0041g

    1

0.0154 g    16
Divide by the smallest value 0.00032

0.00032

0.00509

0.00032

0.0041g

0.00032

0.00096

0.00032

Mole ratios 1 16 13 3

 

Empirical formula = C16H13NO3

 

 

 

 

(d)Molar gas volume

 

The volume occupied by one mole of all gases at the same temperature and pressure is a constant.It is:

(i) 24dm3/24litres/24000cm3 at room temperature(25oC/298K)and pressure(r.t.p).

i.e. 1mole of all gases =24dm3/24litres/24000cm3 at r.t.p

Examples

1mole of O2 = 32g =6.0 x1023 particles= 24dm3/24litres/24000cm3 at r.t.p

1mole of H2 = 2g =6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p

1mole of CO2 = 44g = 6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p

1mole of NH3 = 17g =6.0 x1023 particles = 24dm3/24litres/24000cm3 at r.t.p

1mole of CH4 = 16g =6.0 x1023 particles =24dm3/24litres/24000cm3 at r.t.p

 

(ii)22.4dm3/22.4litres/22400cm3 at standard temperature(0oC/273K) and pressure(s.t.p)

i.e. 1mole of all gases =22.4dm3/22.4litres/22400cm3 at s.t.p

Examples

1mole of O2 = 32g =6.0 x1023 particles= 22.4dm3/22.4litres/22400cm3 at s.t.p

1mole of H2 = 2g =6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p

1mole of CO2 = 44g = 6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p

1mole of NH3 = 17g =6.0 x1023 particles= 22.4dm3/22.4litres/22400cm3 at s.t.p

1mole of CH4 = 16g =6.0 x1023 particles = 22.4dm3/22.4litres/22400cm3 at s.t.p

The volume occupied by one mole of a gas at r.t.p or s.t.p is commonly called the molar gas volume. Whether the molar gas volume is at r.t.p or s.t.p must always be specified.

From the above therefore a less or more volume can be determined as in the examples below.

 

Practice examples

  1. Calculate the number of particles present in:

(Avogadros constant =6.0 x1023mole-1 )

 

  (i) 2.24dm3 of Oxygen.

 

22.4dm3       ->      6.0 x1023

          2.24dm3      ->      2.24 x 6.0 x1023

                              22.4

 

=6.0 x1022   molecules  = 2 x 6.0 x1022.      =   1.2 x1023 atoms

 

 

(ii) 2.24dm3 of Carbon(IV)oxide.

22.4dm3       ->      6.0 x1023

          2.24dm3      ->      2.24 x 6.0 x1023

                              22.4

=6.0 x1022   molecules  = (CO2) = 3 x 6.0 x1022.      =   1.8 x1023 atoms

 

  1. 0.135 g of a gaseous hydrocarbon X on complete combustion produces 0.41g of carbon(IV)oxide and 0.209g of water.0.29g of X occupy 120cm3 at room temperature and 1 atmosphere pressure .Name X and draw its molecular structure.(C=12.0,O= 16.O,H=1.0,1 mole of gas occupies 24dm3 at r.t.p)

Molar mass CO2= 44 gmole-1√ Molar mass H2O = 18 gmole-1

Molar mass X =               0.29 x (24 x 1000)cm3  = 58 gmole-1

120cm3

Since a hydrocarbon is a compound containing Carbon and Hydrogen only. Then:

Mass of carbon in CO2 =                     Mass of C in CO2     x   mass of CO2       =>

                                                  Molar mass of CO2                                                    

12    x   0.41       =   0.1118g

44

Mass of Hydrogen in H2O =      Mass of C in H2O     x   mass of H2O      =>

                                                  Molar mass of H2O                                                    

  2    x   0.209      =   0.0232g

18

 

Element Carbon Hydrogen
Symbol C H
Moles present =    % composition

Molar mass

 0.g118

12

0.0232g

1

Divide by the smallest value 0.0093

0.0093

0.0232

0.0093√

 

Mole ratios 1 x2 2.5×2
  2 5√

Empirical formula   is C2H5

The molecular formular is thus determined :

n  =    Relative formular mass           =   58    =   2

                    Relative empirical formula           29

 

The molecular formula is   (C2H5 )  x  2   =  C4H10.√

Molecule name Butane

Molecula structure

H       H       H       H

 

 

H       C        C        C        C         H√

 

H       H       H       H

 

 

 

 

 

(e)Gravimetric analysis

 

Gravimetric analysis is the relationship between reacting masses and the volumes and /or masses of products. All reactants are in mole ratios to their products in accordance to their stoichiometric equation. Using the mole ration of reactants and products any volume and/or mass can be determined as in the examples:

 

  1. Calculate the volume of carbon(IV)oxide at r.t.p produced when 5.0 g of calcium carbonate is strongly heated.(Ca=40.0, C= 12.0,O = 16.0,1 mole of gas =22.4 at r.t.p)

Chemical equation

CaCO3(s)     ->       CaO(s)         +        CO2(g)

Mole ratios       1:                         1:                            1

 

Molar Mass CaCO3  =100g

 Method 1

100g CaCO3(s)      ->  24dm3 CO2(g) at r.t.p

5.0 g CaCO3(s)      ->       5.0 g x  24dm3      = 1.2dm3/1200cm3

100g

 

Method 2

Moles of 5.0 g CaCO3(s)  =  5.0 g   =  0.05 moles

100 g

Mole ratio 1:1

Moles of CO2(g) = 0.05moles

Volume of CO2(g) = 0.05  x  24000cm3  =1200cm3 /1.2dm3

 

  1. 1.0g of an alloy of aluminium and copper were reacted with excess hydrochloric acid. If 840cm3 of hydrogen at s.t.p was produced, calculate the % of copper in the alloy.(Al =27.0,one mole of a gas at s.t.p =22.4dm3 )

 

Chemical equation

Copper does not react with hydrochloric acid

2Al(s)  +   6HCl(aq)  ->  2AlCl3(aq)  +  3H2(g)

Method 1

 

3H2(g) = 3 moles  x (22.4 x 1000)cm3  => 2 x 27 g Al

840cm3       => 840cm3  x 2 x 27 = 0.675g of Aluminium

3 x 22.4 x 1000

Total mass of alloy – mass of aluminium = mass of copper

=> 1.0g  – 0.675g =0.325g of copper

 

% copper      =    mass of copper x100%       =     32.5%

                                Mass of alloy

Method 2

Mole ratio 2Al: 3H2  = 2:3

Moles of Hydrogen gas =    volume of gas     =>     840cm3 =      0.0375moles

                                         Molar gas volume           22400cm3

 

Moles of Al  =  2/3 moles of H2  =>   2/3x 0.0375moles = 0.025moles

 

Mass of Al = moles x molar mass =>0.025moles x 27 =  0.675g

 

Total mass of alloy – mass of aluminium = mass of copper

=> 1.0g  – 0.675g =  0.325 g of copper

 

% copper      =    mass of copper x100%       =     32.5%

                              Mass of alloy

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(f)Gay Lussac’s  law

 

Gay Lussacs law states that “when gases combine/react they do so in simple volume ratios to each other and to their gaseous products at constant/same temperature and pressure”

Gay Lussacs law thus only apply to gases

Given the volume of one gas reactant, the other gaseous reactants can be deduced thus:

 

Examples

  1. Calculate the volume of Oxygen required to completely react with 50cm3 of Hydrogen.

Chemical equation:   2H2 (g)     +        O2 (g)          ->      2H2O(l)

Volume ratios             2             :           1                :         0

Reacting volumes      50cm3    :         25cm3

50cm3 of Oxygen is used

 

  1. Calculate the volume of air required to completely reacts with 50cm3 of Hydrogen.(assume Oxygen is 21% by volume of air)

 

Chemical equation:   2H2 (g)     +        O2 (g)          ->      2H2O(l)

Volume ratios             2             :           1                :         0

Reacting volumes      50cm3    :         25cm3

50cm3 of Oxygen is used

21%     =     25cm3

100%   =   100 x 25    =

21

 

3.If 5cm3 of a hydrocarbon CxHy burn in 15cm3 of Oxygen to form 10cm3 of Carbon(IV)oxide and 10cm3 of water vapour/steam, obtain the equation for the reaction and hence find the value of x and y in CxHy.

 

Chemical equation:           CxHy (g)     +        O2 (g)          ->      H2O(g)  + CO2(g)

Volumes                            5cm3         :        15cm3          :         10cm3  :  10cm3

Volume ratios                    5cm3         :        15cm3          :         10cm3  :  10cm3   (divide by lowest volume)  5                     5                            5                5

Reacting volume ratios    1volume              3 volume             2 volume  2 volume

 

Balanced chemical equation:       CxHy (g) + 3O2 (g)  ->   2H2O(g) + 2CO2(g)

If “4H” are in 2H2O(g) the y=4

If “2C” are in 2CO2 (g) the x=2

Thus(i) chemical formula of hydrocarbon = C2H4

(ii) chemical name of hydrocarbon = Ethene

 

4.100cm3 of nitrogen (II)oxide NO combine with 50cm3 of Oxygen to form 100cm3 of a single gaseous compound of nitrogen. All volumes measured at the same temperature and pressure. Obtain the equation for the reaction and name the gaseous product.

 

Chemical equation:           NO (g)        +        O2 (g)          ->      NOx

Volumes                          100cm3       :        50cm3          :          100

Volume ratios                    100cm3     :         50cm3         :         100cm3         (divide by lowest volume)   50                      50                          50             

Reacting volume ratios    2volume              1 volume               2 volume

 

Balanced chemical equation:      2 NO (g) +  O2 (g)  ->    2NO x(g)

Thus(i) chemical formula of the nitrogen compound = 2 NO2

(ii) chemical name of compound = Nitrogen(IV)oxide

 

5.When 15cm3 of a gaseous hydrocarbon was burnt in 100cm3 of Oxygen ,the resulting gaseous mixture occupied70cm3 at  room temperature and pressure. When the gaseous mixture was passed through, potassium hydroxide its volume decreased to 25cm3.

(a)What volume of Oxygen was used during the reaction.(1mk)

Volume of Oxygen used =100-25 =75cm3√  

 (P was completely burnt)

 

(b)Determine the molecular formula of the hydrocarbon(2mk)

CxHy  +  O2  ->  xCO2   +   yH2O

          15cm3 : 75cm3

           15          15

            1        :   3

=> 1 atom of C react with 6 (3×2)atoms of Oxygen

Thus x = 1 and y = 2   => P has molecula formula CH4

 

 

 

 

 

 

(g) Ionic equations

An ionic equation is a chemical statement showing the movement of ions (cations and anions ) from reactants to products.

Solids, gases and liquids do not ionize/dissociate into free ions. Only ionic compounds in aqueous/solution or molten state ionize/dissociate into free cations and anions (ions)

An ionic equation is usually derived from a stoichiometric equation by using the following guidelines

 

Guidelines for writing ionic equations

1.Write the balanced stoichiometric equation

2.Indicate the state symbols of the reactants and products

3.Split into cations and anions all the reactants and products that exist in aqueous state.

4.Cancel out any cation and anion that appear on both the product and reactant side.

  1. Rewrite the chemical equation. It is an ionic equation.

 

Practice

(a)Precipitation of an insoluble salt

 

All insoluble salts are prepared in the laboratory from double decomposition /precipitation. This involves mixing two soluble salts to form one soluble and one insoluble salt

  1. When silver nitrate(V) solution is added to sodium chloride solution,sodium nitrate(V) solution and a white precipitate of silver chloride are formed.

 

Balanced stoichiometric equation

AgNO3(aq) +        NaCl(aq)      ->      AgCl (s)       +       NaNO3 (aq)

 

Split reactants product existing in aqueous state as cation/anion

Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)       ->      AgCl(s) +    Na+(aq)+ NO3 (aq)

 

Cancel out ions appearing on reactant and product side

Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq)       ->      AgCl(s) +    Na+(aq)+ NO3 (aq)

 

Rewrite the equation

Ag+(aq) + Cl(aq)  ->      AgCl(s) (ionic equation)

 

  1. When barium nitrate(V) solution is added to copper(II)sulphate(VI) solution, copper(II) nitrate(V) solution and a white precipitate of barium sulphate(VI) are formed.

 

Balanced stoichiometric equation

Ba(NO3)2(aq)         +        CuSO4(aq)   -> BaSO4 (s)         +       Cu(NO3) 2 (aq)

 

Split reactants product existing in aqueous state as cation/anion

Ba2+(aq) + 2NO3(aq) + Cu2+(aq) + SO42-(aq) -> BaSO4 (s) + 2NO3(aq)+ Cu2+(aq)

 

Cancel out ions appearing on reactant and product side

Ba2+(aq) + 2NO3(aq) +Cu2+ (aq) + SO42-(aq)-> BaSO4(s) + 2NO3(aq) + Cu2+(aq)

 

Rewrite the equation

Ba2+(aq) + SO42-(aq) -> BaSO4(s) (ionic equation)

 

3.A yellow precipitate of Potassium Iodide is formed from the reaction of Lead(II)nitrate(v) and potassium iodide.

Balanced stoichiometric equation

Pb(NO3)2(aq)         +        2KI (aq)       -> PbI2 (s)   +       2KNO3  (aq)

 

Split reactants product existing in aqueous state as cation/anion

Pb2+(aq) + 2NO3(aq) + 2K +(aq) + 2I(aq) -> PbI2 (s) + 2NO3(aq)+ 2K +(aq)

 

Cancel out ions appearing on reactant and product side

Pb2+(aq) + 2NO3(aq) + 2K +(aq) + 2I(aq) -> PbI2 (s) + 2NO3(aq)+ 2K +(aq)

 

Rewrite the equation

Pb2+(aq) + 2I (aq)   ->   PbI2 (s) (ionic equation)

 

(b)Neutralization

Neutralization is the reaction of an acid with a soluble base/alkali or insoluble base.

(i)Reaction of alkalis with acids

1.Reaction of nitric(V)acid with potassium hydroxide

 

Balanced stoichiometric equation

HNO3(aq)    +        KOH (aq)     -> H2O (l)   +       KNO3  (aq)

 

Split reactants product existing in aqueous state as cation/anion

H+(aq) + NO3(aq) + K +(aq) + OH(aq) -> H2O (l) + NO3(aq)+  K +(aq)

 

Cancel out ions appearing on reactant and product side

H+(aq) + NO3(aq) + K +(aq) + OH(aq) -> H2O (l) + NO3(aq)+  K +(aq)

 

Rewrite the equation

H+ (aq) + OH(aq)   -> H2O (l)  (ionic equation)

 

2.Reaction of sulphuric(VI)acid with ammonia solution

 

Balanced stoichiometric equation

H2SO4(aq)   +        2NH4OH (aq)        -> H2O (l)   +       (NH4) 2SO4  (aq)

 

Split reactants product existing in aqueous state as cation/anion

2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH(aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)

 

Cancel out ions appearing on reactant and product side

2H+(aq) + SO42- (aq) + 2NH4 +(aq)+ 2OH(aq) ->2H2O (l) +SO42- (aq)+ 2NH4 + (aq)

 

Rewrite the equation

2H+ (aq) + 2OH(aq)   ->        2H2O (l)

 H+ (aq)  + OH(aq)     ->        H2O (l)  (ionic equation)

 

3.Reaction of hydrochloric acid with Zinc hydroxide

 

Balanced stoichiometric equation

2HCl(aq)     +        Zn(OH)2 (s) -> 2H2O (l) +       ZnCl 2  (aq)

 

Split reactants product existing in aqueous state as cation/anion

2H+(aq) + 2Cl(aq) + Zn(OH)2 (s)  ->2H2O (l) + 2Cl (aq)+ Zn 2+ (aq)

 

Cancel out ions appearing on reactant and product side

2H+(aq) + 2Cl(aq) + Zn(OH)2 (s)  ->2H2O (l) + 2Cl (aq)+ Zn 2+ (aq)

 

Rewrite the equation

2H+(aq) + Zn(OH)2 (s)  ->2H2O (l) + Zn 2+ (aq) (ionic equation)

(h)Molar solutions

 

A molar solution is one whose concentration is known. The SI unit of concentration is Molarity denoted M.

Molarity may be defined as the number of moles of solute present in one cubic decimeter of solution.

One cubic decimeter is equal to one litre and also equal to 1000cm3.

The higher the molarity the higher the concentration and the higher/more solute has been dissolved in the solvent  to make one cubic decimeter/ litre/1000cm3 solution.

 

Examples

2M sodium hydroxide means 2 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.

0.02M sodium hydroxide means 0.02 moles of sodium hydroxide solute is dissolved in enough water to make one cubic decimeter/ litre/1000cm3 uniform solution mixture of sodium hydroxide and water.

“2M” is more concentrated than“0.02M”.

 

Preparation of molar solution

 

Procedure

Weigh accurately 4.0 g of sodium hydroxide pellets into a 250cm3 volumetric   flask.

Using a wash bottle add about 200cm3 of distilled water.

Stopper the flask.

Shake vigorously for three minutes.

Remove the stopper for a second then continue to shake for about another two minutes until all  the solid has dissolved.

Add more water slowly upto exactly the 250 cm3 mark.

 

Sample questions

 

1.Calculate the number of moles of sodium hydroxide pellets present in:

(i) 4.0 g.

 

Molar mass of NaOH = (23 + 16 + 1)  =  40g

Moles = Mass                 =>       4.0         =     0.1   /  1.0 x 10 -1 moles

                                Molar mass                      40

 

(ii) 250 cm3 solution in the volumetric flask.

 

                    Moles in 250 cm3 =  0.1   /  1.0 x 10 -1 moles

 

(iii) one decimeter of solution

 

                    Method 1

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.0 x 10 -1 moles  x 1000cm3 =

250cm3

0.4 M  / 0.4 molesdm-3

Method 2

250cm3 solution contain 1.0 x 10 -1 moles

1000cm3 solution = Molarity  contain 1000  x 1.0 x 10 -1 moles 

250 cm3

0.4 M  / 0.4 molesdm-3

 

Theoretical sample practice

 

  1. Calculate the molarity of a solution containing:

  (i) 4.0 g sodium hydroxide dissolved in 500cm3 solution

 

Molar mass of NaOH = (23 + 16 + 1)  =  40g

 

Moles = Mass                 =>       4.0         =     0.1   /  1.0 x 10 -1 moles

                                Molar mass                      40

 

Method 1

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.0 x 10 -1 moles  x 1000cm3

500cm3

0.2 M  / 0.2 molesdm-3

Method 2

500 cm3 solution contain 1.0 x 10 -1 moles

1000cm3 solution = Molarity  contain 1000  x 1.0 x 10 -1 moles 

500 cm3

0.2 M  / 0.2 molesdm-3

(ii) 5.3 g  anhydrous sodium carbonate dissolved in 50cm3 solution

Molar mass of Na2CO3 = (23 x 2 + 12 + 16 x 3)  =  106 g

 

Moles = Mass          =>       5.3         =     0.05 / 5. 0 x 10-2   moles

                                Molar mass              106

 

Method 1

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.0 moles  x 1000cm3 =

50cm3

=1.0 M

 

Method 2

50 cm3 solution contain 5.0 x 10 -2 moles

1000cm3 solution = Molarity  contain 1000  x 5.0 x 10 -2 moles 

50 cm3

= 1.0M /  1.0 molesdm-3

 

(iii) 5.3 g  hydrated sodium carbonate decahydrate dissolved in 50cm3 solution

 

Molar mass of Na2CO3.10H2O  = (23 x 2 + 12 + 16 x 3 + 20 x 1 + 10 x 16)  =286g

 

Moles = Mass          =>       5.3         =     0.0185 / 1.85 x 10 -2   moles

                                 Molar mass              286

Method 1

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.85 x 10 -2 moles  x 1000cm3 =

50cm3

= 0.37 M/0.37 molesdm-3

Method 2

50 cm3 solution contain 1.85 x 10 -2 moles

1000cm3 solution = Molarity  contain 1000  x  1.85 x 10 -2 moles 

50 cm3

= 3.7 x 10-1 M / 3.7 x 10-1 molesdm-3

(iv) 7.1 g of anhydrous sodium sulphate(VI)was dissolved in 20.0 cm3 solution. Calculate the molarity of the solution.

Method 1

                    20.0cm3 solution ->7.1 g

1000cm3 solution ->    1000    x   71   =   3550 g dm-3

20

 

Molar mass Na2SO4  = 142 g

 

Moles dm-3   =    Molarity   =   Mass          3550  = 2.5 M/ molesdm-3

Molar mass    142

 

Method 2

Molar mass Na2SO4  = 142 g

Moles = Mass          =>       7.1         =     0.05 / 5.0 x 10 -2   moles

                                 Molar mass              142

 Method 2(a)

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 5.0 x 10 -2   moles  x 1000cm3

20cm3

= 2.5 M/2.5 molesdm-3

Method 2(b)

20 cm3 solution contain 5.0 x 10 -2 moles

1000cm3 solution = Molarity  contain 1000  x  5.0 x 10 -2 moles 

20 cm3

= 2.5 M/2.5 molesdm-3

 

(iv) The density of sulphuric(VI) is 1.84gcm-3 Calculate the molarity of the acid.

Method 1

                    1.0cm3 solution ->1.84 g

1000cm3 solution ->    1000    x   1.84   =   1840 g dm-3

1

 

Molar mass H2SO4  = 98 g

 

Moles dm-3  = Molarity = Mass                          = 1840

Molar mass                     98

 

= 18.7755 M/ molesdm-3

Method 2

Molar mass H2SO4  = 98 g

 

Moles = Mass          =>       1.84         =   0.0188 / 1.88 x 10 -2   moles

                                 Molar mass               98

 

 Method 2(a)

Moles in decimeters = Molarity = Moles  x  1000cm3/1dm3

Volume of solution

=> 1.88 x 10 -2   moles  x 1000cm3

1.0cm3

= 18.8M/18.8 molesdm-3

Method 2(b)

20 cm3 solution contain 1.88 x 10 -2   moles

1000cm3 solution = Molarity  contain 1000  x  1.88 x 10 -2   moles 

1.0 cm3

= 18.8M/18.8 molesdm-3

 

  1. Calculate the mass of :

(i) 25 cm3 of 0.2M sodium hydroxide solution(Na =23.0.O =16.0, H=1.0)

 

     Molar mass NaOH = 40g

Moles  in 25 cm3  =  Molarity x volume   => 0.2  x   25 = 0.005/5.0 x 10-3moles

                                                1000                      1000

 

Mass of NaOH =Moles x molar mass = 5.0 x 10-3   x   40 = 0.2 g

 

(ii) 20 cm3 of 0.625 M sulphuric(VI)acid (S =32.0.O =16.0, H=1.0)

 

Molar mass H2SO4 = 98g

 

Moles  in 20 cm3 = Molarity x volume=> 0.625 x 20 = 0.0125/1.25.0 x 10-3moles

                                                1000              1000

 

Mass of H2SO4 =Moles x molar mass => 5.0 x 10-3   x   40 = 0.2 g

 

(iii) 1.0 cm3 of 2.5 M Nitric(V)acid (N =14.0.O =16.0, H=1.0)

 

Molar mass HNO3 = 63 g

 

Moles  in 1 cm3 = Molarity x volume => 2.5 x 1 = 0.0025 / 2.5. x 10-3moles

                                         1000                  1000

 

Mass of HNO3 =Moles x molar mass => 2.5 x 10-3   x   40 = 0.1 g

 

  1. Calculate the volume required to dissolve :

(a)(i) 0.25moles of  sodium hydroxide solution to form a 0.8M solution

Volume (in cm3)   =   moles x 1000 =>  0.25  x 1000       = 312.5cm3

                                                                 Molarity                0.8

 

     (ii) 100cm3 was added to the sodium hydroxide solution above. Calculate the concentration of the solution.

 

          C1 x V1 = C2 x V2  where:

C1   =   molarity/concentration before diluting/adding water

C=   molarity/concentration after diluting/adding water

V=   volume before diluting/adding water

                          V=   volume after diluting/adding water

 

=> 0.8M  x 312.5cm3   =  C2    x  (312.5 + 100)

 

                    C= 0.8M  x 312.5cm3 = 0.6061M

412.5

(b)(ii) 0.01M solution  containing 0.01moles of  sodium hydroxide solution .

Volume (in cm3)   =   moles x 1000 =>  0.01  x 1000       = 1000 cm3

                                                                 Molarity                0.01

 

     (ii) Determine the quantity of water which must be added to the sodium hydroxide solution above to form a 0.008M solution.

          C1 x V1 = C2 x V2  where:

C1   =   molarity/concentration before diluting/adding water

C=   molarity/concentration after diluting/adding water

V=   volume before diluting/adding water

                          V=   volume after diluting/adding water

 

=> 0.01M  x 1000 cm3   =  0.008    x V2

 

                    V= 0.01M  x 1000cm3 = 1250cm3

0.008

Volume added  = 1250  – 1000  =   250cm3

 

(c)Volumetric analysis/Titration

 

Volumetric analysis/Titration is the process of determining unknown concentration of one reactant from a known concentration and volume of another.

Reactions take place in simple mole ratio of reactants and products.

Knowing the concentration/ volume of one reactant, the other can be determined from the relationship:

 M1V1  = M2V2   where:

              n1            n2

 

M1  = Molarity of 1st reactant

M2  = Molarity of 2nd  reactant

V1   = Volume of 1st reactant

V1     = Volume of 2nd  reactant

n1     = number of moles of 1st reactant from stoichiometric equation

n2     = number of moles of 2nd reactant from stoichiometric equation

Examples

1.Calculate the molarity of MCO3 if 5.0cm3 of MCO3 react with 25.0cm3 of 0.5M hydrochloric acid.(C=12.0 ,O =16.0)

 

Stoichiometric equation: MCO3(s) + 2HCl(aq)  -> MCl2(aq) + CO2(g) + H2O(l)

Method 1

 

M1V1  = M2V2       ->       M1  x  5.0cm3  = 0.5M  x  25.0cm3

n1            n2                                         1                           2

=> M=         0.5  x  25.0 x1 =   1.25M / 1.25 moledm-3

5.0 x 2

Method 2

Moles of HCl used = molarity x volume  

1000

=> 0.5   x   25.0  =  0.0125 /1.25 x 10-2moles 

                                        1000

 

Mole ratio MCO3 : HCl  =  1:2

Moles MCO3 0.0125 /1.25 x 10-2moles = 0.00625 / 6.25 x 10-3 moles

2

 

Molarity MCO3 = moles x 1000   =>   0.00625 / 6.25 x 10-3   x  1000

Volume                                   5

 

= 1.25M / 1.25 moledm-3

  1. 2.0cm3 of 0.5M hydrochloric acid react with 0.1M of M2CO3. Calculate the volume of 0.1M M2CO3 used.

 

Stoichiometric equation: M2CO3 (aq) + 2HCl(aq)  -> 2MCl (aq) + CO2(g) + H2O(l)

Method 1

 

M1V1  = M2V2       ->       0.5  x  2.0cm3  = 0.1M  x  V2 cm3

n1            n2                                         2                           1

 

=> V=         0.5  x  2.0 x1 =   1.25M / 1.25 moledm-3

0.1 x 2

Method 2

Moles of HCl used = molarity x volume  

1000

=>  0.5     x     2.0  =        0.0125 /1.25 x 10-2moles 

                                          1000

 

Mole ratio M2CO3 : HCl  =  1:2

Moles M2CO3 =    0.0125 /1.25   x   10-2moles =    0.00625 / 6.25 x 10-3 moles

2

 

Molarity M2CO3 = moles x 1000   =>   0.00625 / 6.25 x 10-3   x  1000

Volume                                   5

 

= 1.25M / 1.25 moledm-3

 

  1. 5.0cm3 of 0.1M sodium iodide react with 0.1M of Lead(II)nitrate(V). Calculate(i) the volume of Lead(II)nitrate(V) used.

               (ii)the mass of Lead(II)Iodide formed

(Pb=207.0, I =127.0)

 

Stoichiometric equation:    2NaI(aq)  + Pb(NO3)2(aq) -> 2NaNO3(aq)  + PbI2(s)

 

(i)Volume of Lead(II)nitrate(V) used

 

Method 1

 

M1V1  = M2V2       ->        5 x  0.1cm3  =   0.1M  x  V2 cm3

n1            n2                                         2                           1

 

=> V=    0.1  x  5.0 x 1 =   1.25M / 1.25 moledm-3

0.1  x  2

 

Method 2

Moles of HCl used = molarity x volume  

1000

=>  0.1     x     5.0  =        0.0125 /1.25 x 10-2moles 

                                          1000

 

Mole ratio M2CO3 : HCl  =  1:2

Moles M2CO3 =    0.0125 /1.25   x   10-2moles =    0.00625 / 6.25 x 10-3 moles

2

 

Molarity M2CO3 = moles x 1000   =>   0.00625 / 6.25 x 10-3   x  1000

Volume                                   5

 

= 1.25M / 1.25 moledm-3

 

  1. 0.388g of a monobasic organic acid B required 46.5 cm3 of 0.095M sodium hydroxide for complete neutralization. Name and draw the structural formula of B

Moles of NaOH used = molarity x volume  

1000

=>  0.095     x   46.5 =     0.0044175 /4.4175 x 10-3moles 

                                          1000

 

Mole ratio B : NaOH  =  1:1

Moles B =    0.0044175 /4.4175 x 10-3moles

 

 

Molar mass B                    = mass         =>                0.388

moles                    0.0044175 /4.4175 x 10-3moles

 

= 87.8324 gmole-1

X-COOH = 87.8324 where X is an alkyl group

X =87.8324- 42 = 42.8324=43

By elimination:  CH3  = 15        CH3CH=  29       CH3CH2 CH2  = 43

Molecula formula : CH3CH2 CH2COOH

Molecule name :  Butan-1-oic acid

Molecular structure

H       H       H       O

 

H       C       C       C       C       O       H

 

H       H       H       H

 

  1. 10.5 g of an impure sample containing ammonium sulphate (VI) fertilizer was warmed with 250cm3 of o.8M sodium hydroxide solution.The excess of the alkali was neutralized by 85cm3 of 0.5M hydrochloric acid. Calculate the % of impurities in the ammonium sulphate (VI)fertilizer. (N=14.0,S=32.0,O=16.0, H=1.0)

 

Equation for neutralization

NaOH(aq)    +        HCl(aq)       ->       NaOH(aq)    +        H2O(l)

Mole ratio NaOH(aq):HCl(aq)= 1:1

Moles of HCl = Molarity x volume   =>   0.5  x  85  =     0.0425 moles

1000                                1000

Excess moles of NaOH(aq)= 0.0425 moles

 

Equation for reaction with ammonium salt

2NaOH(aq)  +        (NH4) 2SO4(aq)      -> Na 2SO4(aq) + 2NH3 (g)+     2H2O(l)

Mole ratio NaOH(aq): (NH4) 2SO4(aq)= 2:1

Total moles of  NaOH = Molarity x volume   =>   0.8  x  250  =     0.2 moles

1000                        1000

 

Moles of NaOH that reacted with(NH4) 2SO4  =  0.2  –  0.0425  =  0.1575moles

Moles (NH4) 2SO4   =   ½   x  0.1575moles   =  0. 07875moles

Molar mass (NH4) 2SO4= 132 gmole-1

 

Mass of in impure sample = moles  x  molar mass  =>0. 07875 x 132 = 10.395 g

Mass of impurities =        10.5 -10.395     =   0.105 g

% impurities           =         0.105   x  100  =    1.0 %

10.5

 

 

 

Practically volumetric analysis involves titration.

Titration generally involves filling a burette with known/unknown concentration of a solution then adding the solution to unknown/known concentration of another solution in a conical flask until there is complete reaction. If the solutions used are both colourless, an indicator is added to the conical flask. When the reaction is over, a slight/little excess of burette contents change the colour of the indicator. This is called the end point.

Set up of titration apparatus

 

The titration process involve involves determination of titre. The titre is the volume of burette contents/reading before and after the end point. Burette contents/reading before titration is usually called the Initial burette reading. Burette contents/reading after titration is usually called the Final burette reading. The titre value is thus a sum of the Final less Initial burette readings.

To reduce errors, titration process should be repeated at least once more.

The results of titration are recorded in a titration table as below

 

Sample titration table

 

Titration number       1        2       3
Final burette reading (cm3)     20.0      20.0     20.0
Initial burette reading (cm3)       0.0       0.0      0.0
Volume of solution used(cm3)      20.0      20.0     20.0

 

As evidence of a titration actually done examining body requires the candidate to record their burette readings before and after the titration.

For KCSE candidates burette readings must be recorded in a titration table in the format provided by the Kenya National Examination Council.

As evidence of all titration actually done Kenya National Examination Council require the candidate to record their burette readings before and after the titration to complete the titration table in the format provided.

 

Calculate the average volume of solution used

24.0 + 24.0 + 24.0   =  24.0 cm3

3

As evidence of understanding the degree of accuracy of burettes , all readings must be recorded to a decimal point.

As evidence of accuracy in carrying the out the titration , candidates value should be within 0.2 of the school value .

The school value is the teachers readings presented to the examining body/council based on the concentrations of the solutions s/he presented to her/his candidates.

Bonus mark is awarded for averaged reading within 0.1 school value as Final answer.

 

Calculations involved after the titration require candidates thorough practical and theoretical practice mastery on the:

 

(i)relationship among the mole, molar mass, mole ratios, concentration, molarity.

(ii) mathematical application of 1st principles.

Very useful information which candidates forget appears usually in the beginning of the question paper as:

You are provided with…

 

All calculation must be to the 4th decimal point unless they divide fully to a lesser decimal point.

Candidates are expected to use a non programmable scientific calculator.

 

(a)Sample Titration Practice 1 (Simple Titration)

 

You are provided with:

0.1M sodium hydroxide solution A

Hydrochloric acid solution B

 

You are required to determine the concentration of solution B in moles per litre.

 

Procedure

Fill the burette with solution B. Pipette 25.0cm3 of solution A into a conical flask. Titrate solution A with solution B using phenolphthalein indicator to complete the titration table 1

 

Sample results Titration table 1

Titration number       1        2       3
Final burette reading (cm3)     20.0      20.0     20.0
Initial burette reading (cm3)       0.0       0.0      0.0
Volume of solution B used(cm3)      20.0      20.0     20.0

 

Sample worked questions

 

  1. Calculate the average volume of solution B used

          Average titre = Titre 1 + Titre 2 +Titre 3  => ( 20.0 +20.0 +20.0 ) = 20.0cm3

3                                      3

  1. How many moles of:

(i)solution A were present in 25cm3 solution.

 

Moles of solution A =   Molarity x volume  =  0.1 x 252.5 x 10-3 moles

1000                    1000

 

(ii)solution B were present in the average volume.

Chemical equation:  NaOH(aq) +  HCl(aq)  ->  NaCl(aq) +  H2O(l)

 

Mole ratio 1:1   =>  Moles of A = Moles of B =  2.5 x 10-3 moles

 

(iii) solution B in moles per litre.

Moles of B per litre =    moles x 1000    = 2.5 x 10-3 x 1000  =  0.1M

Volume                      20

 

 

(b)Sample Titration Practice 2 (Redox Titration)

 

You are provided with:

Acidified Potassium manganate(VII) solution A

0.1M of an iron (II)salt  solution B

8.5g of ammonium iron(II)sulphate(VI) crystals(NH4)2 SO4FeSO4.xH2O   solid C

You are required to

(i)standardize acidified potassium manganate(VII)

(ii)determine the value of x in the formula (NH4)2 SO4FeSO4.xH2O.

 

Procedure 1

Fill the burette with solution A. Pipette 25.0cm3 of solution B into a conical flask. Titrate solution A with solution B until a pink colour just appears.

Record your results to complete  table 1.

 

Table 1:Sample results

 

Titration number       1        2       3
Final burette reading (cm3)     20.0      20.0     20.0
Initial burette reading (cm3)       0.0       0.0      0.0
Volume of solution A used(cm3)      20.0      20.0     20.0

 

Sample worked questions

  1. Calculate the average volume of solution A used

          Average titre = Titre 1 + Titre 2 +Titre 3  => ( 20.0 +20.0 +20.0 ) = 20.0cm3

3                                         3

  1. How many moles of:

(i)solution B were present in 25cm3 solution.

 

Moles of solution A =   Molarity x volume  =  0.1 x 252.5 x 10-3 moles

1000                    1000

(ii)solution A were present in the average volume. Assume one mole of B react with five moles of B

 

Mole ratio A : B = 1:5

=>  Moles of A = Moles of B =      2.5 x 10-3 moles     =    5.0 x 10 -4 moles

5                          5

 

(iii) solution B in moles per litre.

Moles of B per litre =    moles x 1000    = 2.5 x 10-3 x 1000

Volume                      20

 0.025 M /moles per litre /moles l-1

Procedure 2

Place all the solid C into the 250cm3 volumetric flask carefully. Add about 200cm3 of distilled water. Shake to dissolve. Make up to the 250cm3 of solution by adding more distilled water. Label this solution C. Pipette 25cm3 of solution C into a conical flask, Titrate solution C with solution A until a permanent pink colour just appears. Complete table 2.

 

Table 2:Sample results

Titration number       1        2       3
Final burette reading (cm3)     20.0      20.0     20.0
Initial burette reading (cm3)       0.0       0.0      0.0
Volume of solution A used(cm3)      20.0      20.0     20.0

 

Sample worked questions

 

  1. Calculate the average volume of solution A used

          Average titre = Titre 1 + Titre 2 +Titre 3  => ( 20.0 +20.0 +20.0 ) = 20.0cm3

3                                          3

  1. How many moles of:

(i)solution A were present inin the average titre.

 

Moles of solution A =   Molarity x volume  =  0.025 x 205.0 x 10-4 moles

1000                     1000

 

(ii)solution C in 25cm3 solution given the equation for the reaction:

MnO4(aq) +  8H+(aq) +  5Fe2+ (aq)  ->   Mn2+(aq)  +  5Fe3+(aq) + 4H2O(l)

 

Mole ratio MnO4(aq): 5Fe2+ (aq) =  1:5     =>                                              Moles of 5Fe2+ (aq) = Moles of MnO4(aq) = 5.0 x 10-4 moles = 1.0 x 10 -4 moles

5                            5

 

(iii) solution B in 250cm3.

Moles of B per litre =  moles x 250   = 1.0 x 10 -4  x 250  = 1.0 x 10 -3 moles                                                 Volume                       25

 

  1. Calculate the molar mass of solid C and hence the value of x in the chemical formula (NH4)2SO4FeSO4.xH2O.

(N=14.0, S=32.0, Fe=56.0, H=1.0 O=16.0)

 

Molar mass =  mass perlitre  =             8.5                 =      8500 g

Moles per litre           1.0 x 10 -3 moles

NH4)2SO4FeSO4.xH2O = 8500

284 + 18x =8500

8500  – 284 =    8216      =        18x      =   454.4444

18                18

x =  454  (whole number)

 

(c)Sample Titration Practice 3 (Back  titration)

 

You are provided with:

(i)an impure calcium carbonate labeled M

(ii)Hydrochloric acid labeled solution N

(iii)solution L containing 20g per litre sodium hydroxide.

You are required to determine the concentration of N in moles per litre and the % of calcium carbonate in mixture M.

 

Procedure 1

Pipette 25.0cm3 of solution L into a conical flask. Add 2-3 drops of phenolphthalein indicator. Titrate with dilute hydrochloric acid solution N and record your results in table 1(4mark)

Sample Table 1

 

       1        2        3
Final burette reading (cm3)      6.5      6.5       6.5
Initial burette reading (cm3)      0.0      0.0        0.0
Volume of N used (cm3)      6.5      6.5        6.5

 

Sample questions

(a) Calculate the average volume of solution N used    

 6.5  +  6.5  +  6.5    =     6.5 cm3

3

 

(b) How many moles of sodium hydroxide are contained in 25cm3of solution L

Molar mass NaOH =40g

Molarity of L   =    mass per litre             =>       20     =   0.5M

Molar mass NaOH                 40

Moles NaOH in 25cm3 = molarity  x volume   =>  0.5M  x 25cm3 =  0.0125 moles

1000                             1000

 

(c)Calculate:

(i)the number of moles of hydrochloric acidthat react with sodium hydroxide in (b)above.

   Mole ratio NaOH : HCl from stoichiometric equation= 1:1

Moles HCl =Moles NaOH => 0.0125 moles

 

(ii)the molarity of hydrochloric acid solution N.

Molarity =   moles x 1000   => 0.0125 moles x 1000 =1.9231M/moledm-3

6.5                             6.5

 

Procedure 2

Place the 4.0 g of M provided into a conical flask and add 25.0cm3 of the dilute hydrochloric acid to it using a clean pipette. Swirl the contents of the flask vigorously until effervescence stop.Using a 100ml measuring cylinder add 175cm3 distilled waterto make up the solution up to 200cm3.Label this solution K.Using a clean pipettetransfer 25.0cm3 of the solution into a clean conical flask and titrate with solution L from the burette using 2-3 drops of methyl orange indicator.Record your observations in table 2.

 

Sample Table 2

 

       1        2        3
Final burette reading (cm3)      24.5      24.5       24.5
Initial burette reading (cm3)      0.0      0.0        0.0
Volume of N used (cm3)      24.5      24.5        24.5

 

Sample calculations

(a)Calculate the average volume of solution L used(1mk)

24.5  +  24.5  +  24.5  = 24.5cm3

3

(b)How many moles of sodium hydroxide are present in the average volume of solution L used?

 

Moles = molarity x average burette volume => 0.5 x 24.5

1000                                          1000

 

= 0.01225 /1.225 x 10-2 moles

 

(c) How many moles of hydrochloric acid are present in the original 200cm3 of solution K?

 

Mole ratio NaOH: HCl = 1:1 => moles of HCl = 0.01225 /1.225 x 10-2 moles

 

Moles in 200cm3 = 200cm3 x 0.01225 /1.225 x 10-2moles

25cm3(volume pipetted)

 

=0.49 /4.9 x 10-1moles

 

(d)How many moles of hydrochloric acid were contained in original 25 cm3 solution N used

 

Original moles = Original molarity x pipetted volume  =>

1000cm3

 

1.9231M/moledm-3 x   25    =   0.04807/4.807 x 10-2 moles

1000

 

(e)How many moles of hydrochloric acid were used to react with calcium carbonate present?

Moles that reacted = original moles –moles in average titre  =>

 

0.04807/4.807 x 10-2moles  –  0.01225 /1.225 x 10-2moles

 

= 0.03582/3.582 x 10 -2 moles

 

(f)Write the equation for the reaction between calcium carbonate and hydrochloric acid.

CaCO3(s)  + 2HCl(aq)   ->   CaCl2(aq)  +   CO2(g)  +   H2O(l)

 

(g)Calculate the number of moles of calcium carbonate that reacted with hydrochloric acid.

From the equation CaCO3(s):2HCl(aq) = 1:2

 

=> Moles CaCO3(s) = 1/2moles HCl

= 1/2 x 0.03582/3.582 x 10 -2 moles

= 0.01791 /1.791 x 10-2moles

 

(h)Calculate the mass of calcium carbonate in 4.0g of mixture M (Ca=40.0,O = 16.0,C=12.0)

 

Molar mass CaCO3 = 100g

Mass CaCO3 = moles x molar mass   =>  0.01791 /1.791 x 10-2moles x 100g

= 1.791g

 

(i)Determine the % of calcium carbonate present in the mixture

 

  % CaCO3 = mass of pure x 100%      =>     1.791g x 100%    =   44.775%

Mass of impure                                 4.0

 

 

(d)Sample titration practice 4 (Multiple titration)

 

You are provided with:

(i)sodium L containing 5.0g per litre of a dibasic organic acid H2X.2H2O.

(ii)solution M which is acidified potassium manganate(VII)

(iii)solution N a mixture of sodium ethanedioate and ethanedioic acid

(iv)0.1M sodium hydroxide solution P

(v)1.0M sulphuric(VI)

You are required to:

(i)standardize solution M using solution L

(ii)use standardized solution M and solution P to determine the % of sodium   ethanedioate in the mixture.

 

Procedure 1

Fill the burette with solution M. Pipette 25.0cm3 of solution L into a conical flask. Heat this solution to about 70oC(but not to boil).Titrate the hot solution L with solution M until a permanent pink colour just appears .Shake thoroughly during the titration. Repeat this procedure to complete table 1.

 

Sample Table 1

 

       1        2        3
Final burette reading (cm3)      24.0      24.0       24.0
Initial burette reading (cm3)      0.0      0.0        0.0
Volume of N used (cm3)      24.0      24.0        24.0

 

Sample calculations

(a)Calculate the average volume of solution L used (1mk)

 

24.0  +  24.0  +  24.0  = 24.0cm3

3

(b)Given that the concentration of the dibasic acid is 0.05molesdm-3.determine the value of x in the formula H2X.2H2O (H=1.0,O=16.0)

 

 Molar mass H2X.2H2O= mass per litre    =>       5.0g/litre           = 100g

Moles/litre                  0.05molesdm-3

 

H2X.2H2O =100

X = 100 – ((2 x1) + 2 x (2 x1)  + (2 x 16) => 100  –  34  =  66

 

  (c) Calculate the number of moles of the dibasic acid H2X.2H2O.

 

Moles = molarity x pipette volume =>    0.5 x 25      =   0.0125/1.25 x10 -2 moles

                          1000                                  1000

 

   (d)Given the mole ratio manganate(VII)(MnO4): acid H2X is 2:5, calculate the number of moles of manganate(VII) (MnO4) in the average titre.

 

Moles H2X  =  2/5 moles of MnO4–  

=>  2/5  x 0.0125/1.25 x10 -2 moles

= 0.005/5.0 x 10 -3moles

 

(e)Calculate the concentration of the manganate(VII)(MnO4) in moles per litre.

      Moles per litre/molarity  =  moles   x   1000

average burette volume

=>0.005/5.0 x 10 -3moles x 1000  =  0.2083 molesl-1/M

24.0

 

Procedure 2

With solution M still in the burette ,pipette 25.0cm3 of solution N into a conical flask. Heat the conical flask containing solution N to about 70oC.Titrate while hot with solution M.Repeat the experiment to complete table 2.

Sample Table 2

 

       1        2        3
Final burette reading (cm3)      12.5      12.5       12.5
Initial burette reading (cm3)      0.0      0.0        0.0
Volume of N used (cm3)      12.5      12.5        12.5

 

Sample calculations

(a)Calculate the average volume of solution L used (1mk)

 

12.5  +  12.5  +  12.5  =12.5cm3

3

(b)Calculations:

(i)How many moles of manganate(VII)ions are contained in the average volume of solution M used?

 

Moles = molarity of solution M x average burette volume

1000

=>    0.2083 molesl-1/ M  x  12.5    =   0.0026 / 2.5 x 10-3 moles

1000

 

(ii)The reaction between manganate(VII)ions and ethanedioate ions that reacted with is as in the equation:

 

2MnO4(aq) +  5C2O42- (aq) + 16H+ (aq) -> 2Mn2+(aq) + 10CO2(g) + 8H2O(l)

 

Calculate the number of moles of ethanedioate ions that reacted with manganate (VII) ions in the average volume of solution M.

 

From the stoichiometric equation,mole ratio MnO4(aq): C2O42- (aq) = 2:5

 

=> moles C2O42-  =  5/2 moles MnO4 => 5/2 x 0.0026 / 2.5 x 10-3 moles

 

= 0.0065 /6.5 x10-3 moles

 

          (iii)Calculate the number of moles of ethanedioate ions contained in 250cm3 solution N.

 

25cm3 pipette volume    ->   0.0065 /6.5 x10-3 moles

250cm3                          ->

 

0.0065 /6.5 x10-3 moles  x  250     =    0.065 / 6.5 x10-2 moles

25

 

 

Procedure 3

Remove solution M from the burette and rinse it with distilled water. Fill the burette with sodium hydroxide solution P. Pipette 25cm3 of solution N into a conical flask and add 2-3 drops of phenolphthalein indicator. Titrate this solution N with solution P from the burette. Repeat the procedure to complete table 3.

Sample Table 2

 

       1        2        3
Final burette reading (cm3)      24.9      24.9       24.9
Initial burette reading (cm3)      0.0      0.0        0.0
Volume of N used (cm3)      24.9      24.9        24.9

 

Sample calculations

(a)Calculate the average volume of solution L used (1mk)

 

24.9  +  24.9  +  24.9  =  24.9 cm3

                                 3

(b)Calculations:

(i)How many moles of sodium hydroxide solution P were contained in the average volume?

 

Moles = molarity of solution P  x average burette volume

1000

=>    0.1 molesl-1   x  24.9    =   0.00249 / 2.49 x 10-3 moles

1000

 

(ii)Given that NaOH solution P reacted with the ethanedioate ions from the acid only and the equation for the reaction is:

 

2NaOH (aq)  +   H2C2O4 (aq)    ->   Na2C2O4(g)   +   2H2O(l)

 

Calculate the number of moles of ethanedioic acid that were used in the reaction

 

From the stoichiometric equation,mole ratio NaOH(aq): H2C2O4 (aq) = 2:1

 

=> moles H2C2O4  =  1/2 moles NaOH => 1/2 x  0.00249 / 2.49 x 10-3 moles

 

= 0.001245/1.245 x10-3 moles.

 

(iii)How many moles of ethanedioic acid were contained in 250cm3 of solution N?

 

25cm3 pipette volume    ->  0.001245/1.245 x10-3 moles

250cm3                          ->

0.001245/1.245 x10-3 moles x  250     =   0.01245/1.245 x10-2 moles

25

(iii)Determine the % by mass of sodium ethanedioate in the micture             (H= 1.0,O=16.0,C=12.0 and total mass of mixture =2.0 g in 250cm3 solution)

 

Molar mass H2C2O4 = 90.0g

Mass of H2C2O4 in 250cm3 = moles in 250cm3 x molar mass H2C2O4

=>0.01245/1.245 x10-2 moles x 90.0

= 1.1205g

 

% by mass of sodium ethanedioate

=(Mass of mixture   –  mass of H2C2O4) x 100%

Mass of mixture

 

=>   2.0  – 1.1205 g   =  43.975%

2.0

 

Note

 (i) L is 0.05M Oxalic acid

(ii) M is 0.01M KMnO4

(iii) N is 0.03M oxalic acid(without sodium oxalate)

 

Practice example 5.(Determining equation for a reaction)

You are provided with

-0.1M hydrochloric acid solution A

-0.5M sodium hydroxide solution B

You are to determine the equation for thereaction between solution A and B

Procedure

Fill the burette  with solution A.Using a pipette and pipette filler transfer 25.0cm3 of solution B into a conical flask.Add 2-3 drops of phenolphthalein indicator.Run solution A into solution B until a permanent pink colour just appears.Record your results in Table 1.Repeat the experiment to obtain three concordant results to complete Table 1

Table 1(Sample results)

Titration 1 2 3
Final  volume(cm3) 12.5 25.0 37.5
Initial volume(cm3) 0.0 12.5 25.0
Volume of solution A used(cm3) 12.5 12.5 12.5

Sample questions

Calculate the average volume of solution A used.

12.5+12.5+12.5  =       12.5cm3

                           3

 

 

 

 

 

 

 

Theoretical Practice examples

 

  1. 1.0g of dibasic acid HOOC(CH2)xCOOH was dissolved in 250cm3 solution. 25.0 cm3 of this solution reacted with 30.0cm3 of 0.06M sodium hydroxide solution. Calculate the value of x in HOOC(CH2)xCOOH. (C=12.0,H=1.0,O=16.)

 

Chemical equation

2NaOH(aq) + H2X(aq) -> Na2X (aq) + 2H2O(aq)

Mole ratio NaOH(aq) :H2X(aq) = 2:1

 

Method 1

Ma Va = na          => Ma  x  25.0     = 1  =>  Ma =0.06  x  30.0 x1

Mb Vb = nb                0.06  x  30.0      2                 25.0 x 2

 

Molarity of acid = 0.036M/Mole l-1

 

Mass of acid per lite =   1.0 x1000       =       4.0 g/l

                           250

0.036M/  Mole l-1     ->  4.0 g /l

1 mole= molar mass of HOOC(CH2)xCOOH =       4.0 x 1         = 111.1111 g

0.036

 

Molar mass (CH2)x =  111.1111 – (HOOCCOOH  = 90.0) = 21.1111

 

(CH2)x = 14x  =  21.1111  =  1.5  = 1 (whole number)

14

 

Method 2

Moles of sodium hydroxide =    Molarity x volume  =  0.06 x 30 = 1.8 x 10 -3moles

1000

Moles of Hydrochloric acid = 1/2  x  1.8 x 10 -3moles = 9.0 x10 -4moles

Molarity of Hydrochloric acid = moles x 1000 = 9.0 x10 -4moles x1000

Volume                       25

Molarity of acid = 0.036M/Mole l-1

 

Mass of acid per lite =   1.0 x1000       =       4.0 g/l

                           250

0.036M/  Mole l-1     ->  4.0 g /l

1 mole= molar mass of HOOC(CH2)xCOOH =       4.0 x 1         = 111.1111 g

0.036

 

Molar mass (CH2)x =  111.1111 – (HOOCCOOH  = 90.0) = 21.1111

 

(CH2)x = 14x  =  21.1111  =  1.5  = 1 (whole number)

14

 

  1. 20.0cm3 of 0.05 M acidified potassium manganate(VII)solution oxidized 25.0cm3 of Fe2+(aq) ions in 40.0g/l of impure Iron (II)sulphate(VI) to Fe3+(aq) ions. Calculate the percentage  impurities in the  Iron (II)sulphate(VI).

 

MnO4 (aq) + 8H+(aq)+ 5Fe2+(aq)-> 5Fe3+(aq)  + Mn2+(aq) + 4H2O(aq)

Fe=56.0,S= 32.0, O=16.0).

 

Moles of MnO4 (aq)   = Molarity  x  volume =   0.05  x  20.0     = 0.001 Moles

1000                       1000

 

Mole ratio MnO4 (aq): 5Fe2+(aq)= 1:5

Moles 5Fe2+(aq)   =   5 x0.001   =   0.005 Moles

 

Moles of 5Fe2+(aq) per litre/molarity = Moles x 1000   =     0005 x  1000

Volume                         25.0

= 0.2 M/ Moles/litre

 

Molar mass =FeSO4=152 g

Mass of  in the mixture  =  Moles x molar mass  => 0.2 x 152  = 30.4 g

 

Mass of impurity = 40.0 – 30.4  =9.6 g

 

% impurity     =      9.6 g x100 = 24.0 % impurity

40.0

3.9.7 g of a mixture of Potassium hydroxide and Potassium chloride was dissolved to make one litre solution.20.0cm3 of this solution required 25.0cm3 of 0.12M hydrochloric acid for completed neutralization. Calculate the percentage by mass of Potassium chloride.(K=39.0,Cl= 35.5)

 

Chemical equation

KOH(aq) +  HCl(aq) ->  KCl(aq) +  H2O(l)

Moles of HCl = Molarity  x  volume => 0.12 x 25.0   = 0.003/3.0 x 10 -3 moles

1000                          1000

Mole ratio KOH(aq) : HCl(aq) -= 1:1

Moles KOH =0.003/3.0 x 10 -3 moles

Method 1

 

Molar mass KOH  =56.0g

Mass KOH in 25cm3 =0.003/3.0 x 10 -3 moles x56.0 = 0.168g

Mass KOH in 1000cm3/1 litre = 0.168 x1000= 8.4 g/l

                                                            20

Mass of KCl =  9.7g  – 8.4g   =  1.3 g

% of KCl =      1.3 x 100       =   13.4021%

                            9.7

 

Method 2

Moles KOH in 1000cm3 /1 litre = Moles in 20cm3 x 1000 =>0.003 x 1000

20                                   20

=0.15M/Moles /litre

Molar mass KOH  =56.0g

Mass KOH in 1000/1 litre = 0.15M/Moles /litre  x 56.0  = 8.4g/l

 

Mass of KCl =  9.7g  – 8.4g   =  1.3 g

% of KCl =      1.3 x 100       =   13.4021%

                            9.7

 

4.A certain carbonate, GCO3, reacts with dilute hydrochloric acid according to the equation given below:

GCO3(s) + 2HCl(aq)              ->      GCl2 (aq)  +   CO2 (g)  + H2O(l)

 

If 1 g of the carbonate reacts completely with 20 cm3 of 1 M hydrochloric acid ,calculate the relative atomic mass of  G (C = 12.0 = 16.0)

 

Moles of HCl = Molarity x volume=> 1 x20  = 0.02 moles

                                        1000                  1000

Mole ratio HCl; GCO3  = 2:1

Moles of GCO3= 0.02 moles = 0.01moles

                                     2

Molar mass of GCO3    =   mass   =>      1         =   100 g

moles       0.01moles

G= GCO3 – CO3 =>100g – (12+ 16 x3 = 60) = 40(no units)

 

 

  1. 46.0g of a metal carbonate MCO3 was dissolved 160cm3 of 0.1M excess hydrochloric acid and the resultant solution diluted to one litre.25.0cm3 of this solution required 20.0cm3 of 0.1M sodium hydroxide solution for complete neutralization. Calculate the atomic mass of ‘M’

Equation

Chemical equation

NaOH(aq) +  HCl(aq) ->  KCl(aq) +  H2O(l)

Moles of NaOH = Molarity x volume=> 0.1 x20  = 0.002 moles

                                        1000                  1000

Mole ratio HCl; NaOH  = 1:1

Excess moles of HCl     =  0.002 moles

 

 

25cm3        ->       0.002 moles

1000cm3     ->          1000 x 0.002      =    0.08moles       

                                                   25cm3

Original moles of HCl   = Molarity x volume  => 1M  x 1litre = 1.0 moles

           

                                              

Moles of HCl reacted with MCO3   =  1.0 – 0.08 moles  =  0.92moles

Chemical equation

MCO3(s) + 2HCl(aq)              ->      MCl2 (aq)  +   CO2 (g)  + H2O(l)

Mole ratio MCO3(s) : HCl(aq) =1:2

                                          

Moles of MCO3    =   0.92moles   => 0.46moles

2

Molar mass of MCO3=       mass     =>              46g           =  100 g

moles                  0.46moles

M= MCO3 – CO3 =>100g – (12+ 16 x3 = 60) = 40

 

 

  1. 25.0cm3 of a mixture of Fe2+ and Fe3+ ions in an aqueous salt was acidified with sulphuric(VI)acid then titrated against potassium manganate(VI).The salt required 15cm3 ofe0.02M potassium manganate(VI) for complete reaction.

A second 25cm3 portion of the Fe2+ and Fe3+ ion salt was reduced by Zinc then titrated against the same concentration of potassium manganate(VI).19.0cm3 of potassium manganate(VI)solution was used for complete reaction. Calculate the concentration of Fe2+ and Fe3+ ion in the solution on moles per litre.

 

Mole ratio Fe2+ :Mn04 = 5:1

Moles Mn04used =        0.02 x 15                        =        3.0 x 10-4 moles

                                          1000

Moles Fe2+            =        3.0 x 10-4 moles              =        6.0 x 10-5 moles

5

Molarity of Fe2+    =        6.0 x 10-4 moles x 1000   =        2.4 x 10-3 moles l-1

                                                           25

Since Zinc reduces  Fe3+  to  Fe2+  in the mixture:

Moles Mn04 that reacted with all  Fe2+=    0.02 x 19       =         3.8 x 10-4 moles

                                                                      1000

Moles of  all  Fe2+                        =   3.8 x 10-4 moles              =        7.6 x 10-5 moles

5

Moles of   Fe3+                             =   3.8 x 10-4 – 6.0 x 10-5       =       1.6 x 10-5 moles

Molarity of Fe3+                   =   1.6 x 10-5 moles x 1000   =        4.0 x 10-4 moles l-1

                                                           25

 

 

 

 

 

 

 

 

 

 

 

 

 

 

     

 

UPGRADE

 

CHEMISTRY

 

                        FORM 3

                 

ORGANIC  Chemistry 1&2

                     

 

 

 

 

 

Comprehensive tutorial notes

 

 MUTHOMI S.G

www.kcselibrary.info

                                     0720096206

 

 

 

 

 

 

 

 

Introduction to Organic chemistry                          

 

Organic chemistry is the branch of chemistry that studies carbon compounds present in living things, once living things or synthetic/man-made.

Compounds that makes up living things whether alive or dead mainly contain carbon. Carbon is tetravalent.

It is able to form stable covalent bonds with itself and many non-metals like hydrogen, nitrogen ,oxygen and halogens to form a variety of compounds. This is because:

(i) carbon uses all the four valence  electrons to form four strong covalent bond.

(ii)carbon can covalently bond to form a single, double  or triple covalent bond with itself.

(iii)carbon atoms can covalently bond to form a very long chain or ring.

When carbon covalently bond with Hydrogen, it forms a group of organic compounds called Hydrocarbons

 

  A.HYDROCARBONS (HCs)

Hydrocarbons are a group of organic compounds containing /made up of hydrogen and carbon atoms only.

Depending on the type of bond that exist between the individual carbon atoms, hydrocarbon are classified as:

(i) Alkanes

(ii) Alkenes

(iii) Alkynes

 

(i) Alkanes

 

(a)Nomenclature/Naming

 

These are hydrocarbons with a general formula CnH2n+2 where n is the number of Carbon atoms in a molecule.

The carbon atoms are linked by single bond to each other and to hydrogen atoms.

 

They include:

 

 

 

 

n General/

Molecular

formula

Structural formula Name

 

1 CH4                   H

 

H       C        H

 

H

 

Methane
2 C2H6              H           H

 

H          C           C         H

 

H           H

 

Ethane

 

 

3 C3H8              H           H        H

 

H          C           C         C       H

 

H           H        H

 

Propane
4 C4H10         H      H     H    H

 

H      C      C    C     C    H

 

H      H    H    H

 

Butane
5 C5H12         H      H     H    H    H

 

H      C      C    C     C    C   H         CH3 (CH2) 6CH3   

 

H      H    H    H     H

 

Pentane
6 C6H14         H      H     H    H    H   H

 

H      C      C    C     C    C    C   H  CH3 (CH2) 6CH3   

 

H      H    H    H     H  H

 

Hexane
7 C7H16         H      H     H    H    H   H   H

 

H      C      C    C     C    C    C   C    H  

 

H      H    H    H     H  H    H

 

Heptane
8 C8H18         H      H     H    H    H   H   H    H

 

H      C      C    C     C    C    C   C    C   H

 

H      H    H    H     H  H    H   H

 

Octane
9 C9H20         H      H     H    H    H   H   H    H   H

 

H      C      C    C     C    C    C   C    C   C   H

 

H      H    H    H     H  H    H   H   H

 

Nonane
10 C10H22         H      H     H    H    H   H   H    H   H    H

 

H      C      C    C     C    C    C   C    C   C    C   H

 

H      H    H    H     H  H    H   H   H    H

 

decane

 

Note

1.The general formula/molecular formular of a compound shows the number of each atoms of elements making the compound e.g.

Decane has a general/molecular formula C10H22 ;this means there are 10 carbon atoms and 22 hydrogen atoms in a molecule of decane.

2.The structural formula shows the arrangement/bonding of atoms of each element making the compound e.g

Decane has the structural formula as in the table above ;this means the 1st carbon from left to right is bonded to three hydrogen atoms and one carbon atom.

The 2nd carbon atom is joined/bonded to two other carbon atoms and two Hydrogen atoms.

3.Since carbon is tetravalent ,each atom of carbon in the alkane MUST always be bonded using four covalent bond /four shared pairs of electrons.

4.Since Hydrogen is monovalent ,each atom of hydrogen in the alkane MUST always be bonded using one covalent bond/one shared pair of electrons.

5.One member of the alkane differ from the next/previous by a CH2  group.

e.g

Propane differ from ethane by one carbon and two Hydrogen atoms form ethane. Ethane differ from methane also by one carbon and two Hydrogen atoms

6.A group of compounds that differ by a CH2  group from the next /previous consecutively is called a homologous series.

7.A homologous series:

(i) differ by a CH2  group from the next /previous consecutively

(ii)have similar chemical properties

(iii)have similar chemical formula that can be represented by a general  formula e.g alkanes have the general formula CnH2n+2.

(iv)the physical properties (e.g.melting/boiling points)show steady gradual change)

8.The 1st four alkanes have the prefix meth_,eth_,prop_ and but_ to represent 1,2,3 and 4 carbons in the compound. All other use the numeral prefix pent_,Hex_,hept_ , etc to show also the number of carbon atoms.

9.If one hydrogen atom in an alkane is removed, an alkyl group is formed.e.g

 

Alkane name molecular structure

CnH2n+2

Alkyl name Molecula structure

CnH2n+1

methane CH4 methyl CH3
ethane CH3CH3 ethyl CH3 CH2
propane CH3 CH2 CH3 propyl CH3 CH2 CH2
butane CH3 CH2 CH2 CH3 butyl CH3 CH2 CH2 CH2

 

(b)Isomers of alkanes

Isomers are compounds with the same molecular general formula but different molecular structural formula.

Isomerism is the existence of a compounds having the same general/molecular formula but different structural formula.

The 1st  three alkanes do not form isomers.Isomers are named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.

The IUPAC system of nomenclature uses the following basic rules/guidelines:

 

1.Identify the longest continuous carbon chain to get/determine the parent alkane.

2.Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible

  1. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens

4.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of branches attached to the parent alkane.

 

Practice on IUPAC nomenclature  of alkanes

(a)Draw the structure of:

 

       (i)2-methylpentane

Procedure

  1. Identify the longest continuous carbon chain to get/determine the parent alkane.

Butane is the parent name         CH3 CH2 CH2 CH3 

 

  1. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible

The methyl group is attached to Carbon “2”

 

  1. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e

 

Position of the branch at carbon “2”

    Number of branches at carbon “1”

Type  of the branch “methyl” hence

Molecular formula

              CH3

 

      CH3 CH CH2 CH3    // CH3 CH (CH3 ) CH2CH3

  

Structural formula  

H      H     H    H

 

H      C      C     C     C    H

 

H             H    H

 

       H     C   H

 

                 H

 

(ii)2,2-dimethylpentane

Procedure

 

  1. Identify the longest continuous carbon chain to get/determine the parent alkane.

Butane is the parent name         CH3 CH2 CH2 CH3 

 

  1. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible

The methyl group is attached to Carbon “2”

 

  1. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e

Position of the branch at carbon “2”

    Number of branches at carbon “2”

Type  of the branch  two“methyl” hence

Molecular formular

    

               CH3

 

       CH3 C CH2 CH3    // CH3 C  (CH3 )2 CH2CH3

 

                       CH3

    Structural formula

H

 

H       C       H

 

H              H    H

 

H      C      C     C     C    H

 

H             H    H

 

       H     C   H

 

                 H

 

(iii) 2,2,3-trimethylbutane

 

Procedure

  1. Identify the longest continuous carbon chain to get/determine the parent alkane.

Butane is the parent name         CH3 CH2 CH2 CH3 

 

  1. Number the longest chain form the end of the chain that is near the branches so as the branch get the lowest number possible

The methyl group is attached to Carbon “2 and 3”

 

  1. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of carbon chains attached to the parent alkane i.e

Position of the branch at carbon “2 and 3”

    Number of branches at carbon “3”

Type  of the branch  three “methyl” hence

Molecular formular

    

                   CH3

 

       CH3     C     CH CH3    // CH3 C  (CH3 )3 CH2CH3

 

                             CH3 CH3

    Structural formula

H

 

H       C       H

 

H              H

 

H      C      C     C   H

 

H             H

                        H

 

 

       H      C    C    H

 

                       H

H      C       H

 

H

(iv) 1,1,1,2,2,2-hexabromoethane

Molecular formula

             

 

      CBr3 CBr3   

  

Structural formula  

Br           Br

 

Br      C           C      Br

 

Br           Br      

 

(v) 1,1,1-tetrachloro-2,2-dimethylbutane

                  CH3  

 

       CCl 3   C      CH3    //   C Cl 3 C  (CH3 )2 CH3

 

                             CH3

    Structural formula

Cl

 

Cl       C       Cl

 

H              H

 

H      C      C     C   H

 

H             H     

       H      C    H

 

                 H

 

(c)Occurrence and extraction

 

Crude oil ,natural gas and biogas are the main sources of alkanes:

(i)Natural gas is found on top of crude oil deposits and consists mainly of methane.

 

(ii)Biogas is formed from the decay of waste organic products like animal dung and cellulose. When the decay takes place in absence of oxygen , 60-75% by volume of the gaseous mixture of methane gas is produced.

 

(iii)Crude oil is a mixture of many flammable hydrocarbons/substances. Using fractional distillation, each hydrocarbon fraction can be separated from the other. The hydrocarbon with lower /smaller number of carbon atoms in the chain have lower boiling point and thus collected first.

 

As the carbon chain increase, the boiling point, viscosity (ease of flow) and colour intensity increase as flammability decrease. Hydrocarbons in crude oil are not pure. They thus have no sharp fixed boiling point.

 

Uses of different crude oil fractions

Carbon atoms in a molecule Common name of fraction Uses of fraction
1-4 Gas L.P.G gas for domestic use
5-12 Petrol Fuel for petrol engines
9-16 Kerosene/Paraffin Jet fuel and domestic lighting/cooking
15-18 Light diesel Heavy diesel engine fuel
18-25 Diesel oil Light diesel engine fuel
20-70 Lubricating oil Lubricating oil to reduce friction.
Over 70 Bitumen/Asphalt Tarmacking roads

 

(d)School laboratory preparation of alkanes

 

In a school laboratory, alkanes may be prepared from the reaction of a sodium alkanoate with solid sodium hydroxide/soda lime.

Chemical equation:

 

Sodium alkanoate            +        soda lime    ->    alkane          +     Sodium carbonate

CnH2n+1COONa(s)                     +        NaOH(s)     ->     C n H2n+2      +       Na2CO3(s)

 

The “H” in NaOH is transferred/moves to the CnH2n+1 in CnH2n+1COONa(s) to form C n H2n+2.

 

Examples

 

  1. Methane is prepared from the heating of a mixture of sodium ethanoate and soda lime and collecting over water

Sodium ethanoate           +        soda lime    ->    methane      +     Sodium carbonate

CH3COONa(s)                 +        NaOH(s)     ->     C H4    +       Na2CO3(s)

 

The “H” in NaOH is transferred/moves to the CH3 in CH3COONa(s) to form CH4.

 

  1. Ethane is prepared from the heating of a mixture of sodium propanoate and soda lime and collecting over water

 

Sodium propanoate         +        soda lime    ->    ethane        +     Sodium carbonate

CH3 CH2COONa(s)         +        NaOH(s)     ->    CH3 CH3    +       Na2CO3(s)

 

The “H” in NaOH is transferred/moves to the CH3 CH2 in CH3 CH2COONa (s) to form CH3 CH3

 

  1. Propane is prepared from the heating of a mixture of sodium butanoate and soda lime and collecting over water

 

Sodium butanoate           +        soda lime    ->    propane     +     Sodium carbonate

CH3 CH2CH2COONa(s) +        NaOH(s)     ->    CH3 CH2CH3  +  Na2CO3(s)

 

The “H” in NaOH is transferred/moves to the CH3 CH2 CH2 in CH3 CH2CH2COONa (s) to form CH3 CH2CH3

 

  1. Butane is prepared from the heating of a mixture of sodium pentanoate and soda lime and collecting over water

Sodium pentanoate         +        soda lime    ->    butane     +     Sodium carbonate

CH3 CH2 CH2CH2COONa(s)+NaOH(s)  ->    CH3 CH2CH2CH3  +  Na2CO3(s)

 

The “H” in NaOH is transferred/moves to the CH3CH2 CH2 CH2 in                      CH3 CH2CH2 CH2COONa (s) to form CH3 CH2 CH2CH3

 

Laboratory set up for the preparation of alkanes 

 

 

 

(d)Properties of alkanes

  1. Physical properties

 

Alkanes are colourless gases, solids and liquids that are not poisonous.

They are slightly soluble in water.

The solubility decrease as the carbon chain and thus the molar mass increase

The melting and boiling point increase as the carbon chain increase.

This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.

The 1st four straight chain alkanes (methane,ethane,propane and butane)are therefore gases ,the nect six(pentane ,hexane, heptane,octane,nonane, and decane) are liquids while the rest from unidecane(11 carbon atoms) are solids .

 

The density of straight chain alkanes increase with increasing carbon chain as the intermolecular forces increases.

This reduces the volume occupied by a given mass of the compound.

 

Summary of physical properties of alkanes

 

Alkane General formula Melting point(K) Boiling point(K) Density gcm-3 State at room(298K) temperature and pressure atmosphere (101300Pa)
Methane CH4 90 112 0.424 gas
Ethane CH3CH3 91 184 0.546 gas
Propane CH3CH2CH3 105 231 0.501 gas
Butane CH3(CH2)2CH3 138 275 0.579 gas
Pentane CH3(CH2)3CH3 143 309 0.626 liquid
Hexane CH3(CH2)4CH3 178 342 0.657 liquid
Heptane CH3(CH2)5CH3 182 372 0.684 liquid
Octane CH3(CH2)6CH3 216 399 0.703 liquid
Nonane CH3(CH2)7CH3 219 424 0.708 liquid
Octane CH3(CH2)8CH3 243 447 0.730 liquid

 

II.Chemical properties         

(i)Burning.

 

Alkanes burn with a blue/non-luminous non-sooty/non-smoky flame in excess air to form carbon(IV) oxide and water.

Alkane         +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

 

Alkanes burn with a blue/non-luminous no-sooty/non-smoky flame in limited air to form carbon(II) oxide and water.

Alkane         +        Air     ->       carbon(II) oxide    +        water  (limited air)

 

Examples

1.(a) Methane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.

Methane       +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

CH4(g)         +        2O2(g)         ->       CO2(g)                   +        2H2O(l/g)

 

(b) Methane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.

Methane       +        Air     ->       carbon(II) oxide    +        water  (excess air/oxygen)

2CH4(g)       +        3O2(g)         ->       2CO(g)                  +        4H2O(l/g)

 

2.(a) Ethane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.

Ethane         +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

2C2H6(g)     +        7O2(g)         ->       4CO2(g)                 +        6H2O(l/g)

 

(b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.

Ethane         +        Air     ->       carbon(II) oxide    +        water  (excess air/oxygen)

2C2H6(g)     +        5O2(g)         ->       4CO(g)                  +        6H2O(l/g)

 

3.(a) Propane when ignited burns with a blue non sooty flame in excess air to form carbon(IV) oxide and water.

Propane       +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

C3H8(g)      +        5O2(g)         ->       3CO2(g)                 +        4H2O(l/g)

 

(b) Ethane when ignited burns with a blue non sooty flame in limited air to form carbon(II) oxide and water.

Ethane         +        Air     ->       carbon(II) oxide    +        water  (excess air/oxygen)

2C3H8(g)     +        7O2(g)         ->       6CO(g)                  +        8H2O(l/g)

 

ii)Substitution

Substitution reaction is one in which a hydrogen atom is replaced by a halogen in presence of ultraviolet light.

Alkanes react with halogens in presence of ultraviolet light to form halogenoalkanes.

During substitution:

(i)the halogen molecule is split into free atom/radicals.

(ii)one free halogen radical/atoms knock /remove one hydrogen from the  alkane leaving an alkyl radical.

(iii) the alkyl radical combine with the other free halogen atom/radical to form halogenoalkane.

(iv)the chlorine atoms substitute repeatedly in the alkane. Each substitution removes a hydrogen atom from the alkane and form hydrogen halide.

(v)substitution stops when all the hydrogen in alkanes are replaced with  halogens.

 

Substitution reaction is a highly explosive reaction in presence of sunlight / ultraviolet light that act as catalyst.

 

Examples of substitution reactions

Methane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of chlorine and methane  explode to form  colourless mixture of chloromethane and hydrogen chloride gas. The pale green colour of chlorine gas fades.

Chemical equation

 

1.(a)Methane         + chlorine    ->       Chloromethane      +        Hydrogen chloride

 

CH4(g)         +        Cl2(g) ->       CH3Cl (g)              +          HCl (g)

 

H                                                           H

 

H       C        H   + Cl         Cl       ->       H       C        Cl      +  H          Cl

 

H                                                           H

 

(b) Chloromethane  + chlorine   ->       dichloromethane    +        Hydrogen chloride

 

CH3Cl (g)                    +        Cl2(g)      ->   CH2Cl2 (g)           +          HCl (g)

 

H                                                           H

 

H       C        Cl   + Cl        Cl       ->       H       C        Cl      +  H          Cl

 

H                                                           Cl

 

(c) dichloromethane  + chlorine           ->       trichloromethane    +          Hydrogen chloride

 

CH2Cl2 (g)         +       Cl2(g)      ->   CHCl3 (g)            +          HCl (g)

 

Cl                                                           H

 

H       C        Cl   + Cl        Cl       ->       Cl      C        Cl      +  H          Cl

 

H                                                           Cl

 

(c) trichloromethane  + chlorine ->      tetrachloromethane + Hydrogen chloride

 

CHCl3 (g)             +     Cl2(g)      ->   CCl4 (g)               +          HCl (g)

H                                                           Cl

 

Cl       C        Cl   + Cl        Cl       ->       Cl      C        Cl      +  H          Cl

 

Cl                                                          Cl

 

Ethane has no effect on bromine or chlorine in diffused light/dark. In sunlight , a mixture of bromine and ethane  explode to form  colourless mixture of bromoethane and hydrogen chloride gas. The red/brown  colour of bromine gas fades.

Chemical equation

 

(a)Ethane     + chlorine    ->       Chloroethane         +        Hydrogen chloride

 

CH3CH3(g)  +        Br2(g)          ->       CH3CH2Br (g)                 +          HBr (g)

 

H           H                                                  H             H

 

H          C           C         H  + Br       Br   ->   H         C            C       H + H      Br

 

H           H                                                            H              Br

                                                                                Bromoethane

 

H           H                                                  H             Br

 

H          C           C         H  + Br       Br   ->  H         C             C       H + H      Br

 

H           Br                                                 H              Br

                                                                      1,1-dibromoethane

 

H           Br                                                 H            Br

 

H          C           C         H  + Br       Br   ->  H         C             C       Br + H      Br

 

H           Br                                                 H              Br

                                                                      1,1,1-tribromoethane

 

H           Br                                                 H             Br

 

H          C           C        Br  + Br       Br   ->  H         C            C       Br + H      Br

 

H           Br                                                 Br             Br

                                                                      1,1,1,2-tetrabromoethane

 

H           Br                                                  H            Br

 

H          C           C        Br  + Br       Br   ->  Br        C            C       Br + H      Br

 

Br           Br                                               Br             Br

                                                                      1,1,1,2,2-pentabromoethane

 

H           Br                                                  Br           Br

 

Br        C            C        Br  + Br       Br   ->  Br       C            C       Br + H      Br

 

Br           Br                                               Br             Br

                                                                      1,1,1,2,2,2-hexabromoethane

 

Uses of alkanes

1.Most alkanes are used as fuel e.g. Methane is used as biogas in homes.Butane is used as the Laboratory gas.

2.On cracking ,alkanes are a major source of Hydrogen for the manufacture of ammonia/Haber process.

3.In manufacture of Carbon black which is a component in printers ink.

4.In manufacture of useful industrial chemicals like methanol, methanol, and chloromethane.

 

 

(ii) Alkenes

 

(a)Nomenclature/Naming

 

 

These are hydrocarbons with a general formula CnH2n  and      C       C      double bond as the functional group . n is the number of Carbon atoms in the molecule.

The carbon atoms are linked by at least one double bond to each other and single bonds to hydrogen atoms.

They include:

 

n General/

Molecular

formula

Structural formula Name

 

1   Does not exist

 

 
2 C2H6

 

 

 

             H           H

 

H          C           C         H

 

CH2 CH2

 

Ethene
3 C3H8              H           H        H

 

H          C           C         C       H

 

H

 

CH2 CH CH3

 

Propene
4 C4H10         H      H     H    H

 

H      C      C    C     C    H

 

H    H

 

CH2 CH CH2CH3

 

Butene
5 C5H12         H      H     H    H    H

 

H      C      C    C     C    C   H         

 

H    H     H

CH2 CH (CH2)2CH3

 

Pentene
6 C6H14         H      H     H    H    H   H

 

H      C      C    C     C    C   C   H      

 

H    H     H  H

 

CH2 CH (CH2)3CH3

 

Hexene
7 C7H16         H      H     H    H    H   H   H

 

H      C      C    C     C    C    C   C    H  

 

H      H    H    H     H  H    H

 

CH2 CH (CH2)4CH3

 

Heptene
8 C8H18         H      H     H    H    H   H   H    H

 

H      C      C    C     C    C    C   C    C   H

 

H    H     H  H    H   H

 

CH2 CH (CH2)5CH3

 

Octene
9 C9H20         H      H     H    H    H   H   H    H   H

 

H      C      C    C     C    C    C   C    C   C   H

 

H    H     H  H    H   H   H

 

CH2 CH (CH2)6CH3

 

Nonene
10 C10H22         H      H     H    H    H   H   H    H   H    H

 

H      C      C    C     C    C    C   C    C   C    C   H

 

H    H     H  H    H   H   H    H

 

CH2 CH (CH2)7CH3

decene

 

Note

1.Since carbon is tetravalent ,each atom of carbon in the alkene MUST always be bonded using four covalent bond /four shared pairs of electrons including at the double bond.

2.Since Hydrogen is monovalent ,each atom of hydrogen in the alkene MUST always be bonded using one covalent bond/one shared pair of electrons.

3.One member of the alkene ,like alkanes,differ from the next/previous by a CH2  group.They also form a homologous series.

e.g

Propene differ from ethene by one carbon and two Hydrogen atoms from ethene. 4.A homologous series of alkenes like that of alkanes:

(i) differ by a CH2  group from the next /previous consecutively

(ii)have similar chemical properties

(iii)have similar chemical formula represented by the general  formula CnH2n

(iv)the physical properties also show steady gradual change

5.The   = C= C = double bond in alkene is the functional group. A functional group is the reacting site of a molecule/compound.

  1. The = C= C = double bond in alkene can easily be broken to accommodate more two more monovalent atoms. The = C= C = double bond in alkenes make it thus unsaturated.
  2. An unsaturated hydrocarbon is one with a double =C=C= or triple – C C – carbon bonds in their molecular structure. Unsaturated hydrocarbon easily reacts to be saturated.

8.A saturated hydrocarbon is one without a double =C=C= or triple – C       C – carbon bonds in their molecular structure.

Most of the reactions of alkenes take place at the  = C = C =bond.

 

(b)Isomers of alkenes

Isomers are alkenes lie alkanes have the same molecular general formula but different molecular structural formula.

Ethene and propene do not form isomers. Isomers of alkenes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.

The IUPAC system of nomenclature of naming alkenes uses the following basic rules/guidelines:

 

1.Identify the longest continuous/straight carbon chain which contains the =C = C= double bond get/determine the parent alkene.

2.Number the longest chain form the end of the chain which contains the =C = C= double bond so he =C = C= double bond lowest number possible.

3 Indicate the positions by splitting “alk-positions-ene” e.g. but-2-ene, pent-1,3-diene.

4.The position indicated must be for the carbon atom at the lower position in the =C = C= double bond.i.e

But-2-ene means the double =C = C=  is between Carbon “2”and “3”

Pent-1,3-diene  means there are two double bond one between carbon “1” and “2”and another between carbon “3” and “4”

  1. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of alkyl carbon chains attached to the alkene. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens

6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of  double C = C  bonds and branches attached to the alkene.

 

7.Position isomers can be formed  when the=C = C= double bond is shifted between carbon atoms e.g.

But-2-ene means the double =C = C=  is between Carbon “2”and “3”

But-1-ene means the double =C = C=  is between Carbon “1”and “2”

Both But-1-ene and But-2-ene are position isomers of Butene

 

8.Position isomers are molecules/compounds having the same general formular but different position of the functional group.i.e.

Butene has the molecular/general formular C4H8 position  but can form both But-1-ene and But-2-ene as position isomers.

 

  1. Like alkanes ,an alkyl group can be attached to the alkene. Chain/branch isomers are thus formed.

10.Chain/branch isomers are molecules/compounds having the same general formula but different structural formula e.g

Butene and 2-methyl propene both have the same general formualr but different branching chain.

 

Practice on IUPAC nomenclature  of alkenes

Name the following isomers of alkene

 

H      H     H    H

 

H     C      C    C     C    H         But-1-ene

 

H    H

H      H    H    H

 

H      C      C    C     C    H         But-2-ene

 

 

H                   H

 

H      H    H    H    H   H

 

H      C       C    C     C    C   C   H       4-methylhex-1-ene      

 

H           H   H

H     C        H

 

H

 

H

 

H       C      H

 

H      H    H          H    H

 

H      C       C    C     C    C   C   H       4,4-dimethylhex-1-ene      

 

H           H   H

H     C        H

 

H

 

  1. H

 

H       C      H

 

H      H    H            H

 

H      C       C    C     C    C   H                          4,4-dimethylpent -1- ene      

 

H           H

H     C        H

 

H

 

  1.  H

 

H       C      H

 

H      H    H            H

 

H      C       C    C     C    C   H                          5,5-dimethylhex-2- ene      

 

H    C     H     H           H

H     C        H

H

H

 

  1. H

 

H       C      H

 

H    H           H

 

H       C    C     C    C   H                2,2-dimethylbut -2- ene      

 

H                 H

H     C        H

 

H

 

8.H2C CHCH2 CH2 CH3                                                pent -1- ene

 

9.H2C C(CH3)CH2 CH2 CH3                       2-methylpent -1- ene

 

10.H2C C(CH3)C(CH3)2 CH2 CH3                          2,3,3-trimethylpent -1- ene

 

11.H2C C(CH3)C(CH3)2 C(CH3)2 CH3                2,3,3,4,4-pentamethylpent -1- ene

 

12.H3C C(CH3)C(CH3) C(CH3)2 CH3                  2,3,4,4-tetramethylpent -2- ene

 

  1. H2C C(CH3)C(CH3) C(CH3) CH3 2,3,4-trimethylpent -1,3- diene

 

  1. H2C CBrCBr CBr CH3 2,3,4-tribromopent -1,3- diene

 

  1. H2C CHCH CH2 But -1,3- diene

 

  1. Br2C CBrCBr CBr2 1,1,2,3,4,4-hexabromobut -1,3- diene

 

  1. I2C CICI CI2 1,1,2,3,4,4-hexaiodobut -1,3- diene

 

  1. H2C C(CH3)C(CH3) CH2 2,3-dimethylbut -1,3- diene

 

(c)Occurrence and extraction

 

At indusrial level,alkenes are obtained from the cracking of alkanes.Cracking is the process of breaking long chain alkanes to smaller/shorter alkanes, an alkene and hydrogen gas at high temperatures.

Cracking is a major source of useful hydrogen gas  for manufacture of ammonia/nitric(V)acid/HCl i.e.

 

Long chain alkane     -> smaller/shorter alkane  + Alkene + Hydrogen gas

 

Examples

1.When irradiated with high energy radiation,Propane undergo cracking to form methane gas, ethene and hydrogen gas.

Chemical equation

CH3CH2CH3 (g)      ->     CH4(g)     +   CH2=CH2(g)     + H2(g)

 

2.Octane undergo cracking to form hydrogen gas, butene and butane gases

Chemical equation

CH3(CH2) 6 CH3 (g)      ->  CH3CH2CH2CH3(g)     +   CH3 CH2CH=CH2(g)   + H2(g)

 

 

(d)School laboratory preparation of alkenes

 

In a school laboratory, alkenes may be prepared from dehydration of alkanols using:

(i) concentrated sulphuric(VI)acid(H2SO4).

(a) aluminium(III)oxide(Al2O3) i.e

 

Alkanol        –Conc. H2SO4 –>           Alkene         +       Water

Alkanol          –Al2O3 –>          Alkene         +       Water e.g.

 

1.(a)At about 180oC,concentrated sulphuric(VI)acid dehydrates/removes water from ethanol to form ethene.

The gas produced contain traces of carbon(IV)oxide and sulphur(IV)oxide gas as impurities.

It is thus passed through concentrated sodium/potassium hydroxide solution to remove the impurities.

Chemical equation

 

CH3CH2OH (l)      –conc H2SO4/180oC–>   CH2=CH2(g)     + H2O(l)

 

(b)On heating strongly aluminium(III)oxide(Al2O3),it dehydrates/removes water from ethanol to form ethene.

Ethanol vapour passes through the hot aluminium (III) oxide which catalyses the dehydration.

Activated aluminium(III)oxide has a very high affinity for water molecules/elements of water and thus dehydrates/ removes water from  ethanol to form ethene.

 

Chemical equation

CH3CH2OH (l)      –(Al2O3/strong heat–>   CH2=CH2(g)     + H2O(l)

 

2(a) Propan-1-ol  and Propan-2-ol(position isomers of propanol)  are dehydrated by conc H2SO4 at about 180oC  to propene(propene has no position isomers).

Chemical equation

CH3CH2 CH2OH (l)      — conc H2SO4/180oC –>   CH3CH2=CH2(g)     + H2O(l)

Propan-1-ol                                                             Prop-1-ene

 

CH3CHOH CH3 (l)      — conc H2SO4/180oC –>   CH3CH2=CH2(g)     + H2O(l)

Propan-2-ol                                                             Prop-1-ene

 

(b) Propan-1-ol  and Propan-2-ol(position isomers of propanol)  are dehydrated by heating strongly aluminium(III)oxide(Al2O3) form propene

Chemical equation

CH3CH2 CH2OH (l)      — Heat/Al2O3 –>   CH3CH2=CH2(g)     + H2O(l)

Propan-1-ol                                                             Prop-1-ene

 

CH3CHOH CH3 (l)      — Heat/Al2O3 –>   CH3CH2=CH2(g)     + H2O(l)

Propan-2-ol                                                             Prop-1-ene

 

3(a) Butan-1-ol  and Butan-2-ol(position isomers of butanol)  are dehydrated by conc H2SO4 at about 180oC  to But-1-ene and But-2-ene respectively

Chemical equation

CH3CH2 CH2 CH2OH (l) — conc H2SO4/180oC –>CH3 CH2CH2=CH2(g)  + H2O(l)

Butan-1-ol                                                               But-1-ene

 

CH3CHOH CH2CH3 (l)– conc H2SO4/180oC –>CH3CH=CH CH2(g) + H2O(l)

Butan-2-ol                                                               But-2-ene

 

(b) Butan-1-ol  and Butan-2-ol   are dehydrated by heating strongly aluminium (III) oxide (Al2O3) form But-1-ene and But-2-ene respectively.

Chemical equation

CH3CH2 CH2 CH2OH (l) — Heat/Al2O3 –>  CH3 CH2CH2=CH2(g)   + H2O(l)

Butan-1-ol                                                               But-1-ene

 

CH3CHOH CH2CH3 (l)   — Heat/Al2O3 –>  CH3CH=CH CH2(g)   + H2O(l)

Butan-2-ol                                                               But-2-ene

 

Laboratory set up for the preparation of alkenes/ethene

 

Caution

(i)Ethanol is highly inflammable

(ii)Conc H2SO4 is highly corrosive on skin contact.

(iii)Common school thermometer has maximum calibration of 110oC and thus cannot be used. It breaks/cracks.

 

(i)Using conentrated sulphuric(VI)acid

 

 

 

Some broken porcelain or sand should be put in the flask when heating to:

          (i)prevent bumping which may break the flask.

(ii)ensure uniform and smooth boiling of the mixture

 

The temperatures should be maintained at above160oC.

At lower temperatures another compound –ether is predominantly formed instead of ethene gas.

 

(ii)Using aluminium(III)oxide

 

 

(e)Properties of alkenes

  1. Physical properties

 

Like alkanes, alkenes are colourles gases, solids and liquids that are not poisonous.

They are slightly soluble in water.

The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene.

The melting and boiling point increase as the carbon chain increase.

This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase.

The 1st four straight chain alkenes (ethene,propane,but-1-ene and pent-1-ene)are gases at room temperature and pressure.

The density of straight chain alkenes,like alkanes, increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkene.

 

 

Summary of physical properties of the 1st five alkenes

 

Alkene General formula Melting point(oC) Boiling point(K) State at room(298K) temperature and pressure atmosphere (101300Pa)
Ethene CH2CH2 -169 -104 gas
Propene CH3 CHCH2 -145 -47 gas
Butene CH3CH2 CHCH2 -141 -26 gas
Pent-1-ene CH3(CH2 CHCH2 -138 30 liquid
Hex-1-ene CH3(CH2) CHCH2 -98 64 liquid

 

  1. Chemical properties

(a)Burning/combustion

 

Alkenes burn with a yellow/ luminous sooty/ smoky flame in excess air to form carbon(IV) oxide and water.

Alkene         +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

 

Alkenes burn with a yellow/ luminous sooty/ smoky flame in limited air to form carbon(II) oxide and water.

Alkene         +        Air     ->       carbon(II) oxide    +        water  (limited air)

 

Burning of alkenes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the  =C=C= double bond because they have  higher C:H ratio.

A homologous series with     C = C    double or     C    C    triple bond is said to be unsaturated.

A homologous series with       C      C      single bond is said to be saturated.Most of the reactions of the unsaturated compound involve trying to be saturated to form a

 

 

C      C      single bond .

 

 

Examples of burning alkenes

 

1.(a) Ethene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.

Ethene         +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

C2H4(g)       +        3O2(g)         ->       2CO2(g)                 +        2H2O(l/g)

 

(b) Ethene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.

Ethene         +        Air     ->       carbon(II) oxide    +        water  (limited air )

C2H4(g)       +        3O2(g)         ->       2CO2(g)                 +        2H2O(l/g)

 

2.(a) Propene when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.

Propene       +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

2C3H6(g)    +        9O2(g)         ->       6CO2(g)                 +        6H2O(l/g)

 

(a) Propene when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.

Propene       +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

C3H6(g)      +        3O2(g)         ->       3CO(g)                  +        3H2O(l/g)

 

(b)Addition reactions

An addition reaction is one which an unsaturated compound reacts to form a saturated compound.Addition reactions of alkenes are named from the reagent used to cause the addtion/convert the double  =C=C= to  single C-C bond.

 

(i)Hydrogenation

Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at high temperatures react with alkenes to form alkanes.

Examples

1.When Hydrogen gas is passed through liquid vegetable and animal oil at about 180oC in presence of Nickel catalyst,solid fat is formed.

Hydrogenation is thus used to harden oils to solid fat especially margarine.

During hydrogenation, one hydrogen atom in the hydrogen molecule attach itself to one carbon and the other hydrogen to the second carbon breaking the double bond to single bond.

 

Chemical equation

          H2C=CH2    +     H2     -Ni/Pa->     H3C  –  CH3

 

H       H                                                        H    H

 

C   =  C        +        H – H  – Ni/Pa ->      H – C – C – H

 

H       H                                                         H    H

 

2.Propene undergo  hydrogenation to form Propane

 

Chemical equation

          H3C CH=CH2    +     H2     -Ni/Pa->     H3C CH –  CH3

 

H       H     H                                                 H   H    H

 

 

H      C       C  = C     +   H – H  – Ni/Pa->      H – C –  C –  C- H

 

H               H                                                 H    H     H

 

3.Both But-1-ene and But-2-ene undergo  hydrogenation to form Butane

Chemical equation

But-1-ene      +    Hydrogen      –Ni/Pa->      Butane

          H3C CH2 CH=CH2    +     H2     -Ni/Pa->     H3C CH2CH –  CH3

 

H       H     H    H                                                H   H    H     H

 

 

H      C      C –   C  = C     +   H – H  – Ni/Pa->      H –  C-  C –  C –  C- H

 

H       H             H                                                H    H   H    H

 

 

But-2-ene      +    Hydrogen      –Ni/Pa->      Butane

          H3C CH2 =CH CH2    +     H2     -Ni/Pa->     H3C CH2CH –  CH3

 

H       H     H    H                                                     H   H    H     H

 

H      C      C =   C – C -H    +   H – H  – Ni/Pa->      H –  C-  C –  C –   C- H

 

H                      H                                                     H    H             H    H

 

  1. But-1,3-diene should undergo hydrogenation to form Butane. The reaction uses two moles of hydrogen molecules/four hydrogen atoms to break the two double bonds.

But-1,3-diene          +    Hydrogen      –Ni/Pa->      Butane

          H2C CH CH=CH2    +     2H2     -Ni/Pa->     H3C CH2CH –  CH3

 

 

H       H     H    H                                                     H   H    H     H

 

H      C      C –   C = C -H    +   2(H – H)  – Ni/Pa->  H –  C-  C –  C –   C- H

 

H    H             H    H

 

(ii) Halogenation.

Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkene to form an alkane.

The double bond in the alkene break and form a single bond.

The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases/reduces.

One bromine atom bond at the 1st carbon in the double bond while the other goes to the 2nd carbon.

 

Examples

1Ethene reacts with bromine to form 1,2-dibromoethane.

Chemical equation

          H2C=CH2    +     Br2                            H2 Br C  –  CH2 Br

 

H       H                                                        H    H

 

C   =  C        +        Br – Br                     Br – C – C – Br

 

H       H                                                         H    H

Ethene   + Bromine                              1,2-dibromoethane

 

2.Propene reacts with chlorine to form 1,2-dichloropropane.

Chemical equation

          H3C CH=CH2    +     Cl2                             H3C CHCl –  CH2Cl

Propene       +        Chlorine                     1,2-dichloropropane

H       H     H                                                 H   H    H

 

 

H      C       C  = C     + Cl – Cl                       H – C –  C –  C- Cl

 

H               H                                                 H    Cl    H

 

 

 

 

H       H     H    H                                                H    H    H     H

 

 

H      C      C –   C  = C     +   I – I                         H –  C-  C –  C –  C- I

 

H       H     H    H                                                 H   H    H     H

 

3.Both But-1-ene and But-2-ene undergo  halogenation with iodine to form 1,2-diiodobutane and 2,3-diiodobutane

Chemical equation

But-1-ene      +    iodine                               1,2 diiodobutane

          H3C CH2 CH=CH2    +     I2                      H3C CH2CH I –  CH2I

 

But-2-ene      +    Iodine                          2,3-diiodobutane

          H3C CH= CH-CH2    +     F2                         H3C CHICHI –  CH3

 

H       H     H    H                                                     H   H    H     H

 

H      C      C =   C – C -H    +   I – I                                     H –  C-  C –  C –   C- H

 

H                      H                                                     H    I    I    H

  1. But-1,3-diene should undergo halogenation to form Butane. The reaction uses two moles of  iodine  molecules/four iodine atoms to break the two double bonds.

But-1,3-diene          +    iodine                  1,2,3,4-tetraiodobutane

          H2C= CH CH=CH2    +     2I2                      H2CI CHICHI –  CHI

 

H       H     H    H                                                     H   H    H     H

 

H      C      C –   C = C -H    +   2(I – I)                   H –  C-  C –  C –   C- H

 

I    I     I        I

 

(iii) Reaction with hydrogen halides.

Hydrogen halides reacts with alkene to form a halogenoalkane. The double bond in the alkene break and form a single bond.

The main compound is one which the hydrogen atom bond at the carbon with more hydrogen .

Examples

  1. Ethene reacts with hydrogen bromide to form bromoethane.

Chemical equation

 

H2C=CH2    +     HBr                            H3 C  –  CH2 Br

 

H       H                                                        H    H

 

C   =  C        +        H – Br                      H – C – C – Br

 

H       H                                                         H    H

Ethene         +    Bromine                       bromoethane

 

  1. Propene reacts with hydrogen iodide to form 2-iodopropane.

Chemical equation

          H3C CH=CH2    +     HI                             H3C CHI –  CH3

Carbon atom with more Hydrogen atoms gets extra hydrogen

Propene       +        Chlorine                     2-chloropropane

H       H     H                                                 H   H    H

 

H      C       C  = C     + H – Cl                        H – C –  C –  C– H

 

H               H                                                 H    Cl    H

 

  1. Both But-1-ene and But-2-ene reacts with hydrogen bromide to form 2- bromobutane

Chemical equation

But-1-ene      + hydrogen bromide                                    2-bromobutane

          H3C CH2 CH=CH2    +   HBr                    H3C CH2CHBr -CH3

 

H       H     H    H                                                H   H    H     H

 

 

H      C      C –   C  = C     + H – Br                                 H –  C-  C –  C –  C- H

 

H       H             H                                                H    H   Br    H

 

But-2-ene      + Hydrogen bromide                       2-bromobutane

          H3C CH= CH-CH2    +  HBr                         H3C CHBrCH2 –  CH3

 

H       H     H    H                                                     H   H    H     H

 

H      C      C =   C – C -H    +   Br – H                      H –  C-  C –  C –   C- H

 

H                      H                                                     H   Br     H    H

  1. But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses two moles of hydrogen iodide molecules/two iodine atoms and two hydrogen atoms to break the two double bonds.

But-1,3-diene          +    iodine                  2,3-diiodobutane

          H2C= CH CH=CH2    +     2HI2                      H3CCHICHI –  CH3

 

H       H     H    H                                                     H   H    H     H

 

H      C      C –   C = C -H  +   2(H – I)                    H –  C-   C –  C –   C- H

 

H     I     I      H

 

(iv) Reaction with bromine/chlorine water.

Chlorine and bromine water is formed when the halogen is dissolved in distilled water.Chlorine water has the formular HOCl(hypochlorous/chloric(I)acid) .Bromine water has the formular HOBr(hydrobromic(I)acid).

During the addition reaction .the halogen move to one carbon and the OH to the other carbon in the alkene  at the =C=C= double bond to form a halogenoalkanol.

 

Bromine water  + Alkene -> bromoalkanol

Chlorine water + Alkene -> bromoalkanol

Examples

 

1Ethene reacts with bromine water to form  bromoethanol.

Chemical equation

          H2C=CH2    +     HOBr                                 H2 Br C  –  CH2 OH

 

H       H                                                        H    H

 

C   =  C        +        Br – OH                    Br – C – C – OH

 

H       H                                                         H    H

Ethene   + Bromine  water                             bromoethanol

 

2.Propene reacts with chlorine water to form  chloropropan-2-ol / 2-chloropropan-1-ol.

Chemical equation

 

          I.H3C CH=CH2    +     HOCl                                H3C CHCl –  CH2OH

Propene       + Chlorine water                                    2-chloropropane

H       H     H                                                 H   H    H

 

 

H      C       C  = C     + HO – Cl                     H – C –  C –  C- OH

 

H               H                                                 H    Cl    H

 

II.H3C CH=CH2    +     HOCl                              H3C CHOH –  CH2Cl

Propene       +        Chlorine                                     chloropropan-2-ol

H       H     H                                                 H   H    H

 

 

H      C       C  = C     + HO – Cl                     H – C –  C –  C- Cl

 

H               H                                                 H    OH  H

 

3.Both But-1-ene and But-2-ene react with bromine water to form  2-bromobutan-1-ol /3-bromobutan-2-ol respectively

 

Chemical equation

I.But-1-ene              +    bromine water                             2-bromobutan-1-ol

 

          H3C CH2 CH=CH2    +    HOBr                          H3C CH2CH Br –  CH2OH

 

H       H     H    H                                                H   H    H     H

 

 

H      C      C –   C  = C     + HO– Br                    H –  C-  C –  C –  C- OH

 

H       H             H                                                H    H   Br    H

 

II.But-2-ene            +    bromine water                             3-bromobutan-2-ol

          H3C CH= CHCH3    +    HOBr                            H3C CH2OHCH Br CH3

 

H       H     H    H                                                H   H    H     H

 

 

H      C      C –   C  = C     + HO– Br                    H –  C-  C –  C –  C- OH

 

H       H             H                                                H    H   Br    H

 

  1. But-1,3-diene reacts with bromine water to form Butan-1,3-diol.

The reaction uses two moles of  bromine water molecules  to break the two double bonds.

But-1,3-diene  +    bromine water                 2,4-dibromobutan-1,3-diol

 

          H2C= CH CH=CH2    + 2HOBr                      H2COH CHBrCHOH CHBr

H       H     H    H                                                   H   H    H     H

 

H      C      C –   C = C -H    + 2(HO – Br)                        H –  C-  C –  C –   C- H

 

HO   Br HO    Br

 

(v) Oxidation.

 Alkenes are oxidized to alkanols with duo/double functional groups by oxidizing agents.

When an alkene is bubbled into orange acidified potassium/sodium dichromate (VI) solution,the colour of the oxidizing agent changes to green.

When an alkene is bubbled into purple acidified potassium/sodium manganate(VII) solution, the oxidizing agent is decolorized.

Examples

 

1Ethene is oxidized to ethan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution.

The purple acidified potassium/sodium manganate(VII) solution is decolorized.

The orange acidified potassium/sodium dichromate(VI) solution turns to green.

Chemical equation

          H2C=CH2             [O] in H+/K2Cr2O7                   HO CH2  –  CH2 OH

 

H       H                                                        H    H

 

C   =  C+ [O] in H+/KMnO4                   H – C – C – H

 

H       H                                                         OH   OH

Ethene   + [O] in H+/KMnO4                         ethan-1,2-diol

 

  1. Propene is oxidized to propan-1,2-diol by acidified potassium/sodium manganate(VII) solution/ acidified potassium/sodium dichromate(VI) solution.

The purple acidified potassium/sodium manganate(VII) solution is decolorized.

The orange acidified potassium/sodium dichromate(VI) solution turns to green.

Chemical equation

 

  H3C CH=CH2            [O] in H+/KMnO4                    H3C CHOH –  CH2OH

Propene              [O] in H+/KMnO4                      propan-1,2-diol

H       H     H                                                 H   H    H

 

 

H      C       C  = C      [O] in H+/KMnO4        H – C –  C –  C- OH

 

H               H                                                 H    OH  H

 

 

3.Both But-1-ene and But-2-ene react with bromine water to form  butan-1,2-diol and butan-2,3-diol

 

Chemical equation

I.But-1-ene   +  [O] in H+/KMnO4                                                      butan-1,2-diol

 

          H3C CH2 CH=CH2   +     [O]                            H3C CH2CHOH –  CH2OH

 

H       H     H    H                                                H   H    H     H

 

 

H      C      C –   C  = C     + [O]                            H –  C-  C –  C –  C- OH

 

H       H             H                                                H    H   OH    H

 

(v) Hydrolysis.

Hydrolysis is the reaction of a compound with water/addition of H-OH to a compound.

Alkenes undergo hydrolysis to form alkanols .

This takes place in two steps:

(i)Alkenes react with concentrated sulphuric(VI)acid  at room temperature and pressure to form alkylhydrogen sulphate(VI).

 

Alkenes + concentrated sulphuric(VI)acid -> alkylhydrogen sulphate(VI)

 

(ii)On adding water to alkylhydrogen sulphate(VI) then warming, an alkanol is formed.

alkylhydrogen sulphate(VI) + water  -warm-> Alkanol.

Examples

(i)Ethene reacts with cold concentrated sulphuric(VI)acid to form ethyl hydrogen sulphate(VII)

Chemical equation

          H2C=CH2         +   H2SO4                                   CH3  –  CH2OSO3H

 

H       H                                                       H    O-SO3H

 

C   =  C          +      H2SO                     H – C – C – H

 

H       H                                                         H    H

Ethene        +         H2SO                             ethylhydrogen sulphate(VI)

 

(ii) Ethylhydrogen sulphate(VI) is hydrolysed by water to ethanol

Chemical equation

  CH3  –  CH2OSO3H         +      H2O                                      CH3  –  CH2OH  + H2SO4

 

H       OSO3H                                               H   OH

 

H –   C   –    C – H           +  H2O                     H – C – C – H  + H2SO4

 

H       H                                                         H    H

ethylhydrogen sulphate(VI)  H2O                        Ethanol

 

 

  1. Propene reacts with cold concentrated sulphuric(VI)acid to form propyl hydrogen sulphate(VII)

Chemical equation

          CH3H2C=CH2         +       H2SO4                                   CH3CH2  – CH2OSO3H

 

H       H    H                                                           H     H   O-SO3H

 

C   =  C –  C – H  + H2SO                    H –  C – C – C – H

 

H       H    H                                                  H    H    H

Propene       +        H2SO                             propylhydrogen sulphate(VI)

 

(ii) Propylhydrogen sulphate(VI) is hydrolysed by water to propanol

Chemical equation

  CH3  –  CH2OSO3H         +      H2O                                      CH3  –  CH2OH  + H2SO4

 

H     H        OSO3H                                       H   H   OH

 

H – C –   C   –    C – H         +  H2O                  H – C – C – C – H  + H2SO4

 

H     H       H                                                 H   H    H

propylhydrogen sulphate(VI)  H2O                                propanol

 

(vi) Polymerization/self addition

 

Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule.

Only alkenes undergo addition polymerization.

Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene

 

During addition polymerization

(i)the double bond in alkenes break

(ii)free radicals are formed

(iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule.

 

Examples of addition polymerization

 

1.Formation of Polyethene

Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles)

 

H    H                     H    H                   H    H                    H    H

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

H    H                     H    H                   H    H                    H    H

 

Ethene         +        Ethene         +        Ethene         +        Ethene    + 

 

(ii)the double bond joining the ethane molecule break to free readicals

 

H    H                     H    H                   H    H                    H    H

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

H    H                     H    H                   H    H                    H    H

 

Ethene radical     +  Ethene radical    +   Ethene radical   +   Ethene radical    + 

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

H    H    H    H  H    H   H    H             lone pair of electrons

 

  • C – C – C – C – C – C – C – C• +  

 

H    H    H    H  H    H   H    H

Lone pair of electrons can be used to join more monomers to form longer polyethene.

Polyethene molecule can be represented as:

 

 

 

H    H    H    H  H    H   H    H             extension of

molecule/polymer

– C – C – C – C – C – C – C – C-

 

H    H    H    H  H    H   H    H

Since the molecule is a repetition of one monomer, then the polymer is:

H    H

 

( C – C  )n

 

H    H

Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship:

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

 

Examples

Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 )

 

Number of monomers/repeating units in polyomer =  Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760

 

Substituting           4760           =                 170 ethene molecules

28

The commercial name of polyethene is polythene.

It is an elastic, tough, transparent and durable plastic.

Polythene is used:

(i)in making plastic bag

(ii)bowls and plastic bags

(iii)packaging materials

 

2.Formation of Polychlorethene

Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 

H    H                     H    H                   H    H                    H    H

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

H    Cl                     H    Cl                   H    Cl                     H    Cl

 

chloroethene          +   chloroethene     +    chloroethene    +    chloroethene    + 

 

(ii)the double bond joining the chloroethene molecule break to free radicals

 

H    H                     H    H                   H    H                    H    H

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

H    Cl                     H    Cl                   H    Cl                    H    Cl

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

H    H    H    H  H    H   H    H             lone pair of electrons

 

  • C – C – C – C – C – C – C – C• +  

 

H   Cl    H   Cl  H    Cl   H    Cl

Lone pair of electrons can be used to join more monomers to form longer polychloroethene.

Polychloroethene molecule can be represented as:

 

H    H    H    H  H    H   H    H             extension of

molecule/polymer

– C – C – C – C – C – C – C – C-     +  

 

H   Cl    H   Cl  H   Cl   H   Cl

 

Since the molecule is a repetition of one monomer, then the polymer is:

H    H

 

( C – C  )n

 

H    Cl

 

Examples

Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 )

 

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760

 

Substituting 4760   =  77.16   => 77  polychloroethene molecules(whole number)

62.5

The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used:

(i)in making plastic rope

(ii)water pipes

(iii)crates  and boxes

 

 

3.Formation of Polyphenylethene

Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 

H    H                     H    H                   H    H                    H    H

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

H    C6H5                H    C6H5              H    C6H5                H    C6H5

 

phenylethene         +   phenylethene    +    phenylethene   +    phenylethene    + 

 

(ii)the double bond joining the phenylethene molecule break to free radicals

 

H    H                     H    H                   H    H                    H    H

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

H    C6H5                H    C6H5              H    C6H5               H    C6H5

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

H         H      H       H       H       H       H        H             lone pair of electrons

 

  • C –   C   –   C   –   C   –   C   –   C   –   C   –   C  •     +  

 

H         C6H5   H    C6H5  H       C6H5   H       C6H5

Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.

Polyphenylethene molecule can be represented as:

H         H      H       H       H       H       H        H

 

– C   –    C   –    C   –   C   –   C   –   C   –   C   –   C  –

 

H         C6H5   H    C6H5  H       C6H5   H       C6H5

 

Since the molecule is a repetition of one monomer, then the polymer is:

H    H

 

( C – C  )n

 

H    C6H5

 

Examples

Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, )

 

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

 

=> Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760

 

Substituting 476045.7692  =>45  polyphenylethene molecules(whole number)

104

The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:

(i)in making packaging material for carrying delicate items like computers, radion,calculators.

(ii)ceiling tiles

(iii)clothe linings

 

4.Formation of Polypropene

Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

 

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 

H    H                     H    H                   H    H                    H    H

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

H    CH3                 H    CH3               H    CH3                 H    CH3

 

propene       +   propene  +    propene +    propene   + 

 

(ii)the double bond joining the phenylethene molecule break to free radicals

 

H    H                     H    H                   H    H                    H    H

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

H    CH3                 H    CH3               H    CH3                H    CH3

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

H         H      H       H       H       H       H        H             lone pair of electrons

 

  • C –   C   –   C   –   C   –   C   –   C   –   C   –   C  •     +  

 

H         CH3   H        CH3  H        CH3   H        CH3

 

Lone pair of electrons can be used to join more monomers to form longer propene.

propene molecule can be represented as:

H         H      H       H       H       H       H        H

 

– C   –    C   –    C   –   C   –   C   –   C   –   C   –   C  –

 

H         CH3   H        CH3  H        CH3   H        CH3

 

Since the molecule is a repetition of one monomer, then the polymer is:

H    H

 

( C – C  )n

 

H    CH3

Examples

Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, )

 

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

=> Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760

 

Substituting 4760108.1818  =>108  propene molecules(whole number)

44

The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:

(i)in making packaging material for carrying delicate items like computers, radion,calculators.

(ii)ceiling tiles

(iii)clothe linings

 

 

 

 

5.Formation of Polytetrafluorothene

Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 

F    F                       F     F                    F     F                      F    F

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

F    F                       F     F                    F     F                      F    F

 

tetrafluoroethene    +  tetrafluoroethene+  tetrafluoroethene+ tetrafluoroethene + 

 

(ii)the double bond joining the tetrafluoroethene molecule break to free radicals

 

F    F                       F    F                    F    F                      F    F

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

F    F                       F    F                     F    F                      F    F

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

F    F     F     F   F     F   F    F             lone pair of electrons

 

  • C – C – C – C – C – C – C – C• +  

 

F     F     F    F   F    F     F    F

Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.

polytetrafluoroethene molecule can be represented as:

 

F    F     F     F   F     F   F    F             extension of

molecule/polymer

– C – C – C – C – C – C – C – C-     +  

 

F     F     F    F   F    F     F    F

 

Since the molecule is a repetition of one monomer, then the polymer is:

F    F

 

( C – C  )n

 

F     F

Examples

Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 )

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760

 

Substituting 4760   =  77.16   => 77  polychloroethene molecules(whole number)

62.5

The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used:

(i)in making plastic rope

(ii)water pipes

(iii)crates  and boxes

 

 

6.Formation of rubber from Latex

Natural rubber is obtained from rubber trees.

During harvesting an incision is made on the rubber tree to produce a milky white substance called latex.

Latex is a mixture of rubber and lots of water.

The latex is then added an acid to coagulate the rubber.

Natural rubber is a polymer of 2-methylbut-1,3-diene ;

 

H         CH3       H        H

 

CH2=C (CH3) CH = CH2       H   –   C    =    C    –    C    =    C    –   H

 

During natural polymerization to rubber, one double  C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;

 

H         CH3      H          H           H        CH3       H        H

 

–   C    –    C     =   C    –    C       –   C    –    C    =    C        C    –

 

H                                            H           H                                 H

Generally the structure of rubber is thus;

 

H         CH3      H          H

 

-(-   C    –    C    =    C    –    C   -)n

 

 

H                                 H

Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.

During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer.

 

Sulphur atoms make cross link between polymers

H         CH3      H        H           H        CH3       H        H

 

–   C    –    C     –   C    –    C       –   C    –    C        C        C    –

 

H                      S         H           H                                S         H

 

 

 

H         CH3       S        H           H        CH3       S         H

 

–   C    –    C     –    C    –    C       –  C    –    C         C        C    –

 

H                       H          H          H                     H        H

Vulcanized rubber is used to make tyres, shoes  and valves.

 

7.Formation of synthetic rubber

Synthetic rubber is able to resist action of oil,abrasion and organic solvents which  rubber cannot.

Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ;

H         Cl         H          H

 

CH2=C (Cl CH = CH2            H   –   C    =    C    –    C    =    C    –   H

 

During  polymerization to synthetic rubber, one double  C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus;

 

H         Cl         H          H           H        Cl       H           H

 

–   C    –    C     =   C    –    C       –   C    –    C    =    C        C    –

 

H                                            H           H                                 H

Generally the structure of rubber is thus;

 

H         Cl         H          H

 

-(-   C    –    C    =    C    –    C   -)n

 

 

H                                 H

Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.

 

 

(c)Test for the presence of    – C = C – double bond.

 

(i)Burning/combustion

 

All unsaturated hydrocarbons with  a   – C = C – or   – C  =  C – bond burn with a yellow sooty flame.

 

Experiment

Scoop a sample of the substance provided in a clean metallic spatula. Introduce it on a Bunsen burner.

 

               Observation                  Inference
Solid melt then burns with a yellow sooty flame

 

– C = C –,

 

– C  =  C – bond

 

(ii)Oxidation by acidified KMnO4/K2Cr2O7

 

Bromine water ,Chlorine water and Oxidizing agents acidified KMnO4/K2Cr2O7 change to unique colour in presence of  – C = C –

 

or   – C =  C – bond.

 

Experiment

Scoop a sample of the substance provided into a clean test tube. Add 10cm3 of distilled water. Shake. Take a portion of the solution mixture. Add three drops of acidified KMnO4/K2Cr2O7 .

 

              Observation                     Inference
Acidified KMnO4 decolorized

 

Orange colour of acidified K2Cr2O7 turns green

 

Bromine water is decolorized

 

Chlorine water is decolorized

– C = C –

 

– C = C – bond

 

 

(d)Some uses of Alkenes

  1. In the manufacture of plastic
  2. Hydrolysis of ethene is used in industrial manufacture of ethanol.
  3. In ripening of fruits.
  4. In the manufacture of detergents.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(iii) Alkynes

 

(a)Nomenclature/Naming

 

These are hydrocarbons with a general formula CnH2n2  and        C      C      double bond as the functional group . n is the number of Carbon atoms in the molecule.

The carbon atoms are linked by at least one triple bond to each other and single bonds to hydrogen atoms.

 

They include:

n General/

Molecular

formula

Structural formula Name
1   Does not exist

 

 

2 C2H2

 

 

 

 

 

H          C           C         H

CH CH

 

Ethyne
3 C3H4                                       H

 

H          C           C         C       H

 

H

CH  C  CH3

 

Propyne
4 C4H6                         H    H

 

H      C      C    C     C    H

 

H    H

CH  C  CH2CH3

 

Butyne
5 C5H8                         H    H    H

 

H      C      C    C     C    C   H         

 

H    H     H

CH  C  (CH2)2CH3

 

Pentyne
6 C6H10                         H    H    H   H

 

H      C      C    C     C    C   C   H      

 

H    H     H  H

CH   C  (CH2)3CH3

 

Hexyne

 

7 C7H12                         H    H    H   H   H

 

H      C      C    C     C    C    C   C    H  

 

H      H    H    H     H  H    H

CH  C  (CH2)4CH3

Heptyne
8 C8H14                          H    H    H   H   H    H

 

H      C      C    C     C    C    C   C    C   H

 

H    H     H  H    H   H

CH  C  (CH2)5CH3

Octyne
9 C9H16                         H    H    H   H   H    H   H

 

H      C      C    C     C    C    C   C    C   C   H

 

H    H     H  H    H   H   H

CH  C  (CH2)6CH3

Nonyne
10 C10H18                         H    H    H   H   H    H   H    H

 

H      C      C    C     C    C    C   C    C   C    C   H

 

H    H     H  H    H   H   H    H

CH  C  (CH2)7CH3

Decyne

 

Note

  1. Since carbon is tetravalent ,each atom of carbon in the alkyne MUST always be bonded using four covalent bond /four shared pairs of electrons including at the triple bond.
  2. Since Hydrogen is monovalent ,each atom of hydrogen in the alkyne MUST always be bonded using one covalent bond/one shared pair of electrons.
  3. One member of the alkyne ,like alkenes and alkanes, differ from the next/previous by a CH2 group(molar mass of 14 atomic mass units).They thus form a homologous series.

e.g

Propyne differ from ethyne by (14 a.m.u) one carbon and two Hydrogen atoms from ethyne.

4.A homologous series of alkenes like that of alkanes:

(i) differ by a CH2  group from the next /previous consecutively

(ii) have similar chemical properties

(iii)have similar chemical formula with general  formula CnH2n-2

(iv)the physical properties also show steady gradual change

 

5.The   – C = C – triple bond in alkyne is the functional group. The functional group is the reacting site of the alkynes.

  1. The – C = C – triple bond in alkyne can easily be broken to accommodate more /four more monovalent atoms. The – C = C – triple bond in alkynes make it thus unsaturated like alkenes.
  2. Most of the reactions of alkynes like alkenes take place at the – C = C- triple bond.

 

(b)Isomers of alkynes

Isomers of alkynes have the same molecular general formula but different molecular structural formula.

Isomers of alkynes are also named by using the IUPAC(International Union of Pure and Applied Chemistry) system of nomenclature/naming.

The IUPAC system of nomenclature of naming alkynes uses the following basic rules/guidelines:

 

1.Identify the longest continuous/straight carbon chain which contains the  – C = C- triple bond to get/determine the parent alkene.

  1. Number the longest chain form the end of the chain which contains the -C = C- triple bond so as  – C = C- triple bond get lowest number possible.

 

3 Indicate the positions by splitting “alk-positions-yne” e.g. but-2-yne, pent-1,3-diyne.

4.The position indicated must be for the carbon atom at the lower position in the

-C = C- triple bond. i.e

But-2-yne means the triple -C = C-  is between Carbon “2”and “3”

Pent-1,3-diyne  means there are two triple bonds; one between carbon “1” and “2”and another between carbon “3” and “4”

 

  1. Determine the position, number and type of branches. Name them as methyl, ethyl, propyl e.tc. according to the number of alkyl carbon chains attached to the alkyne. Name them fluoro-,chloro-,bromo-,iodo- if they are halogens

 

6.Use prefix di-,tri-,tetra-,penta-,hexa- to show the number of  triple  – C = C-  bonds and branches attached to the alkyne.

7.Position isomers can be formed  when the – C = C- triple bond is shifted between carbon atoms e.g.

But-2-yne means the double  – C = C- is between Carbon “2”and “3”

But-1-yne means the double – C = C-  is between Carbon “1”and “2”

Both But-1-yne and But-2-yne are position isomers of Butyne.

 

  1. Like alkanes and alkynes , an alkyl group can be attached to the alkyne. Chain/branch isomers are thus formed.

Butyne and 2-methyl propyne both have the same general formular but different branching chain.

 

(More on powerpoint)

 

(c)Preparation of Alkynes.

 

Ethyne is prepared from the reaction of water on calcium carbide. The reaction is highly exothermic and thus a layer of sand should be put above the calcium carbide to absorb excess heat to prevent the reaction flask from breaking. Copper(II)sulphate(VI) is used to catalyze the reaction

 

Chemical equation

CaC2(s)   +  2 H2O(l)   ->   Ca(OH) 2 (aq)  +  C2H2 (g)

 

 

(d)Properties of alkynes

  1. Physical properties

 

Like alkanes and alkenes,alkynes  are colourles gases, solids and liquids that are not poisonous.

They are slightly soluble in water. The solubility in water decrease as the carbon chain and as the molar mass increase but very soluble in organic solvents like tetrachloromethane and methylbenzene. Ethyne has a pleasant taste when pure.

The melting and boiling point increase as the carbon chain increase.

This is because of the increase in van-der-waals /intermolecular forces as the carbon chain increase. The 1st three straight chain alkynes (ethyne,propyne and but-1-yne)are gases at room temperature and pressure.

The density of straight chain alkynes increase with increasing carbon chain as the intermolecular forces increases reducing the volume occupied by a given mass of the alkyne.

 

Summary of physical properties of the 1st five alkenes

Alkyne General formula Melting point(oC) Boiling point(oC) State at room(298K) temperature and pressure atmosphere (101300Pa)
Ethyne CH CH -82 -84 gas
Propyne CH3 C CH -103 -23 gas
Butyne CH3CH2 CCH -122  8 gas
Pent-1-yne CH3(CH2) 2 CCH -119 39 liquid
Hex-1-yne CH3(CH2) 3C CH -132 71 liquid

 

  1. Chemical properties

(a)Burning/combustion

 

Alkynes burn with a yellow/ luminous very sooty/ smoky flame in excess air to form carbon(IV) oxide and water.

Alkyne         +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

 

Alkenes burn with a yellow/ luminous verysooty/ smoky flame in limited air to form carbon(II) oxide/carbon and water.

Alkyne         +        Air     ->       carbon(II) oxide /carbon +        water  (limited air)

Burning of alkynes with a yellow/ luminous sooty/ smoky flame is a confirmatory test for the presence of the  – C = C – triple bond because they have  very high   C:H ratio.

 

Examples of burning alkynes

1.(a) Ethyne when ignited burns with a yellow very sooty flame in excess air to form carbon(IV) oxide and water.

Ethyne         +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

2C2H2(g)     +        5O2(g)         ->       4CO2(g)                 +        2H2O(l/g)

 

(b) Ethyne when ignited burns with a yellow sooty flame in limited air to form a mixture of unburnt carbon and carbon(II) oxide and water.

Ethyne         +        Air     ->       carbon(II) oxide    +        water  (limited air )

C2H2(g)       +        O2(g) ->       2CO2(g)   +   C  +  2H2O(l/g)

 

2.(a) Propyne when ignited burns with a yellow sooty flame in excess air to form carbon(IV) oxide and water.

Propyne       +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

C3H4(g)      +        4O2(g)         ->       3CO2(g)                 +        2H2O(l/g)

 

(a) Propyne when ignited burns with a yellow sooty flame in limited air to form carbon(II) oxide and water.

Propene       +        Air     ->       carbon(IV) oxide   +        water  (excess air/oxygen)

2C3H4(g)    +        5O2(g)         ->       6CO(g)                  +        4H2O(l/g)

 

(b)Addition reactions

An addition reaction is one which an unsaturated compound reacts to form a saturated compound. Addition reactions of alkynes are also named from the reagent used to cause the addition/convert the triple   – C  =  C- to  single C- C bond.

 

(i)Hydrogenation

Hydrogenation is an addition reaction in which hydrogen in presence of Palladium/Nickel catalyst at 150oC temperatures react with alkynes to form alkenes then alkanes.

Examples

1.During hydrogenation, two hydrogen atom in the hydrogen molecule attach itself to one carbon and the other two  hydrogen to the second carbon breaking the triple bond to double the single.

Chemical equation

   HC = CH  +  H2     -Ni/Pa ->  H2C = CH2 +  H2     -Ni/Pa -> H2C  –  CH2

 

H       H                                    H    H                                             H    H

 

 

C   = C + H – H  – Ni/Pa -> H – C = C – H + H – H  – Ni/Pa -> H – C  –  C – H

 

H      H                                      H    H                                            H     H

 

2.Propyne undergo  hydrogenation to form Propane

 

Chemical equation

          H3C CH = CH2    +     2H2     -Ni/Pa->     H3C CH –  CH3

 

H       H     H                                                 H   H    H

 

 

H      C       C  = C     +   2H – H  – Ni/Pa->      H – C –  C –  C- H

 

H               H                                                 H    H     H

 

3(a) But-1-yne undergo  hydrogenation to form Butane

Chemical equation

But-1-yne      +    Hydrogen      –Ni/Pa->      Butane

          H3C CH2 C = CH    +     2H2     -Ni/Pa->     H3C CH2CH –  CH3

 

H       H         H                                                  H   H    H     H

 

 

H      C      C –   C  = C     + 2H – H  – Ni/Pa->      H –  C-  C –  C –  C- H

 

H       H                                                                H    H   H    H

(b) But-2-yne undergo  hydrogenation to form Butane

Chemical equation

But-2-yne      +    Hydrogen      –Ni/Pa->      Butane

          H3C C = C CH2    +     2H2     -Ni/Pa->     H3C CH2CH –  CH3

 

H                     H                                                 H   H    H     H

 

 

H      C      C =   C  – C   H   +   2H – H- Ni/Pa->  H –  C-  C –  C –  C- H

 

H                       H                                                H    H   H    H

 

 

(ii) Halogenation.

Halogenation is an addition reaction in which a halogen (Fluorine, chlorine, bromine, iodine) reacts with an alkyne to form an alkene then alkane.

The reaction of alkynes with halogens with alkynes is faster than with alkenes. The triple bond in the alkyne break and form a double then single bond.

The colour of the halogen fades as the number of moles of the halogens remaining unreacted decreases.

Two bromine atoms bond at the 1st carbon in the triple bond while the other two goes to the 2nd carbon.

 

Examples

1Ethyne reacts with brown bromine vapour to form 1,1,2,2-tetrabromoethane.

Chemical equation

     HC =  CH   +     2Br2                                    H Br2 C  –  CH Br2

 

H       H                                                        H    H

 

C   =  C        +        2Br – Br                   Br – C – C – Br

 

Br    Br

Ethyne   + Bromine                              1,1,2,1-tetrabromoethane

 

2.Propyne reacts with chlorine to form 1,1,2,2-tetrachloropropane.

Chemical equation

          H3C C = CH    +     2Cl2                             H3C CHCl2 –  CHCl2

Propyne       +        Chlorine                     1,1,2,2-tetrachloropropane

H                                                                            H   Cl    H

 

 

H      C       C  = C     + 2Cl – Cl                     H – C –  C –  C- Cl

 

H               H                                                 H    Cl    Cl

Propyne       +        Iodine                        1,1,2,2-tetraiodopropane

H3C C = CH    +     2I2                               H3C CHI2 –  CHI2

 

H       H            H                                                H   H      I     H

 

 

H      C      C –   C  = C     +   2I – I                                 H –  C-  C –  C –  C- I

 

H       H                                                                H   H     I       I

 

 

 

3(a)But-1-yne undergo  halogenation  to form 1,1,2,2-tetraiodobutane with iodine

Chemical equation

But-1-yne      +    iodine                               1,1,2,2-tetrabromobutane

          H3C CH2 C = CH    +     2I2                      H3C CH2C I2 –  CHI2

H       H                                                                  H   H      I      I

 

H      C      C –   C = C -H    +   2I – I                        H –  C-  C –  C –   C- H

 

H       H              H                                                   H    H             I       I

 

(b) But-2-yne undergo  halogenation  to form 2,2,3,3-tetrafluorobutane with fluorine

But-2-yne      +    Fluorine                           2,2,3,3-tetrafluorobutane

          H3C C = C -CH2    +     2F2                           H3C CF2CF2 –  CH3

 

H       H     H    H                                                     H   H    H     H

 

H      C      C =   C – C -H    +   F – F                        H –  C-  C –  C –   C- H

 

H                      H                                                     H    H             H    H

  1. But-1,3-diyne should undergo halogenation to form 1,1,2,3,3,4,4 octaiodobutane. The reaction uses four moles of  iodine  molecules/eight iodine atoms to break the two(2) triple double bonds at carbon “1” and “2”.

But-1,3-diene          +    iodine                  1,2,3,4-tetraiodobutane

          H C = C  C = C H    +     4I2                      H C I2 C I2 C I2 C H I2

 

I    I      I      I

 

H      C      C –   C = C -H    +   4(I – I)                   H –  C-  C –  C –   C- H

 

I    I      I      I

 

(iii) Reaction with hydrogen halides.

Hydrogen halides reacts with alkyne to form a halogenoalkene then halogenoalkane. The triple bond in the alkyne break and form a double then single bond.

The main compound is one which the hydrogen atom bond at the carbon with more hydrogen .

Examples

  1. Ethyne reacts with hydrogen bromide to form bromoethane.

Chemical equation

 

H C = C H    +     2HBr                                   H3 C  –  CH Br2

 

H       H                                                        H    H

 

C   =  C        +        2H – Br                    H – C – C – Br

 

H    Br

Ethyne         +    Bromine                       1,1-dibromoethane

 

  1. Propyne reacts with hydrogen iodide to form 2,2-diiodopropane (as the main product )

Chemical equation

          H3C C = CH    +     2HI                             H3C CHI2 –  CH3

Carbon atom with more Hydrogen atoms gets extra hydrogen

Propene       +        Chlorine                     2,2-dichloropropane

H                                                                  H    I    H

 

H      C       C  = C     + 2H – I                        H – C –  C –  C– H

 

H               H                                                 H     I     H

 

  1. Both But-1-yne and But-2-yne reacts with hydrogen bromide to form 2,2- dibromobutane

 

Chemical equation

But-1-ene      + hydrogen bromide                                    2,2-dibromobutane

 

          H3C CH2 C = CH    +  2HBr                     H3C CH2CHBr -CH3

 

H       H                                                              H   H    Br     H

 

 

H      C      C –   C  = C    + 2H – Br                      H –  C-  C –  C –  C- H

 

H       H             H                                                H    H   Br    H

 

But-2-yne      + Hydrogen bromide                       2,2-dibromobutane

 

          H3C C = C -CH3    +  2HBr                           H3C CBr2CH2 –  CH3

 

H                     H                                                     H   Br    H     H

 

H      C      C =   C – C -H    +   2Br – H                    H –  C-  C –  C –   C- H

 

H                      H                                                     H   Br    H    H

 

  1. But-1,3-diene react with hydrogen iodide to form 2,3- diiodobutane. The reaction uses four moles of hydrogen iodide molecules/four iodine atoms and two hydrogen atoms to break the two double bonds.

But-1,3-diyne          +    iodine                  2,2,3,3-tetraiodobutane

          H C = C C = C H    +     4HI                     H3C C I2 C I2 CH3

 

H           H                                                     H   I      I      H

 

H      C      C –   C = C -H  +   4(H – I)                    H –  C-   C –  C –   C- H

 

H     I     I      H

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.ALKANOLS(Alcohols)

 

(A) INTRODUCTION.

 

Alkanols belong to a homologous series of organic compounds with a general formula CnH2n +1 OH and thus  -OH  as the functional group .The 1st ten alkanols include

n General / molecular formular Structural formula IUPAC name
1 CH3OH  

H – C –O – H

H

 

 

Methanol

2 CH3 CH2OH

C2H5 OH

       H     H

 

H     C  – C –O – H

H     H

 

Ethanol
3 CH3 (CH2)2OH

C3H7 OH

       H     H     H

 

H     C  – C  – C –O – H

H     H     H

 

Propanol
4 CH3 (CH2)3OH

C4H9 OH

       H     H     H    H

 

H     C  – C –  C –  C –O – H

H     H    H     H

Butanol
5 CH3(CH2)4OH

C5H11 OH

       H     H     H  H   H

 

H     C  – C –  C- C-  C –O – H

H     H    H   H   H

 

Pentanol
6 CH3(CH2)5OH

C6H13 OH

       H     H     H  H   H   H

 

H     C  – C –  C- C-  C–  C – O – H

H     H    H   H   H    H

 

Hexanol
7 CH3(CH2)6OH

C7H15 OH

       H     H     H  H   H   H    H

 

H     C  – C –  C- C-  C–  C –C- O – H

H     H    H   H   H    H   H

 

Heptanol
8 CH3(CH2)7OH

C8H17 OH

       H     H     H  H   H   H    H  H

 

H     C  – C –  C- C-  C–  C –C- C -O – H

H     H    H   H   H    H   H  H

 

Octanol
9 CH3(CH2)8OH

C9H19 OH

       H     H     H  H   H   H    H  H  H

 

H     C  – C –  C- C-  C–  C –C- C –C- O – H

H     H    H   H   H    H   H  H   H

 

Nonanol
10 CH3(CH2)9OH

C10H21 OH

       H     H     H  H   H   H    H  H  H  H

 

H     C  – C –  C- C-  C–  C –C- C –C- C-O – H

H     H    H   H   H    H   H  H   H  H

 

Decanol

 

Alkanols like Hydrocarbons( alkanes/alkenes/alkynes) form a homologous series where:

(i)general name is derived from the alkane name then ending with “-ol

(ii)the members have –OH as the fuctional group

(iii)they have the same general formula represented by R-OH where R is an alkyl group.

(iv) each member differ by –CH2 group from the next/previous.

(v)they show a similar and gradual change in their physical properties e.g. boiling and melting points.

(vi)they show similar and gradual change in their chemical properties.

 

  1. ISOMERS OF ALKANOLS.

Alkanols exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines:

(i)Like alkanes  , identify the longest carbon chain to be the parent name.

(ii)Identify the position of the  -OH functional group to give it the smallest /lowest position.

(iii) Identify the type and position of the side branches.

 

Practice examples of isomers of alkanols

 

(i)Isomers of propanol C3H7OH

CH3CH2CH2OH  – Propan-1-ol

OH

 

CH3CHCH3  – Propan-2-ol

 

Propan-2-ol and Propan-1-ol are position isomers because only the position of the –OH functional group changes.

 

(ii)Isomers of  Butanol   C4H9OH

 

CH3 CH2 CH3 CH2 OH   Butan-1-ol

 

CH3 CH2 CH CH3

                

OH                   Butan-2-ol

 

CH3

 

CH3 CH3 CH3

 

OH                          2-methylpropan-2-ol

Butan-2-ol and Butan-1-ol are position isomers because only the position of the         -OH functional group changes.

2-methylpropan-2-ol is both a structural and position isomers because both the position of the functional group and the arrangement of the atoms in the molecule changes.

 

(iii)Isomers of  Pentanol   C5H11OH

 

CH3 CH2 CH2CH2CH2 OH   Pentan-1-ol (Position isomer)

 

CH3 CH2 CH CH3

                

OH                   Pentan-2-ol (Position isomer)

 

CH3 CH2 CH CH2 CH3

                

OH                   Pentan-3-ol (Position isomer)

 

 

 

CH3

 

 

CH3 CH2 CH2 C CH3

                

OH                    2-methylbutan-2-ol (Position /structural isomer)

 

CH3

 

 

CH3 CH2 CH2 C CHOH

                

CH3                   2,2-dimethylbutan-1-ol (Position /structural isomer)

 

CH3

 

 

CH3 CH2 CH   C CH3

                

CH3  OH           2,3-dimethylbutan-1-ol (Position /structural isomer)

 

(iv)1,2-dichloropropan-2-ol

 

CClH2 CCl CH3

                

OH

(v)1,2-dichloropropan-1-ol

 

CClH2 CHCl CH2

                

OH

 

 (vi) Ethan1,2-diol

                                                           H   H

                             

          HOCH2CH2OH              H-O – C – C – O-H

 

H    H

 

  (vii) Propan1,2,3-triol                                  H   OH H                      

                                                                            

HOCH2CHOHCH2OH                 H-O – C- C – C – O-H

H  H    H

  1. LABORATORY PREPARATION OF ALKANOLS.

 

For decades the world over, people have been fermenting grapes juice, sugar, carbohydrates and starch to produce ethanol as a social drug for relaxation.

In large amount, drinking of ethanol by mammals /human beings causes mental and physical lack of coordination.

Prolonged intake of ethanol causes permanent mental and physical lack of coordination because it damages vital organs like the liver.

Fermentation is the reaction where sugar is converted to alcohol/alkanol using biological catalyst/enzymes in yeast.

It involves three processes:

(i)Conversion of starch to maltose using the enzyme diastase.

 

(C6H10O5)n (s) + H2O(l)    –diastase enzyme –> C12H22O11(aq)

           (Starch)                                                                   (Maltose)

 

(ii)Hydrolysis of Maltose to glucose using the enzyme maltase.

 

C12H22O11(aq)+ H2O(l)    — maltase enzyme –>2 C6H12O6(aq)

          (Maltose)                                                            (glucose)

 

(iii)Conversion of glucose to ethanol and carbon(IV)oxide gas using the enzyme zymase.

 

C6H12O6(aq)     — zymase enzyme –> 2 C2H5OH(aq)  + 2CO2(g)

(glucose)                                              (Ethanol)

 

At concentration greater than 15% by volume, the ethanol produced kills the yeast enzyme stopping the reaction.

To increases the concentration, fractional distillation is done to produce spirits (e.g. Brandy=40% ethanol).

Methanol is much more poisonous /toxic than ethanol.

Taken large quantity in small quantity it causes instant blindness and liver, killing the consumer victim within hours.

 

School laboratory preparation of ethanol from fermentation of glucose

Measure 100cm3 of pure water into a conical flask.

Add about five spatula end full of glucose.

Stir the mixture to dissolve.

Add about one spatula end full of yeast.

Set up the apparatus as below.

 

 

 

Preserve the mixture for about three days.

 

D.PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOLS

Use the prepared sample above for the following experiments that shows the characteristic properties of alkanols

 

  • Role of yeast

Yeast is a single cell fungus which contains the enzyme maltase and zymase that catalyse the fermentation process.

 

  • Observations in lime water.

A white precipitate is formed that dissolve to a colourless solution later. Lime water/Calcium hydroxide reacts with carbon(IV)0xide produced during the fermentation to form insoluble calcium carbonate and water.

More carbon (IV)0xide produced  during fermentation react with  the insoluble calcium carbonate and water to form soluble calcium hydrogen carbonate.

 

Ca(OH)2(aq)               +        CO2 (g)        ->       CaCO3(s)

H2O(l)    +   CO2 (g)  +    CaCO3(s) ->           Ca(HCO3) 2 (aq)

 

 (c)Effects on litmus paper

Experiment

Take the prepared sample and test with both blue and red litmus papers.

Repeat the same with pure ethanol and methylated spirit.

Sample Observation table

    

Substance/alkanol Effect on litmus paper
Prepared sample Blue litmus paper remain blue

Red litmus paper remain red

Absolute ethanol Blue litmus paper remain blue

Red litmus paper remain red

Methylated spirit Blue litmus paper remain blue

Red litmus paper remain red

Explanation 

Alkanols are neutral compounds/solution that have characteristic sweet smell and taste.

They have no effect on both blue and red litmus papers.

 

(d)Solubility in water.

Experiment

Place about 5cm3 of prepared sample into a clean test tube Add equal amount of distilled water.

Repeat the same with pure ethanol and methylated spirit.

Observation

No layers formed between the two liquids.

Explanation

Ethanol is miscible in water.Both ethanol and water are polar compounds .

The solubility of alkanols decrease with increase in the alkyl chain/molecular mass.

The alkyl group is insoluble in water while –OH functional group is soluble in water.

As the molecular chain becomes longer ,the effect of the alkyl group increases as the effect of the functional group decreases.

 

e)Melting/boiling point.

Experiment

Place pure ethanol in a long boiling tube .Determine its boiling point.

Observation

Pure ethanol has a boiling point of 78oC at sea level/one atmosphere pressure.

Explanation

The melting and boiling point of alkanols increase with increase in molecular chain/mass .

This is because the intermolecular/van-der-waals forces of attraction between the molecules increase.

More heat energy is thus required to weaken the longer chain during melting and break during boiling.

 

f)Density

Density of alkanols increase with increase in the intermolecular/van-der-waals forces of attraction between the molecule, making it very close to each other.

This reduces the volume occupied by the molecule and thus increase the their mass per unit volume (density).

 

Summary table showing the trend in physical properties of alkanols

 

Alkanol Melting point

(oC)

Boiling point

(oC)

Density

gcm-3

Solubility in water
Methanol -98 65 0.791 soluble
Ethanol -117 78 0.789 soluble
Propanol -103 97 0.803 soluble
Butanol -89 117 0.810 Slightly soluble
Pentanol -78 138 0.814 Slightly soluble
Hexanol -52 157 0.815 Slightly  soluble
Heptanol -34 176 0.822 Slightly soluble
Octanol -15 195 0.824 Slightly soluble
Nonanol -7 212 0.827 Slightly soluble
Decanol 6 228 0.827 Slightly soluble

 

g)Burning

 

Experiment

Place the prepared sample in a watch glass. Ignite. Repeat with pure ethanol and methylated spirit.

Observation/Explanation

Fermentation produce ethanol with a lot of water(about a ratio of 1:3)which prevent the alcohol from igniting.

Pure ethanol and methylated spirit easily catch fire / highly flammable.

They burn with an almost colourless non-sooty/non-smoky blue flame to form carbon(IV) oxide (in excess air/oxygen)or carbon(II) oxide (limited  air) and water.

Ethanol is thus a saturated compound like alkanes.

Chemica equation

C2 H5OH(l)    +   3O2 (g)      -> 3H2O(l)    +         2CO2 (g)  ( excess air)

C2 H5OH(l)    + 2O2 (g)      -> 3H2O(l)    +         2CO (g)  ( limited air)

 

2CH3OH(l)    + 3O2 (g)      -> 4H2O(l)    +         2CO2 (g)  ( excess air)

2 CH3OH(l)    +           2O2 (g)      -> 4H2O(l)    +         2CO (g)  ( limited air)

 

2C3 H7OH(l)    +    9O2 (g)      -> 8H2O(l)    +         6CO2 (g)  ( excess air)

C3 H7OH(l)    + 3O2 (g)      -> 4H2O(l)    +         3CO (g)  ( limited air)

 

2C4 H9OH(l)    +    13O2 (g)      -> 20H2O(l)    +     8CO2 (g)  ( excess air)

C4 H9OH(l)    + 3O2 (g)      -> 4H2O(l)    +         3CO (g)  ( limited air)

 

Due to its flammability, ethanol is used;

  • as a fuel in spirit lamps
  • as gasohol when blended with gasoline

 

(h)Formation of alkoxides

Experiment

Cut a very small piece of sodium. Put it in a beaker containing about 20cm3 of the prepared sample in a beaker.

Test the products with litmus papers. Repeat with absolute ethanol and methylated spirit.

Sample observations

Substance/alkanol Effect of adding sodium
Fermentation prepared sample (i)effervescence/fizzing/bubbles

(ii)colourless gas produced that extinguish burning splint with explosion/ “Pop” sound

(iii)colourless solution formed

(iv)blue litmus papers remain blue

(v)red litmus papers turn blue

Pure/absolute ethanol/methylated spirit (i)slow effervescence/fizzing/bubbles

(ii)colourless gas slowly  produced that extinguish burning splint with explosion/ “Pop” sound

(iii)colourless solution formed

(iv)blue litmus papers remain blue

(v)red litmus papers turn blue

Explanations

Sodium/potassium reacts slowly with alkanols to form basic solution called alkoxides and producing hydrogen gas.

If  the alkanol has some water the metals react faster with the water to form soluble hydroxides/alkalis i.e.

 

Sodium        +        Alkanol        ->      Sodium alkoxides        + Hydrogen gas

Potassium    +       Alkanol        ->       Potassium alkoxides    + Hydrogen gas

 

Sodium        +        Water            ->      Sodium hydroxides             + Hydrogen gas

Potassium    +       Water           ->       Potassium hydroxides        + Hydrogen gas

 

Examples

1.Sodium metal reacts with ethanol to form sodium ethoxide

Sodium metal reacts with water to form sodium Hydroxide

2CH3CH2OH(l)     +        2Na(s)         ->       2CH3CH2ONa (aq)    + H2 (s)

2H2O(l)       +        2Na(s)         ->       2NaOH (aq)    + H2 (s)

 

2.Potassium metal reacts with ethanol to form Potassium ethoxide

Potassium metal reacts with water to form Potassium Hydroxide

2CH3CH2OH(l)     +        2K(s)           ->       2CH3CH2OK (aq)    + H2 (s)

2H2O(l)       +        2K(s)           ->       2KOH (aq)    + H2 (s)

 

3.Sodium metal reacts with propanol to form sodium propoxide

Sodium metal reacts with water to form sodium Hydroxide

2CH3CH2 CH2OH(l)        +        2Na(s) -> 2CH3CH2 CH2ONa (aq)    + H2 (s)

2H2O(l)       +        2Na(s)         ->       2NaOH (aq)    + H2 (s)

 

4.Potassium metal reacts with propanol to form Potassium propoxide

Potassium metal reacts with water to form Potassium Hydroxide

2CH3CH2 CH2OH(l)        +        2K(s)           -> 2CH3CH2 CH2OK (aq)    + H2 (s)

2H2O(l)       +        2K(s)           ->       2KOH (aq)    + H2 (s)

 

5.Sodium metal reacts with butanol to form sodium butoxide

Sodium metal reacts with water to form sodium Hydroxide

2CH3CH2 CH2 CH2OH(l)  + 2Na(s) -> 2CH3CH2 CH2 CH2ONa (aq)    + H2 (s)

2H2O(l)       +        2Na(s)         ->       2NaOH (aq)    + H2 (s)

 

6.Sodium metal reacts with pentanol to form sodium pentoxide

Sodium metal reacts with water to form sodium Hydroxide

2CH3CH2 CH2 CH2 CH2OH(l)+2Na(s) -> 2CH3CH2 CH2 CH2 CH2ONa (aq) + H2 (s)

2H2O(l)       +        2Na(s)         ->       2NaOH (aq)    + H2 (s)

 

(i)Formation of Esters/Esterification

Experiment

Place 2cm3 of ethanol in a boiling tube.

Add equal amount of ethanoic acid.To the mixture add carefully 2drops of concentrated sulphuric(VI)acid.

Warm/Heat gently.

Pour the mixture into a beaker containing about 50cm3 of cold water.

Smell the products.

Repeat with methanol

Sample observations

 

Substance/alkanol Effect on adding equal amount of ethanol/concentrated sulphuric(VI)acid
Absolute ethanol Sweet fruity smell
Methanol Sweet fruity smell

Explanation

Alkanols react with alkanoic acids to form a group of homologous series of sweet smelling compounds called esters and water. This reaction is catalyzed by concentrated sulphuric(VI)acid in the laboratory.

 

Alkanol        +    Alkanoic acid   –Conc. H2SO4->    Ester   +  water

 

Naturally esterification is catalyzed by sunlight. Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids that create a variety of known natural(mostly in fruits) and synthetic(mostly in juices) esters .

Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g.

 

Ethanol        +        Ethanoic acid         ->       Ethylethanoate      +    Water

Ethanol        +        Propanoic acid       ->       Ethylpropanoate    +    Water

Ethanol        +        Methanoic acid      ->       Ethylmethanoate   +     Water

Ethanol        +        butanoic acid         ->       Ethylbutanoate      +     Water

Propanol      +        Ethanoic acid         ->       Propylethanoate     +     Water

Methanol     +        Ethanoic acid         ->       Methyethanoate     +     Water

Methanol     +        Decanoic acid        ->       Methyldecanoate    +     Water

Decanol       +        Methanoic acid      ->       Decylmethanoate   +      Water

 

During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol.

R-COOH   +    R2 –OH     ->    R-COO –R2    +  H2O

e.g.

  1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water.

Ethanol        + Ethanoic acid      –Conc. H2SO4 –>Ethylethanoate + Water

C2H5OH (l)   + CH3COOH(l) –Conc. H2SO4 –>  CH3COO C2H5(aq)   +H2O(l)

CH3CH2OH (l)+ CH3COOH(l) –Conc. H2SO4 –>  CH3COOCH2CH3(aq)   +H2O(l)

 

  1. Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water.

Ethanol        + Propanoic acid    –Conc. H2SO4 –>Ethylethanoate + Water

C2H5OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C2H5(aq)   +H2O(l)

CH3CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>

CH3 CH2COOCH2CH3(aq)   +H2O(l)

 

  1. Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water.

Methanol     + Ethanoic acid      –Conc. H2SO4 –>Methylethanoate + Water

CH3OH (l)    + CH3COOH(l) –Conc. H2SO4 –>  CH3COO CH3(aq)   +H2O(l)

 

  1. Methanol reacts with propanoic acid to form the ester methyl propanoate and water.

Methanol     + propanoic acid    –Conc. H2SO4 –>Methylpropanoate + Water

CH3OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –> CH3 CH2COO CH3(aq)   +H2O(l)

 

  1. Propanol reacts with propanoic acid to form the ester propylpropanoate and water.

Propanol      + Propanoic acid    –Conc. H2SO4 –>Ethylethanoate + Water

C3H7OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C3H7(aq)   +H2O(l)

CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>

CH3 CH2COOCH2 CH2CH3(aq)   +H2O(l)

 

(j)Oxidation

Experiment

Place 5cm3 of absolute ethanol in a test tube.Add three drops of acidified potassium manganate(VII).Shake thoroughly for one minute/warm.Test the solution mixture using pH paper. Repeat by adding acidified potassium dichromate(VII).

Sample observation table

Substance/alkanol Adding acidified KMnO4/K2Cr2O7 pH of resulting solution/mixture Nature of resulting solution/mixture
Pure ethanol (i)Purple colour of KMnO4decolorized

 

(ii) Orange colour of K2Cr2O7turns green.

pH= 4/5/6

 

 

pH = 4/5/6

Weakly acidic

 

 

Weakly acidic

 

 

Explanation

Both acidified KMnO4 and K2Cr2O7 are oxidizing agents(add oxygen to other compounds. They oxidize alkanols to a group of homologous series called alkanals then further oxidize them to alkanoic acids.The oxidizing agents are themselves reduced hence changing their colour:

(i) Purple KMnO4 is reduced to colourless Mn2+

(ii)Orange K2Cr2O7is reduced to green Cr3+

The pH of alkanoic acids show they have few H+ because they are weak acids i.e

 

Alkanol        +        [O]    ->       Alkanal       +        [O]     ->       alkanoic acid

 

    NB The [O] comes from the oxidizing agents acidified KMnO4 or K2Cr2O7

Examples

1.When ethanol is warmed with three drops of acidified KMnO4 there is decolorization of KMnO4

 

Ethanol    +     [O]    ->     Ethanal      +    [O]     ->   Ethanoic acid

CH3CH2OH   +    [O]    ->   CH3CH2O     +    [O]     ->   CH3COOH

 

2.When methanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.

methanol    +     [O]    ->     methanal      +    [O]     ->   methanoic acid

CH3OH     +     [O]    ->         CH3O        +    [O]     ->   HCOOH

 

3.When propanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.

Propanol    +     [O]    ->     Propanal      +    [O]     ->   Propanoic acid

CH3CH2 CH2OH   +    [O]    ->  CH3CH2 CH2O   +  [O]   -> CH3 CH2COOH

4.When butanol is warmed with three drops of acidified K2Cr2O7 ,the orange colour of acidified K2Cr2O7 changes to green.

Butanol    +     [O]    ->     Butanal      +    [O]     ->   Butanoic acid

CH3CH2 CH2 CH2OH  + [O]  ->CH3CH2 CH2CH2O +[O] -> CH3 CH2COOH

 

Air slowly oxidizes ethanol to dilute ethanoic acid commonly called vinegar. If beer is not tightly corked, a lot of carbon(IV)oxide escapes and there is slow oxidation of the beer making it “flat”.

 

(k)Hydrolysis /Hydration and Dehydration

  1. Hydrolysis/Hydration is the reaction of a compound/substance with water.

Alkenes react with water vapour/steam at high temperatures and high pressures in presence of phosphoric acid catalyst to form alkanols.i.e.

Alkenes       +        Water           – H3PO4 catalyst->  Alkanol

Examples

(i)Ethene is mixed with steam over a phosphoric acid catalyst at 300oC temperature and 60 atmosphere pressure to form ethanol

Ethene        + water    —60 atm/300oC/ H3PO4      –> Ethanol

H2C =CH2 (g)  +  H2O(l)   –60 atm/300oC/ H3PO4        –>   CH3 CH2OH(l)

This is the main method of producing large quantities of ethanol instead of fermentation

(ii) Propene        + water    —60 atm/300oC/ H3PO4      –> Propanol

CH3C =CH2 (g)  +  H2O(l)   –60 atm/300oC/ H3PO4        –>   CH3 CH2 CH2OH(l)

(iii) Butene        + water    —60 atm/300oC/ H3PO4      –> Butanol

CH3 CH2 C=CH2 (g) + H2O(l)   –60 atm/300oC/ H3PO4   –> CH3 CH2 CH2 CH2OH(l)

 

  1. Dehydration is the process which concentrated sulphuric(VI)acid (dehydrating agent) removes water from a compound/substances.

Concentrated sulphuric(VI)acid dehydrates alkanols to the corresponding alkenes at about 180oC. i.e

Alkanol  –Conc. H2 SO4/180oC–> Alkene  + Water

Examples

  1. At 180oC and in presence of Concentrated sulphuric(VI)acid, ethanol undergoes dehydration to form ethene.

Ethanol              —180oC/ H2SO4      –>    Ethene              +   Water

CH3 CH2OH(l)         –180oC/ H2SO4        –>    H2C =CH2 (g)  +  H2O(l)

  1. Propanol undergoes dehydration to form propene.

Propanol              —180oC/ H2SO4      –>    Propene              +   Water

CH3 CH2 CH2OH(l)  –180oC/ H2SO4        –>    CH3CH =CH2 (g)  +  H2O(l)

  1. Butanol undergoes dehydration to form Butene.

Butanol              —180oC/ H2SO4      –>    Butene              +   Water

CH3 CH2 CH2CH2OH(l)         –180oC/ H2SO4        –>    CH3 CH2C =CH2 (g)  +  H2O(l)

  1. Pentanol undergoes dehydration to form Pentene.

Pentanol              —180oC/ H2SO4     –>    Pentene              +   Water

CH3 CH2 CH2 CH2 CH2OH(l)–180oC/ H2SO4–>CH3 CH2 CH2C =CH2 (g)+H2O(l)

 

(l)Similarities of alkanols with Hydrocarbons

  1. Similarity with alkanes

Both alkanols and alkanes burn with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air)  and water. This shows they are saturated with high C:H ratio. e.g.

Both ethanol and ethane ignite and burns in air with a blue non-sooty flame to form carbon(IV)oxide(in excess air/oxygen)/carbon(II)oxide(in limited air)  and water.

 

CH2 CH2OH(l)   +  3O2(g)   -Excess air->    2CO2 (g)  +  3H2 O(l)

CH2 CH2OH(l)   +  2O2(g)   -Limited air->    2CO (g)  +  3H2 O(l)

 

CH3 CH3(g)   +    3O2(g)   -Excess air->     2CO2 (g)  +  3H2 O(l)

2CH3 CH3(g)   +  5O2(g)   -Limited air->    4CO (g)   +  6H2 O(l)

 

  1. Similarity with alkenes/alkynes

Both alkanols(R-OH) and alkenes/alkynes(with = C = C = double and  – C = C- triple ) bond:

(i)decolorize acidified KMnO4

(ii)turns Orange acidified K2Cr2O7  to green.

Alkanols(R-OH) are oxidized to alkanals(R-O) ant then alkanoic acids(R-OOH).

Alkenes are oxidized to alkanols with duo/double functional groups.

Examples

1.When ethanol is warmed with three drops of acidified K2Cr2O7 the orange of acidified K2Cr2O7 turns to green. Ethanol is oxidized to ethanol and then to ethanoic acid.

 

Ethanol          +     [O]    ->     Ethanal      +    [O]     ->   Ethanoic acid

CH3CH2OH   +     [O]    ->   CH3CH2O     +    [O]     ->   CH3COOH

 

2.When ethene is bubbled in a test tube containing acidified K2Cr2O7 ,the orange of acidified K2Cr2O7 turns to green. Ethene is oxidized to ethan-1,2-diol.

Ethene       +    [O]        ->       Ethan-1,2-diol.

H2C=CH2      +    [O]        ->      HOCH2 -CH2OH

 

III. Differences with alkenes/alkynes

Alkanols do not decolorize bromine and chlorine water.

Alkenes decolorizes bromine and chlorine water to form halogenoalkanols

Example

When ethene is bubbled in a test tube containing bromine water,the bromine water is decolorized. Ethene is oxidized to bromoethanol.

Ethene       +   Bromine water           ->       Bromoethanol.

H2C=CH2      +    HOBr                       ->       BrCH2 -CH2OH

 

  1. Differences in melting and boiling point with Hydrocarbons

Alkanos have higher melting point than the corresponding hydrocarbon (alkane/alkene/alkyne)

This is because most alkanols exist as dimer.A dimer is a molecule made up of two other molecules joined usually by van-der-waals forces/hydrogen bond or dative bonding.

Two alkanol molecules form a dimer joined by hydrogen bonding.

Example

In Ethanol the oxygen atom attracts/pulls the shared electrons in the covalent bond more to itself than Hydrogen.

This creates a partial negative charge (δ-) on oxygen and partial positive charge(δ+) on hydrogen.

Two ethanol molecules attract each other at the partial charges through Hydrogen bonding forming a dimmer.

 

 

Hydrogen bonds

H       H           H

Covalent bonds

 

 

H       C        C        O                 H       H

 

H       H           H     O     C        C        H

 

H       H

Dimerization of alkanols means more energy is needed to break/weaken the Hydrogen bonds before breaking/weakening the intermolecular forces joining the molecules of all organic compounds during boiling/melting.

 

E.USES OF SOME ALKANOLS

(a)Methanol is used as industrial alcohol and making methylated spirit

(b)Ethanol is used:

  1. as alcohol in alcoholic drinks e.g Beer, wines and spirits.

2.as antiseptic to wash woulds

3.in manufacture of vanishes, ink ,glue and paint because it is volatile and thus easily evaporate

4.as a fuel when blended with petrol to make gasohol.

 

 

 

 

B.ALKANOIC ACIDS (Carboxylic acids)

 

(A) INTRODUCTION.

 

Alkanoic acids belong to a homologous series of organic compounds with a general formula CnH2n +1 COOH and thus  -COOH  as the functional group .The 1st ten alkanoic acids include:

n General /molecular

formular

Structural formula IUPAC name
0 HCOOH  

H – C –O – H

O

Methanoic acid

 

1 CH3 COOH

 

        H

 

H – C  – C – O – H

H     O

Ethanoic acid
2 CH3 CH2 COOH

C2 H5 COOH

                                    H    H

 

H-C – C  – C – O – H

 

H    H     O

Propanoic acid
3 CH3 CH2 CH2 COOH

C3 H7 COOH

                             H     H    H

 

H- C –  C – C  – C – O – H

 

H    H    H     O

Butanoic acid
4 CH3CH2CH2CH2 COOH

C4 H9 COOH

                       H   H     H    H

 

H –  C – C –  C – C  – C – O – H

 

H    H    H    H     O

Pentanoic acid
5 CH3CH2 CH2CH2CH2 COOH

C5 H11 COOH

                H     H   H    H    H

 

H     C –  C – C –  C – C  – C – O – H

 

H    H    H    H    H     O

 

Hexanoic acid
6 CH3CH2 CH2 CH2CH2CH2 COOH

C6 H13 COOH

       H     H      H    H    H    H

 

H  C      C –  C – C –  C – C – C – O – H

 

H    H    H    H    H    H     O

 

Pentanoic acid

 

Alkanoic acids like alkanols /alkanes/alkenes/alkynes form a homologous series where:

(i)the general name of an alkanoic acids is derived from the alkane name then ending with “–oic”  acid as the table above shows.

(ii) the members have R-COOH/R    C-O-H    as the functional  group.

 

O

(iii)they have the same general formula represented by R-COOH where R is an alkyl group.

(iv)each member differ by –CH2– group from the next/previous.

(v)they show a similar and gradual change in their physical properties e.g. boiling and melting point.

(vi)they show similar and gradual change in their chemical properties.

(vii) since they are acids they show similar properties with mineral acids.

 

(B) ISOMERS OF ALKANOIC ACIDS.

 

Alkanoic acids exhibit both structural and position isomerism. The isomers are named by using the following basic guidelines

(i)Like alkanes. identify the longest carbon chain to be the parent name.

(ii)Identify the position of the   -C-O-H      functional  group to give it the smallest

 

O

/lowest position.

(iii)Identify the type and position of the side group branches.

 

Practice examples on isomers of alkanoic acids

1.Isomers of butanoic acid C3H7COOH

CH3 CH2 CH2 COOH

Butan-1-oic acid

CH3

 

H2C    C      COOH   2-methylpropan-1-oic acid

 

2-methylpropan-1-oic acid and Butan-1-oic acid are structural isomers because the position of the functional group does not change but the arrangement of the atoms in the molecule does.

 

 2.Isomers of pentanoic acid C4H9COOH

 

CH3CH2CH2CH2 COOH        pentan-1-oic acid

 

CH3

 

CH3CH2CH COOH   2-methylbutan-1-oic acid

 

CH3

 

H3C     C    COOH        2,2-dimethylpropan-1-oic acid

 

CH3

 

3.Ethan-1,2-dioic acid

O    O

 

HOOC- COOH      //                  H – O – C – C – O – H

 

4.Propan-1,3-dioic acid

O    H   O

 

HOOC- CH2COOH      //          H – O – C – C – C – O – H

 

H

 

 

5.Butan-1,4-dioic acid

O    H    H   O

 

HOOC CH2 CH2 COOH        H- O – C – C – C – C –O – H

 

H    H

 

 

6.2,2-dichloroethan-1,2-dioic acid

HOOCCHCl2                                                    Cl                                       

                             

H – O –  C – C – Cl

 

O   H

 

 

(C) LABORATORY AND INDUSTRIAL PREPARATIONOF ALKANOIC ACIDS.

 

In a school laboratory, alkanoic acids can be prepared by adding an oxidizing agent (H+/KMnO4 or H+/K2Cr2O7)to the corresponding alkanol then warming.

The oxidation converts the alkanol first to an alkanal the alkanoic acid.

NB Acidified KMnO4 is a stronger oxidizing agent than acidified K2Cr2O7

General equation:

R- CH2OH        + [O]            –H+/KMnO4–>     R- CH –O    +        H2O(l)

(alkanol)                                                             (alkanal)

 

R- CH – O   +    [O]         –H+/KMnO4–>     R- C –OOH

(alkanal)                                                  (alkanoic acid)

 

Examples

1.Ethanol on warming in acidified KMnO4 is oxidized to ethanal then ethanoic acid .

 

CH3– CH2OH    + [O]            –H+/KMnO4–>     CH3– CH –O          +           H2O(l)

(ethanol)                                                             (ethanal)

 

CH3– CH – O         +    [O]         –H+/KMnO4–>     CH3– C –OOH

(ethanal)                                                  (ethanoic acid)

 

2Propanol on warming in acidified KMnO4 is oxidized to propanal then propanoic acid

CH3– CH2 CH2OH       + [O]  –H+/KMnO4–>     CH3– CH2 CH –O +           H2O(l)

(propanol)                                                              (propanal)

 

CH3– CH – O         +    [O]         –H+/KMnO4–>     CH3– C –OOH

(propanal)                                                          (propanoic acid)

 

Industrially,large scale manufacture of alkanoic acid like ethanoic acid is obtained from:

(a)Alkenes reacting with steam at high temperatures and pressure in presence of phosphoric(V)acid catalyst and undergo hydrolysis to form alkanols. i.e.

 

Alkenes + Steam/water    — H2PO4 Catalyst–> Alkanol

The alkanol is then oxidized by air at 5 atmosphere pressure  with Manganese (II)sulphate(VI) catalyst to form the alkanoic acid.

Alkanol  +  Air      — MnSO4 Catalyst/5 atm pressure–> Alkanoic acid

 

Example

Ethene is mixed with steam over a phosphoric(V)acid catalyst,300oC temperature and 60 atmosphere pressure to form ethanol.

 

CH2=CH2     +    H2O   ->  CH3 CH2OH

(Ethene)                           (Ethanol)

 

This is the industrial large scale method of manufacturing ethanol

Ethanol is then oxidized  by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid.

 

CH3 CH2OH          +        [O] — MnSO4 Catalyst/5 atm pressure–>  CH3 COOH

(Ethanol)                                                                                          (Ethanoic acid)

 

(b)Alkynes react with liquid water at high temperatures and pressure in presence of Mercury(II)sulphate(VI)catalyst and 30% concentrated sulphuric(VI)acid to form alkanals.

Alkyne         +        Water — Mercury(II)sulphate(VI)catalyst–> Alkanal

 

The alkanal is then oxidized by air at 5 atmosphere pressure with Manganese (II) sulphate(VI) catalyst to form the alkanoic acid.

Alkanal        + air/oxygen — Manganese(II)sulphate(VI)catalyst–> Alkanoic acid

 

Example

Ethyne react with liquid water at high temperature and pressure with Mercury (II) sulphate (VI)catalyst and 30% concentrated sulphuric(VI)acid to form ethanal.

 

 

CH   =   CH           +    H2O   –HgSO4–>  CH3 CH2O

(Ethyne)                                                        (Ethanal)

 

This is another industrial large scale method of manufacturing ethanol from large quantities of ethyne found in natural gas.

Ethanal is then oxidized  by air at 5 atmosphere pressure with Manganese (II)sulphate(VI) catalyst to form the ethanoic acid.

 

CH3 CH2O   +        [O] — MnSO4 Catalyst/5 atm pressure–>  CH3 COOH

(Ethanal)   (Oxygen from air)                                                (Ethanoic acid)

 

 

(D) PHYSICAL AND CHEMICAL PROPERTIES OF ALKANOIC ACIDS.

 

I.Physical properties of alkanoic acids

 

The table below shows some physical properties of alkanoic acids

Alkanol Melting point(oC) Boiling point(oC) Density(gcm-3) Solubility in water
Methanoic acid 18.4 101 1.22 soluble
Ethanoic acid 16.6 118 1.05 soluble
Propanoic acid -2.8 141 0.992 soluble
Butanoic acid -8.0 164 0.964 soluble
Pentanoic acid -9.0 187 0.939 Slightly soluble
Hexanoic acid -11 205 0.927 Slightly soluble
Heptanoic acid -3 223 0.920 Slightly soluble
Octanoic acid 11 239 0.910 Slightly soluble
Nonanoic acid 16 253 0.907 Slightly soluble
Decanoic acid 31 269 0.905 Slightly soluble

 

From the table note the following:

  • Melting and boiling point decrease as the carbon chain increases due to increase in intermolecular forces of attraction between the molecules requiring more energy to separate the molecules.

 

  • The density decreases as the carbon chain increases as the intermolecular forces of attraction increases between the molecules making the molecule very close reducing their volume in unit mass.

 

  • Solubility decreases as the carbon chain increases as the soluble –COOH end is shielded by increasing insoluble alkyl/hydrocarbon chain.
  • Like alkanols ,alkanoic acids exist as dimmers due to the hydrogen bonds within the molecule. i.e..

 

II Chemical properties of alkanoic acids

The following experiments shows the main chemical properties of ethanoic (alkanoic) acid.

(a)Effect on litmus papers

Experiment

Dip both blue and red litmus papers in ethanoic acid. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute nitric(V)acid.

Sample observations

Solution/acid Observations/effect on litmus papers Inference
Ethanoic acid Blue litmus paper turn red

Red litmus paper remain red

H3O+/H+(aq)ion
Succinic acid Blue litmus paper turn red

Red litmus paper remain red

H3O+/H+(aq)ion
Citric acid Blue litmus paper turn red

Red litmus paper remain red

H3O+/H+(aq)ion
Oxalic acid Blue litmus paper turn red

Red litmus paper remain red

H3O+/H+(aq)ion
Tartaric acid Blue litmus paper turn red

Red litmus paper remain red

H3O+/H+(aq)ion
Nitric(V)acid Blue litmus paper turn red

Red litmus paper remain red

H3O+/H+(aq)ion

 

Explanation

All acidic solutions  contains H+/H3O+(aq) ions. The H+ /H3O+ (aq) ions is responsible for turning blue litmus paper/solution to red

(b)pH

Experiment

Place 2cm3 of ethaoic acid in a test tube. Add 2 drops of universal indicator solution and determine its pH. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI)acid.

Sample observations

Solution/acid pH Inference
Ethanoic acid 4/5/6 Weakly acidic
Succinic acid 4/5/6 Weakly acidic
Citric acid 4/5/6 Weakly acidic
Oxalic acid 4/5/6 Weakly acidic
Tartaric acid 4/5/6 Weakly acidic
Sulphuric(VI)acid 1/2/3 Strongly acidic

Explanations

Alkanoic acids are weak acids that partially/partly dissociate to release few H+ ions in solution. The pH of their solution is thus 4/5/6 showing they form weakly acidic solutions when dissolved in water.

All alkanoic acid dissociate to releases the “H” at the functional group in -COOH to form the alkanoate ion; –COO

Mineral acids(Sulphuric(VI)acid, Nitric(V)acid and Hydrochloric acid) are strong acids that wholly/fully dissociate to release many H+ ions in solution. The pH of their solution is thus 1/2/3 showing they form strongly acidic solutions when dissolved in water.i.e

Examples

  1. CH3COOH(aq) CH3COO(aq)         +       H+(aq)

(ethanoic acid)                           (ethanoate ion)                 (few H+ ion)

 

  1. CH3 CH2COOH(aq) CH3 CH2COO(aq)  +      H+(aq)

(propanoic acid)                                  (propanoate ion)           (few H+ ion)

 

  1. CH3 CH2 CH2COOH(aq)    CH3 CH2 CH2COO(aq)  + H+(aq)

(Butanoic acid)                                        (butanoate ion)               (few H+ ion)

 

  1. HOOH(aq)                                             HOO(aq)           +       H+(aq)

(methanoic acid)                          (methanoate ion)            (few H+ ion)

 

  1. H2 SO4 (aq)                     SO42- (aq)     +       2H+(aq)

(sulphuric(VI) acid)                             (sulphate(VI) ion)  (many  H+ ion)

 

  1. HNO3 (aq)                     NO3 (aq)      +       H+(aq)

(nitric(V) acid)                                (nitrate(V) ion)        (many  H+ ion)

 

(c)Reaction with metals

Experiment

Place about 4cm3 of ethanoic acid in a test tube. Put about 1cm length of polished magnesium ribbon. Test any gas produced using a burning splint. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.

Sample observations

Solution/acid Observations Inference
Ethanoic acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that burn with “pop” sound/explosion

H3O+/H+(aq)ion
Succinic acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that burn with “pop” sound/explosion

H3O+/H+(aq)ion
Citric acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that burn with “pop” sound/explosion

H3O+/H+(aq)ion
Oxalic acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that burn with “pop” sound/explosion

H3O+/H+(aq)ion
Tartaric acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that burn with “pop” sound/explosion

H3O+/H+(aq)ion
Nitric(V)acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that burn with “pop” sound/explosion

H3O+/H+(aq)ion

Explanation

Metals higher in the reactivity series displace the hydrogen in all acids to evolve/produce hydrogen gas and form a salt. Alkanoic acids react with metals with metals to form alkanoates salt and produce/evolve hydrogen gas .Hydrogen extinguishes a burning splint with a pop sound/explosion. Only the “H”in the functional group  -COOH is /are displaced and not in the alkyl hydrocarbon chain.

 

Alkanoic acid  +  Metal   ->   Alkanoate   +   Hydrogen gas. i.e.

Examples

  1. For a monovalent metal with monobasic acid

2R – COOH    +  2M       ->   2R- COOM   +   2H2(g)

 

2.For a divalent metal with monobasic acid

2R – COOH    +   M       ->   (R- COO) 2M   +   H2(g)

3.For a divalent metal with dibasic acid

HOOC-R-COOH+   M       ->  MOOC-R-COOM  +   H2(g)

4.For a monovalent metal with dibasic acid

HOOC-R-COOH+   2M       ->  MOOC-R-COOM  +   H2(g)

5 For mineral acids

(i)Sulphuric(VI)acid is a dibasic acid

H2 SO4 (aq)   + 2M     ->  M2 SO4 (aq) + H2(g)

H2 SO4 (aq)   + M     ->    MSO4 (aq)    + H2(g)

 

(ii)Nitric(V) and hydrochloric acid are  monobasic acid

HNO3 (aq)   + 2M     ->  2MNO3 (aq) + H2(g)

HNO3 (aq)   +  M     ->   M(NO3 ) 2 (aq) + H2(g)

 

Examples

1.Sodium reacts with ethanoic acid to form sodium ethanoate and produce. hydrogen gas.

Caution: This reaction is explosive.

CH3COOH (aq)   +   Na(s)  ->  CH3COONa (aq)  + H2(g)

(Ethanoic acid)                      (Sodium ethanoate)

 

2.Calcium reacts with ethanoic acid to form calcium ethanoate and produce. hydrogen gas.

2CH3COOH (aq)   +   Ca(s)  ->  (CH3COO) 2Ca (aq)  + H2(g)

(Ethanoic acid)                      (Calcium ethanoate)

 

3.Sodium reacts with ethan-1,2-dioic acid to form sodium ethan-1,2-dioate and produce. hydrogen gas.

HOOC-COOH+   2Na       ->  NaOOC – COONa  +   H2(g)

(ethan-1,2-dioic acid)                (sodium ethan-1,2-dioate)

 

Commercial name of ethan-1,2-dioic acid is oxalic acid. The salt is sodium oxalate.

 

4.Magnesium reacts with ethan-1,2-dioic acid to form magnesium ethan-1,2-dioate and produce. hydrogen gas.

 

HOOC-R-COOH+   Mg       ->  ( OOC – COO) Mg  +   H2(g)

(ethan-1,2-dioic acid)                (magnesium ethan-1,2-dioate)

 

5.Magnesium reacts with

(i)Sulphuric(VI)acid to form Magnesium sulphate(VI)

H2 SO4 (aq)   + Mg     ->  MgSO4 (aq) + H2(g)

 

(ii)Nitric(V) and hydrochloric acid are  monobasic acid

2HNO3 (aq)   +  Mg     ->   M(NO3 ) 2 (aq) + H2(g)

 

(d)Reaction with hydrogen carbonates and carbonates

Experiment

Place  about 3cm3 of ethanoic acid in a test tube. Add about 0.5g/ ½ spatula end full of sodium hydrogen carbonate/sodium carbonate. Test the gas produced using lime water. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.

Sample observations

Solution/acid Observations Inference
Ethanoic acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that forms a white precipitate with lime water

H3O+/H+(aq)ion
Succinic acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that forms a white precipitate with lime water

H3O+/H+(aq)ion
Citric acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that forms a white precipitate with lime water

H3O+/H+(aq)ion
Oxalic acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that forms a white precipitate with lime water

H3O+/H+(aq)ion
Tartaric acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that forms a white precipitate with lime water

H3O+/H+(aq)ion
Nitric(V)acid (i)effervescence, fizzing, bubbles

(ii)colourless gas produced that forms a white precipitate with lime water

H3O+/H+(aq)ion

 

All acids react with hydrogen carbonate/carbonate to form salt ,water and evolve/produce bubbles of carbon(IV)oxide and water.

Carbon(IV)oxide forms a white precipitate when bubbled in lime water/extinguishes a burning splint.

Alkanoic acids react with hydrogen carbonate/carbonate to form alkanoates ,water and evolve/produce bubbles of carbon(IV)oxide and water.

 

Alkanoic acid + hydrogen carbonate -> alkanoate + water + carbon(IV)oxide

 

Alkanoic acid +   carbonate      ->     alkanoate   +   water   + carbon(IV)oxide

Examples

  1. Sodium hydrogen carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.

CH3COOH (aq)   +   NaHCO3 (s)  ->  CH3COONa (aq)  + H2O(l) + CO2 (g)

(Ethanoic acid)                      (Sodium ethanoate)

 

2.Sodium carbonate reacts with ethanoic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.

2CH3COOH (aq)   +   Na2CO3 (s)  -> 2CH3COONa (aq)  + H2O(l) + CO2 (g)

(Ethanoic acid)                           (Sodium ethanoate)

 

3.Sodium carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.

HOOC-COOH+   Na2CO3 (s)       ->  NaOOC – COONa  + H2O(l) + CO2 (g)

(ethan-1,2-dioic acid)                (sodium ethan-1,2-dioate)

 

4.Sodium hydrogen carbonate reacts with ethan-1,2-dioic acid to form sodium ethanoate ,water and carbon(IV)oxide gas.

HOOC-COOH+   2NaHCO3 (s)       ->  NaOOC – COONa  + H2O(l) + 2CO2 (g)

(ethan-1,2-dioic acid)                (sodium ethan-1,2-dioate)

 

(e)Esterification

 

Experiment

Place 4cm3 of ethanol acid in a boiling tube.

Add equal volume of ethanoic acid. To the mixture, add 2 drops of concentrated sulphuric(VI)acid carefully. Warm/heat gently on Bunsen flame.

Pour the mixture into a beaker containing 50cm3 of water. Smell the products. Repeat with a solution of succinic acid, citric acid, oxalic acid, tartaric acid and dilute sulphuric (VI) acid.

 

Sample observations

Solution/acid Observations
Ethanoic acid Sweet fruity smell
Succinic acid Sweet fruity smell
Citric acid Sweet fruity smell
Oxalic acid Sweet fruity smell
Tartaric acid Sweet fruity smell
Dilute sulphuric(VI)acid No sweet fruity smell

 

Explanation

Alkanols react with alkanoic acid to form the sweet smelling homologous series of esters and water.The reaction is catalysed by concentrated sulphuric(VI)acid in the laboratory but naturally by sunlight /heat.Each ester has a characteristic smell derived from the many possible combinations of alkanols and alkanoic acids.

 

Alkanol        +        Alkanoic acids       ->       Ester  +       water

Esters derive their names from the alkanol first then alkanoic acids. The alkanol “becomes” an alkyl group and the alkanoic acid “becomes” alkanoate hence alkylalkanoate. e.g.

 

Ethanol        +        Ethanoic acid         ->       Ethylethanoate      +    Water

Ethanol        +        Propanoic acid       ->       Ethylpropanoate    +    Water

Ethanol        +        Methanoic acid      ->       Ethylmethanoate   +     Water

Ethanol        +        butanoic acid         ->       Ethylbutanoate      +     Water

Propanol      +        Ethanoic acid         ->       Propylethanoate     +     Water

Methanol     +        Ethanoic acid         ->       Methyethanoate     +     Water

Methanol     +        Decanoic acid        ->       Methyldecanoate    +     Water

Decanol       +        Methanoic acid      ->       Decylmethanoate   +      Water

 

During the formation of the ester, the “O” joining the alkanol and alkanoic acid comes from the alkanol.

R-COOH   +    R2 –OH     ->    R-COO –R2    +  H2O

 

Examples

  1. Ethanol reacts with ethanoic acid to form the ester ethyl ethanoate and water.

Ethanol        + Ethanoic acid      –Conc. H2SO4 –>Ethylethanoate + Water

C2H5OH (l)   + CH3COOH(l) –Conc. H2SO4 –>  CH3COO C2H5(aq)   +H2O(l)

CH3CH2OH (l)+ CH3COOH(l) –Conc. H2SO4 –>  CH3COOCH2CH3(aq)   +H2O(l)

 

  1. Ethanol reacts with propanoic acid to form the ester ethylpropanoate and water.

Ethanol        + Propanoic acid    –Conc. H2SO4 –>Ethylethanoate + Water

C2H5OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C2H5(aq)   +H2O(l)

CH3CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>

CH3 CH2COOCH2CH3(aq)   +H2O(l)

 

  1. Methanol reacts with ethanoic acid to form the ester methyl ethanoate and water.

Methanol     + Ethanoic acid      –Conc. H2SO4 –>Methylethanoate + Water

CH3OH (l)    + CH3COOH(l) –Conc. H2SO4 –>  CH3COO CH3(aq)   +H2O(l)

 

  1. Methanol reacts with propanoic acid to form the ester methyl propanoate and water.

Methanol     + propanoic acid    –Conc. H2SO4 –>Methylpropanoate + Water

CH3OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –> CH3 CH2COO CH3(aq)   +H2O(l)

 

  1. Propanol reacts with propanoic acid to form the ester propylpropanoate and water.

Propanol      + Propanoic acid    –Conc. H2SO4 –>Ethylethanoate + Water

C3H7OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>CH3CH2COO C3H7(aq)   +H2O(l)

CH3CH2 CH2OH (l)+ CH3 CH2COOH(l) –Conc. H2SO4 –>

CH3 CH2COOCH2 CH2CH3(aq)   +H2O(l)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

                             

  1. DETERGENTS

Detergents are cleaning agents that improve the cleaning power /properties of water.A detergent therefore should be able to:

(i)dissolve substances which water can not e.g grease ,oil, fat

(ii)be washed away after cleaning.

There are two types of detergents:

(a)Soapy detergents

(b)Soapless detergents

 

  • SOAPY DETERGENTS

Soapy detergents usually called soap is long chain salt of organic alkanoic acids.Common soap is sodium octadecanoate .It is derived from reacting concentrated sodium hydroxide solution with octadecanoic acid(18 carbon alkanoic acid) i.e.

Sodium hydroxide  +  octadecanoic acid   ->   Sodium octadecanoate   +   water

NaOH(aq) + CH3 (CH2) 16 COOH(aq)           -> CH3 (CH2) 16 COO –  Na(aq) +H2 O(l)

Commonly ,soap can thus be represented ;

R- COO –  Na+      where;

R is a long chain alkyl group and   -COO –  Na+   is the alkanoate ion.

In a school laboratory and at industrial and domestic level,soap is made by reacting concentrated sodium hydroxide solution with esters from (animal) fat and oil. The process of making soap is called saponification. During saponification ,the ester is hydrolyzed by the alkali to form sodium salt /soap and glycerol/propan-1,2,3-triol is produced.

 

Fat/oil(ester)+sodium/potassium hydroxide->sodium/potassium salt(soap)+ glycerol

 

Fats/Oils are esters with fatty acids and glycerol parts in their structure;

C17H35COOCH2

 

C17H35COOCH

 

C17H35COOCH2

 

When boiled with concentrated sodium hydroxide solution NaOH;

(i)NaOH ionizes/dissociates into Naand  OHions

(ii)fat/oil split into three C17H35COOand one CH2 CH  CH2

(iii) the three Na+ combine with the three C17H35COOto form the salt     C17H35COO Na+

(iv)the three OHions combine with the CH2 CH  CH2 to form an alkanol with  three functional groups CH2 OH CH OH  CH2 OH(propan-1,2,3-triol)

C17H35COOCH2                                                                                                     CH2OH                                                                                                             

C17H35COOCH    +NaOH -> 3 C17H35COO Na+  +                    CHOH

 

 

C17H35COOCH2                                                                                                CH2OH

Ester                     Alkali Soap                                glycerol

 

Generally:

CnH2n+1COOCH2                                                                                                   CH2OH                                                                                                             

CnH2n+1COOCH    +NaOH -> 3 CnH2n+1COO Na+  +       CHOH

 

CnH2n+1COOCH2                                                                                              CH2OH

Ester                     Alkali Soap                                glycerol

 

R – COOCH2                                                                                                   CH2OH                                                                                                             

R – COOCH    +NaOH -> 3R-COO Na+           +                     CHOH

 

R- COOCH2                                                                                              CH2OH

         Ester              Alkali         Soap                                glycerol

 

During this process a little sodium chloride is added to precipitate the soap by reducing its solubility. This is called salting out.

The soap is then added colouring agents ,perfumes and herbs of choice.

 

School laboratory preparation of soap

Place about 40 g of fatty (animal fat)beef/meat in 100cm3  beaker .Add about 15cm3 of 4.0M sodium hydroxide solution. Boil the mixture for about 15minutes.Stir the mixture .Add about 5.0cm3 of distilled water as you boil to make up for evaporation. Boil for about another 15minutes.Add about four spatula end full of pure sodium chloride crystals. Continue stirring for another five minutes. Allow to cool. Filter of /decant and wash off the residue with distilled water .Transfer the clean residue into a dry beaker. Preserve.

The action of soap

Soapy detergents:

(i)act by reducing the surface tension of water by forming a thin layer on top of the water.

(ii)is made of a non-polar alkyl /hydrocarbon tail and a polar  -COONa+ head. The non-polar alkyl /hydrocarbon tail is hydrophobic (water hating) and thus does not dissolve in water .It dissolves in non-polar solvent like grease, oil and fat. The polar  -COONa+ head is hydrophilic (water loving)and thus dissolve in water. When washing with soapy detergent, the non-polar tail of the soapy detergent surround/dissolve in the dirt on the garment /grease/oil while the polar head dissolve in water.

Through mechanical agitation/stirring/sqeezing/rubbing/beating/kneading, some  grease is dislodged/lifted of the surface of the garment. It is immediately surrounded by more soap molecules It float and spread in the water as tiny droplets that scatter light in form of emulsion making the water cloudy and shinny. It is  removed from the garment by rinsing with fresh water.The repulsion of the soap head prevent /ensure the droplets do not mix.Once removed, the dirt molecules cannot be redeposited back because it is surrounded by soap molecules.

 

Advantages and disadvantages of using soapy detergents

Soapy detergents are biodegradable. They are acted upon by bacteria and rot.They thus do not cause environmental pollution.

Soapy detergents have the diadvatage in that:

(i)they are made from fat and oils which are better eaten as food than make soap.

(ii)forms an insoluble precipitate with hard water called scum. Scum is insoluble calcium octadecanoate and Magnesium octadecanoate  formed when soap reacts with Ca2+ and Mg2+ present in hard water.

Chemical equation

2C17H35COO Na+ (aq)    +   Ca2+(aq)    ->      (C17H35COO )Ca2+ (s)  +     2Na+(aq)

(insoluble Calcium octadecanote/scum)

2C17H35COO Na+ (aq)    +   Mg2+(aq)    ->     (C17H35COO )Mg2+ (s)  +     2Na+(aq)

(insoluble Magnesium octadecanote/scum)

This causes wastage of soap.

Potassium soaps are better than Sodium soap. Potassium is more expensive than sodium and thus its soap is also more expensive.

 

(b)SOAPLESS DETERGENTS

Soapless detergent usually called detergent is a long chain salt fromed from by-products of fractional distillation of crude oil.Commonly used soaps include:

(i)washing agents

(ii)toothpaste

(iii)emulsifiers/wetting agents/shampoo

Soapless detergents are derived from reacting:

(i)concentrated sulphuric(VI)acid with a long chain alkanol e.g. Octadecanol(18 carbon alkanol) to form alkyl hydrogen sulphate(VI)

Alkanol   +  Conc sulphuric(VI)acid  ->  alkyl hydrogen sulphate(VI)  + Water

R –OH  +         H2SO4           -> R –O-SO3H     +      H2O

 

(ii)the alkyl hydrogen sulphate(VI) is then neutralized with sodium/potassium hydroxide to form sodium/potassium alkyl hydrogen sulphate(VI)

Sodium/potassium alkyl hydrogen sulphate(VI) is the soapless detergent.

 

alkyl hydrogen         +    Potassium/sodium  ->        Sodium/potassium      +      Water

sulphate(VI)                     hydroxide            alkyl hydrogen sulphate(VI)

R –O-SO3H              +      NaOH                   ->          R –O-SO3 Na+        +      H2O

 

Example

Step I : Reaction of Octadecanol with Conc.H2SO4

C17H35CH2OH (aq)  +  H2SO4          ->  C17H35CH2O- SO3 H+ (aq)   +   H2O (l)

 

octadecanol + sulphuric(VI)acid -> Octadecyl hydrogen sulphate(VI) + water

 

Step II: Neutralization  by an alkali

C17H35CH2O- SO3 H+ (aq) +  NaOH       ->  C17H35CH2O- SO3 Na+ (aq)  +   H2O (l)

 

Octadecyl hydrogen  + sodium/potassium -> sodium/potassium octadecyl+Water

sulphate(VI)                    hydroxide                   hydrogen sulphate(VI)

 

School laboratory preparation of soapless detergent

Place about 20g of olive oil in a 100cm3 beaker. Put  it in a trough containing ice cold water.

Add dropwise carefully 18M concentrated sulphuric(VI)acid stirring continuously into the olive oil until the oil turns brown.Add 30cm3 of 6M sodium hydroxide solution.Stir.This is a soapless detergent.

 

The action of soapless detergents

 

The action of soapless detergents is similar to that of soapy detergents.The soapless detergents contain the hydrophilic head and a long hydrophobic tail. i.e.

 

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-COONa+

 

vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv-O-SO3 Na+

(long hydrophobic /non-polar alkyl tail)          (hydrophilic/polar/ionic head)

 

The tail dissolves in fat/grease/oil while the ionic/polar/ionic head dissolves in water.

The tail stick to the dirt which is removed by the attraction of water molecules and the polar/ionic/hydrophilic head by mechanical agitation /squeezing/kneading/ beating/rubbing/scrubbing/scatching.

The suspended dirt is then surrounded by detergent molecules and repulsion of the anion head preventing the dirt from sticking on the material garment.

The tiny droplets of dirt emulsion makes the water cloudy. On rinsing the cloudy emulsion is washed away.

 

Advantages and disadvantages of using soapless detergents

Soapless detergents are non-biodegradable unlike soapy detergents.

They persist in water during sewage treatment by causing foaming in rivers ,lakes and streams leading to marine /aquatic death.

Soapless detergents have the advantage in that they:

(i)do not form scum with hard water.

(ii)are cheap to manufacture/buying

(iii)are made from petroleum products but soapis made from fats/oil for    human consumption.

 

Sample revision questions

 

  1. Study the scheme below
KOH
Fat/oil
Residue X
Filtrate Y
Filtration
Sodium Chloride
Boiling

 

(a)Identify the process

Saponification

(b)Fats and oils are esters. Write the formula of the a common structure of ester

C17H35COOCH2

 

C17H35COOCH

 

C17H35COOCH2

 

(c)Write a balanced equation for the reaction taking place during boiling

C17H35COOCH2                                                                                                      CH2OH                                                                                                              

C17H35COOCH    +3NaOH   -> 3 C17H35COO Na+       CHOH

 

 

C17H35COOCH2                                                                                                 CH2OH

Ester                      Alkali              Soap                             glycerol

 

(d)Give the IUPAC name of:

          (i)Residue X

                    Potassium octadecanoate

(ii)Filtrate Y

                    Propan-1,2,3-triol

 

(e)Give one use of fitrate Y

Making paint

 

(f)What is the function of sodium chloride

          To reduce the solubility of the soap hence helping in precipitating it out

 

(g)Explain how residue X helps in washing.

          Has a non-polar hydrophobic tail that dissolves in dirt/grease /oil/fat

Has a polar /ionic hydrophilic head that dissolves in water.

From mechanical agitation,the dirt is plucked out of the garment and surrounded by the tail end preventing it from being deposited back on the garment.

 

(h)State one:

          (i)advantage of continued use of residue X on the environment

                    Is biodegradable and thus do not pollute the environment

(ii)disadvantage of using residue X

Uses fat/oil during preparation/manufacture which are better used for human consumption.

(i)Residue X was added dropwise to some water.The number of drops used before lather forms is as in the table below.

 

  Water sample

 

 

A

 

B

 

C

Drops of residue X 15 2 15
Drops of residue X in boiled water 2 2 15

 

(i)State and explain which sample of water is:

  1. Soft

Sample B .Very little soap is used and no effect on amount of soap even on boiling/heating.

  1. Permanent hard

Sample C . A lot of soap is used and no effect on amount of soap even on boiling/heating. Boiling does not remove permanent hardness of water.

III. Temporary hard

Sample A . A lot of soap is used before boiling. Very little soap is used on boiling/heating. Boiling remove temporary hardness of water.

 

(ii)Write the equation for the reaction at water sample C.

 

Chemical equation

2C17H35COO K+ (aq)    +   CaSO4(aq)    ->    (C17H35COO )Ca2+ (s)  +     K2SO4(aq)

(insoluble Calcium octadecanote/scum)

Ionic equation

2C17H35COO K+ (aq)    +   Ca2+(aq)    ->        (C17H35COO )Ca2+ (s)  +     2K+(aq)

(insoluble Calcium octadecanote/scum)

Chemical equation

2C17H35COO K+ (aq)    +   MgSO4(aq)    ->   (C17H35COO )Mg2+ (s)  +     K2SO4(aq)

(insoluble Calcium octadecanote/scum)

Ionic equation

2C17H35COO K+ (aq)    +   Mg2+(aq)    ->      (C17H35COO )Mg2+ (s)  +     2K+(aq)

(insoluble Magnesium octadecanote/scum)

 

(iii)Write the equation for the reaction at water sample A before boiling.

Chemical equation

2C17H35COO K+ (aq)    Ca(HCO3)(aq)   ->(C17H35COO )Ca2+ (s)  +  2KHCO3 (aq)

(insoluble Calcium octadecanote/scum)

Ionic equation

2C17H35COO K+ (aq)    +   Ca2+(aq)    ->        (C17H35COO )Ca2+ (s)  +     2K+(aq)

(insoluble Calcium octadecanote/scum)

Chemical equation

2C17H35COO K+ (aq)    Mg(HCO3)(aq) ->(C17H35COO )Mg2+ (s)  +   2KHCO3 (aq)

(insoluble Calcium octadecanote/scum)

Ionic equation

2C17H35COO K+ (aq)    +   Mg2+(aq)    ->      (C17H35COO )Mg2+ (s)  +     2K+(aq)

(insoluble Magnesium octadecanote/scum)

 

(iv)Explain how water becomes hard

Natural or rain water flowing /passing through rocks containing calcium (chalk, gypsum, limestone)and magnesium compounds (dolomite)dissolve them to form soluble Ca2+  and  Mg2+ ions that causes water hardness.

 

(v)State two useful benefits of hard water

          -Used in bone and teeth formation

-Coral polyps use hard water to form coral reefs

-Snails use hard water to make their shells

 

2.Study the scheme below and use it to answer the questions that follow.

 

Conc. H2SO4
Substance B
6M sodium hydroxide
Brown solid A
Ice cold water
Olive oil

                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                               

(a)Identify :

          (i)brown solid A

                    Alkyl hydrogen sulphate(VI)

(ii)substance B

                    Sodium alkyl hydrogen sulphate(VI)

(b)Write a general formula of:

          (i)Substance A.

                                             O

 

R-O-S O3  H  // R- O – S –  O  – H 

 

                                   O

(ii)Substance B                                             O

 

R-O-S O3 –  Na+                                                     R- O – S –  O –  Na+

 

                                                                       O

(c)State one

(i) advantage of continued use of substance B

          –Does not form scum with hard water

-Is cheap to make

-Does not use food for human as a raw material.

 

(ii)disadvantage of continued use of substance B.

          Is non-biodegradable therefore do not pollute the environment

 

(d)Explain the action of B during washing.

          Has a non-polar hydrocarbon long tail that dissolves in dirt/grease/oil/fat.

Has a polar/ionic hydrophilic head that dissolves in water

Through mechanical agitation the dirt is plucked /removed from the garment and surrounded by the tail end preventing it from being deposited back on the garment.

(e) Ethene was substituted for olive oil in the above process. Write the equation and name of the new products A and B.

Product A

Ethene         +        Sulphuric(VI)acid            ->       Ethyl hydrogen sulphate(VI)

H2C=CH2             +        H2SO4                   –>      H3C – CH2 –O-SO3H

 

Product B

Ethyl hydrogen sulphate(VI)  + sodium hydroxide  ->  sodium Ethyl    +        Water

hydrogen sulphate(VI)

 

H3C – CH2 –O-SO3H        +     NaOH     ->  H3C – CH2 –O-SO3Na+  + H2O

(f)Ethanol can also undergo similar reactions forming new products A and B.Show this using a chemical equation.

Product A

Ethanol        +        Sulphuric(VI)acid            ->Ethyl hydrogen sulphate(VI) + water

H3C-CH2OH        +        H2SO4                   –>      H3C – CH2 –O-SO3H  + H2O

 

Product B

Ethyl hydrogen sulphate(VI)  + sodium hydroxide  ->  sodium Ethyl    +        Water

hydrogen sulphate(VI)

 

H3C – CH2 –O-SO3H        +     NaOH     ->  H3C – CH2 –O-SO3Na+  + H2O

 

 

3.Below is part of a detergent

          H3C  – (CH2 )16    –  O  –  SO–  K +

(a)Write the formular of the polar and non-polar end

Polar end

H3C  – (CH2 )16   

Non-polar end

–  O  –  SO–  K +

(b)Is the molecule a soapy or saopless detergent?

          Soapless detergent

(c)State one advantage of using the above detergent

          -does not form scum with hard water

-is cheap to manufacture

 

4.The structure of a detergent is

 

H    H    H    H H   H  H   H   H   H  H    H     H

 

 

 

H- C-  C-  C-C-  C-  C-   C- C-   C- C -C-  C-   -C- COONa+

 

 

H   H   H    H   H   H   H   H   H   H  H   H   H

 

  1. a) Write the molecular formula of the detergent. (1mk)

CH3(CH2)12COONa+

 

  1. b) What type of detergent is represented by the formula? (1mk)

Soapy detergent

  1. c) When this type of detergent is used to wash linen in hard water, spots (marks) are left on the linen. Write the formula of the substance responsible for the spots (CH3(CH2)12COO)2Ca2+       /  CH3(CH2)12COO)2Mg2+

 

 

 

 

 

 

 

 

 

  1. POLYMERS AND FIBRES

Polymers and fibres are giant molecules of organic compounds. Polymers and fibres are formed when small molecules called monomers join together to form large molecules called polymers at high temperatures and pressures. This process is called polymerization.

Polymers and fibres are either:

(a)Natural polymers and fibres

(b)Synthetic polymers and fibres

Natural polymers and fibres are found in living things(plants and animals) Natural polymers/fibres include:

-proteins/polypeptides making amino acids in animals

-cellulose that make cotton,wool,paper and silk

-Starch that come from glucose

-Fats and oils

-Rubber from latex in rubber trees.

 

Synthetic polymers and fibres are man-made. They include:

-polyethene

-polychloroethene

-polyphenylethene(polystyrene)

-Terylene(Dacron)

-Nylon-6,6

-Perspex(artificial glass)

Synthetic polymers and fibres have the following characteristic advantages over natural polymers

  1. They are light and portable
  2. They are easy to manufacture.
  3. They can easily be molded into shape of choice.
  4. They are resistant to corrosion, water, air , acids, bases and salts.
  5. They are comparatively cheap, affordable, colourful and aesthetic

 

Synthetic polymers and fibres however have the following disadvantages over natural polymers

  1. They are non-biodegradable and hence cause environmental pollution during disposal
  2. They give out highly poisonous gases when burnt like chlorine/carbon(II)oxide
  3. Some on burning produce Carbon(IV)oxide. Carbon(IV)oxide is a green house gas that cause global warming.
  4. Compared to some metals, they are poor conductors of heat,electricity and have lower tensile strength.

To reduce environmental pollution from synthetic polymers and fibres, the followitn methods of disposal should be used:

1.Recycling: Once produced all synthetic polymers and fibres should be recycled to a new product. This prevents accumulation of the synthetic polymers and fibres in the environment.

2.Production of biodegradable synthetic polymers and fibres that rot away.

 

There are two types of polymerization:

(a)addition polymerization

(b)condensation polymerization

 

(a)addition polymerization

Addition polymerization is the process where a small unsaturated monomer (alkene ) molecule join together to form a large saturated molecule. Only alkenes undergo addition polymerization.

Addition polymers are named from the alkene/monomer making the polymer and adding the prefix “poly” before the name of monomer to form a polyalkene

 

During addition polymerization

(i)the double bond in alkenes break

(ii)free radicals are formed

(iii)the free radicals collide with each other and join to form a larger molecule. The more collisions the larger the molecule.

 

Examples of addition polymerization

 

1.Formation of Polyethene

Polyethene is an addition polymer formed when ethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting paticles)

 

H    H                     H    H                   H    H                    H    H

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

H    H                     H    H                   H    H                    H    H

 

Ethene         +        Ethene         +        Ethene         +        Ethene    + 

 

(ii)the double bond joining the ethane molecule break to free readicals

 

H    H                     H    H                   H    H                    H    H

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

H    H                     H    H                   H    H                    H    H

 

Ethene radical     +  Ethene radical    +   Ethene radical   +   Ethene radical    + 

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

H    H    H    H  H    H   H    H             lone pair of electrons

 

  • C – C – C – C – C – C – C – C• +  

 

H    H    H    H  H    H   H    H

Lone pair of electrons can be used to join more monomers to form longer polyethene.

Polyethene molecule can be represented as:

 

 

 

H    H    H    H  H    H   H    H             extension of

molecule/polymer

– C – C – C – C – C – C – C – C-     +  

 

H    H    H    H  H    H   H    H

Since the molecule is a repetition of one monomer, then the polymer is:

H    H

 

( C – C  )n

 

H    H

Where n is the number of monomers in the polymer. The number of monomers in the polymer can be determined from the molar mass of the polymer and monomer from the relationship:

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

 

Examples

Polythene has a molar mass of 4760.Calculate the number of ethene molecules in the polymer(C=12.0, H=1.0 )

Number of monomers/repeating units in polyomer =  Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C2H4 )= 28 Molar mass polyethene = 4760

 

Substituting           4760           =                 170 ethene molecules

28

The commercial name of polyethene is polythene. It is an elastic, tough, transparent and durable plastic. Polythene is used:

(i)in making plastic bag

(ii)bowls and plastic bags

(iii)packaging materials

 

2.Formation of Polychlorethene

Polychloroethene is an addition polymer formed when chloroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 

H    H                     H    H                   H    H                    H    H

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

H    Cl                     H    Cl                   H    Cl                     H    Cl

 

chloroethene          +   chloroethene     +    chloroethene    +    chloroethene    + 

 

(ii)the double bond joining the chloroethene molecule break to free radicals

 

H    H                     H    H                   H    H                    H    H

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

H    Cl                     H    Cl                   H    Cl                    H    Cl

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

H    H    H    H  H    H   H    H             lone pair of electrons

 

  • C – C – C – C – C – C – C – C•     +  

 

H   Cl    H   Cl  H    Cl   H    Cl

Lone pair of electrons can be used to join more monomers to form longer polychloroethene.

Polychloroethene molecule can be represented as:

 

H    H    H    H  H    H   H    H             extension of

molecule/polymer

– C – C – C – C – C – C – C – C-     +  

 

H   Cl    H   Cl  H   Cl   H   Cl

 

Since the molecule is a repetition of one monomer, then the polymer is:

H    H

 

( C – C  )n

 

H    Cl

 

Examples

Polychlorothene has a molar mass of 4760.Calculate the number of chlorethene molecules in the polymer(C=12.0, H=1.0,Cl=35.5 )

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C2H3Cl )= 62.5 Molar mass polyethene = 4760

 

Substituting 4760   =  77.16   => 77  polychloroethene molecules(whole number)

62.5

The commercial name of polychloroethene is polyvinylchloride(PVC). It is a tough, non-transparent and durable plastic. PVC is used:

(i)in making plastic rope

(ii)water pipes

(iii)crates  and boxes

 

 

3.Formation of Polyphenylethene

Polyphenylethene is an addition polymer formed when phenylethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 

H    H                     H    H                   H    H                    H    H

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

H    C6H5                H    C6H5              H    C6H5                H    C6H5

 

phenylethene         +   phenylethene    +    phenylethene   +    phenylethene    + 

 

(ii)the double bond joining the phenylethene molecule break to free radicals

 

H    H                     H    H                   H    H                    H    H

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

H    C6H5                H    C6H5              H    C6H5               H    C6H5

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

H         H      H       H       H       H       H        H             lone pair of electrons

 

  • C –   C   –   C   –   C   –   C   –   C   –   C   –   C  •     +  

 

H         C6H5   H    C6H5  H       C6H5   H       C6H5

Lone pair of electrons can be used to join more monomers to form longer polyphenylethene.

Polyphenylethene molecule can be represented as:

H         H      H       H       H       H       H        H

 

– C   –    C   –    C   –   C   –   C   –   C   –   C   –   C  –

 

H         C6H5   H    C6H5  H       C6H5   H       C6H5

 

Since the molecule is a repetition of one monomer, then the polymer is:

H    H

 

( C – C  )n

 

H    C6H5

 

Examples

Polyphenylthene has a molar mass of 4760.Calculate the number of phenylethene molecules in the polymer(C=12.0, H=1.0, )

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C8H8 )= 104 Molar mass polyethene = 4760

 

Substituting 476045.7692  =>45  polyphenylethene molecules(whole number)

104

The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:

(i)in making packaging material for carrying delicate items like computers, radion,calculators.

(ii)ceiling tiles

(iii)clothe linings

 

4.Formation of Polypropene

Polypropene is an addition polymer formed when propene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 

H    H                     H    H                   H    H                    H    H

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

H    CH3                 H    CH3               H    CH3                 H    CH3

 

propene       +   propene  +    propene +    propene   + 

 

(ii)the double bond joining the phenylethene molecule break to free radicals

 

H    H                     H    H                   H    H                    H    H

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

H    CH3                 H    CH3               H    CH3                H    CH3

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

H         H      H       H       H       H       H        H             lone pair of electrons

 

  • C –   C   –   C   –   C   –   C   –   C   –   C   –   C  •     +  

 

H         CH3   H        CH3  H        CH3   H        CH3

Lone pair of electrons can be used to join more monomers to form longer propene.

propene molecule can be represented as:

H         H      H       H       H       H       H        H

 

– C   –    C   –    C   –   C   –   C   –   C   –   C   –   C  –

 

H         CH3   H        CH3  H        CH3   H        CH3

 

Since the molecule is a repetition of one monomer, then the polymer is:

H    H

 

( C – C  )n

 

H    CH3

Examples

Polypropene has a molar mass of 4760.Calculate the number of propene molecules in the polymer(C=12.0, H=1.0, )

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

=> Molar mass propene (C3H8 )= 44 Molar mass polyethene = 4760

 

Substituting 4760108.1818  =>108  propene molecules(whole number)

44

The commercial name of polyphenylethene is polystyrene. It is a very light durable plastic. Polystyrene is used:

(i)in making packaging material for carrying delicate items like computers, radion,calculators.

(ii)ceiling tiles

(iii)clothe linings

 

 

 

 

5.Formation of Polytetrafluorothene

Polytetrafluorothene is an addition polymer formed when tetrafluoroethene molecule/monomer join together to form a large molecule/polymer at high temperatures and pressure.

During polymerization:

(i)many molecules are brought nearer to each other by the high pressure(which reduces the volume occupied by reacting particles)

 

F    F                       F     F                    F     F                      F    F

 

C = C           +        C = C          +        C = C           +        C = C     +  

 

F    F                       F     F                    F     F                      F    F

 

tetrafluoroethene    +  tetrafluoroethene+  tetrafluoroethene+ tetrafluoroethene + 

 

(ii)the double bond joining the tetrafluoroethene molecule break to free radicals

 

F    F                       F    F                     F    F                      F    F

 

  • C – C•      +        •C – C•         +       •C – C•          +       •C – C•     +  

 

F    F                       F    F                     F    F                      F    F

 

(iii)the free radicals collide with each other and join to form a larger molecule

 

F    F     F     F   F     F   F    F             lone pair of electrons

 

  • C – C – C – C – C – C – C – C• +  

 

F     F     F    F   F    F     F    F

Lone pair of electrons can be used to join more monomers to form longer polytetrafluoroethene.

polytetrafluoroethene molecule can be represented as:

 

F    F     F     F   F     F   F    F             extension of

molecule/polymer

– C – C – C – C – C – C – C – C-     +  

 

F     F     F    F   F    F     F    F

 

Since the molecule is a repetition of one monomer, then the polymer is:

F    F

 

( C – C  )n

 

F     F

 

Examples

Polytetrafluorothene has a molar mass of 4760.Calculate the number of tetrafluoroethene molecules in the polymer(C=12.0, ,F=19 )

Number of monomers/repeating units in monomer =  Molar mass polymer

Molar mass monomer

=> Molar mass ethene (C2F4 )= 62.5 Molar mass polyethene = 4760

 

Substituting 4760   =  77.16   => 77  polychloroethene molecules(whole number)

62.5

The commercial name of polytetrafluorethene(P.T.F.E) is Teflon(P.T.F.E). It is a tough, non-transparent and durable plastic. PVC is used:

(i)in making plastic rope

(ii)water pipes

(iii)crates  and boxes

 

5.Formation of rubber from Latex

Natural rubber is obtained from rubber trees.

During harvesting an incision is made on the rubber tree to produce a milky white substance called latex.

Latex is a mixture of rubber and lots of water.

The latex is then added an acid to coagulate the rubber.

Natural rubber is a polymer of 2-methylbut-1,3-diene ;

 

H         CH3       H        H

 

CH2=C (CH3) CH = CH2       H   –   C    =    C    –    C    =    C    –   H

 

During natural polymerization to rubber, one double  C=C bond break to self add to another molecule.The double bond remaining move to carbon “2” thus;

 

H         CH3      H          H           H        CH3       H        H

 

–   C    –    C     =   C    –    C       –   C    –    C    =    C        C    –

 

H                                            H           H                                 H

Generally the structure of rubber is thus;

 

H         CH3      H          H

 

-(-   C    –    C    =    C    –    C   -)n

 

 

H                                 H

Pure rubber is soft and sticky.It is used to make erasers, car tyres. Most of it is vulcanized.Vulcanization is the process of heating rubber with sulphur to make it harder/tougher.

During vulcanization the sulphur atoms form a cross link between chains of rubber molecules/polymers. This decreases the number of C=C double bonds in the polymer.

 

Sulphur atoms make cross link between polymers

H         CH3      H        H           H        CH3       H        H

 

–   C    –    C     –   C    –    C       –   C    –    C        C        C    –

 

H                      S         H           H                                S         H

 

 

 

H         CH3       S        H           H        CH3       S         H

 

–   C    –    C     –    C    –    C       –  C    –    C         C        C    –

 

H                       H          H          H                     H        H

Vulcanized rubber is used to make tyres, shoes  and valves.

 

6.Formation of synthetic rubber

Synthetic rubber is able to resist action of oil,abrasion and organic solvents which  rubber cannot.

Common synthetic rubber is a polymer of 2-chlorobut-1,3-diene ;

H         Cl         H          H

 

CH2=C (Cl CH = CH2            H   –   C    =    C    –    C    =    C    –   H

 

During  polymerization to synthetic rubber, one double  C=C bond is broken to self add to another molecule. The double bond remaining move to carbon “2” thus;

 

H         Cl         H          H           H        Cl       H           H

 

–   C    –    C     =   C    –    C       –   C    –    C    =    C        C    –

 

H                                            H           H                                 H

Generally the structure of rubber is thus;

 

H         Cl         H          H

 

-(-   C    –    C    =    C    –    C   -)n

 

 

H                                 H

Rubber is thus strengthened through vulcanization and manufacture of synthetic rubber.

 

 

 

 

 

 

(b)Condensation polymerization

Condensation polymerization is the process where two or more small monomers join together to form a larger molecule by elimination/removal of a simple molecule. (usually water).

Condensation polymers acquire a different name from the monomers because the two monomers are two different compounds

During condensation polymerization:

(i)the two  monomers are brought together by high pressure to reduce distance between them.

(ii)monomers realign themselves at the functional group.

(iii)from each functional group an element is removed so as to form simple molecule (of usually H2O/HCl)

(iv)the two monomers join without the simple molecule of H2O/HCl

 

Examples of condensation polymerization

 

1.Formation of Nylon-6,6

Method 1: Nylon-6,6 can be made from the  condensation polymerization of hexan-1,6-dioic acid with hexan-1,6-diamine.Amines are a group of homologous series with a general formula R-NH2 and thus  -NH2 as the functional group.

During the formation of Nylon-6,6:

(i)the two  monomers are brought together by high pressure to reduce distance between them  and realign themselves at the functional groups.

O                    O                           H                  H

 

H- O –  C –  (CH2 ) 4 – C – O – H   +   H –N – (CH2) 6 – N – H

 

 

(iii)from each functional group an element is removed so as to form a  molecule  of H2O and the two monomers join at the linkage .

O                    O      H                    H

 

H- O –  C –  (CH2 ) 4 –  C    N  – (CH2) 6 – N – H    + H 2O

 

.

Polymer bond linkage

 

Nylon-6,6  derive its name from the two monomers each with six carbon chain

 

Method 2: Nylon-6,6 can be made from the  condensation polymerization of hexan-1,6-dioyl dichloride with hexan-1,6-diamine.

Hexan-1,6-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus  -OCl  as the functional group.

The R-OCl is formed when the “OH” in R-OOH/alkanoic acid is replaced by Cl/chlorine/Halogen

 

During the formation of Nylon-6,6:

(i)the two  monomers are brought together by high pressure to reduce distance between them  and realign themselves at the functional groups.

O                    O                    H                  H

 

 

Cl –    C –  (CH2 ) 4  – C – Cl   +  H –N – (CH2) 6 – N – H

 

 

(iii)from each functional group an element is removed so as to form a  molecule  of HCl and the two monomers join at the linkage .

O                    O      H                   H

 

Cl –  C –  (CH2 ) 4 –  C    N  – (CH2) 6 – N – H    + HCl

 

.

Polymer bond linkage

The two monomers each has six carbon chain hence the name “nylon-6,6”

The commercial name of Nylon-6,6 is Nylon It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and carpets.

 

2.Formation of Terylene

Method 1: Terylene can be made from the  condensation polymerization of ethan-1,2-diol with benzene-1,4-dicarboxylic acid.

Benzene-1,4-dicarboxylic acid a group of homologous series with a general formula R-COOH where R is a ring of six carbon atom called Benzene ring .The functional group is -COOH.

During the formation of Terylene:

(i)the two  monomers are brought together by high pressure to reduce distance between them  and realign themselves at the functional groups.

O                    O

 

H- O –  C –  C6H5     – C – O – H   +   H –O –  CH2 CH2 – O – H

 

(iii)from each functional group an element is removed so as to form a  molecule  of H2O and the two monomers join at the linkage .

O                    O

 

H- O –  C –  C6H5       –  C    O  – (CH2) 6 – N – H      +   H 2O

 

.

Polymer bond linkage of terylene

 

 

 

 

Method 2: Terylene can be made from the  condensation polymerization of benzene-1,4-dioyl dichloride with ethan-1,2-diol.

Benzene-1,4-dioyl dichloride belong to a group of homologous series with a general formula R-OCl and thus  -OCl  as the functional group and R as a benzene ring.

The R-OCl is formed when the “OH” in R-OOH  is replaced by Cl/chlorine/Halogen

 

During the formation of Terylene

(i)the two  monomers are brought together by high pressure to reduce distance between them  and realign themselves at the functional groups.

O                    O

 

Cl  –    C –   C5H5   –  C – Cl   +  H –O – CH2 CH2 – O – H

 

 

(iii)from each functional group an element is removed so as to form a  molecule  of HCl and the two monomers join at the linkage .

O                    O

 

Cl –   C –  C5H5    –   C    O  –  CH2 CH2 – O – H    +   HCl

 

.

Polymer bond linkage of terylene

 

 

The commercial name of terylene is Polyester /polyster It is a a tough, elastic and durable plastic. It is used to make clothes, plastic ropes and sails and  plastic model kits.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Practice questions Organic chemistry

 

  1. A student mixed equal volumes of Ethanol and butanoic acid. He added a few drops of concentrated Sulphuric (VI) acid and warmed the mixture

(i) Name and write the formula of the main products

Name………………………………….

Formula……………………………………..

(ii) Which homologous series does the product named in (i) above belong?

 

HC = CH2
O
  1. The structure of the monomer phenyl ethene is given below:-

 

 

 

 

  1. a) Give the structure of the polymer formed when four of the monomers are added together
  2. b) Give the name of the polymer formed in (a) above
  3. Explain the environmental effects of burning plastics in air as a disposal method
  4. Write chemical equation to represent the effect of heat on ammonium carbonate
  5. Sodium octadecanoate has a chemical formula CH3(CH2)6 COONa+, which is used as soap.

Explain why a lot of soap is needed when washing with hard water

                        O

 

CH3CH2CH    C    OH

 

NH2

  1. A natural polymer is made up of the monomer:

 

 

 

 

 

(a) Write the structural formula of the repeat unit of the polymer

(b) When 5.0 x 10-5 moles of the polymer were hydrolysed, 0.515g of the monomer

were obtained.

Determine the number of the monomer molecules in this polymer.

(C = 12; H = 1; N = 14; O =16)

  1. The formula below represents active ingredients of two cleansing agents A and B

 

 

 

 

 

 

 

 

Which one of the cleansing agents would be suitable to be used in water containing magnesium hydrogen carbonate? Explain

 

 

 

H            H         H         H

 

C             C             C             C

 

O            H         O              H        n

 

 

  1. Study the polymer below and use it to answer the questions that follow:

 

 

 

 

 

 

 

(a) Give the name of the monomer and draw its structures

(b) Identify the type of polymerization that takes place

(c) State one advantage of synthetic polymers

  1. Ethanol and Pentane are miscible liquids. Explain how water can be used to separate a mixture

of ethanol and pentane

 

GLUCOSE SOLUTION
CRUDE ETHANOL
95% ETHANOL
ABSOLUTE ETHANOL
G
H

 

 

(a) What is absolute ethanol?

(b) State two conditions required for process G to take place efficiently

 

  1. (a) (i) The table below shows the volume of oxygen obtained per unit time when hydrogen

peroxide was decomposed in the presence of manganese (IV) Oxide. Use it to answer

the questions that  follow:-

Time in seconds Volume of Oxygen evolved (cm3)
0

30

60

90

120

150

180

210

240

270

300

0

10

19

27

34

38

43

45

45

45

45

(i) Plot a graph of volume of oxygen gas against time

(ii) Determine the rate of reaction at time 156 seconds

(iii) From the graph, find the time taken for 18cm3 of oxygen to be produced

(iv) Write a chemical equation to show how hydrogen peroxide decomposes in the presence

of manganese (IV) Oxide

(b) The diagram below shows how a Le’clanche (Dry cell) appears:-

 

 

 

 

 

 

 

 

 

 

 

(i) What is the function of MnO2 in the cell above?

(ii) Write the equation of a reaction that occurs at the cathode

(iii) Calculate the mass of Zinc that is consumed when a current of 0.1amperes flows

through the above cell for 30minutes (1F =96500c Zn =65)

  1. (a) Give the IUPAC names of the following compounds:

(i) CH3COOCH2CH3                                                                                    *

CH2 = C – CHCH3

 

Br

 

(ii)

 

 

(b) The structure below shows some reactions starting with ethanol. Study it and answer

CH3COOH
CH3COONa
CH3CH2OH
CH2=CH2
CH3CH3
CH2      CH2
n
S
P
CH4
T
Na Metal
Compound U
Step II
Step I
Step III
CH3COOH

 

Drops of Conc. H2SO4

Reagent R
NaOH(aq)
Heat
Excess Cl2 /U.V

the questions  that follow:

 

 

 

 

 

 

 

 

 

 

 

(i) Write the formula of the organic compounds P and S                                                 *

(ii) Name the type of reaction, the reagent(s) and condition for the reactions in the following steps :-

(I) Step I                                                                                                *

(II) Step II                                                                                             *

(III) Step III                                                                                           *

(iii) Name reagent R          ……………………………………………………………  *

(iv) Draw the structural formula of T and give its name                                                  *

(v) (I) Name compound U………………………………………………………..              

(II) If the relative molecular mass of U is 42000, determine the value of n (C=12, H=1)

(c) State why C2H4 burns with a more smoky flame than C2H6                                 *

  1. a) State two factors that affect the properties of a polymer
  2. b) Name the compound with the formula below :

CH3CH2CH2ONa

CH3CH2CH3
P
Step

 

 V

CH3CH = CH2
Step

 

 W

CH3CH2CH2OH
CH3CH2COOH
Step R
         Gas F
CH3CH2COONa
NaOH

 

Step J

K2CO3
Step X
NaOH

Heat

CH3CH2COOCH3

                 G

Product T

+ Na2CO3

  H      H

 

C –  C

 

CH3  H

n
K
  1. c) Study the scheme below and use it to answer the questions that follow:-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. i) Name the following compounds:-
  2. Product T …………………………          II. K    ………
  3. ii) State one common physical property of substance G

iii) State the type of reaction that occurred in step J                                                        

  1. iv) Give one use of substance K
  2. v) Write an equation for the combustion of compound P
  3. vi) Explain how compounds CH3CH2COOH and CH3CH2CH2OH can be distinguished chemically

vii) If a polymer K has relative molecular mass of 12,600, calculate the value of n (H=1 C =12)

  1. Study the scheme given below and answer the questions that follow:-

 

H2 (g)     Ni

High temp

Polymer Q
Polymerization
Compound P
CH3CH2CH3
CH3CH2CH2ONa + H2

 

Na(s)
Propan-l-ol
Step I
Propylethanoate
CH3CH2COOH

 

Solution T + CO2 (g)

 

Step III
Na2CO3(aq)
Conc. H2SO4 180oC
Step II

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(a) (i) Name compound P ……………………………………………………………………

(ii) Write an equation for the reaction between CH3CH2COOH and Na2CO3                                        

(b) State one use of polymer Q

(c) Name one oxidising agent that can be used in step II        …………………………………..

(d) A sample of polymer Q is found to have a molecular mass of 4200. Determine the number of

monomers in the polymer  (H = 1, C = 12)

(e) Name the type of reaction in step I …………………………………………………………..

(f) State one industrial application of step III

(g)State how burning can be used to distinguish between propane and propyne. Explain your

answer

(h) 1000cm3 of ethene (C2H4) burnt in oxygen to produce Carbon (II) Oxide and water vapour.

Calculate the minimum volume of air needed for the complete combustion of ethene

(Air contains 20% by volume of oxygen)

  1. (a) Study the schematic diagram below and answer the questions that follow:-
CH3CH2COOCH2CH2CH3
CH3CHCH2
CH3CH2CH2ONa + Gas P
CH3CH2CH2OH
X
V
HCl Step 5
 Step 1      R
Na

Step2

H+
Step 3
Q + H2O
MnO4

Step 4

Ni    H2

 

 

 

 

 

 

 

 

 

 

 

 

(i) Identify the following:

Substance Q ………………………………………………………………………………………………..

Substance R…………………………………………………………………………………………………

Gas P…………………………………………………………………………………………………………..

(ii) Name:

                    Step 1…………………………………………………………………………………….

                    Step 4…………………………………………………………………………………….

(iii) Draw the structural formula of the major product of step 5

(iv) State the condition and reagent in step 3                                                            

  1. Study the flow chart below and answer the questions that follow
Ethyl Ethanoate
CH2CH2OH
L
Reagent Q

Step 3

KMnO4/H+(aq)
M
KMnO4/H+
CH2CH2
J
K
CO2 (g)
STEP 2
Ni/H2(g)

     Step 4

Reagent P

 

 

 

 

 

 

 

 

 

 

 

(a) (i) Name the following organic compounds:

               M……………………………………………………………..……..

               L…………………………………………………………………..

(ii) Name the process in step:

                         Step 2 ………………………………………………………….….

                         Step 4 ………………………………………………………….…

(iii) Identify the reagent P and Q

(iv) Write an equation for the reaction between CH3CH2CH2OH and sodium

 

  1. a) Give the names of the following compounds:
  2. i) CH3CH2CH2CH2OH ……………………………………………………………………
  3. ii) CH3CH2COOH …………………………………………………………………

iii) CH3C – O- CH2CH3 ……………………………………………………………………

 

 

Step I
  1. Study the scheme given below and answer the questions that follow;
Step V

Complete combustion

Products
CH ºCH
C2H5COONa
Step II
Step IV + Heat
CH2 = CH2
C2H6
Step III
 

CH2 = CHCl

n
  1. i) Name the reagents used in:

Step I:                    ………………………………………………………………………

Step II                    ……………………………………………………………………

Step III          ………………………………………………………………………

  1. ii) Write an equation to show products formed for the complete combustion of CH = CH

iii) Explain one disadvantage of continued use of items made form the compound formed

in step III

  1. A hydrated salt has the following composition by mass. Iron 20.2 %, oxygen 23.0%,

sulphur 11.5%, water 45.3%

  1. i) Determine the formula of the hydrated salt (Fe=56, S=32, O=16, H=11)
  2. ii) 6.95g of the hydrated salt in c(i) above were dissolved in distilled water and the total

volume made to 250cm3 of solution. Calculate the concentration of the resulting salt solution

in moles per litre. (Given that the molecula mass of the salt is 278)

  1. Write an equation to show products formed for the complete combustion of CH = CH

iii) Explain one disadvantage of continued use of items made form the compound formed

in step III

 

 

  1. Give the IUPAC name for each of the following organic compounds;
  2.   i) CH3 – CH – CH2 – CH3

 

OH

ii)CH3 – CH – CH2 – CH2 – CH3

 

 

C2H5

iii)CH3COOCH2CH2CH3

 

  1. The structure below represents a cleansing agent.

O

 

R – S – ONa+

 

O

  1. a) State the type of cleansing agent represented above
  2. b) State one advantage and one disadvantage of using the above cleansing agent.

 

  1. The structure  below  shows part of  polymer .Use it to  answer the  questions  that follow.

CH3                 CH3         CH3

ï                     ï              ï

― CH  – CH2 – CH- CH2 – CH – CH2

  1. a) Derive the structure of the monomer
  2. b) Name the type of polymerization represented above
  3. The flow chart below represents a series of reactions starting with ethanoic acid:-
Ethanol
B

process I

Ethanoic acid

 

Na2CO3
Salt A + CO2 + H2O

 

 

 

 

 

(a) Identify substances A and B

(b) Name the process I

 

  1. a) Write an equation showing how ammonium nitrate may be prepared starting with

ammonia gas

(b) Calculate the maximum mass of ammonium nitrate that can be prepared using 5.3kg of

ammonia (H=1, N=14, O=16)

 

  1. (a) What is meant by the term, esterification?

(b) Draw the structural formulae of two compounds that may be reacted to form ethylpropanoate

 

  1. (a) Draw the structure of pentanoic acid

(b) Draw the structure and give the name of the organic compound formed when ethanol

reacts with   pentanoic acid in presence of concentrated sulphuric acid

 

 

 

 

 

 

 

 

  1. The scheme below shows some reactions starting with ethanol. Study it and answer the questions
Q
P
CH3COONa
Ethanol
CH3CH2ONa
C
H2SO4(l)

 

170oC step 1

Cr2O7(aq) / H+(aq)
Na(s)
Step 2
Step  4
CH3CH2OH/H2SO4
Step 3
2-

that follow:-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(i) Name and draw the structure of substance Q                                             

(ii) Give the names of the reactions that take place in steps 2 and 4

(iii) What reagent is necessary for reaction that takes place in step 3

  1. Substances A and B are represented by the formulae ROH and RCOOH respectively.

They belong to two different homologous series of organic compounds. If both A and B

react with potassium metal:

(a) Name the common product produced by both

(b) State the observation made when each of the samples A and B are reacted with sodium

hydrogen  carbonate

(i) A

(ii) B

  1. Below are structures of particles. Use it to answer questions that follow. In each case only

electrons in the outermost energy level are shown

W
 
 
 
 
U
V
 
19P

20N

Z
Y

key

P = Proton

N = Neutron

X = Electron

 

 

 

 

 

 

 

 

 

 

(a) Identify the particle which is an anion

  1. Plastics and rubber are extensively used to cover electrical wires.

(a) What term is used to describe plastic and rubbers used in this way?

(b) Explain why plastics and rubbers are used this way

 

 

 

 

  1. The scheme below represents the manufacture of a cleaning agent X

 

R
Conc.

 

H2SO4

R
 

SO3H

Cleaning agent X

 

 

 

 

 

 

(a) Draw the structure of X and state the type of cleaning agent to which X belong

(b) State one disadvantage of using X as a cleaning agent

  1. Y grams of a radioactive isotope take 120days to decay to 3.5grams. The half-life period

of the isotope is 20days

(a) Find the initial mass of the isotope

(b) Give one application of radioactivity in agriculture

       H     H

 

-C – C –

 

CH3   H

n

n

  1. The structure below represents a polymer. Study and answer the questions that follow:-

 

 

 

 

 

(i) Name the polymer above……………………………………………………………………….

(ii) Determine the value of n if giant molecule had relative molecular mass of 4956

  1. RCOONa+ and RCH2OSO3Na+ are two types of cleansing agents;
  2. i) Name the class of cleansing agents to which each belongs
  3. ii) Which one of these agents in (i) above would be more suitable when washing with water

from the Indian ocean. Explain

iii) Both sulphur (IV) oxide and chlorine are used bleaching agents.  Explain the difference

in their bleaching  properties

 

               H        H        H        H

 

C         C        C        C

 

O        H        O        H

n
  1. The formula given below represents a portion of a polymer

 

 

 

 

 

 

 

 

(a)  Give the name of the polymer

(b) Draw the structure of the monomer used to manufacture the polymer

 

 

 

 

 

 

 

 

 

 

     

 

UPGRADE

 

 

CHEMISTRY

                

FORM 3

                 

The Chemistry of NITROGEN

                       

 

 

 

 

 

Comprehensive tutorial notes

 

 

 MUTHOMI S.G

www.kcslibrary.info

                                     0720096206

 

 

 

 

 

 

Contents

 

 

 A NITROGEN …………………………………………………………………….3

          Occurrence………………………………………………………..……..…..3

Isolation from air…………………………………………………………….3

Fractional distillation …………………………………………………….….4

Preparation of Nitrogen…………………………………………………..….6

Properties of  Nitrogen…………………………………………………….…..6

 

 

B.OXIDES OFNITROGEN……………………………………………….……..7

          Nitrogen(I)Oxide………………………..……………………………………7

Nitrogen(II) oxide……………………….…………………………………..9

Nitrogen(IV) oxide …………………………………………………………11

 

  C.AMMONIA…..………………………….…………………………….……..13

          Occurrence…………………………………………………………………15

Preparation………………………………………………………………….15

Properties……………………………………………………………………16

Haber process……………………………….………………………………12

 

  1. NITRIC(V) ACID………..………………………….………………………28

          Preparation…………………………………….……………………………13

Properties…………………………………………………………………..33

Ostwalds process…………………………………………………………..12

 

E.NO3 and NO2salts ……………………….……………………………….……27

 

 

  1. COMPREHENSIVE REVISION QUESTIONS…………………………..…32

 

 

 

 

 

 

 

 

 

A.NITROGEN

  1. a) Occurrence:

Nitrogen is found in the atmosphere occupying about 78% by volume of air.

Proteins, amino acids, polypeptides in living things contain nitrogen.

 

  1. b) Isolation of nitrogen from the air.

 

Nitrogen can be isolated from other gases present in air like oxygen, water (vapour), carbon (IV) oxide and noble gases as in the school laboratory as in the flow chart below:

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Water is added slowly into an “empty flask” which forces the air out into another flask containing concentrated sulphuric (VI) acid. Concentrated sulphuric (VI) acid is hygroscopic. It therefore absorb/remove water present in the air sample.

 

More water forces the air into the flask containing either concentrated sodium hydroxide or potassium hydroxide solution. These alkalis react with carbon IV) oxide to form the carbonates and thus absorbs/remove carbon IV) oxide present in the air sample.

 

Chemical equation 2NaOH (aq) + CO2 (g)     ->       Na2CO3 (aq)          +           H2O(l)

Chemical equation 2KOH (aq)   + CO2 (g)     ->       K2CO3 (aq)  +        H2O(l)

 

More water forces the air through a glass tube packed with copper turnings. Heated brown copper turnings react with oxygen to form black copper (II) oxide.

 

Chemical equation         2Cu (s) +        O2 (g)          ->       CuO (s)

(brown)                              (black)

 

The remaining gas mixture is collected by upward delivery/downward displacement of water/over water. It contains about 99% nitrogen and 1% noble gases.

 

On a large scale for industrial purposes, nitrogen is got from fractional distillation of air.

 

c)Nitrogen from fractional distillation of air.

 

For commercial purposes nitrogen is got from the fractional of air.

Air is first passed through a dust precipitator/filter to remove dust particles.

 

The air is then bubbled through either concentrated sodium hydroxide or potassium hydroxide solution to remove/absorb Carbon(IV) oxide gas.

 

Chemical equation         2NaOH (aq)   + CO2 (g)    ->       Na2CO3 (aq)          + H2O(l)

Chemical equation         2KOH (aq)     + CO2 (g)     ->       K2CO3 (aq)  + H2O(l)

 

Air mixture is the cooled to -25oC.At this temperature, water (vapour ) liquidifies and then solidify to ice and thus removed.

 

The air is further cooled to -200oC during which it forms a blue liquid.

 

The liquid is then heated. Nitrogen with a boiling point of -196oC distils first then Argon at-186oC and then finally Oxygen at -183oC boils last.

 

 

  1. c) School laboratory preparation of Nitrogen.

 

The diagram below shows the set up of the school laboratory preparation of nitrogen gas.

 

 

d.Properties of Nitrogen gas(Questions)

 

1.Write the equation for the reaction for the school laboratory preparation of nitrogen gas.

                    Chemical equation   NH4Cl (s) + NaNO2(s)->NaCl (g)+ NH4NO2 (s)

Chemical equation         NH4NO2 (s)   ->       N2 (g)          +           H2O (l)

 

 

  1. State three physical properties of nitrogen gas.

                    – colourless, odourless, less dense than air ,neutral and slightly soluble in water

 

  1. State and explain the observation made when a burning magnesium ribbon is lowered in a gas jar containing nitrogen gas.

 

Observation; It continues burning with a blight blindening flame forming white ash.

ExplanationMagnesium burns to produce enough heat /energy to reacts with nitrogen to form white magnesium nitride.

Chemical equation3Mg (s)         +        N2 (g)          ->       Mg3N2 (s)

(white ash/solid)

  1. State two main uses of nitrogen gas

-manufacture of ammonia from Haber process

– As a refrigerant in storage of semen for Artificial insemination.

 

 

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  1. OXIDES OF NITROGEN

Nitrogen forms three main oxides:

i)Nitrogen(I) oxide(N2O)

  1. ii) Nitrogen(II) oxide (NO)

iii) Nitrogen (IV) oxide( NO2)

 

  1. i) Nitrogen (I) oxide(N2O)

 

  1. a) Occurrence

Nitrogen (I) oxide does not occur naturally but prepared in a laboratory.

 

b)Preparation                         

The set up below shows the set up of apparatus that can be used to prepare Nitrogen (I) oxide in a school laboratory.

 

 

 

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  1. c) Properties of nitrogen (I) oxide (Questions)

 

  1. Write the equation for the reaction for the school laboratory preparation of Nitrogen (I) oxide.

 

Chemical equation NH4NO2(s) ->       H2O (l)        + N2O (g)

 

2.a) State and explain three errors made in the above set up

          –Oxygen is being generated instead of Nitrogen (I) oxide.

Ammonium Nitrate(V) should be used instead of potassium manganate(VI) and manganese(IV)oxide.

  1. b) State three physical properties of Nitrogen (I) oxide.

-slightly soluble in water.

-colourless

-odourless

-less dense than air

-slightly sweet smell

  1. State and explain the observation made when a burning magnesium ribbon is lowered in a gas jar containing Nitrogen (I) oxide.

Observation – Continues to burn with a bright flame

-White solid/residue is formed

Explanation-Magnesium burns in air to produce enough heat/energy split/break   Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.

Chemical equation

Mg(s)           + N2O (g)-> MgO (s)       + N2(g)

 

  1. State and explain the observation made when the following non metals are burnt then lowered in a gas jar containing Nitrogen (I) oxide.

 

  1. a) Carbon/charcoal

Observation – Continues to burn with an orange glow

-colorless gas is formed that forms white precipitate with lime water.

Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break   Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas. Carbon (IV) oxide gas reacts to form a white precipitate with lime water.

Chemical equation      C(s)         + 2N2O (g)->         CO2 (g)        + 2N2(g)

 

  1. b) sulphur powder

 

          Observation – Continues to burn with a blue flame

-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.

Explanation-Sulphur burns in air to produce enough heat/energy split/break   Nitrogen (I) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.

Chemical equation S(s)    + 2N2O (g)->         SO2 (g)        + 2N2(g)

 

  1. State two uses of nitrogen (I) oxide

-As laughing gas because as anesthesia the patient regain consciousness laughing hysterically after surgery.

-improves engine efficiency.

 

  1. State three differences between nitrogen (I) oxide and oxygen

          –Oxygen is odourless while nitrogen (I) oxide has faint sweet smell

-Both relight/rekindle a glowing wooden splint but Oxygen can relight a feeble glowing splint while nitrogen (I) oxide relights well lit splint.

-Both are slightly soluble in water but nitrogen (I) oxide is more soluble.

 

 

  1. ii) Nitrogen (II) oxide (NO)

 

  1. a) Occurrence

Nitrogen (II) oxide does not occur naturally but prepared in a laboratory.

 

b)Preparation

                             

The set up below shows the set up of apparatus that can be used to prepare Nitrogen (II) oxide in a school laboratory.

 

  1. c) Properties of nitrogen (II) oxide (Questions)

 

  1. Write the equation for the reaction for the school laboratory preparation of Nitrogen (II) oxide.

 

Chemical equation 3Cu(s)  + 8HNO3(aq)   ->  4H2O (l)+2NO (g)  +2Cu(NO3)2(aq)

Chemical equation 3Zn(s)  + 8HNO3(aq)   ->  4H2O (l)+2NO (g)  +2Zn(NO3)2(aq)

Chemical equation 3Mg(s)  + 8HNO3(aq)   ->  4H2O (l)+2NO (g)+2Mg(NO3)2(aq)

 

  1. State three physical properties of Nitrogen (II) oxide.

-insoluble in water.

-colourless

-odourless

-denser dense than air

-has no effect on both blue and red litmus papers

 

  1. State and explain the observation made when a burning magnesium ribbon is lowered in a gas jar containing Nitrogen (II) oxide.

 

Observation – Continues to burn with a bright flame

-White solid/residue is formed

Explanation-Magnesium burns in air to produce enough heat/energy split/break   Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.

Chemical equation 2Mg(s)         + 2NO (g)->          2MgO (s)     + N2(g)

 

  1. State and explain the observation made when the following non metals are burnt then lowered in a gas jar containing Nitrogen (II) oxide.

 

  1. a) Carbon/charcoal

Observation – Continues to burn with an orange glow

-colorless gas is formed that forms white precipitate with lime water.

Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break   Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas.Carbon (IV) oxide gas reacts to form a white precipitate with lime water.

Chemical equation      C(s)         +  2NO (g)->         CO2 (g)        + N2(g)

 

  1. b) sulphur powder

 

          Observation – Continues to burn with a blue flame

-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.

Explanation-Sulphur burns in air to produce enough heat/energy split/break   Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.

Chemical equation      S(s)         + N2O (g)-> SO2 (g)        + N2(g)

 

  1. c) Phosphorus

Observation – Continues to produce dense white fumes

Explanation-Phosphorus burns in air to produce enough heat/energy split/break   Nitrogen (II) oxide gas into free Nitrogen and oxygen then continues to burn in oxygen to form dense white fumes of phosphorus (V) oxide gas.

Chemical equation      4P(s)       + 10NO (g)->        2P2O5(g)      + 5N2(g)

 

  1. State one use of nitrogen (II) oxide

As an intermediate gas in the Ostwalds process for manufacture of nitric(V) gas.

 

  1. State and explain the observation made when nitrogen (II) oxide is exposed to the atmosphere.

Observation–brown fumes produced/evolved that turn blue litmus paper red.

Explanation– Nitrogen (II) oxide gas on exposure to air is quickly oxidized by the air/ oxygen to brown nitrogen (IV) oxide gas. Nitrogen (IV) oxide gas is an acidic gas.

Chemical equation 2NO (g)+      O2(g)->      2NO2 (g)

(colorless)                                  (brown)

 

  1. ii) Nitrogen (IV) oxide (NO2)

 

  1. a) Occurrence

Nitrogen (IV) oxide occurs -naturally from active volcanic areas.

-formed from incomplete combustion of the internal      combustion engine of motor vehicle exhaust fumes.

-from lightening

 

b)Preparation                         

The set up below shows the set up of apparatus that can be used to prepare Nitrogen (IV) oxide in a school laboratory.

 

 

  1. c) Properties of nitrogen (IV)oxide (Questions)

 

  1. Write the equation for the reaction for the school laboratory preparation of Nitrogen (II) oxide.

Chemical equation Cu(s)  + 4HNO3(aq)   ->  2H2O (l)+2NO 2(g)  +Cu(NO3)2(aq)

Chemical equation Zn(s)  + 4HNO3(aq)   ->  2H2O (l)+2NO 2(g)  +Zn(NO3)2(aq)

Chemical equation Fe(s)  + 4HNO3(aq)   ->  2H2O (l)+2NO 2(g)  +Fe(NO3)2(aq)

 

  1. State three physical properties of Nitrogen (IV) oxide.

-soluble/dissolves in water.

-brown in colour

-has pungent irritating poisonous odour/smell

-denser dense than air

-turns blue litmus papers to red

 

  1. State and explain the observation made when Nitrogen (IV) oxidegas is bubbled in water.

Observation–The gas dissolves and thus brown colour of the gas fades

-A colourless solution is formed

-solution formed turns blue litmus papers to red

-solution formed  has no effect on red

Explanation-Magnesium burns in air to produce enough heat/energy split/break   Nitrogen (IV) oxide gas  dissolves then react with water to form an acidic mixture of  nitric(V) acid andnitric(III) acid.

 

Chemical equationH2O (l) +  2NO 2(g)->HNO3(aq)    + HNO2(aq)

(nitric(V) acid)         (nitric(III) acid)

 

  1. State and explain the observation made when a test tube containing Nitrogen (IV) oxide is cooled then heated gently then strongly.

 

Observation on cooling

-Brown colour fades

-Yellow liquid formed

Observation on gentle heating

            –Brown colour reappears

            –Yellow liquid formed changes to brown fumes/gas

Observation on gentle heating

            –Brown colour fades

            –brown fumes/gas changes to a colourless gas

Explanation-Brown nitrogen (IV) oxide gas easily liquefies to yellow dinitrogen tetraoxide liquid.When the yellow dinitrogen tetraoxide liquid is gently heated it changes back to the brown nitrogen (IV) oxidegas.When the brown nitrogen (IV) oxide gas is strongly heated it decomposes to colourless mixture of Nitrogen (II) oxide gas and Oxygen.

 

Chemical equation  O2(s) +  2NO (g)   =====  2NO2 (g)   ===== N2O4(l)

                              (colourless gases)           (brown gas)                    (yellow liquid)

 

  1. State and explain the observation made when a burning magnesium ribbon is lowered in a gas jar containing Nitrogen (IV) oxide.

Observation – Continues to burn with a bright flame

-White solid/residue is formed

-Brown fumes/colour fades

Explanation-Magnesium burns in air to produce enough heat/energy split/break   brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form white solid/ash of Magnesium oxide.

 

Chemical equation      4Mg(s)    +  2NO 2(g)->        4MgO (s)     + N2(g)

 

  1. State and explain the observation made when the following non metals are burnt then lowered in a gas jar containing Nitrogen (IV) oxide.
  2. a) Carbon/charcoal

Observation – Continues to burn with an orange glow

-Brown fumes/colour fades

-colorless gas is formed that forms white precipitate with lime water.

Explanation-Carbon/charcoal burns in air to produce enough heat/energy split/break   brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form carbon (IV) oxide gas.Carbon (IV) oxide gas reacts to form a white precipitate with lime water.

 

Chemical equation2C(s)   +  2NO 2(g)->        2CO2 (g)      + N2(g)

 

  1. b) sulphur powder

Observation – Continues to burn with a blue flame

-Brown fumes/colour fades

-colorless gas is formed that turn orange acidified potassium dichromate (VI) to green.

Explanation-Sulphur burns in air to produce enough heat/energy split/break   brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form sulphur (IV) oxide gas.Sulphur (IV) oxide gas turns orange acidified potassium dichromate (VI) to green.

 

Chemical equation2S(s)   + 2NO2 (g)->         2SO2 (g)      + N2(g)

 

  1. c) Phosphorus

Observation– Continues to produce dense white fumes

-Brown fumes/colour fades

Explanation-Phosphorus burns in air to produce enough heat/energy split/break   brown Nitrogen (IV) oxide gas into free colourless Nitrogen and oxygen then continues to burn in oxygen to form dense white fumes of phosphorus (V) oxide gas.

Chemical equation      8P(s)       + 10NO2 (g)->       4P2O5(g)      + 5N2(g)

 

  1. State two uses of nitrogen (IV) oxide

-In theOstwald process for industrial manufacture of nitric (V) gas.

-In the manufacture of T.N.T explosives

  1. State and explain the observation made when nitrogen (II) oxide is exposed to the atmosphere.

Observation–brown fumes produced/evolved that turn blue litmus paper red.

Explanation– Nitrogen (II) oxide gas on exposure to air is quickly oxidized by the air/ oxygen to brown nitrogen (IV) oxide gas. Nitrogen (IV) oxide gas is an acidic gas.

Chemical equation   2NO (g)  +      O2(g)  ->        2NO2 (g)

(colourless)                                (brown)

 

 

 

 

 

 

 

 

 

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  1. AMMONIA (NH3)

Ammonia is a compound of nitrogen and hydrogen only.  It is therefore a hydride of nitrogen.

  1. a) Occurrence

Ammonia gas occurs -naturally from  urine of mammals and excretion of birds

-formed in the kidney of human beings

 

b)Preparation                         

The set up below shows the set up of apparatus that can be used to prepare dry Ammonia gas in a school laboratory.

Set up method 1

  1. c) P
  2. Write the equation for the reaction taking place in:
  3. Method 1

 

Chemical equation

Ca (OH)2(s) + NH4 Cl(s)->CaCl2 (aq)  + H2O(l)      + 2NH3(g)

 

b)Method 2

Chemical equation

NaOH (aq)   + NH4 Cl(aq)   -> NaCl (aq)  + H2O(l) + NH3(g)

 

  1. State three physical properties of ammonia.

-has a pungent choking smell of urine

-Colourless

-Less dense than air hence collected by upward delivery

-Turns blue litmus paper blue thus is the only naturally occurring basic gas    (at this level)

  1. Calcium oxide is used as the drying agent. Explain why calcium chloride and concentrated sulphuric(VI) acid cannot be used to dry the gas.

-Calcium chloride reacts with ammonia forming the complex compound  CaCl2.8H2O.

Chemical equation  CaCl2 (s)       + 8NH3(g)     ->  CaCl2 .8NH3(g)

-Concentrated sulphuric(VI) acid reacts  with ammonia forming  ammonium sulphate(VI) salt compound

Chemical equation  2NH3(g)      +H2SO4(l)    ->(NH4)2SO4(aq)

 

4.Describe the test for the presence of ammonia gas.

Using litmus paper:

Dip moist/damp/wet blue and red litmus papers in a gas jar containing a gas suspected to be ammonia.The blue litmus paper remain blue and the red litmus paper turns blue.Ammonia is the only basic gas.(At this level)

 

 

Using hydrogen chloride gas

Dip a glass rod in concentrated hydrochloric acid. Bring the glass rod near the mouth of a gas jar suspected to be ammonia. White fumes (of ammonium chloride)are produced/evolved.

 

  1. Describe the fountain experiment to show the solubility of ammonia.

Ammonia is very soluble in water.

When a drop of water is introduced into flask containing ammonia, it dissolves all the ammonia in the flask. If water is subsequently allowed into the flask through a small inlet, atmospheric pressure forces it very fast to occupy the vacuum forming a fountain. If the water contains three/few drops of litmus solution,the litmus solution  turns blue because ammonia is an alkaline/basic gas. If the water contains three/few drops of phenolphthalein indicator,the indicator turns pink because ammonia is an alkaline/basic gas. Sulphur(IV)oxide and hydrogen chloride gas are also capable of the fountain experiment . If the water contains three/few drops of  phenolphthalein indicator, the indicator turns colourless because both Sulphur(IV) oxide and hydrogen chloride gas are acidic gases.

 

 

6.State and explain the observation made when hot platinum /nichrome wire is placed over concentrated ammonia solution with Oxygen gas bubbled into the mixture.

 

Observations

Hot platinum /nichrome wire continues to glow red hot.

Brown fumes of a gas are produced.

Explanation

Ammonia reacts with Oxygen on the surface of the wire .This reaction is exothermic producing a lot of heat/energy that enables platinum wire to glow red hot. Ammonia is oxidized to Nitrogen(II)oxide gas and water. Hot platinum /nichrome wire acts as catalyst to speed up the reaction. Nitrogen(II)oxide gas is further  oxidized to brown Nitrogen(IV)oxide gas on exposure to air.

 

Chemical equation

(i)4NH3(g)   +        5O2(g)  -Pt-> 4NO(g) + 6H2O(l)

 

(ii)2NO(g)     +   O2(g)    ->     2NO2(g)

 

  1. Ammonia gas was ignited in air enriched with Oxygen gas. State and explain the observations made

Observations

             – Ammonia gas burns with a green flame

-Colourless gas produced

Explanation

Ammonia gas burns with a green flame in air enriched with Oxygen to from Nitrogen gas and water.

Chemical equation

2NH3(g)   +        O2(g)  -> N2(g) + 3H2O(l)

  1. Dry ammonia was passed through heated copper(II)Oxide as in the set up below.

 

 

(a)State the observations made in tube K

-Colour changes from black to brown

-Colourless liquid droplet form on the cooler parts of tube K

 

(b)(i)Identify liquid L.

-Water/ H2O(l)

 

(ii)Explain a chemical and physical test that can be used to identify liquid L.

Chemical test

(i) Add three/few drops of liquid L into anhydrous copper(II)sulphate(VI).

Colour changes from white to blue.

Explanation-Water changes white anhydrous copper(II)sulphate(VI) to blue hydrated copper(II)sulphate(VI)

 

(ii) Add three/few drops of liquid L into anhydrous cobalt(II)Chloride.

Colour changes from blue to pink.

Explanation-Water changes blue anhydrous cobalt(II)Chloride to pink  hydrated cobalt(II)Chloride.

 

Physical test

(i)Heat the liquid. It boils at 100oC at sea level (1atmosphere  pressure/760mmHg pressure, 101300Pa,101300Nm-2).

(ii)Cool the liquid. It freezes  at 0.0oC .

(iii)Determine the density. It is 1.0gcm-3

 

(c)Write the equation for the reaction that take place.

2NH3(g)      +        3CuO(s)    ->  N2(g)  +  3H2O(l) + 3Cu(s)

(black)                                      (brown)

 

2NH3(g)      +        3PbO(s)    ->  N2(g)  +  3H2O(l) + 3Pb(s)

                                        (brown when hot)                         (grey)

 

8.(a)What is aqueous ammonia

    Aqueous ammonia is formed when ammonia gas is dissolved in water.

NH3(g) + (aq)        ->   NH3(aq)

 

A little NH3(aq) reacts with ammonia water to form ammonia solution(NH4OH)

 

NH3 (aq)   +  H2O(l)                           OH(aq)   + NH4+(aq)

 

This makes a solution of aqueous ammonia is a weak base /alkali unlike other two alkalis.

 

9.Using dot and cross to represent outer electrons show the bonding in:

 

(a) NH3

         

  

(b) NH4+

 

 

(c)NH4Cl

 

10.Name four uses of ammonia

(i)In the manufacture of nitrogenous fertilizers.

(ii) In the manufacture of nitric(V)acid from Ostwalds process.

(iii)As a refrigerant in ships and warehouses.

(iv)In softening hard water.

(v)In the solvay process for the manufacture of sodium carbonate.

(vi)In the removal of grease and stains.

 

11.(a)Calculate the percentage of Nitrogen in the following fertilizers:

          (i) (NH4)2SO4

                              Molar mass of  (NH4)2SO4 = 132g

Mass of N in (NH4)2SO4= 28g

%  of N => 28  x  100   = 21.2121%

132

(ii) (NH4)3PO4

                              Molar mass of  (NH4)3PO4 = 149g

Mass of N in (NH4)3PO4= 42g

%  of N => 42  x  100   = 28.1879%

149

 

(b)State two advantages of fertilizer a (i) over a (ii) above.

                    (i)Has higher % of Nitrogen

(ii)Has phosphorus which is necessary for plant growth.

 

 

 

(c) Calculate the mass of Nitrogen in a 50kg bag of:

(i) (NH4)2SO4

                              % of N in  (NH4)2SO4 = 21.2121%

Mass of N in  50 kg (NH4)2SO4= 21.2121 x 50  =  10.6 kg

100

(ii) NH4NO3

                              Molar mass of  NH4NO3 = 80g

Mass of N in (NH4)3PO4= 28g

%  of N => 28  x  100   = 35%

80

                              % of N in  NH4NO3 = 35%

Mass of N in  50 kg (NH4)2SO4= 35 x 50  =  17.5 kg

100

                    NH4NO3 therefore has a higher mass of Nitrogen than (NH4)2SO4

 

 

d).Manufacture of Ammonia /Haber process

 

Most of the Ammonia produced for industrial purposes uses the Haber process developed by the German Scientist Fitz Haber.

 

(i)Raw materials

The raw materials include:

(i)Nitrogen from Fractional distillation of air from the atmosphere.

(ii)Hydrogen from:

  1. Water gas-passing steam through heated charcoal

C(s) + H2O(l) -> CO(g) + H2 (g)

II .Passing natural gas /methane through steam.

CH4(g)+ H2O(l) -> CO(g) + 3H2 (g)

 

(ii)Chemical process

Hydrogen and Nitrogen are passed through a purifier to remove unwanted gases like Carbon(IV)oxide,Oxygen,sulphur(IV)oxide, dust, smoke which would poison the catalyst.

Hydrogen and Nitrogen are then mixed in  the ratio of 3:1 respectively. The mixture is compressed to 200-250atmoshere pressure to liquidify. The liquid mixture is then heated to 400- 450oC.The hot compressed gases are then passed over finely divided Iron catalyst promoted/impregnated with Al2O3 /K2O .Promoters increase the efficiency of the catalyst.

 

 

 

 

Optimum conditions in Haber processs

Chemical equation

 

N2 (g)  +   3H2 (g)   ===Fe/Pt===  2NH3 (g)     ΔH = -92kJ

 

Equilibrium/Reaction rate considerations

(i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained.

(ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 200atmospheres is normally used.

 

(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical.

An optimum temperature of about 450oC is normally used.

 

(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.

 

  1. e) Nitric(V)acid (HNO3)

a)Introduction.

Nitric(V)acid is one of the mineral acids .There are three mineral acids; Nitric(V)acid, sulphuric(VI)acid and hydrochloric acid.Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level.

  1. b) School laboratory preparation

Nitric(V)acid is prepared in a school laboratory from the reaction  of Concentrated sulphuric(VI)acid and potassium nitrate(V) below.

(c)Properties of Concentrated Nitric (V)acid(Questions)

1.Write an equation for the school laboratory preparation of nitric(V)acid.

             KNO3(s) + H2SO4(l)     -> KHSO4(s)  + HNO3(l)

 

2.Sodium nitrate(V)can also be used to prepare nitric(V)acid. State two reasons why potassium nitrate(V) is preferred over Sodium nitrate(V).

 

(i) Potassium nitrate(V) is more volatile than sodium nitrate(V) and therefore readily displaced from the less volatile concentrated sulphuric(VI)acid

 

(ii) Sodium nitrate(V) is hygroscopic and thus absorb water . Concentrated sulphuric(VI)acid dissolves in water. The dissolution is a highly exothermic process.

  1. An all glass apparatus /retort is used during the preparation of nitric(V) acid. Explain.

Hot concentrated nitric(V) acid vapour is highly corrosive and attacks rubber cork apparatus if used.

  1. Concentrated nitric(V) acid is colourless . Explain why the prepared sample in the school laboratory appears yellow.

 

Hot concentrated   nitric(V) acid decomposes to brown nitrogen(IV)oxide  and Oxygen gases.

 

4HNO3(l/g)     -> 4NO2(g)   + H2O (l)  +O2(g)

 

Once formed  the brown nitrogen(IV)oxide dissolves in the acid forming a yellow solution .

 

  1. State and explain the observation made when concentrated nitric (V) acid is heated.

Observation

Brown fumes are produced.

Colourless gas that relights/rekindles glowing splint

Explanation

Hot concentrated   nitric(V) acid decomposes to water, brown nitrogen(IV)oxide  and Oxygen gases. Oxygen gas is not visible in the brown fumes of nitrogen (IV) oxide.

4HNO3(g)     -> 4NO2(g)   + H2O (l)  +O2(g)

 

  1. Explain the observations made when:

     (a) About 2cm3 of Iron(II)sulphate(VI) solution is added about 5 drops of concentrated   nitric(V) acid and the mixture then heated/warmed in a test tube.

Observation

(i)Colour changes from green to brown.

(ii)brown fumes /gas produced on the upper parts of the test tube.

 

Explanation

Concentrated   nitric(V) acid is a powerful/strong oxidizing agent. It oxidizes  green Fe2+ ions in FeSO4 to brown/yellow Fe3+ .The acid is reduced to colourless Nitrogen(II)oxide.

 

Chemical equation:

6FeSO4(aq) + 3H2SO4 (aq) + 2HNO3(aq) -> 3Fe2(SO4) 3 (aq)+ 4H2O + 2NO(g)

 

Colourless  Nitrogen(II)oxide is rapidly further oxidized to brown Nitrogen(IV)oxide by atmospheric oxygen.

 

Chemical equation:

2NO(g)     +        O(g)  ->      2NO2 (g)

(colourless)                             (brown)

 

 (b) A spatula full of sulphur powder in a claen dry beaker was added to 10cm3 concentrated nitric (V) acid and then heated gently/warmed.

Observation

(i)Yellow colour of sulphur fades .

(ii)brown fumes /gas produced.

 

Explanation

Concentrated   nitric(V) acid is a powerful/strong oxidizing agent. It oxidizes yellow sulphur to colourless concentrated sulphuric(VI)acid. The acid is reduced to brown Nitrogen(IV)oxide gas.

 

Chemical equation:

S(s)  + 6HNO3(l)     -> 4NO2(g)   + H2O (l)  +H2SO4(l)

(c) A few/about 1.0g pieces of copper turnings/Zinc granules/ Magnesium ribbon  are added 10cm3 of concentrated   nitric(V) acid in a beaker.

 

 Observation

(i) brown fumes /gas produced.

(ii) blue solution formed with copper turnings

(iii) colourless solution formed with Zinc granules/Magnesium ribbon

 

Explanation

Concentrated   nitric (V) acid is a powerful/strong oxidizing agent. It oxidizes metals to their metal nitrate (VI) salts. The acid is reduced to brown Nitrogen (IV) oxide gas.

Chemical equation:

Cu(s)   +  4HNO3(l)     ->  2NO2(g)   + H2O (l)   +   Cu(NO3) 2 (aq)

             Zn(s)   +  4HNO3(l)     ->  2NO2(g)   +  H2O (l)   +   Zn(NO3) 2 (aq)

             Mg(s)  +  4HNO3(l)     ->  2NO2(g)   +  H2O (l)   +    Mg(NO3) 2 (aq)

Pb(s)   +  4HNO3(l)     ->  2NO2(g)   +   H2O (l)   +   Pb(NO3) 2 (aq)

Ag(s)  +  2HNO3(l)     ->  NO2(g)      +   H2O (l)   +    AgNO3 (aq)

 

 

 

(d)Properties of Dilute Nitric (V)acid(Questions)

 

(i)What is dilute nitric(v)acid

When concentrated nitric(v)acid is added to over half portion of water ,it is relatively said to be dilute. A dilute solution is one which has more solvent/water than solute/acid. The number of moles of the acid are present in a large amount/volume of the solvent.This makes the molarity /number of moles present in one cubic decimeter of the solution to be low e.g. 0.02M.

If more water is added to the acid until the acid is too dilute to be diluted further then an infinite dilute solution if formed.

 

(ii))1cm length of polished Magnesium ribbon was put is a test tube containing 0.2M dilute nitric(v)acid. State and explain the observation made.

 

Observation

-Effervescence/bubbling/fizzing

-Colourless gas produced that extinguish burning splint with an explosion/pop sound

-Colourless solution formed

-Magnesium ribbon dissolves/decrease in size

 

Explanation

Dilute dilute nitric(v)acid reacts with Magnesium to form hydrogen gas.

Mg(s)  +  2HNO3(aq)     ->    H2 (g)   +    Mg(NO3) 2 (aq)

 

With other reactive heavy metals, the hydrogen gas produced is rapidly oxidized to water.

 

Chemical equation 3Pb(s)  + 8HNO3(aq)   ->  4H2O (l)+2NO (g)  +2Pb(NO3)2(aq)

Chemical equation 3Zn(s)  + 8HNO3(aq)   ->  4H2O (l)+2NO (g)  +2Zn(NO3)2(aq)

Chemical equation 3Fe(s)  + 8HNO3(aq)   ->  4H2O (l)+2NO (g)   +2Fe(NO3)2(aq)

 

Hydrogen gas therefore is usually not prepared in a school laboratory using dilute nitric (v)acid.

 

(iii)A half spatula full of sodium hydrogen carbonate and Copper(II) carbonate were separately to separate test tubes containing 10cm3 of 0.2M dilute nitric (V) acid. 

Observation

-Effervescence/bubbling/fizzing

-Colourless gas produced that forms a white precipitate with lime water.

-Colourless solution formed with sodium hydrogen carbonate.

– Blue solution formed with Copper(II) carbonate.

Explanation

Dilute dilute nitric (v)acid reacts with Carbonates and hydrogen carbonates to form Carbon(IV)oxide, water and nitrate(V)salt

CuCO3 (s)  +  2HNO3(aq)     ->    H2O (l)   +    Cu(NO3) 2 (aq) + CO2 (g)

ZnCO3 (s)  +  2HNO3(aq)     ->    H2O (l)   +    Zn(NO3) 2 (aq) + CO2 (g)

 

CaCO3 (s)  +  2HNO3(aq)     ->    H2O (l)   +    Ca(NO3) 2 (aq) + CO2 (g)

PbCO3 (s)  +  2HNO3(aq)     ->    H2O (l)   +    Pb(NO3) 2 (aq) + CO2 (g)

 

FeCO3 (s)  +  2HNO3(aq)     ->    H2O (l)   +    Fe(NO3) 2 (aq) + CO2 (g)

NaHCO3 (s)  +  HNO3(aq)     ->    H2O (l)   +    NaNO3 (aq) + CO2 (g)

 

KHCO3 (s)  +  HNO3(aq)     ->    H2O (l)   +    KNO3 (aq) + CO2 (g)

NH4HCO3 (aq)  +  HNO3(aq)     ->    H2O (l)   +    NH4NO3 (aq) + CO2 (g)

 

Ca(HCO3) 2 (aq)  +  2HNO3(aq)     ->    2H2O (l)   +    Ca(NO3) 2 (aq) + 2CO2 (g)

Mg(HCO3) 2 (aq)  +  2HNO3(aq)     ->    2H2O (l)   +    Mg(NO3) 2 (aq) + 2CO2 (g)

 

(iii) 25.0cm3 of 0.1M Nitric(V) acid was titrated with excess 0.2M sodium hydroxide solution using phenolphthalein indicator.

  1. State the colour change at the end point

                               Colourless

 

  1. What was the pH of the solution at the end point. Explain.

                    pH 1/2/3

A little of the acid when added to the base changes the colour of the indicator to show the end point. The end point therefore is acidic with low pH of Nitric(V) acid. Nitric(V) acid is a strong acid with pH 1/2/3.

 

III. Calculate the number of moles of acid used.

Number of moles   =    molarity  x volume     =>        0.1 x 25    =   2.5 x 10-3moles

1000                             1000

 

  1. Calculate the volume of sodium hydroxide used

Volume of sodium hydroxide in cm3

1000 x Number of moles  =>            1000x 2.5 x 10-3             =        12.5cm3

Molarity                                            0.2

 

(e)Industrial large scale manufacture of Nitric (V)acid

(i)Raw materials

  1. Air/Oxygen

Oxygen is got from fractional distillation of air

Ammonia from Haber process.

 

  1. 2. Chemical processes

 

Air from the atmosphere is passes through electrostatic precipitators/filters to remove unwanted gases like Nitrogen,Carbon(IV)oxide,dust,smoke which may poison the catalyst.The ammonia -air mixture  is compressed to 9 atmospheres to reduce the distance between reacting gases.

The mixture is passed through the heat exchangers where a temperature of 850oC-900oC is maintained.

The first reaction take place in the catalytic chamber where Ammonia reacts with the air to form Nitrogen(II)Oxide and water.

Optimum condition in Ostwalds process

Chemical equation                                                                              

4NH3(g)  + 5O2(g)  === Pt/Rh === 4NO(g) + 6H2O(g)  ΔH = 950kJ

The reaction is reversible and exist in dynamic equilibrium where the products reform back the reactants. The following factors are used to increase the yield/amount of  Notrogen(II)oxide:

(i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II) oxide.

More/higher yield of Nitrogen(II) oxide  is attained as reactants try to return the equilibrium balance.

(ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules .

Less/lower yield of Nitrogen(II)oxide is attained.

Very low pressures increases the distance between reacting NH3 and O2 molecules.

An optimum pressure of about 9 atmospheres is normally used.

Cooling the mixture condenses the water vapour to liquid water

 

(iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ).

Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained.

Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical.

An optimum temperature of about 900oC is normally used.

(iv)Platinum can be used as catalyst.

Platinum is very expensive. It is:

-promoted with Rhodium to increase the surface area/area of contact.

-added/coated on the surface of asbestos to form platinized –asbestos to  reduce the amount/quantity used.

The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation.

Nitrogen(II)oxide formed is passed through an oxidation reaction chamber where more air oxidizes the Nitrogen(II)Oxide   to  Nitrogen(IV)Oxide gas.

 

Chemical equation

2NO(g)        +        O2(g)    ->   2NO2(g)

Nitrogen(IV)Oxide gas is passed up to meet a downward flow of water in the absorption chamber. The gas react with water to form a mixture of Nitric(V) and Nitric(III)acids

Chemical equation.

2NO2(g)      +        H2O (l)    ->   HNO2(aq) + HNO3(aq)

Excess air is bubbled through the mixture to oxidize Nitric(III)/ HNO2(aq) to Nitric(V)/HNO3(aq)

Chemical equation.

O2(g)    +     2HNO2(aq)  ->   2HNO3(aq)

Overall chemical equation in the absorption chamber.

O2(g)    +     4NO2(g) +   2H2O (l)  ->   4HNO3(aq)

The acid is 65% concentrated.It is made 100% concentrated by either:

(i)fractional distillation or

(ii)added to concentrated sulphuric(VI) acid to remove the 35% of water.

 

 

A factory uses  63.0 kg of  68% pure nitric(V)acid per day to produce an ammonium fertilizer for an agricultural county. If the density of the acid is 1.42 gcm-3 , calculate :

 

(i)the concentration  of the acid used in moles per litre.

  Molar mass HNO3               =  63

Method 1

Moles of HNO3 in 1cm3 = Mass in 1cm3 1.42         =>    1.42    = 0.0225 moles

Molar mass HNO3                             63

Molarity = Moles x 1000=>0.0225 moles x10000 = 22.5molesdm-3/M

1 cm3

 

100%        =       22.5molesdm-3/M

68%           =               68  x 22.5                          =  15.3M/ molesdm-3

100

 

Method 2

Moles of HNO3 in 1000cm3 = Mass in 1000cm3 =>1.42 x1000

Molar mass HNO3                   63

=22.5397  molesdm-3/M

 

100%        =       22.5397  molesdm-3/M

68%           =               68  x  22.5397                   =  15.327 molesdm-3

100

 

(ii)the volume of ammonia gas at r.t.p used. (H=1.0,N=14.0,O=16.0,one mole of gas = 24 dm-3 at r.t.p)

Chemical equation

HNO3 (aq)    + NH3 (g)   ->       NH4NO3(aq)

 

Mole ratio HNO3 (aq) :  NH3 (g)  = 1 : 1

1 mole  HNO3 (aq) -> 24dm3 NH3 (g)

15.327 mole HNO3 (aq) ->15.327 mole x 24 dm3 = 367.848dm3

1dm3

(iii)the number of crops which can be applied the fertilizer if each crop require 4.0g.

 

HNO3 (aq)    + NH3 (g)      ->    NH4NO3(aq)

Molar mass NH4NO3 =80 g

Mole ratio HNO3  : NH4NO3   = 1 : 1

Mass of HNO3 in 63.0 kg = 68% x 63 =42.84kg

1 mole HNO3 (aq)=63g    -> 80g NH4NO3

  (42.84×1000)g HNO3 (aq) -> (42.84×1000)g  x 80

63

                                         = 54400g

Mass of fertilizer =           54400g        =  13600 crops

Mass  per crop                    4.0

 

 

 

  1. NITRATE(V) NO3 and NITRATE(III) NO2 Salts

 

Nitrate(V) /NO3 and Nitrate(III) /NO2  are salts derived from Nitric(V)/HNO3 and Nitric(III)/HNO2 acids. Both HNO3 and HNO2 are monobasic acids with only one ionizable hydrogen in a molecule.

Only KNO2 ,NaNO2  and NH4NO2 exist. All metallic nitrate(V)salts exist.

All Nitrate(V) /NO3 and Nitrate(III) /NO2  are soluble/dissolve in water.

 

(a)Effect of heat on Nitrate(V) /NO3 and Nitrate(III) /NO2 salts(Test for presence of Nitrate(V) /NO3  ions in solid state)

 

  1. All Nitrate(III) /NO2 salts are not affected by gentle or strong heating except ammonium nitrate(III) NH4NO2 .

Ammonium nitrate(III) NH4NOis a colourless solid that decompose to form Nitrogen gas and water.

 

Chemical equation

NH4NO2 (s) ->   H2O(l)   + N2(g)

This reaction is used to prepare small amounts of Nitrogen in a school laboratory.

  1. All Nitrate(V) /NO3 salts decompose on strong heating:

 

Experiment

Put ½ spatula full of sodium nitrate(V) into a a test tube. Place moist blue/red litmus papers on the mouth of the test tube.  Heat strongly when test tube is slanted.

Test the gases produced using glowing splint

Caution (i) Wear safety gas mask and hand gloves

              (ii)Lead(II)nitrate(V)decomposes to Lead(II)oxide that react and fuses with the test tube permanently.

 

Repeat with potassium nitrate(V), copper(II) nitrate(V), Lead(II)nitrate(V), silver nitrate(V), Zinc nitrate(V), Magnesium nitrate(V) and Ammonium nitrate(V).

 

Observations

Cracking sound

Brown fumes/gas produced except in potassium nitrate(V) and Sodium nitrate(V)

Glowing splint relights/rekindles but feebly in Ammonium nitrate(V).

Black solid residue with copper(II) nitrate(V)

White residue/solid with sodium nitrate(V), potassium nitrate(V),silver nitrate(V), Magnesium nitrate(V)

Yellow residue/solid when hot but white on cooling with Zinc nitrate(V)

Brown residue/solid when hot but yellow on cooling with Lead(II)nitrate(V)

Explanation

  1. Potassium nitrate(V) and Sodium nitrate(V) decomposes on strong heating to form potassium nitrate(III) and Sodium nitrate(III) producing Oxygen gas. Oxygen gas relights/rekindles a glowing splint.

Chemical equation.

2KNO3(s)    ->       2KNO2(s)   +     O2 (g)

2NaNO3(s) ->       2NaNO2(s) +     O2 (g)

 

2.Heavy metal nitrate(V)salts decomposes to form the oxide, brown nitrogen (IV) oxide and  Oxygen gas.

Copper(II)oxide is black.Zinc oxide is yellow when hot and white when cool/cold. Lead(II)oxide is yellow when cold/cool and brown when hot/heated.

Hydrated copper(II)nitrate is blue. On heating it melts and dissolves in its water of crystallization to form a green solution. When all the water of crystallization has evaporated,the nitrate(V)salt decomposes to black Copper(II)oxide and a mixture of brown nitrogen(IV)oxide gas and colourless Oxygen gas.

Chemical equation

2Cu(NO3)2(s)        ->     2CuO (s)       +    4NO2(g)    +    O2(s)

2Ca(NO3)2(s)        ->     2CaO (s)       +    4NO2(g)    +    O2(s)

2Zn(NO3)2(s)        ->     2ZnO (s)       +    4NO2(g)    +    O2(s)

2Mg(NO3)2(s)        ->     2MgO (s)     +    4NO2(g)    +    O2(s)

2Pb(NO3)2(s)          ->     2PbO (s)      +    4NO2(g)    +    O2(s)

2Fe(NO3)2(s)          ->     2FeO (s)      +    4NO2(g)    +    O2(s)

 

Silver nitrate(V)and Mercury(II)nitrate decomposes to the corresponding metal and a mixture of brown nitrogen(IV)oxide gas and colourless Oxygen gas.

Chemical equation

2AgNO3 (s)          ->     2Ag (s)      +    2NO2(g)    +    O2(s)

Hg(NO3)2(s)        ->     Hg(l)          +    2NO2(g)    +    O2(s)

The production/evolution of brown fumes of Nitrogen(IV)oxide gas on heating a salt is a confirmatory test for presence of NO3 ions of heavy metals

 

(b)Brown ring test (Test for presence of Nitrate(V) /NO3  ions in aqueous/ solution state)

 

Experiment

Place 5cm3 of Potassium nitrate(V)solution onto a clean test tube. Add 8 drops of freshly prepared Iron(II)sulphate(VI)solution.  Swirl/ shake.

Using a test tube holder  to firmly slant and hold the test tube, carefully add 5cm3 of Concentrated sulphuric (VI) acid down along the side of test tube.Do not shake the test tube contents.

Caution: Concentrated sulphuric (VI) acid is highly corrosive.

 

Observation.

Both Potassium nitrate(V)and freshly prepared Iron(II)sulphate (VI)do not form layers

On adding Concentrated Sulphuric(VI)acid,two layers are formed.

A brown ring is formed between the layers.

 

Explanation

All nitrate(V)salts are soluble. They form a miscible mixture when added freshly prepared Iron(II)sulphate(VI)solution. Concentrated sulphuric(VI)acid is denser than the miscible mixture thus settle at the bottom.

At the junction of the layers, the acid reacts with nitrate(V)salts to form Nitric(V)acid/HNO3. Nitric(V)acid/HNO3 is   reduced to Nitrogen (II)oxide by the Iron(II)sulphate(VI) salt to form the complex compound Nitroso-iron(II)sulphate(VI)/FeSO4.NO . Nitroso-iron(II)sulphate(VI) is brown in colour.It forms a thin layer at the junction between concentrated sulphuric (VI)acid and the miscible mixture of freshly prepared Iron(II) sulphate(VI) and the nitrate(V)salts as a brown ring.

Chemical equation

FeSO4(aq)      +      NO(g)    ->    FeSO4.NO(aq)

(Nitroso-iron(II)sulphate(VI)complex)

 

The brown ring disappear if shaken because concentrated sulphuric (VI)acid mixes with the aqueous solution generating a lot of heat which decomposes Nitroso-iron(II)sulphate(VI)/FeSO4.NO to iron(II)sulphate(VI) and Nitrogen(II)oxide.

Chemical equation

      FeSO4.NO(aq)                 ->FeSO4(aq)     +      NO(g)    ->

 

Iron(II)sulphate(VI) solution is easily/readily oxidized to iron(III)sulphate(VI) on exposure to air/oxygen. The brown ring test thus require freshly prepared Iron(II) sulphate(VI) solution

 

(c)Devardas alloy  test (Test for presence of Nitrate(V) /NO3  ions in aqueous/ solution state)

 

Experiment

Place 5cm3 of Potassium nitrate(V)solution onto a clean test tube. Add 5 drops of sodium hydroxide solution.  Swirl/ shake. Add a piece of aluminium foil to the mixture.Heat.Test any gases produced using both blue and red litmus papers.

 

Observation.                                                                                          Inference

Effervescence/bubbles/fizzing

colourless gas that has a pungent smell of urine              NO3

Blue limus paper remain blue

Red litmus paper turn red.

 

 

Explanation

The Devardas alloy test for NO3 ions in solution was developed by the Italian scientist Artulo Devarda(1859-1944)

When a NO3salt is added sodium hydroxide and aluminium foil, effervescence of ammonia gas is a confirmatory test for NO3ions.

 

  1. ENVIRONMENTAL EFFECTS OF NITROGENOUS COMPOUNDS

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

     

 

UPGRADE

CHEMISTRY

                

FORM 3

                 

Chemistry of SULPHUR

      

 

                

 

 

 

Comprehensive tutorial notes

 

 MUTHOMI S.G

www.kcselibrary.info

                                     0720096206

 

 

 

 

 

 

 

 

 

 

 

 

                       A.SULPHUR (S)

 

Sulphur is an element in Group VI Group 16)of the Periodic table . It has atomic number 16 and electronic configuration 16 and valency 2 /divalent and thus forms the ion S2-

 

  1. Occurrence.

 

Sulphur mainly occurs :

(i) as free element in Texas and Louisiana in USA and Sicily in Italy.

(ii)Hydrogen sulphide gas in active volcanic areas e.g. Olkaria near           Naivasha in Kenya

(iii)as copper pyrites(CuFeS2) ,Galena (PbS,Zinc blende(ZnS))and iron pyrites(FeS2) in other parts of  the world.

 

  1. Extraction of Sulphur from Fraschs process

 

Suphur occurs about 200 metres underground. The soil structure in these areas is usually weak and can easily cave in.

 

Digging of tunnels is thus discouraged in trying to extract the mineral.

Sulphur is extracted by drilling three concentric /round pipes of diameter of ratios 2:8: 18 centimeters.

 

Superheated water at 170oC and 10atmosphere pressure is forced through the outermost pipe.

 

The high pressures ensure the water remains as liquid at high temperatures instead of vapour of vapour /gas.

 

The superheated water melts the sulphur because the melting point of sulphur is lower at about at about 115oC.

 

A compressed air at 15 atmospheres is forced /pumped through the innermost pipe.

 

The hot air forces the molten sulphur up the middle pipe where it is collected and solidifies in a large tank.

 

It is about 99% pure.

 

Diagram showing extraction of Sulphur from Fraschs Process

 

 

 

 

  1. Allotropes of Sulphur.

 

  1. Sulphur exist as two crystalline allotropic forms:

(i)Rhombic sulphur

(ii)Monoclinic sulphur

 

Rhombic sulphur Monoclinic sulphur
Bright yellow crystalline solid

Has a melting point of 113oC

Has a density of 2.06gcm-3

Stable below 96oC

Has octahedral structure

Pale yellow crystalline solid

Has a melting point of 119oC

Has a density of 1.96gcm-3

Stable above 96oC

Has a needle-like structure

 

Rhombic sulphur and Monoclinic sulphur have a transition temperature of 96oC.This is the temperature at which one allotrope changes to the other.

 

 

 

 

 

 

 

 

 

  1. Sulphur exists in non-crystalline forms as:

(i)Plastic sulphur-

Plastic sulphur is prepared from heating powdered sulphur to boil then pouring a thin continuous stream in a beaker with cold water. A long thin elastic yellow thread of plastic sulphur is formed .If left for long it turn to bright yellow crystalline rhombic sulphur.

 

(ii)Colloidal sulphur-

Colloidal sulphur is formed when sodium thiosulphate (Na2S2O3) is added hydrochloric acid to form a yellow precipitate.

 

  1. Heating Sulphur.

 

A molecule of sulphur exists as puckered ring of eight atoms joined by covalent bonds as S8.

 

 

 

 

 

On heating the yellow sulphur powder melts at 113oC to clear amber liquid with low viscosity and thus flows easily.

On further heating to 160oC the molten liquid darkens to a brown very viscous liquid that does not flow easily.

This is because the S8 rings break into S8 chain that join together to form very long chains made of over 100000 atoms of Sulphur.

The long chains entangle each other reducing their mobility /flow and hence increases their viscosity.

 

On continued further heating to above 160oC, the viscous liquid darkens but becomes more mobile/flows easily and thus less viscous.

 

This is because the long chains break to smaller/shorter chains.

At 444oC, the liquid boils and forms brown vapour of a mixture of S8 ,S6 ,S2 molecules that solidifies to S8 ring of “flowers of sulphur” on the cooler parts.

 

Summary of changes on heating sulphur

 

Observation on heating Explanation/structure of Sulphur
Solid sulphur

 

 

Heat to 113oC

Amber yellow liquid

 

 

Heat to 160oC

Liquid darkens

 

 

 

Heat to 444oC

Liquid boils to brown vapour

 

 

Cool to room temperature

Yellow sublimate

(Flowers of Sulphur)

Puckered S8 ring

 

 

Puckered S8 ring in liquid form (low viscosity/flow easily)

 

Puckered S8 ring break/opens then join to form long chains that entangle (very high viscosity/very low rate of flow)

 

 

Mixture of S8 ,S6 ,S2 vapour

 

 

Puckered S8 ring

 

  1. Physical and Chemical properties of Sulphur.(Questions)

 

  1. State three physical properties unique to Sulphur

Sulphur is a yellow solid, insoluble in water, soluble in carbon disulphide/tetrachloromethane/benzene, poor conductor of heat and electricity. It has a melting point of 115oC and a boiling point of 444oC.

 

  1. Moist/damp/wet blue and red litmus papers were put in a gas jar containing air/oxygen. Burning sulphur was then lowered into the gas jar. State and explain the observation made.

 

Observations

-Sulphur melts then burns with a blue flame

Colourless gas produced that has a pungent smell

Red litmus paper remains red. Blue litmus paper turns red.

 

Explanation

Sulphur burns in air and faster in Oxygen to form Sulphur(IV)Oxide gas and traces/small amount of Sulphur(VI)Oxide gas. Both oxides react with water to form the corresponding acidic solution i.e

 

(i) Sulphur(IV)Oxide gas reacts with water to form sulphuric(IV)acid

(ii) Sulphur(VI)Oxide gas reacts with water to form sulphuric(VI)acid

 Chemical equation

S(s)              + O2(g)        ->       SO2(g) (Sulphur(IV)Oxide gas)

2S(s)            + 3O2(g)      ->       2SO3(g) (Sulphur(VI)Oxide gas traces)

SO2(g)         +  H2O(l)     ->       H2 SO3 (aq) ( sulphuric(IV)acid)                  SO3(g)         +  H2O(l)     ->       H2 SO4 (aq) ( sulphuric(VI)acid).

 

  1. Iron filings were put in a test tube containing powdered sulphur then heated on a Bunsen flame. Stop heating when reaction starts. State and explain the observations made. Test the effects of a magnet on the mixture before and after heating. Explain.

 

Observations

Before heating, the magnet attracts iron filings leaving sulphur

After heating, the magnet does not attract the mixture.

After heating, a red glow is observed that continues even when heating is stopped..

Black solid is formed.

 

Explanation                                                                                                     Iron is attracted to a magnet because it is ferromagnetic.

When a mixture of iron and sulphur is heated, the reaction is exothermic giving out heat energy that makes the mixture to continue glowing even after stopping heating.

Black Iron(II)sulphide is formed which is a compound and thus not ferromagnetic.

 

Chemical equation                                                                                           Fe(s) + S(s)                      -> FeS(s) (Exothermic reaction/ ∆H)

 

Heated powdered heavy metals  combine with sulphur to form black sulphides.

Cu(s)           + S(s)           ->       CuS(s)

Zn(s)            + S(s)           ->       ZnS(s)

Pb(s)            + S(s)           ->       PbS(s)

 

4.The set up below show the reaction of sulphur on heated concentrated sulphuric(VI)acid.

 

                                                              

(i)State and explain the observation made.

Observation

Yellow colour of sulphur fades

Orange colour of potassium dichromate(VI)paper turns to green.

 

Explanation

Hot concentrated sulphuric(VI)acid oxidizes sulphur to sulphur (IV)oxide gas. The oxide is also reduced to water. Traces of sulphur (VI)oxide is formed.

 

Chemical equation

S(s)   +    3H2 SO4 (l)     ->    3SO2(g) +  3H2O(l) +SO3(g)

 

Sulphur (IV)oxide gas turns Orange potassium dichromate(VI)paper to green.

 

(ii)State and explain the observation made if concentrated sulphuric (VI) acid is replaced with concentrated Nitric (V) acid in the above set up.

 

          Observation

Yellow colour of sulphur fades

Colurless solution formed

Brown fumes/gas produced.

 

Explanation

Hot concentrated Nitric(V)acid oxidizes sulphur to sulphuric (VI)acid. The Nitric (V) acid  is reduced to brown nitrogen(IV)oxide gas.

 

Chemical equation

S(s)   +    6HNO3 (l)     ->    6NO2(g) +  2H2O(l) +H2SO4 (l)

 

NB:

Hydrochloric acid is a weaker oxidizing agent and thus cannot oxidize sulphur like the other mineral acids.

 

5.State three main uses of sulphur                                                     .

Sulphur is mainly used in:

(i)Contact process for the manufacture/industrial/large scale production of concentrated sulphuric(VI)acid.

(ii)Vulcanization of rubber to make it harder, tougher, stronger, and more durable.

(iii)Making gun powder and match stick heads

(iv) As ointments to treat fungal infections

 

 

 

  1. Revision Practice

The diagram below represents the extraction of sulphur by Fraschs process. Use it to answer the questions that follow.

N
M
L

 

 

 

 

 

 

 

 

(a)Name the substances that passes through:

M      Superheated water at 170oC and 10 atmosphere pressure

L        Hot compressed air

        N       Molten sulphur

 

(b)What is the purpose of the substance that passes through L and M?

M- Superheated water at 170oC and 10 atmosphere pressure is used to melt the sulphur

L- Hot compressed air is used to force up the molten sulphur.

 

(c) The properties of the two main allotropes of sulphur represented by letters A and B are given in the table below. Use it to answer the questions that follow.

 

  A B
Appearance Bright yellow Pale yellow
Density(gcm-3) 1.93 2.08
Melting point(oC) 119 113
Stability Above 96oC Below 96oC

 

I.What are allotropes?

          Different forms of the same element existing at the same temperature and pressure without change of state.

 

  1. Identify allotrope:
  2. Monoclinic sulphur

        B . Rhombic sulphur

 

III. State two main uses of sulphur.

          -Manufacture of sulphuric(VI)acid

-as fungicide

-in vulcanization of rubber to make it harder/tougher/ stronger

-manufacture of dyes /fibres

 

(d)Calculate the volume of sulphur (IV)oxide produced when 0.4 g of sulphur is completely burnt in excess air.(S = 32.0 ,I mole of a gas occupies 24 dm3 at room temperature)

 

Chemical equation

S(s)   +  O2(g)   ->  SO2(g)

Mole ratio S: SO2  =  1:1

 

Method 1

32.0 g of sulphur -> 24 dm3 of SO2(g)

0.4 g of sulphur ->  0.4 g x 24 dm3  =   0.3 dm3

32.0 g

Method 2

Moles of sulphur used = Mass of sulphur         =>  0.4 = 0.0125 moles

Molar mass of sulphur                    32

 

Moles of sulphur used = Moles of sulphur(IV)oxide used=>0.0125 moles

Volume of sulphur(IV)oxide used = Moles of sulphur(IV)oxide x volume of one mole of gas =>0.0125 moles x 24 dm3 =     0.3 dm3

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

B.COMPOUNDS OF SULPHUR

The following are the main compounds of sulphur:

(i) Sulphur(IV)oxide

          (ii) Sulphur(VI)oxide                                              .

(iii) Sulphuric(VI)acid

(iv) Hydrogen Sulphide

(v) Sulphate(IV)/SO32- and Sulphate(VI)/ SO42- salts

 

                  (i) Sulphur(IV)oxide(SO2)

 

(a) Occurrence

Sulphur (IV)oxide is found in volcanic areas as a gas or dissolved in water from geysersand hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya.

 

(b) School laboratory preparation

 

In a Chemistry school laboratory Sulphur (IV)oxide is prepared from the reaction of

 

Method 1:Using Copper and  Sulphuric(VI)acid.

 

 

 

Method 2:Using Sodium Sulphate(IV) and hydrochloric acid.

 

 

(c)Properties of Sulphur(IV)oxide(Questions)

 

  1. Write the equations for the reaction for the formation of sulphur (IV)oxide using:

(i)Method 1

 

Cu(s)  + 2H2SO4(l)  ->  CuSO4(aq)  + SO2(g)  + 2H2O(l)

Zn(s)  + 2H2SO4(l)  ->  ZnSO4(aq)  + SO2(g)  + 2H2O(l)

 

Mg(s)  + 2H2SO4(l)  ->  MgSO4(aq)  + SO2(g)  + 2H2O(l)

Fe(s)  + 2H2SO4(l)  ->  FeSO4(aq)  + SO2(g)  + 2H2O(l)

 

Calcium ,Lead and Barium will form insoluble sulphate(VI)salts that will cover unreacted metals stopping further reaction thus producing very small amount/quantity of sulphur (IV)oxide gas.

 

(ii)Method 2

Na2SO3(aq) + HCl(aq) -> NaCl(aq ) + SO2(g)  + 2H2O(l)

K2SO3(aq) + HCl(aq) -> KCl(aq ) + SO2(g)  + 2H2O(l)

 

BaSO3(s) + 2HCl(aq) -> BaCl2(aq ) + SO2(g)  + H2O(l)

CaSO3(s) + 2HCl(aq) -> CaCl2(aq ) + SO2(g)  + H2O(l)

PbSO3(s) + 2HCl(aq) -> PbCl2(s ) + SO2(g)  + H2O(l)

 

Lead(II)chloride is soluble on heating thus reactants should be heated to prevent it coating/covering unreacted PbSO3(s)

 

2.State the physical properties unique to sulphur (IV)oxide gas.

Sulphur (IV)oxide gas is a colourless gas with a pungent irritating and  choking smell which liquidifies easily. It is about two times denser than air.

 

  1. The diagram below show the solubility of sulphur (IV)oxide gas. Explain.

 

 

 

Sulphur(IV) oxide is very soluble in water.

One drop of water dissolves all the Sulphur (IV) oxide in the flask leaving a vacuum.

If the clip is removed, atmospheric pressure forces the water up through the narrow tube to form a fountain to occupy the vacuum.

An acidic solution of sulphuric (IV)acid is formed which turns litmus solution red.

Chemical equation

SO2(g) +  H2O(l)             ->     H2 SO3 (aq) ( sulphuric(IV)acid turn litmus red)

 

4.Dry litmus papers and wet/damp/moist litmus papers were put in a gas jar containing sulphur(IV) oxide gas. State and explain the observations made.

Observations

(i)Dry Blue litmus paper remains blue.

Dry red litmus paper remains red.

(ii) Wet/damp/moist blue litmus paper turns red.

Moist/damp/wet red litmus paper remains red.

Both litmus papers are then bleached /decolorized.

 

Explanation

Dry sulphur(IV) oxide gas is a molecular compound that does not dissociate/ionize to release H+(aq)ions and thus has no effect on dry blue/red litmus papers.

Wet/damp/moist litmus papers contain water that dissolves /react with dry sulphur(IV) oxide gas to form a solution of weak sulphuric(IV)acid (H2 SO3 (aq)).

Weak sulphuric(IV)acid(H2 SO3 (aq)) dissociates /ionizes into free H+(aq)ions:

H2 SO3 (aq) -> 2H+(aq)   +  SO32- (aq)

 

The free H+(aq)ions are responsible for turning blue litmus paper turns red showing the gas is acidic.

The SO32- (aq) ions in wet/damp/moist sulphur(IV) oxide gas is responsible for many reactions of the gas.

It is easily/readily oxidized to sulphate(VI) SO42- (aq) ions making sulphur(IV) oxide gas act as a reducing agent as in the following examples:

 

(a)Bleaching agent

 

Wet/damp/moist coloured flowers/litmus papers are bleached/decolorized when put in sulphur(IV) oxide gas.

This is because sulphur(IV) oxide removes atomic oxygen from the coloured dye/ material to form sulphuric(VI)acid.

Chemical equations

 

(i)Formation of sulphuric(IV)acid

SO2(g) +  H2O(l)             ->     H2 SO3 (aq)

 

(ii)Decolorization/bleaching of the dye/removal of atomic oxygen.

Method I. H2 SO3 (aq)  + (dye + O)   -> H2 SO4 (aq)   + dye

(coloured)                            (colourless)

 

Method II. H2 SO3 (aq)  + (dye)   -> H2 SO4 (aq)   + (dye – O)

(coloured)                         (colourless)

 

Sulphur(IV) oxide gas therefore bleaches by reduction /removing oxygen from a dye unlike chlorine that bleaches by oxidation /adding oxygen.

The bleaching by removing oxygen from Sulphur(IV) oxide gas is temporary.

 

This is because the bleached dye regains the atomic oxygen from the atmosphere/air in presence of sunlight as catalyst thus regaining/restoring its original colour. e.g.

 

Old newspapers turn brown on exposure to air on regaining the atomic oxygen.

The bleaching through adding oxygen by chlorine gas is permanent.

 

(b)Turns Orange acidified potassium dichromate(VI) to green

 

Experiment:

 

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium dichromate(VI) solution. or;

 

(ii)Dip a filter paper soaked in acidified potassium dichromate(VI) into a gas jar containing Sulphur(IV) oxide gas.

 

Observation:

Orange acidified potassium dichromate(VI) turns to green.

 

Explanation:

Sulphur(IV) oxide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in  sulphuric(VI) acid.

 

Chemical/ionic equation:

(i)Reaction of Sulphur(IV) oxide gas with water

SO2(g) +  H2O(l)             ->     H2 SO3 (aq)

 

(ii)Dissociation /ionization of Sulphuric(IV)acid.

H2 SO3 (aq) -> 2H+(aq)   +  SO32- (aq)

 

(iii)Oxidation of SO32- (aq)and reduction of Cr2O72-(aq)

3SO32-(aq)  +   Cr2O72-(aq) +8H+(aq)   -> 3SO42-(aq) +  2Cr3+(aq) +  4H2O(l)

 

This is a confirmatory test for the presence of Sulphur(IV) oxide gas.

 

Hydrogen sulphide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a yellow residue.

 

 

(c)Decolorizes acidified potassium manganate(VII)

 

Experiment:

 

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or;

 

(ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Sulphur(IV) oxide gas.

 

Observation:

Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized.

 

Explanation:

Sulphur(IV) oxide gas reduces acidified potassium manganate(VII) from purple MnO4 ions to green Mn2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in  sulphuric(VI) acid.

 

Chemical/ionic equation:

(i)Reaction of Sulphur(IV) oxide gas with water

SO2(g) +  H2O(l)             ->     H2 SO3 (aq)

 

(ii)Dissociation /ionization of Sulphuric(IV)acid.

H2 SO3 (aq) -> 2H+(aq)   +  SO32- (aq)

 

(iii)Oxidation of SO32- (aq)and reduction of MnO4 (aq)

5SO32-(aq)  +   2MnO4 (aq) +6H+(aq)   -> 5SO42-(aq) +  2Mn2+(aq) +  3H2O(l)

(purple)                                                     (colourless)

 

This is another test for the presence of Sulphur(IV) oxide gas.

 

Hydrogen sulphide also decolorizes acidified potassium manganate(VII) from purple MnO4 ions to colourless Mn2+ ions leaving a yellow residue.

 

(d)Decolorizes bromine water

 

Experiment:

 

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing bromine water . or;

 

(ii)Put three drops of bromine water into a gas jar containing Sulphur(IV) oxide gas. Swirl.

 

Observation:

Yellow bromine water turns to colourless/ bromine water is decolorized.

 

Explanation:

Sulphur(IV) oxide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) without leaving a residue itself oxidized from SO32- ions in sulphuric (IV) acid to SO42- ions in  sulphuric(VI) acid.

 

Chemical/ionic equation:

(i)Reaction of Sulphur(IV) oxide gas with water

SO2(g) +  H2O(l)             ->     H2 SO3 (aq)

 

(ii)Dissociation /ionization of Sulphuric(IV)acid.

H2 SO3 (aq) -> 2H+(aq)   +  SO32- (aq)

 

(iii)Oxidation of SO32- (aq)and reduction of MnO4 (aq)

SO32-(aq)  +   Br2 (aq) + H2O(l)   ->  SO42-(aq) +    2HBr(aq)

(yellow)                                           (colourless)

 

This can also be used as another test for the presence of Sulphur(IV) oxide gas.

 

Hydrogen sulphide also decolorizes yellow bromine water to colourless leaving a yellow residue.

 

(e)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+

Experiment:

 

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or;

 

(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Sulphur(IV) oxide gas.Swirl.

 

Observation:

Yellow/brown Iron (III)chloride solution turns to green

 

Explanation:

Sulphur(IV) oxide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions without leaving a residue itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in  sulphuric(VI) acid.

 

Chemical/ionic equation:

(i)Reaction of Sulphur(IV) oxide gas with water

SO2(g) +  H2O(l)             ->     H2 SO3 (aq)

 

(ii)Dissociation /ionization of Sulphuric(IV)acid.

H2 SO3 (aq) -> 2H+(aq)   +  SO32- (aq)

 

(iii)Oxidation of SO32- (aq)and reduction of Fe3+ (aq)

SO32-(aq)  +   2Fe3+ (aq) +3H2O(l)  -> SO42-(aq) +  2Fe2+(aq) +  2H+(aq)

(yellow)                                               (green)

 

 

(f)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas

Experiment:

 

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or;

 

(ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Sulphur(IV) oxide gas. Swirl.

 

Observation:

Brown fumes of a gas evolved/produced.

 

Explanation:

Sulphur(IV) oxide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in  sulphuric(VI) acid.

 

Chemical/ionic equation:

SO2(g) +  2HNO3 (l)       ->     H2 SO4 (l)  + NO2 (g)

(brown fumes/gas)

 

(g)Reduces Hydrogen peroxide to water

Experiment:

 

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide. Add four drops of Barium nitrate(V)or Barium chloride followed by five drops of  2M hydrochloric acid/ 2M nitric(V) acid.

 

Observation:

A white precipitate is formed that persist /remains on adding 2M hydrochloric acid/ 2M nitric(V) acid.

 

Explanation:

Sulphur(IV) oxide gas reduces 20 volume hydrogen peroxide and  itself oxidized from SO32- ions in sulphuric(IV) acid to SO42- ions in  sulphuric(VI) acid.

When Ba2+ ions in Barium Nitrate(V) or Barium chloride solution is added, a white precipitate of insoluble Barium salts is formed showing the presence of of either  SO32- ,SO42- ,CO32- ions. i.e.

 

Chemical/ionic equation:

SO32-(aq)  +   Ba2+ (aq)  -> BaSO3(s)

white precipitate

SO42-(aq)  +   Ba2+ (aq)  -> BaSO4(s)

white precipitate

CO32-(aq)  +   Ba2+ (aq)  -> BaCO3(s)

white precipitate

 

If nitric(V)/hydrochloric acid is added to the three suspected insoluble white precipitates above, the white precipitate:

(i) persist/remains if SO42-(aq)ions (BaSO4(s)) is present.

(ii)dissolves if SO32-(aq)ions (BaSO3(s)) and CO32-(aq)ions (BaCO3(s))is present. This is because:

 

  1. BaSO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic SO2 gas that turns Orange moist filter paper dipped in acidified Potassium dichromate to green.

 

Chemical equation

BaSO3(s) +2H+(aq)  -> Ba2+ (aq)  + SO2(g)  + H2O(l)

 

  1. BaCO3(s) reacts with Nitric(V)/hydrochloric acid to produce acidic CO2 gas that forms a white precipitate when bubbled in lime water.

Chemical equation

 

BaCO3(s) +2H+(aq)  -> Ba2+ (aq)  + CO2(g)  + H2O(l)

 

5.Sulphur(IV)oxide also act as an oxidizing agent as in the following examples.

 

(a)Reduction by burning Magnesium

 

Experiment

Lower a burning Magnesium ribbon into agas jar containing Sulphur(IV)oxide gas

 

Observation

Magnesium ribbon continues to burn with difficulty.

White ash and yellow powder/speck

 

Explanation

Sulphur(IV)oxide does not support burning/combustion. Magnesium burns to produce enough heat energy to decompose Sulphur(IV)oxide to sulphur and oxygen.

The metal continues to burn on Oxygen forming white Magnesium oxide solid/ash.

Yellow specks of sulphur residue form on the sides of reaction flask/gas jar.

During the reaction, Sulphur(IV)oxide is reduced(oxidizing agent)while the metal is oxidized (reducing agent)

 

Chemical equation

SO2(g)  +  2Mg(s)          ->       2MgO(s)      +       S(s)

(white ash/solid)     (yellow speck/powder)

 

(b)Reduction by Hydrogen sulphide gas

 

Experiment

Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas

Bubble hydrogen sulphide gas into the gas jar containing Sulphur(IV)oxide gas.

Or

Put two drops of water into a gas jar containing dry Sulphur(IV)oxide gas

Invert a gas jar full of hydrogen sulphide gas over the gas jar containing Sulphur(IV)oxide gas. Swirl

 

Observation

Yellow powder/speck

Explanation

Sulphur(IV)oxide  oxidizes hydrogen sulphide to yellow specks of sulphur residue and itself reduced to also sulphur that form on the sides of reaction flask/gas jar.

A little moisture/water act as catalyst /speeds up the reaction.

Chemical equation

SO2(g)  +  2H2S(g)         ->       2H2O(l)       +       3S(s)

(yellow speck/powder)

 

6.Sulphur(IV)oxide has many industrial uses. State three.

 

(i)In the contact process for the manufacture of Sulphuric(VI)acid

(ii)As a bleaching agent of pulp and paper.

(iii)As a fungicide to kill microbes’

(iv)As a preservative of jam, juices to prevent fermentation

 

(ii) Sulphur(VI)oxide(SO3)

 

(a) Occurrence

 

Sulphur (VI)oxide is does not occur free in nature/atmosphere

 

(b) Preparation

 

In a Chemistry school laboratory Sulphur (VI)oxide may  prepared from:

 

Method 1;Catalytic oxidation  of sulphur(IV)oxide gas.

 

Sulphur(IV)oxide gas and oxygen mixture are first dried by being passed through Concentrated Sulphuric(VI)acid .

The dry mixture is then passed through platinised asbestos to catalyse/speed up the combination to form Sulphur (VI)oxide gas.

Sulphur (VI)oxide gas readily solidify as silky white needles if passed through a freezing mixture /ice cold water.

The solid fumes out on heating to a highly acidic poisonous gas.

 

Chemical equation 

2SO2(g) +  O2(g)   –platinised asbestos–>      2SO3 (g)

 

Method 2; Heating Iron(II)sulphate(VI) heptahydrate

 

When green hydrated Iron(II)sulphate(VI) heptahydrate crystals are heated in a boiling tube ,it loses the water of crystallization and colour changes from green to white.

 

Chemical equation 

FeSO4.7H2O(s)    ->    FeSO4(s)    +   7H2O(l)

(green solid)             (white solid)

 

On further heating ,the white anhydrous Iron(II)sulphate(VI) solid decomposes to a mixture of Sulphur (VI)oxide and  Sulphur (IV)oxide gas.

Sulphur (VI) oxide readily / easily solidify as white silky needles when the mixture is passed through a freezing mixture/ice cold water.

Iron(III)oxide is left  as a brown residue/solid.

 

Chemical equation 

 

2FeSO4 (s)    ->         Fe2O3(s)    +   SO2 (g) +   SO3(g)

(green solid)             (brown solid)

 

Caution

On exposure to air Sulphur (VI)oxide gas produces highly corrosive poisonous fumes of concentrated sulphuric(VI)acid and thus its preparation in a school laboratory is very risky.

 

(c) Uses of sulphur(VI)oxide

 

One of the main uses of sulphur(VI)oxide gas is as an intermediate product in the contact process for industrial/manufacture/large scale/production of sulphuric(VI)acid.

 

 

(iii) Sulphuric(VI)acid(H2SO4)

 

(a) Occurrence

 

Sulphuric (VI)acid(H2SO4) is one of the three mineral acids. There are three mineral acids;

Nitric(V)acid

Sulphuric(VI)acid

Hydrochloric acid.

Mineral acids do not occur naturally but are prepared in a school laboratory and manufactured at industrial level.

 

(b)The Contact process for industrial manufacture of H2SO4 .

 

  1. Raw materials

 

The main raw materials for industrial preparation of Sulphuric(VI)acid include:

(i)Sulphur from Fraschs process or from heating metal sulphide ore like Galena(PbS),Zinc blende(ZnS)

(ii)Oxygen from fractional distillation of air

(iii)Water from rivers/lakes

 

  1. Chemical processes

 

The contact process involves four main chemical processes:

 

(i)Production of Sulphur (IV)oxide

 

As one of the raw materials, Sulphur (IV)oxide gas is got from the following sources;

 

  1. Burning/roasting sulphur in air.

 

Sulphur from Fraschs process is roasted/burnt in air to form Sulphur (IV)oxide gas in the burners

Chemical equation

S(s) +  O2(g)              –>      SO2 (g)

 

  1. Burning/roasting sulphide ores in air.

 

Sulphur (IV)oxide gas is produced as a by product in extraction of some metals like:

– Lead from Lead(II)sulphide/Galena,(PbS)

– Zinc from zinc(II)sulphide/Zinc blende, (ZnS)

– Copper from Copper iron sulphide/Copper pyrites, (CuFeS2)

 

On roasting/burning, large amount /quantity of sulphur(IV)oxide is generated/produced.

Chemical equation

 

(i)2PbS (s)             + 3O2 (g)    ->     2PbO(s)    +   2SO2 (g)

(ii)2ZnS (s)            + 3O2 (g)    ->     2ZnO(s)    +   2SO2 (g)

(ii)2CuFeS2 (s)      + 4O2 (g)    ->     2FeO(s)    +   3SO2 (g)   +   Cu2O(s)

 

Sulphur(IV)oxide easily/readily liquefies and thus can be transported to a far distance safely.

 

(ii)Purification of Sulphur(IV)oxide

 

Sulphur(IV)oxide gas contain dust particles and Arsenic(IV)oxide as impurities. These impurities “poison”/impair the catalyst by adhering on/covering its surface.

The impurities are removed by electrostatic precipitation method .

In the contact process Platinum or Vanadium(V)oxide may be used. Vanadium(V)oxide is preferred because it is :

(i) cheaper/less expensive

(ii) less easily poisoned by impurities

 

(iii)Catalytic conversion of Sulphur(IV)oxide to Sulphur(VI)oxide

 

Pure and dry mixture of Sulphur (IV)oxide gas and Oxygen is heated to 450oC in a heat exchanger.

The heated mixture is passed through long pipes coated with pellets of Vanadium (V)oxide catalyst.

The close “contact” between the reacting gases and catalyst give the process its name.

Vanadium (V)oxide catalyse the conversion/oxidation of Sulphur(IV)oxide to Sulphur(VI)oxide gas.

Chemical equation

2SO2 (g) +  O2(g)      — V2O5 –>      2SO2 (g)

 

This reaction is exothermic (-∆H) and the temperatures need to be maintained at around 450oC to ensure that:

(i)reaction rate/time taken for the formation of Sulphur(VI)oxide is not too slow/long at lower temperatures below 450oC

(ii) Sulphur(VI)oxide gas does not decompose back to Sulphur(IV)oxide gas and Oxygen gas at higher temperatures than 450oC.

 

(iv)Conversion of Sulphur(VI)oxide of Sulphuric(VI)acid

 

Sulphur(VI)oxide is the acid anhydride of concentrated Sulphuric(VI)acid. Sulphur(VI)oxide reacts with water to form thick mist of fine droplets of very/highly corrosive concentrated Sulphuric(VI)acid because the reaction is highly exothermic.

To prevent this, Sulphur (VI)oxide is a passed up  to meet downward flow of 98% Sulphuric(VI)acid  in the absorption chamber/tower.

The reaction forms a very viscous oily liquid called Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid.

 

Chemical equation

H2SO4 (aq)  + SO3 (g)    ->        H2S2O7 (l)

 

Oleum/fuming Sulphuric (VI) acid/ pyrosulphuric (VI) acid is diluted carefully with distilled water to give concentrated sulphuric (VI) acid .

Chemical equation

H2S2O7 (l)  + H2O (l)     ->        2H2SO4 (l)

 

The acid is stored ready for market/sale.

 

 III. Environmental effects of contact process

 

Sulphur(VI)oxide and Sulphur(IV)oxide gases are atmospheric pollutants that form acid rain if they escape to the atmosphere.

In the Contact process, about 2% of these gases do not form sulphuric (VI) acid.

The following precautions prevent/minimize pollution from Contact process:

(i)recycling back any unreacted Sulphur(IV)oxide gas back to the heat exchangers.

(ii)dissolving Sulphur(VI)oxide gas in concentrated sulphuric (VI) acid instead of water.

This prevents the formation of fine droplets of the corrosive/ toxic/poisonous fumes of concentrated sulphuric (VI) acid.

 

(iii)scrubbing-This involves passing the exhaust gases through very tall chimneys lined with quicklime/calcium hydroxide solid.

This reacts with Sulphur (VI)oxide gas forming harmless calcium(II)sulphate (IV) /CaSO3

 

Chemical equation

Ca(OH)2 (aq) +  SO2(g)        –>      CaSO3 (aq) + H2O (g)

 

  1. Uses of Sulphuric(VI)acid

 

Sulphuric (VI) acid is used:

(i) in making dyes and paint

(ii)as acid in Lead-acid accumulator/battery

(iii) for making soapless detergents

(iv) for making sulphate agricultural fertilizers

 

 

  1. Sketch chart diagram showing the Contact process

 

(c) Properties of Concentrated sulphuric(VI)acid       (i)Concentrated sulphuric(VI)acid is a colourless oily liquid with a density of 1.84gcm-3.It has a boiling point of 338oC.

 

(ii) Concentrated sulphuric(VI)acid is very soluble in water.

The solubility /dissolution of the acid very highly exothermic.

The concentrated acid should thus be diluted slowly in excess water.

Water should never be added to the acid because the hot acid scatters highly corrosive fumes out of the container.

 

(iii) Concentrated sulphuric (VI)acid is a covalent compound. It has no free H+ ions.

Free H+ ions are responsible for turning the blue litmus paper red. Concentrated sulphuric (VI) acid thus do not change  the blue litmus paper red.

 

(iv) Concentrated sulphuric (VI)acid is hygroscopic. It absorbs water from the atmosphere and do not form a solution.

This makes concentrated sulphuric (VI) acid very suitable as drying agent during preparation of gases.

(v)The following are some chemical properties of concentrated sulphuric (VI) acid:

  1. As a dehydrating agent

 

Experiment I;

Put about  four spatula end fulls of brown sugar and glucose in separate 10cm3 beaker.

Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Allow to stand for about 10 minutes.

Observation;

Colour( in brown sugar )change from brown to black.

Colour (in glucose) change from white to black.

10cm3 beaker becomes very hot.

Explanation

Concentrated sulphuric (VI) acid is strong dehydrating agent.

It removes chemically and physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds.

When added to sugar /glucose a vigorous reaction that is highly exothermic take place.

The sugar/glucose is charred to black mass of carbon because the acid dehydrates the sugar/glucose leaving carbon.

Caution

This reaction is highly exothermic that start slowly but produce fine particles of carbon that if inhaled cause quick suffocation by blocking the lung villi.

 

Chemical equation

Glucose:    C6H12O6(s)     –conc.H2SO4–>   6C (s)  +      6H2O(l)

(white)                                     (black)

 

Sugar:    C12H22O11(s)      –conc.H2SO4–> 12C (s)  +11H2O(l)

(brown)                                  (black)

 

Experiment II;

Put about two spatula end full of hydrated copper(II)sulphate(VI)crystals in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid .Warm .

Observation;

Colour change from blue to white.

Explanation

 

Concentrated sulphuric (VI) acid is strong dehydrating agent.It removes physically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from hydrated compounds.

The acid dehydrates blue copper(II)sulphate to white anhydrous copper(II)sulphate .

 

Chemical equation

CuSO4.5H2O(s) –conc.H2SO4–>   CuSO4 (s)  +      5H2O(l)

(blue)                                         (white)

 

Experiment III;

Put about 4cm3 of absolute ethanol in a boiling tube .Carefully add about 10cm3 of concentrated sulphuric (VI) acid.

Place moist/damp/wet filter paper dipped in acidified potassium dichromate(VI)solution on the mouth of the boiling tube. Heat strongly.

Caution:

Absolute ethanol is highly flammable.

Observation;

Colourless gas produced.

Orange acidified potassium dichromate (VI) paper turns to green.

 

Explanation

Concentrated sulphuric (VI) acid is strong dehydrating agent.

It removes chemically combined elements of water(Hydrogen and Oxygen in ratio 2:1)from compounds.

The acid dehydrates ethanol to ethene gas at about 170oC.

Ethene with  =C=C= double bond turns orange acidified potassium dichromate (VI) paper turns to green.

 

Chemical equation

C2H5OH(l) –conc.H2SO4/170oC –>   C2H4 (g)  +      H2O(l)

NB: This reaction is used for the school laboratory preparation of ethene gas

 

Experiment IV;

Put about 4cm3 of methanoic acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Heat gently

 

Caution:

This should be done in a fume chamber/open

Observation;

Colourless gas produced.

Explanation

Concentrated sulphuric (VI) acid is strong dehydrating agent. It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds.

The acid dehydrates methanoic acid to poisonous/toxic carbon(II)oxide gas.

 

Chemical equation

HCOOH(l) –conc.H2SO4 –>   CO(g)  +      H2O(l)

NB: This reaction is used for the school laboratory preparation of small amount carbon (II)oxide gas

 

Experiment V;

Put about 4cm3 of ethan-1,2-dioic/oxalic  acid in a boiling tube .Carefully add about 6 cm3 of concentrated sulphuric (VI) acid. Pass any gaseous product through lime water.Heat gently

 

Caution:

This should be done in a fume chamber/open

 

Observation;

Colourless gas produced.

Gas produced forms a white precipitate with lime water.

Explanation

Concentrated sulphuric (VI) acid is strong dehydrating agent.

It removes chemically combined elements of water (Hydrogen and Oxygen in ratio 2:1)from compounds.

The acid dehydrates ethan-1,2-dioic/oxalic  acid to a mixture of  poisonous/toxic carbon(II)oxide and carbon(IV)oxide gases.

 

Chemical equation

HOOCCOOH(l) –conc.H2SO4 –>   CO(g)  +  CO2(g)  +   H2O(l)

 

NB: This reaction is also used for the school laboratory preparation of small amount carbon (II) oxide gas.

Carbon (IV) oxide gas is removed by passing the mixture through concentrated sodium/potassium hydroxide solution.

 

  1. As an Oxidizing agent

 

Experiment I

Put about 2cm3 of Concentrated sulphuric (VI) acid into three separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube. Put about 0.5g of Copper turnings, Zinc granule and Iron filings to each boiling tube separately.

 

Observation;

Effervescence/fizzing/bubbles

Blue solution formed with copper,

Green solution formed with Iron

Colourless solution formed with Zinc

Colourless gas produced that has a pungent irritating choking smell.

Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.

 

Explanation

Concentrated sulphuric (VI) acid is strong oxidizing agent.

It oxidizes metals to metallic sulphate(VI) salts and itself reduced to sulphur(IV)oxide gas.

Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.

CuSO4(aq) is a blue solution. ZnSO4(aq) is a colourless solution. FeSO4(aq) is a green solution.

 

Chemical equation

Cu(s) +  2H2SO4(aq) –>   CuSO4(aq)  +  SO2(g)  +   2H2O(l)

Zn(s) +  2H2SO4(aq) –>   ZnSO4(aq)  +  SO2(g)  +   2H2O(l)

Fe(s) +  2H2SO4(aq) –>   FeSO4(aq)  +  SO2(g)  +   2H2O(l)

 

Experiment II

Put about 2cm3 of Concentrated sulphuric (VI) acid into two separate boiling tubes. Place a thin moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution on the mouth of the boiling tube.

Put about 0.5g of powdered charcoal and sulphur powder to each boiling tube separately.

Warm.

Observation;

Black solid charcoal dissolves/decrease

Yellow solid sulphur dissolves/decrease

Colourless gas produced that has a pungent irritating choking smell.

Gas produced turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.

 

Explanation

Concentrated sulphuric (VI) acid is strong oxidizing agent. It oxidizes non-metals to non metallic oxides and itself reduced to sulphur(IV)oxide gas. Sulphur (IV) oxide gas turn orange moist/damp/wet filter paper dipped in acidified potassium dichromate (VI)solution to green.

Charcoal is oxidized to carbon(IV)oxide. Sulphur is oxidized to Sulphur(IV)oxide .

Chemical equation

C(s) +  2H2SO4(aq)  –>   CO2(aq)  +  2SO2(g)  +   2H2O(l)

S(s) +  2H2SO4(aq)  –>     3SO2(g)  +   2H2O(l)

 

III. As the least volatile acid

 

Study the table below showing a comparison in boiling points of the three mineral acids

 

Mineral acid Relative molecula mass Boiling point(oC)
Hydrochloric acid(HCl) 36.5 35.0
Nitric(V)acid(HNO3) 63.0 83.0
Sulphuric(VI)acid(H2SO4) 98.0 333

 

1.Which is the least volatile acid? Explain

Sulphuric(VI)acid(H2SO4) because it has the largest molecule and joined by Hydrogen bonds making it to have the highest boiling point/least volatile.

 

  1. Using chemical equations, explain how sulphuric(VI)acid displaces the less volatile mineral acids.

 

(i)Chemical equation

KNO3(s) +  H2SO4(aq) –>   KHSO4(l)  +  HNO3(g)

             NaNO3(s) +  H2SO4(aq) –>   NaHSO4(l)  +  HNO3(g)

 

This reaction is used in the school laboratory preparation of Nitric(V) acid (HNO3).

 

(ii)Chemical equation

KCl(s) +  H2SO4(aq) –>   KHSO4(s)  +  HCl(g)

             NaCl(s) +  H2SO4(aq) –>   NaHSO4(s)  +  HCl(g)

 

This reaction is used in the school laboratory preparation of Hydrochloric acid (HCl).

 

(d) Properties of dilute sulphuric(VI)acid.

 

Dilute sulphuric(VI)acid is made when about 10cm3 of concentrated  sulphuric

(VI) acid is carefully added to about 90cm3 of distilled water.

Diluting concentrated sulphuric (VI) acid should be done carefully because the reaction is highly exothermic.

Diluting concentrated sulphuric (VI) acid decreases the number of moles present in a given volume of solution which makes the acid less corrosive.

On  diluting concentrated  sulphuric(VI) acid, water ionizes /dissociates the acid fully/wholly into two(dibasic)free H+(aq) and SO42-(aq)ions:

 

H2SO4 (aq)   ->   2H+(aq)  +  SO42-(aq)

 

The presence of free H+(aq)ions is responsible for ;

(i)turn litmus red because of the presence of free H+(aq)ions

(ii)have pH 1/2/3 because of the presence of many free H+(aq)ions hence a strongly acidic solution.

(iii)Reaction with metals

 

Experiment:

Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean test tubes. Add about 0.1g of Magnesium ribbon to one test tube. Cover the mixture with a finger as stopper. Introduce a burning splint on top of the finger and release the finger “stopper”. Repeat by adding Zinc, Copper and Iron instead of the Magnesium ribbon.

 

Observation:

          No effervescence/ bubbles/ fizzing with copper

Effervescence/ bubbles/ fizzing with Iron ,Zinc and Magnesium

Colourless gas produced that extinguishes burning splint with a “pop” sound.

Colourless solution formed with Zinc and Magnesium.

Green solution formed with Iron

Explanation:

When a metal higher than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of Hydrogen gas.

Impure hydrogen gas extinguishes burning splint with a “pop” sound.

A sulphate (VI) salts is formed. Iron, Zinc and Magnesium are higher than hydrogen in the reactivity/electrochemical series.

They form Iron (II)sulphate(VI), Magnesium sulphate(VI) and Zinc sulphate(VI).

.         When a metal lower than hydrogen in the reactivity/electrochemical series is put in a test tube containing dilute sulphuric(VI)acid, there is no effervescence/ bubbling/ fizzing that take place.

Copper thus do not react with dilute sulphuric(VI)acid.

Chemical/ionic equation

Mg(s)        +   H2SO4(aq)         –>   MgSO4(aq)   +        H2(g)

Mg(s)       +   2H+(aq)            –>   Mg2+ (aq)       +        H2(g)

 

Zn(s)         +   H2SO4(aq)         –>   ZnSO4(aq)    +        H2(g)

Zn(s)        +   2H+(aq)             –>   Zn2+ (aq)       +        H2(g)

 

              Fe(s)        +   H2SO4(aq)        –>   FeSO4(aq)     +        H2(g)

               Fe(s)        +   H+(aq)              –>   Fe2+ (aq)        +       H2(g)

 

NB:(i) Calcium,Lead and Barium forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.

(ii)Sodium and Potassium react explosively with dilute sulphuric(VI)acid

 

(iv)Reaction with metal carbonates and hydrogen carbonates

 

Experiment:

Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of sodium carbonate to one boiling tube. Introduce a burning splint on top of the boiling tube. Repeat by adding Zinc carbonate, Copper (II)carbonate and Iron(II)Carbonate in place of the sodium hydrogen carbonate.

Observation:

 

          Effervescence/ bubbles/ fizzing.

Colourless gas produced that extinguishes burning splint.

Colourless solution formed with Zinc carbonate, sodium hydrogen carbonate and sodium carbonate.

Green solution formed with Iron(II)Carbonate

Blue solution formed with Copper(II)Carbonate

Explanation:

 

When a metal carbonate or a hydrogen carbonates is put in a test tube containing dilute sulphuric(VI)acid, effervescence/ bubbling/ fizzing takes place with evolution of carbon(IV)oxide gas. carbon(IV)oxide gas extinguishes a burning splint and forms a white precipitate when bubbled in lime water.

A sulphate (VI) salts is formed.

Chemical/ionic equation

 

ZnCO3(s) +   H2SO4(aq)  –>   ZnSO4(aq)  + H2O(l) + CO2(g)

ZnCO3(s) +   2H+(aq)  –>   Zn2+ (aq)  + H2O(l) + CO2(g)

 

CuCO3(s) +   H2SO4(aq)  –>   CuSO4(aq)  + H2O(l) + CO2(g)

CuCO3(s) +   2H+(aq)  –>   Cu2+ (aq)  + H2O(l) + CO2(g)

 

              FeCO3(s) +   H2SO4(aq)  –>   FeSO4(aq)  + H2O(l) + CO2(g)

FeCO3(s) +   2H+(aq)  –>   Fe2+ (aq)  + H2O(l) + CO2(g)

 

2NaHCO3(s) +   H2SO4(aq)  –>   Na2SO4(aq)  + 2H2O(l) + 2CO2(g)

NaHCO3(s) +   H+(aq)  –>   Na+ (aq)  + H2O(l) + CO2(g)

 

Na2CO3(s) +   H2SO4(aq)  –>   Na2SO4(aq)  + H2O(l) + CO2(g)

NaHCO3(s) +   H+(aq)  –>   Na+ (aq)  + H2O(l) + CO2(g)

 

(NH4)2CO3(s) +   H2SO4(aq)  –>   (NH4)2SO4 (aq)  + H2O(l) + CO2(g)

(NH4)2CO3 (s) +   H+(aq)  –>   NH4+ (aq)  + H2O(l) + CO2(g)

 

2NH4HCO3(aq) +   H2SO4(aq)  –>   (NH4)2SO4 (aq)  + H2O(l) + CO2(g)

NH4HCO3(aq) +   H+(aq)  –>   NH4+ (aq)  + H2O(l) + CO2(g)

 

NB:

Calcium, Lead and Barium  carbonates forms insoluble sulphate(VI)salts that cover/coat the unreacted metals.

(v)Neutralization-reaction of metal oxides and alkalis/bases

 

Experiment I:

Place 5cm3 of 0.2M dilute sulphuric(VI)acid into four separate clean boiling tubes. Add about 0.1g of copper(II)oxide to one boiling tube. Stir.

Repeat by adding Zinc oxide, calcium carbonate and Sodium (II)Oxide in place of the Copper(II)Oxide.

Observation:         

Blue solution formed with Copper(II)Oxide

Colourless solution formed with other oxides

Explanation:

 

When a metal oxide is put in a test tube containing dilute sulphuric(VI)acid, the oxide dissolves forming a sulphate (VI) salt.

Chemical/ionic equation

 

ZnO(s) +   H2SO4(aq)  –>   ZnSO4(aq)  + H2O(l)

ZnO(s) +   2H+(aq)  –>   Zn2+ (aq)  + H2O(l)

 

CuO(s) +   H2SO4(aq)  –>   CuSO4(aq)  + H2O(l)

CuO(s) +   2H+(aq)  –>   Cu2+ (aq)  + H2O(l)

 

              MgO(s) +   H2SO4(aq)  –>   MgSO4(aq)  + H2O(l)

MgO(s) +   2H+(aq)  –>   Mg2+ (aq)  + H2O(l)

 

Na2O(s) +   H2SO4(aq)  –>   Na2SO4(aq)  + H2O(l)

Na2O(s) +   2H+(aq)  –>   2Na+ (aq)  + H2O(l)

 

K2CO3(s) +   H2SO4(aq)  –>   K2SO4(aq)  + H2O(l)

K2O(s) +   H+(aq)  –>   2K+ (aq)  + H2O(l)

 

NB:

Calcium, Lead and Barium oxides forms insoluble sulphate(VI)salts that cover/coat the unreacted metals oxides.

 

Experiment II:

Fill a burette wuth 0.1M dilute sulphuric(VI)acid. Pipette 20.0cm3 of 0.1Msodium hydroxide solution into a 250cm3 conical flask. Add three drops of phenolphthalein indicator.Titrate the acid  to get a permanent colour change.Repeat with0.1M potassium hydroxide solution inplace of 0.1Msodium hydroxide solution

Observation:

 

Colour of phenolphthalein changes from pink to colourless at the end point.

 

Explanation

Like other (mineral) acids dilute sulphuric(VI)acid neutralizes bases/alkalis to a sulphate salt and water only.

Colour of the indicator used changes when a slight excess of acid is added to the base at the end point

Chemical equation:

2NaOH(aq) +   H2SO4(aq)  –>   Na2SO4(aq)  + H2O(l)

OH(s) +   H+(aq)  –>   H2O(l)

 

2KOH(aq) +   H2SO4(aq)  –>   K2SO4(aq)  + H2O(l)

OH(s) +   H+(aq)  –>   H2O(l)

 

2NH4OH(aq) +   H2SO4(aq)  –>   (NH4)2SO4(aq)  + H2O(l)

OH(s) +   H+(aq)  –>   H2O(l)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(iv) Hydrogen sulphide(H2S)

 

(a) Occurrence

Hydrogen sulphide  is found in volcanic areas as a gas or dissolved in water from geysers and hot springs in active volcanic areas of the world e.g. Olkaria and Hells gate near Naivasha in Kenya.

It is present in rotten eggs and human excreta.

 

(b) Preparation

 

Hydrogen sulphide is prepared in a school laboratory by heating Iron (II) sulphide with dilute hydrochloric acid.

 

 

 

 

 

 

(c) Properties of Hydrogen sulphide(Questions)

  1. Write the equation for the reaction for the school laboratory preparation of Hydrogen sulphide.

Chemical equation: FeS (s)  + 2HCl (aq)       ->  H2S (g)  FeCl2 (aq)

 

  1. State three physical properties unique to Hydrogen sulphide.

Hydrogen sulphide is a colourless gas with characteristic pungent poisonous smell of rotten eggs. It is soluble in cold water but insoluble in warm water. It is denser than water and turns blue litmus paper red.

 

  1. Hydrogen sulphide exist as a dibasic acid when dissolved in water. Using a chemical equation show how it ionizes in aqueous state.

                    H2S(aq)       ->    H+(aq)   +   HS(aq)

H2S(aq)       ->   2H+(aq)  +   S2- (aq)

Hydrogen sulphide therefore can form both normal and acid salts  e.g

Sodium hydrogen sulphide and sodium sulphide both exist

 

  1. State and explain one gaseous impurity likely to be present in the gas jar containing hydrogen sulphide above.

 

Hydrogen/ H2

Iron(II)sulphide contains Iron as impurity .The iron will react with dilute hydrochloric acid to form iron(II)chloride and produce hydrogen gas that mixes with hydrogen sulphide  gas.

  1. State and explain the observations made when a filter paper dipped in Lead(II) ethanoate /Lead (II) nitrate(V) solution is put in a gas jar containing hydrogen sulphide gas.

Observations

Moist Lead(II) ethanoate /Lead (II) nitrate(V) paper turns black.

 

Explanation

When hydrogen sulphide is bubbled in a metallic salt solution, a metallic sulphide is formed.

All sulphides are insoluble black salts except sodium sulphide, potassium sulphide and ammonium sulphides.

Hydrogen sulphide gas blackens moist Lead (II) ethanoate /Lead (II) nitrate(V) paper .

The gas reacts with Pb2+ in the paper to form black Lead(II)sulphide.

This is the chemical test for the presence of H2S other than the physical smell of rotten eggs.

Chemical equations

Pb2+(aq)       + H2S           -> PbS         + 2H+(aq)

(black)

Fe2+(aq)       + H2S           -> FeS          + 2H+(aq)

(black)

Zn2+(aq)       + H2S           -> ZnS         + 2H+(aq)

(black)

Cu2+(aq)       + H2S           -> CuS         + 2H+(aq)

(black)

2Cu+(aq)      + H2S           -> Cu2S        + 2H+(aq)

(black)

  1. Dry hydrogen sulphide was ignited as below.
Dry Hydrogen sulphide gas
Flame A

 

(i) State the observations made in flame A

Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water.

Chemical equation:           2H2S(g)      +    3O2(g)  ->    2H2O(l) +   2SO2(g)

 

Hydrogen sulphide burns in limited air with a blue flame to form sulphur solid and water.

Chemical equation:           2H2S(g)      +    O2(g)    ->    2H2O(l) +   2S(s)

 

  1. Hydrogen sulphide is a strong reducing agent that is oxidized to yellow solid sulphur as precipitate. The following experiments illustrate the reducing properties of Hydrogen sulphide.

 

(a)Turns Orange acidified potassium dichromate(VI) to green

 

Experiment:

 

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing acidified potassium dichromate (VI) solution. or;

 

(ii)Dip a filter paper soaked in acidified potassium dichromate (VI) into a gas jar containing Hydrogen sulphide gas.

 

Observation:

Orange acidified potassium dichromate (VI) turns to green.

Yellow solid residue.

 

Explanation:

Hydrogen sulphide gas reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions leaving a  yellow solid residue as itself is oxidized to sulphur.

 

Chemical/ionic equation:

4H2S(aq)  +   Cr2O72-(aq) +6H+(aq)   -> 4S(aq) +  2Cr3+(aq) +  7H2O(l)

 

This test is used for differentiating Hydrogen sulphide and sulphur (IV)oxide gas.

 

Sulphur(IV)oxide also reduces acidified potassium dichromate(VI) from orange Cr2O72- ions to green Cr3+ ions without leaving a yellow residue.

 

(b)Decolorizes acidified potassium manganate(VII)

 

Experiment:

 

(i)Pass a stream of Sulphur(IV) oxide gas in a test tube containing acidified potassium manganate(VII) solution. or;

 

(ii)Dip a filter paper soaked in acidified potassium manganate(VII) into a gas jar containing Hydrogen Sulphide gas.

 

Observation:

Purple acidified potassium manganate(VII) turns to colourless/ acidified potassium manganate(VII) is decolorized.

Yellow solid residue.

 

Explanation:

Hydrogen sulphide gas reduces acidified potassium manganate(VII) from purple MnO4 ions to green Mn2+ ions leaving a residue as the gas itself is oxidized to sulphur.

Chemical/ionic equation:

5H2S(g)  +   2MnO4 (aq) +6H+(aq)   -> 5S (s) +  2Mn2+(aq) +  8H2O(l)

(purple)                                       (colourless)

 

This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas.

Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4 ions to colourless Mn2+ ions leaving no  yellow residue.

 

(c)Decolorizes bromine water

 

Experiment:

 

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing bromine water . or;

 

(ii)Put three drops of bromine water into a gas jar containing Hydrogen sulphide gas. Swirl.

 

Observation:

Yellow bromine water turns to colourless/ bromine water is decolorized.

Yellow solid residue

 

Explanation:

Hydrogen sulphide gas reduces yellow bromine water to colourless hydrobromic acid (HBr) leaving a yellow residue as the gas  itself is oxidized to sulphur.

 

Chemical/ionic equation:

H2 S(g)            +   Br2 (aq)            ->       S (s)    +       2HBr(aq)

(yellow solution)    (yellow solid)        (colourless)

 

This is another test for differentiating Hydrogen sulphide and Sulphur(IV) oxide gas.

Sulphur(IV) oxide also decolorizes acidified potassium manganate(VII) from purple MnO4 ions to colourless Mn2+ ions leaving no  yellow residue.

 

(d)Reduces Iron(III) Fe3+ salts to Iron(II) salts Fe2+

Experiment:

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of Iron (III)chloride solution. or;

 

(ii)Place about 3cm3 of Iron (III)chloride solution into a gas jar containing Hydrogen sulphide gas. Swirl.

 

Observation:

Yellow/brown Iron (III)chloride solution turns to green.

Yellow solid solid

 

Explanation:

Hydrogen sulphide gas reduces Iron (III)chloride solution from yellow/brown Fe3+ ions to green Fe2+ ions leaving a yellow  residue.The gas is  itself oxidized to sulphur.

 

Chemical/ionic equation:

H2S(aq)      +    2Fe3+ (aq) ->       S (s)   +        Fe2+(aq)       +  2H+(aq)

(yellow solution)   (yellow residue)    (green)

 

(e)Reduces Nitric(V)acid to Nitrogen(IV)oxide gas

Experiment:

 

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated nitric(V)acid. or;

 

(ii)Place about 3cm3 of concentrated nitric(V)acid into a gas jar containing Hydrogen sulphide gas. Swirl.

 

Observation:

Brown fumes of a gas evolved/produced.

Yellow solid residue

 

Explanation:

Hydrogen sulphide gas reduces concentrated nitric(V)acid to brown nitrogen(IV)oxide gas itself oxidized to yellow sulphur.

 

Chemical/ionic equation:

H2S(g) +  2HNO3 (l)       ->     2H2O(l)         + S (s)           + 2NO2 (g)

(yellow residue)  (brown fumes)

 

(f)Reduces sulphuric(VI)acid to Sulphur

Experiment:

 

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of concentrated sulphuric(VI)acid. or;

 

(ii)Place about 3cm3 of concentrated sulphuric (VI) acid into a gas jar containing Hydrogen sulphide gas. Swirl.

 

Observation:

Yellow solid residue

 

Explanation:

Hydrogen sulphide gas reduces concentrated sulphuric(VI)acid to yellow sulphur.

 

Chemical/ionic equation:

3H2S(g) +  H2SO4 (l)       ->     4H2O(l)         +        4S (s)

(yellow residue)

 

(g)Reduces Hydrogen peroxide to water

Experiment:

 

(i)Pass a stream of Hydrogen sulphide gas in a test tube containing about 3 cm3 of 20 volume hydrogen peroxide.

Observation:

Yellow solid residue

 

Explanation:

Hydrogen sulphide gas reduces 20 volume hydrogen peroxide to water and  itself oxidized to yellow sulphur

Chemical/ionic equation:

H2S(g) +  H2O2 (l) ->     2H2O(l)         +        S (s)

(yellow residue)

8.Name the salt formed when:

(i)equal volumes of equimolar hydrogen sulphide neutralizes sodium hydroxide solution:

    Sodium hydrogen sulphide

Chemical/ionic equation:

H2S(g) +  NaOH (l)         ->     H2O(l)           +        NaHS (aq)

 

(ii) hydrogen sulphide neutralizes excess concentrated sodium hydroxide solution:

    Sodium sulphide

Chemical/ionic equation:

H2S(g) +  2NaOH (l)       ->     2H2O(l)         +        Na2S (aq)

 

Practice

Hydrogen sulphide gas was bubbled into a solution of metallic nitrate(V)salts as in the flow chart below

Hydrogen sulphide
Blue solution
Brown solution
Black solid
Green solution

 

 

 

 

 

 

 

 

 

(a)Name the black solid                     Copper(II)sulphide

 

(b)Identify the cation responsible for the formation of:

  1. Blue solution Cu2+(aq)
  2. Green solution                     Fe2+(aq)

III. Brown solution                  Fe3+(aq)

 

(c)Using acidified potassium dichromate(VI) describe how you would differentiate between sulphur(IV)Oxide and hydrogen sulphide

-Bubble the gases in separate test tubes containing acidified Potassium dichromate(VI) solution.

-Both changes the Orange colour of acidified Potassium dichromate(VI) solution to green.

-Yellow solid residue/deposit is formed with Hydrogen sulphide

Chemical/ionic equation:

4H2S(aq)  +   Cr2O72-(aq) +6H+(aq)   -> 4S(aq) +  2Cr3+(aq) +  7H2O(l)

3SO32-(aq)  +   Cr2O72-(aq) +8H+(aq)   -> 3SO42-(aq) +  2Cr3+(aq) +  4H2O(l)

 

(d)State and explain the observations made if a burning splint is introduced at the mouth of a hydrogen sulphide generator.

          ObservationGas continues burning with a blue flame

Explanation: Hydrogen sulphide burns in excess air with a blue flame to form sulphur(IV)oxide gas and water.

Chemical equation: 2H2S(g)+ 3O2(g) -> 2H2O(l) + 2SO2 (g)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(v)Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts

  1. Sulphate (VI) (SO42-) salts are normal and acid salts derived from Sulphuric (VI)acid H2SO4.
  2. Sulphate(IV) (SO32-) salts are normal and acid salts derived from Sulphuric (IV)acid H2SO3.
  3. Sulphuric (VI)acid H2SO4 is formed when sulphur(VI)oxide gas is bubbled in water.

The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (VI) (SO42-) and hydrogen sulphate (VI) (HSO4) salts.

i.e.

H2SO4 (aq)   ->   2H+(aq) + SO42-(aq)

H2SO4 (aq)   ->    H+(aq) + HSO4(aq)

 

All Sulphate (VI) (SO42-) salts dissolve in water/are soluble except Calcium (II) sulphate (VI) (CaSO4), Barium (II) sulphate (VI) (BaSO4) and Lead (II) sulphate (VI) (PbSO4)

All Hydrogen sulphate (VI) (HSO3) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (VI) (NaHSO4), Potassium (I) hydrogen sulphate (VI) (KHSO4) and Ammonium hydrogen sulphate (VI) (NH4HSO4) exist also as solids.

Other Hydrogen sulphate (VI) (HSO4) salts do not exist except those of Calcium (II) hydrogen sulphate (VI) (Ca (HSO4)2) and Magnesium (II) hydrogen sulphate (VI) (Mg (HSO4)2).

 

  1. Sulphuric (IV)acid H2SO3 is formed when sulphur(IV)oxide gas is bubbled in water.

The acid exist as a dibasic acid with two ionisable hydrogen. It forms therefore the Sulphate (IV) (SO32-) and hydrogen sulphate (VI) (HSO4) salts.

i.e.

H2SO3 (aq)   ->   2H+(aq) + SO32-(aq)

H2SO3 (aq)   ->    H+(aq) + HSO3(aq)

 

All Sulphate (IV) (SO32-) salts dissolve in water/are soluble except Calcium (II) sulphate (IV) (CaSO3), Barium (II) sulphate (IV) (BaSO3) and Lead (II) sulphate (IV) (PbSO3)

All Hydrogen sulphate (IV) (HSO3) salts exist in solution/dissolved in water. Sodium (I) hydrogen sulphate (IV) (NaHSO3), Potassium (I) hydrogen sulphate (IV) (KHSO3) and Ammonium hydrogen sulphate (IV) (NH4HSO3) exist also as solids.

Other Hydrogen sulphate (IV) (HSO3) salts do not exist except those of Calcium (II) hydrogen sulphate (IV) (Ca (HSO3)2) and Magnesium (II) hydrogen sulphate (IV) (Mg (HSO3)2).

 

5.The following experiments show the effect of heat on sulphate(VI) (SO42-)and sulphate(IV) (SO32-) salts:

 

Experiment:

In a clean dry test tube place separately about 1.0g of :

Zinc(II)sulphate (VI), Iron(II)sulphate(VI), Copper(II)sulphate(VI),Sodium (I) sulphate (VI), Sodium (I) sulphate (IV).Heat gently then strongly. Test any gases produced using litmus papers.

 

Observations:

-Colourless droplets of liquid forms on the cooler parts of the test tube in all cases.

-White solid residue is left in case of Zinc (II)sulphate(VI),Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV).

-Colour changes from green to brown /yellow in case of Iron (II)sulphate(VI)

 -Colour changes from blue to white then black in case of Copper (II) sulphate (VI)

-Blue litmus paper remain and blue and red litmus paper remain red in case of Zinc(II)sulphate(VI), Sodium (I) sulphate (VI) and Sodium (I) sulphate (IV)

-Blue litmus paper turns red and red litmus paper remain red in case of Iron (II)sulphate(VI) and Copper (II) sulphate (VI).

 

Explanation

(i)All Sulphate (VI) (SO42-) salts exist as hydrated salts with water of crystallization that condenses and collects on cooler parts of test tube as a colourless liquid on gentle heating. e.g.

 

K2SO4.10H2O(s)    ->       K2SO4(s)    +   10H2O(l)

                    Na2SO4.10H2O(s)  ->       Na2SO4(s)    +   10H2O(l)

MgSO4.7H2O(s)    ->       MgSO4(s)    +   7H2O(l)

CaSO4.7H2O(s)     ->       CaSO4(s)    +   7H2O(l)

ZnSO4.7H2O(s)     ->       ZnSO4(s)    +   7H2O(l)

FeSO4.7H2O(s)      ->       FeSO4(s)    +   7H2O(l)

Al2(SO4)3.6H2O(s) ->       Al2(SO4)3 (s)    +   6H2O(l)

CuSO4.5H2O(s)     ->       CuSO4(s)    +   5H2O(l)

All Sulphate (VI) (SO42-) salts do not decompose on heating except Iron (II) sulphate (VI) and Copper (II) sulphate (VI).

(i)Iron (II) sulphate (VI) decomposes on strong heating to produce acidic sulphur (IV)oxide and sulphur(VI)oxide gases. Iron(III)oxide is formed as a brown /yellow residue.

Chemical equation

2FeSO4 (s)   ->       Fe2O3(s)    +   SO2(g) +   SO3(g)

This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas.

Sulphur (VI) oxide readily /easily solidifies as white silky needles when the mixture is passed through freezing mixture/ice cold water.

Sulphur (IV) oxide does not.

(ii) Copper(II)sulphate(VI) decomposes on strong heating to black copper (II) oxide and Sulphur (VI) oxide gas.

Chemical equation

2CuSO4 (s)  ->       CuO(s)    +   SO3(g)

This reaction is used for the school laboratory preparation of small amount of sulphur(VI)oxide gas.

 

  1. The following experiments show the test for the presence of sulphate (VI) (SO42-)and sulphate(IV) (SO32-) ions in a sample of a salt/compound:

 

Experiments/Observations:

 

(a)Using Lead(II)nitrate(V)

 

  1. To about 5cm3 of a salt solution in a test tube add four drops of Lead(II)nitrate(V)solution. Preserve.

 

Observation Inference
White precipitate/ppt SO42- , SO32- , CO32- , Clions

 

  1. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve.

 

Observation 1

Observation Inference
White precipitate/ppt persists SO42- , Clions

 

Observation 2

Observation Inference
White precipitate/ppt dissolves SO32- , CO32- , ions

 

III.(a)To the preserved sample observation 1 in (II) above, Heat to boil.

 

Observation 1

 

Observation Inference
White precipitate/ppt persists on boiling SO42-  ions

 

Observation 2

Observation Inference
White precipitate/ppt dissolves on boiling Cl  ions

 

.(b)To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).

 

Observation 1

Observation Inference
(i)acidified potassium manganate(VII)decolorized

(ii)Orange colour of acidified potassium

dichromate(VI) turns to green

SO32- ions

 

 

Observation 2

Observation Inference
(i)acidified potassium manganate(VII) not decolorized

(ii)Orange colour of acidified potassium

dichromate(VI) does not turns to green

 

CO32- ions

 

 

 

Experiments/Observations:

(b)Using Barium(II)nitrate(V)/ Barium(II)chloride

 

  1. To about 5cm3 of a salt solution in a test tube add four drops of Barium(II) nitrate (V) / Barium(II)chloride. Preserve.

 

 

Observation Inference
White precipitate/ppt SO42- , SO32- , CO32- ions

 

 

  1. To the preserved sample in (I) above, add six drops of 2M nitric(V) acid . Preserve.

 

 

Observation 1

Observation Inference
White precipitate/ppt persists SO42- , ions

 

 

Observation 2

Observation Inference
White precipitate/ppt dissolves SO32- , CO32- , ions

 

 

III.To the preserved sample observation 2 in (II) above, add 4 drops of acidified potassium manganate(VII) /dichromate(VI).

 

 

Observation 1

Observation Inference
(i)acidified potassium manganate(VII)decolorized

(ii)Orange colour of acidified potassium

dichromate(VI) turns to green

SO32- ions

 

 

 

Observation 2

Observation Inference
(i)acidified potassium manganate(VII) not decolorized

(ii)Orange colour of acidified potassium

dichromate(VI) does not turns to green

 

CO32- ions

 

 

 

 

 

Explanations

 

Using Lead(II)nitrate(V)

(i)Lead(II)nitrate(V) solution reacts with chlorides(Cl), Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Lead(II)chloride, Lead(II)sulphate(VI), Lead(II) sulphate (IV) and Lead(II)carbonate(IV).

 

Chemical/ionic equation:

 

Pb2+(aq)       +        Cl (aq)       ->       PbCl2(s)

Pb2+(aq)       +        SO42+ (aq)   ->       PbSO4 (s)

Pb2+(aq)       +        SO32+ (aq)   ->       PbSO3 (s)

Pb2+(aq)       +        CO32+ (aq)  ->       PbCO3 (s)

 

(ii)When the insoluble precipitates are acidified with nitric(V) acid,

– Lead(II)chloride and Lead(II)sulphate(VI) do not react with the acid and thus their white precipitates remain/ persists.

 

– Lead(II) sulphate (IV) and Lead(II)carbonate(IV) reacts with the acid to form soluble Lead(II) nitrate (V) and produce/effervesces/fizzes/bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.

 

. Chemical/ionic equation:

PbSO3 (s)   +   2H+(aq)       ->    H2 O (l)    +   Pb2+(aq)    +        SO2 (g)

PbCO3 (s)   +   2H+(aq)      ->    H2 O (l)    +   Pb2+(aq)    +        CO2 (g)

 

(iii)When Lead(II)chloride and Lead(II)sulphate(VI) are heated/warmed;

 

– Lead(II)chloride dissolves in hot water/on boiling(recrystallizes on cooling)

 

– Lead(II)sulphate(VI) do not dissolve in hot water thus its white precipitate persists/remains on heating/boiling.

 

(iv)When sulphur(IV)oxide and carbon(IV)oxide gases are produced;

 

sulphur(IV)oxide will decolorize acidified potassium manganate(VII)  and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.

 

Chemical equation:

5SO32-(aq)  +   2MnO4 (aq) +6H+(aq)   -> 5SO42-(aq) +  2Mn2+(aq) +  3H2O(l)

(purple)                                                     (colourless)

 

3SO32-(aq)  +   Cr2O72-(aq) +8H+(aq)   -> 3SO42-(aq) +  2Cr3+(aq) +  4H2O(l)

(Orange)                                                 (green)

 

Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.

 

Chemical equation:

Ca(OH)2(aq)    +   CO2 (g)    ->  CaCO3(s)  +  H2O(l)

 

These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.

 

Using Barium(II)nitrate(V)/ Barium(II)Chloride

(i)Barium(II)nitrate(V) and/ or Barium(II)chloride solution reacts with Sulphate (VI) salts (SO42- ), Sulphate (IV)salts (SO32-) and carbonates(CO32-) to form the insoluble white precipitate of Barium(II)sulphate(VI), Barium(II) sulphate (IV) and Barium(II)carbonate(IV).

 

Chemical/ionic equation:

Ba2+(aq)       +        SO42+ (aq)   ->       BaSO4 (s)

 

Ba2+(aq)       +        SO32+ (aq)   ->       BaSO3 (s)

 

Ba2+(aq)       +        CO32+ (aq)  ->       BaCO3 (s)

 

(ii)When the insoluble precipitates are acidified with nitric(V) acid,

 

– Barium (II)sulphate(VI) do not react with the acid and thus its white precipitates remain/ persists.

 

– Barium(II) sulphate (IV) and Barium(II)carbonate(IV) reacts with the acid to form soluble Barium(II) nitrate (V) and produce /effervesces /fizzes/ bubbles out sulphur(IV)oxide and carbon(IV)oxide gases respectively.

 

. Chemical/ionic equation:

BaSO3 (s)   +   2H+(aq)       ->    H2 O (l)    +   Ba2+(aq)    +        SO2 (g)

 

BaCO3 (s)   +   2H+(aq)      ->    H2 O (l)    +   Ba2+(aq)    +        CO2 (g)

 

(iii) When sulphur(IV)oxide and carbon(IV)oxide gases are produced;

 

sulphur(IV)oxide will decolorize acidified potassium manganate(VII)  and / or Orange colour of acidified potassium dichromate(VI) will turns to green. Carbon(IV)oxide will not.

 

Chemical equation:

5SO32-(aq)  +   2MnO4 (aq) +6H+(aq)   -> 5SO42-(aq) +  2Mn2+(aq) +  3H2O(l)

(purple)                                                     (colourless)

 

3SO32-(aq)  +   Cr2O72-(aq) +8H+(aq)   -> 3SO42-(aq) +  2Cr3+(aq) +  4H2O(l)

(Orange)                                                 (green)

 

Carbon(IV)oxide forms an insoluble white precipitate of calcium carbonate if three drops of lime water are added into the reaction test tube when effervescence is taking place. Sulphur(IV)oxide will not.

 

Chemical equation:

Ca(OH)2(aq)    +   CO2 (g)    ->  CaCO3(s)  +  H2O(l)

 

These tests should be done immediately after acidifying to ensure the gases produced react with the oxidizing agents/lime water.

 

 

 

Summary test for Sulphate (VI) (SO42-)and Sulphate(IV) (SO32-) salts

Heat to boil
White ppt persist

on heating in SO42-

in CO32-

 

White ppt dissolves on heating in Cl

 

Acidified KMnO4 decolorized in SO32-
White ppt with lime water in CO32-
Acidified KMnO4

K2Cr2O7 / Lime water

white ppt persist /remains in SO32- and CO32-

 

white ppt dissolves in SO32- and CO32-
Dilute nitric(V) acid
Lead(II)nitrate(V)
White precipitates of Cl, SO42- , SO32- and CO32-
Unknown salt

 

 

Practice revision question

 

  1. Study the flow chart below and use it to answer the questions that follow
Sodium salt solution

 

 

Acidified K2Cr2O7
Gas G and colour of solution

changes orange to green

Colourless solution B
Barium nitrate(VI) (VI)(aq)
White precipitate
Dilute HCl

 

 

(a)Identify the:

          I: Sodium salt solution  

          Sodium sulphate(IV)/Na2SO3

 

II: White precipitate               

Barium sulphate(IV)/BaSO3

                             

III: Gas G                                

Sulphur (IV)Oxide /SO2

 

IV: Colourless solution H        

Barium chloride /BaCl2

 

(b)Write an ionic equation for the formation of:                                                  I.White precipitate

 

          Ionic equation        Ba2+(aq)        +       SO32-(aq)    ->    BaSO3(s)

II.Gas G

          Ionic equation        BaSO3(s)+ 2H+(aq)  -> SO2 (g)  +  H2O (l) + Ba2+(aq)

III. Green solution from the orange solution

 

3SO32-(aq)  +   Cr2O72-(aq) +8H+(aq)   -> 3SO42-(aq) +  2Cr3+(aq) +  4H2O(l)

(Orange)                                                 (green)

 

  1. Study the flow chart below and answer the questions that follow.

 

(i)Write equation for the reaction taking place at:

I.The roasting furnace                                                                            (1mk)

          2FeS2 (s)  +  5O2 (g)       ->  2FeO(s)  + 4SO2 (g)

II.The absorption tower                                                                         (1mk)

H2SO4 (l)  +  SO3 (g)      ->  H2S2O7(l)

III.The diluter                                                                                        (1mk)

H2S2O7(l)  +  H2 O(l)       -> 2H2SO4 (l)

(ii)The reaction taking place in chamber K is

SO2 (g) + 1/2O2 (g)                        SO3 (g)

  1. Explain why it is necessary to use excess air in chamber K

To ensure all the SO2 reacts

II.Name another substance used in chamber K

Vanadium(V)oxide

3.(a)Describe a chemical test that can be used to differentiate between sodium sulphate (IV) and sodium sulphate (VI).

 

Add acidified Barium nitrate(V)/chloride.

White precipitate formed with sodium sulphate (VI)

No white precipitate formed with sodium sulphate (IV)

 

(b)Calculate the volume of sulphur (IV) oxide formed when 120 kg of  copper is reacted with excess concentrated sulphuric(VI)acid.(Cu = 63.5 ,1 mole of a gas at s.t.p =22.4dm3)

 

Chemical equation

Cu(s) + 2H2SO4(l) -> CuSO4(aq)+ H2O(l) + SO2 (g)

Mole ratio Cu(s: SO2 (g) = 1:1

 

Method 1

1 Mole Cu =63.5 g          ->                 1 mole SO2 =  22.4dm3

(120 x 1000) g                ->                 (120 x 1000) g  x 22.4.dm3)

63.5 g

 

42330.7087

 

Method 2

Moles  of Cu =      ( 120  x  1000 ) g =1889.7639 moles

63.5

Moles SO2   =   Moles  of Cu    = 1889.7639 moles

Volume of  SO2 = Mole x molar gas volume = (1889.7639 moles x 22.4)

= 42330.7114

 

(a)Identify the:

(i)cation responsible for the green solution T

Cr3+

(ii)possible anions present in white precipitate R

CO32-,  SO32-,  SO42-

(b)Name gas V

Sulphur (IV)oxide

(c)Write a possible ionic equation for the formation of white precipitate R.

Ba2+ (aq)  +   CO32- (aq)    ->  BaCO3(s)

Ba2+ (aq)  +   SO32- (aq)    ->  BaSO3(s)

Ba2+ (aq)  +   SO42- (aq)    ->  BaSO4(s)

 

 

 

     

UPGRADE

CHEMISTRY

                

FORM 3

                 

Chemistry of CHLORINE

      

 

 

 

 

 

 

 

 

 

 

Comprehensive tutorial notes

 

 MUTHOMI S.G

www.kcselibrary.info

                                     0720096206

 

 

 

 

 

 

CONTENTS

 

  1. Chlorine.
  2. The Halogens.
  3. Hydrogen chloride.
  4. Chloride (Cl) salts.
  5. Comprehensive revision questions.

 

 

 

 

 

 

 

 

 

 

 

 

                                     A.CHLORINE

Chlorine is a non-metallic element in group VII (Group 17) of the periodic table. It has electronic configuration 2:8:7. It gains one valence election to form stable Clion, it belongs to the chemical family of halogens.

Occurrence

-As Brine-concentration sodium chloride solution dissolved in salty seas water, oceans and lakes e.g. Lake Magadi in Kenya is very salty.

-As rock-salt solid sodium chloride crystals in the earths crust all over the world.

  1. B) Preparation

Chlorine gas may be prepared in the school laboratory from the following:

a)Heating solid Manganese (iv) Oxide and Concentrated Hydrochloric acid.

  1. b) Heating Lead (IV) Oxide and concentrated hydrochloric acid.

c)Reacting Potassium Manganate (VII) with concentrated Hydrochloric acid

d)Reacting Potassium /sodium Dichromate (VI) Acid with Concentrated Hydrochloric acid.

Set up of school laboratory preparation of chlorine.

 

 

  1. c) Properties of chlorine. (Questions)
  2. What is the colour of chlorine?

Pale green.

  1. Describe the smell of chlorine.

Pungent irritating smell.

  1. What method is used in collection of chlorine gas explain.

-Downward delivery.

-Chlorine is 11/2 denser than air.

4.(i) What is the purpose of concentrated sulphuric (VI) acid.

-To dry the gas.

(ii) Name two other substances that can be used in place of concentrated sulphuric (VI) acid.

-Calcium chloride

          -Silica gel

(iii) Name a substance that cannot be used in place of concentrated sulphuric (VI) acid explain.

-Calcium oxide reacts with chlorine.

 

5.(a)Write three possible reactions between concentrated hydrochloric acid and the oxidizing agents.

 

  1. 2KMnO4(s) +16HCl(aq) → 2KCl(aq)+2MnCl2(aq) + 8H2O(l)  +  5Cl2(g)

 

               2.K2Cr2O7(s)  +14HCl(aq) →  2KCl(aq) +  2CrCl3(aq)  +  7H­2O(l)  +  3Cl2(g)

 

3.NaCr2O7(s) + 14HCl(aq) → 2NaCl(aq) + CrCl3(aq) + 7H2O(l) + 3Cl2(g)

 

          4.PbO2(s)  + 4HCl(aq)  → PbCl2(aq)  +  Cl2(g)  +  2H2O(l)

         

          5.MnO2(s)+  4HCl(aq)  →  MnCl2(aq) +  Cl­2(g)  + 2H2O(l)

 

(b) Why is Hydrochloric acid used in all the above cases?

Oxidizing agents KMnO4/PbO2/MnO2/K2Cr2O/Na2Cr­2O7 readily oxidize hydrochloric acid to chlorine themselves reduced to their chlorides.

Generally:

          2HCl (aq) + [O] → Cl2 (g) + H2O (l)

                           (From oxidizing agent)

 

  1. State and explain the observation made when chlorine is bubbled in water.

Observation

-Pale yellow colour of chlorine fades.

          -yellow solution formed.

Explanation

Chlorine dissolves then reacts with water to form yellow chlorine water. Chlorine water is chemically a mixture of hydrochloric acid and chloric(I)acid (hypochlorous acid).

 A mixture of hydrochloric acid and chloric(I)acid (hypochlorous acid) is commonly called Chlorine water

 

Chemical equation:

Cl2(g)  +  H2O(l)  →  HCl(aq)  +  HClO(aq)

 

  1. Chlorine water in a boiling tube inverted into a trough was exposed to sunlight for two hours. Using a well labeled diagram show and explain the observations made.

 

 

 

 

Chlorine (I) acid is an unstable compound.

          After two hours the chloric (I) acid in chlorine water decomposes to hydrochloric acid and releases oxygen gas. This reaction takes place in sunlight.

Chemical equation

2HOCl(aq)   →  2HCl(aq)  +  O2 (g)

 

  1. State and explain the observation made when chlorine gas is bubbled in gas jar containing damp/wet/moist litmus papers.

Observation

The blue litmus turns red then both the red/blue litmus papers are bleached/decolourized.

Explanation

Chlorine reacts with water in the litmus papers to form acidic hydrochloric acid and chloric (l) acid that turns blue litmus papers red.

Chemical Equation

Cl2(g)    +    H2O(l)  →  HCl(aq)  +  HClO(aq)

Explanation

Unstable chloric (I) acid oxidizes the dye/colured litmus paper to colourless material

Chemical Equation

HClO(aq)   +    dye    →  HCl(aq)  +   (dye +  O)

(coloured)                         (colourless)

Or:

HClO(aq)   +    dye-O    →  HCl(aq)  +   dye

 (coloured)                         (colourless)

 

NB Chlorine does not therefore bleach/decolourize dry litmus paper/dye because chloric(I) acid cannot be formed in absence of water.

 

  1. Blue litmus papers were put in a flask containing cold dilute sodium hydroxide. Chlorine gas was bubbled into the solution. State and explain the observations made.

Observation

blue litmus papers were bleached /decolorized.

           Pale green colour of chlorine fades.

 

Explanation

-Sodium hydroxide reacts with chlorine to form sodium chloride and sodium hypochlorite. Sodium hypochlorite bleaches dyes by oxidation.

 

Chemical Equation

   Cl2  +  2NaOH(aq)  →  NaCl(aq)  +  NaClO(aq)  +  H2O

              NaClO(aq)      +      dye        →  NaCl(aq)  +  (dye + O)

                                        (coloured)                             (Colourless)

 

     NaClO(aq)   +    (dye-O)  →       NaCl(aq)   +      dye

                                        (Coloured)                                (Colourless)

 

10.Blue litmus papers were put in flask containing hot concentrated sodium hydroxide. Chlorine gas was bubbled into the solution. State and explain the observations made.

          Observation.

     blue litmus papers were bleached.

               Pale green colour of chlorine fades.

 

          Explanation

Hot concentrated sodium hydroxide reacts with chlorine to form sodium chloride and sodium chloride (V).Sodium chlorate (V) bleaches by oxidation.

 

Chemical equation

          2Cl2(g) + 4NaOH(aq)  →  3NaCl(aq)   +  NaClO3(aq)  + H2O(l)

 

          NaClO3(aq)  +  3(dyes)  →  NaCl(aq)  +  3(dye + O)

NaClO3(aq)  +  3(dyes-O)  →  NaCl(aq)  +  3 dyes

 

                    NaClO3 is also a weed killer

 

  1. State three main use of chlorine gas.

-Manufacture of polyvinyl chloride (P.V.C) // polychloroethene pipes.

-Manufacture of hydrochloric acid used in “Pickling” of metals.

-Manufacture of bleaching agents

-Chlorination of water to kill germs.

 

 

 

  1. The diagram below shows the effect of chlorine on heated iron wool.

Method I

 

          Method II

Suction pump
HEAT
Chlorine gas
Concentrated sodium/potassium hydroxide.
Iron/Aluminium
Aluminium(III)oxide/iron(III)oxide

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. a) Identify a suitable drying agent to dry chlorine gas.

-Conc. H2SO4 / Concentrated sulphuric (VI) acid.

-Anhydrous Calcium (II) Chloride.

-Silica gel

  1. b) State and explain the observations made in combustion tube in method I and II

    Observation

          Iron glows red hot

 Brown crystals are formed

 

    Explanation  

Iron reacts with chlorine to form dark brown crystals of iron (III) Chloride.

          This reaction is exothermic and requires no farther heating once started.

Iron (III) Chloride sublimes away ensuring the unreacted Iron completely reacts with chlorine gas.

Chemical equation

 2Fe(s)    +  3Cl2 (g)    →  2FeCl3(g)

 

  1. c) (i) Why is the brown solid collected at the point as shown in method I and II.

-Heated iron (III) Chloride crystals sublime to gas and solidify on the cooler parts.

 

(ii) Name another metal that can be used in place of iron to react with chlorine and collected  at similar point on heating explain.

Metal  Aluminum

 

          Explanation

Aluminum reacts with chlorine to form a while sublimate of aluminum (III) chloride at the cooler parts

Chemical equation

  2Al(s)        +    3Cl2(g)    →    2AlCl3(s/g)

 

  1. d) What is the purpose of suction pump?

          To pull the gaseous products into the set up.      

 

  1. e) What is the function of:

(i) Sodium hydroxide in method II. Explain.

To absorb poisonous/toxic excess unreacted chlorine gas.

Sodium hydroxide reacts with chlorine to form sodium chloride, Sodium hypochlorite and water.

Chemical equation:

 2NaOH(aq)   +   Cl2(g)   →   NaCl(aq)   +   NaClO(aq)   +   H2O(l)

2KOH(aq)      +   Cl2(g)   →   KCl(aq)    +   KClO(aq)     +   H2O(l)

 

(ii) Anhydrous calcium chloride/calcium oxide in method I. Explain.

To absorb moisture/water in the set up to prevent it from hydrolyzing iron (III) chloride/aluminium oxide.

 

Explanation

Iron (III) chloride and Aluminium chloride fumes and reacts with small traces of water to form a solution of iron (III) hydroxide/aluminium hydroxide and hydrogen chloride gas.

Chemical equation

FeCl3(s)   +  3HCl(aq) → Fe(OH)3(aq) + 3HCl(g)

         AlCl3(s)     +  3HCl(aq) → Al(OH)3(aq) + 3HCl(g)

 

  1. f) Based on e (i) and (ii) above what precaution should be made in:

(i) method II to ensure correct results.

-Tube B should be completely dry to prevent hydrolysis of iron (III) Chloride to iron (III)   hydroxide.

 

(ii) Carrying out method I

-Should be done in a fume chamber or in the open because chlorine gas is poisonous/toxic.

 

(g) Name another substance that can be used place of Sodium hydroxide in method I

Potassium hydroxide

         

(h) Calcium oxide cannot be used in place of calcium chloride during preparation of chlorine. Explain.

Calcium oxide is a base. It reacts /absorbs water to form calcium hydroxide solution.

          Calcium hydroxide reacts with chlorine to form a mixture of calcium chloride and calcium hypochlorite.

          Chemical equation

          2Ca (OH)2(aq) + 2Cl2(g) → CaCl2(aq) + CaOCl2(aq) + H2O(l)

 

  1. (a)State and explain the observation made when a piece of burning magnesium ribbon is lowered in a gas jar containing chlorine gas.

-Magnesium ribbon continues burning with a bright flame.

          -White solid formed.

          -Pale yellow colour of chlorine fades

 

          Explanation:

 Magnesium reacts with chlorine forming a white solid of magnesium chloride.

          Chemical equation

                    Mg(s)    +    Cl2(g) → MgCl2(s)

 

(b) Write the equation for the reaction that takes place if zinc is used.

 

Zn(s)   +   Cl2(g)   →   ZnCl2(s)

 

  1. Burning phosphorus was lowered in a gas jar containing chlorine gas.
  2. a) State the observations made.

-Phosphorus continues to burn.

          -Dense white fumes formed.

          -Pale green colour of chlorine fades.

 

  1. b) Write two possible equations that take place.

P4(s)   +   6Cl2(g)     →   4 PCl3(s)

          P4(s)   +   10Cl2(g)  →    4 PCl3(s)

         

(c) State two reasons why the deflagrating spoon with rid/cover should be used.

-Chlorine in the gas jar is poisonous/toxic.

          -Burning phosphorus produces poisonous/toxic phosphorus (III) chloride //     phosphorus (V) chloride.

          -Ensure the reaction is not affected by air/oxygen from the atmosphere.

 

(d) After the reaction is complete, 2cm3 of distilled water were added. The solution formed was tested with both blue and red litmus papers.

(i) State the observations made.

-Blue litmus paper turns red

           -Red litmus paper remain red

 

     (ii) Explain the observation made in d(i) above

-Phosphoric (V) Chloride hydrolyze in water to phosphoric (V) acid and produce hydrogen chloride gas. Both hydrogen chloride and phosphoric (V) acid are acidic.

Chemical equation

                    PCl5 (l)  +    4H2O(l)    →   H3PO4 (aq)     +     5HCl(g)

 

  1. State and explain the observations made when gas jar containing chlorine is inverted over another containing hydrogen sulphide gas.

          Observation

                    Yellow solid formed.

                    Pale colour of chlorine fades

          Explanation

          Chlorine oxidizes hydrogen sulphide to sulphur itself reduced to hydrogen chloride gas. A little water catalyzes the reaction.

    Chemical equation

            H2S(g)            +           Cl2(g)         → S(s)            +   HCl(g)

                                                                 (yellow solid)        (White Fume)

 

  1. Chlorine was bubbled in aqueous ammonia solution in a beaker state and explain the observation made.

Observation:

                  White fumes evolved.

 Pale green colour of chlorine fades.

Explanation

Chlorine reacts with ammonia gas to form a dense white fume of ammonia chloride and Nitrogen gas is produced.

Chemical equation

8NH3(g)      +        3Cl2(g)   → 6NH4Cl(s)    +    N2(g)

 

  1. (a) Dry gas was bubbled in cold dilute sodium hydroxide solution. Explain the observations made:

 

Observation

                 Pale green colour of chlorine fades.

                 Pale yellow solution is formed.

Explanation

     Chlorine reacts withhot concentrated sodiumsodium hydroxide / Potassium hydroxide solution to form pale yellow solution of metal chlorate (V) and chlorides of the metal

Chemical equation

   Cl2(g)     +  2NaOH       →         NaClO(aq)     +  NaCl(aq)   + H2O(l)

              (sodium hydroxide)  (Sodium Chlorate (I))

Cl2(g)    +  2KOH          →                  KClO(aq)       +     NaCl(aq) + H2O(l)

(Potassium  hydroxide)  (Potassium Chlorate (I))

 

(b)The experiment in 17(a) was repeated with hot concentrated sodium hydroxide solution. Explain the observation made.

Observation

                Pale green colour of chlorine fades.

                      Pale yellow solution is formed.

    Explanation

Chlorine reacts with hot concentrated Sodium hydroxide/Potassium hydroxide solution to form pale yellow solution of metal chlorate (v) and chlorides of metals.

Chemical equation

     3Cl2(g) + 6NaOH(aq)    →    NaClO3 (aq)     +     5NaCl(aq)  + 3H2O(l) 

                (Sodium hydroxide)  (Sodium Chlorate (V))

 

     3Cl2(g) + 6KOH(aq)    →      KClO3 (aq)     +     5KCl(aq)  + 3H2O(l) 

              (Potassium hydroxide) (Potassium Chlorate (V)) 

 

    The products formed when chlorine reacts with alkalis depend thus on temperature and the concentration of alkalis.

 

(c) (i) Write the equation for the formation of calcium chlorite (I) and calcium chlorate (V).

 

2Ca (OH)2(aq)   +   2Cl2(g)   →  CaCl2(aq)   + CaOCl2(aq)    + H2O(l)

    (Calcium hydroxide)                                     (Calcium Chlorate(I))   

        (Cold/dilute)

Ca (OH)2(aq) + Cl2(g)     →     CaCl2(aq)    +     Ca(ClO3)2(aq)   + H2O(l)

(Calcium Chlorate(V))       

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

                             

 

                              B: THE HALOGENS

 

  1. a) What are halogens?

These are elements in group VII of the periodic table. They include:

 

Element Symbol Atomic number Electric configuration Charge of ion Valency State at Room Temperature
Fluorine

 

Chlorine

 

Bromine

 

Iodine

 

Astatine

F

 

Cl

 

Br

 

I

 

At

9

 

17

 

35

 

53

 

85

2:7

 

2:8:7

 

2:8:18:7

 

2:8:18:18:7

 

2:8:18:32:18:7

F

 

Cl

 

Br

 

I

 

At

1

 

1

 

1

 

1

 

1

Pale yellow gas

Pale green gas

Red liquid

 

Grey Solid

 

Radioactive

 

 

  1. b) Compare the atomic radius and ionic radius of chloride ion and chlorine. Explain.

The radius of chlorine is smaller than the ionic radius o the chloride ion.

          Effective nucleus attraction on outer energy level in chloride ion is less than chlorine atom because of extra gained electron gained electron that repelled thus causes the outer energy level to expand/increase.

 

  1. c) Compare the atomic radius of chlorine and fluorine Explain.

Atomic radius of Fluorine is smaller than that of chlorine.

          Chlorine has more energy levels than fluorine occupied by more electrons.

 

  1. d) Chlorine is a gas, Bromine is a liquid, Iodine is a solid. Explain the above observations.

Bromine, Chlorine and iodine exists as diatomic molecules bonded by strong covalent bond. Each molecule is joined to the other by weak intermolecular forces/ Van-der-waals forces.

 

          -The strength of intermolecular/Van-der-waals forces of attraction increase with increase in molecular size/atomic radius.Iodine has therefore the largest atomic radius and thus strongest intermolecular forces to make it a solid.

 

  1. e) (i) What is electronegativity?

Electronegativity is the tendency/ease of acquiring /gaining electrons by an element during chemical reaction.

          It is measured using Pauling’s scale.

           Fluorine with Pauling scale 4.0 is the most electronegative element in the periodic table and thus the highest tendency to acquire/gain extra electron.

 

(ii) The table below shows the electronegativity of the halogens.

 

Halogen F Cl Br I At
Electronegativity (Pauling’s scale) 4.0 3.0 2.8 2.5 2.2

 

Explain the trend in electronegativity of the halogens.

Decrease down the group from fluorine to Astatine

Atomic radius increase down the group decreasing electron – attracting power down the group from fluorine to astatine.

 

(f)  (i)What is electron affinity

          Electron affinity is the energy required to gain an electron in an atom of an element in its gaseous state.

 

(ii) Study the table below showing the election affinity of halogens for the process                  x      +    e       →    x

 

Halogen F Cl Br I
Electron affinity kJmole-1 -333 -364 -342 -295

 

 

 

(iii) Explain the trend in electron affinity of the halogens.

-Decrease down the group

-Atomic radius of halogens increase down the group thus incoming/gained electron is attracted less strongly by the progressively larger atoms with a decreasing effective nuclear charge on outer energy level

 

(iv) Which is a move stable ion Cl or Br explain?

-Cl ion.

      -Has a more negative/exothermic electron affinity than Br

 

(v) Differentiate between electron affinity and:

  1. Ionization energy.

 

Ionization energy is the energy required to lose /donate an electron in an atom of an element in its gaseous state while electron affinity is the energy required to gain/acquire extra electron by an atom of an element in its gaseous state.

      Both are measured in kilojoules per mole.

     

  1. Electronegativity.

 

-Electron affinity is the energy required to gain an electron in an atom of an element in gaseous state. It involves the process:

               X(g)    + e → X(g)

 

Electronegativity is the ease/tendency of gaining/ acquiring electrons by an element during chemical reactions.

  It does not involve use of energy but theoretical arbitrary Pauling’s scale of measurements.

 

(g) (i) 5cm3 of sodium chloride, Sodium bromide and Sodium iodide solutions were put separately in test tubes. 5 drops of chlorine water was added to each  test tube: state and explain the observation made.

 

Observation

 Yellow colour of chlorine water fades in all test tubes expect with sodium chloride.

-Coloured Solution formed.

 

Explanation 

Chlorine is more electronegative than bromine and iodine. On adding chlorine water, bromine and Iodine are displaced from their solutions by chlorine.

 

(ii) The experiment in g (i) was repeated with 5 drops of bromine water instead of chlorine water .explain the observation made.

 

Observation

 

 Yellow colour of bromine water fades in test tube containing sodium iodide.

  • Brown solution formed in test tube containing sodium iodide

 

Explanation

 Bromine is more electronegative than iodide but less 6than chlorine.

On adding Bromine water, iodide displaced from its solution but not chlorine.

 

(iii) Using the knowledge in g(i) and (ii) above,

  1. Complete the table below using (X) to show no reaction and (√) to show a reaction.

 

 

 

 

 

 

                   Halide ion                 Halogen

ion in                                    solution

Halogen

F Cl Br I
F2 X

 

Cl2 X X

 

Br2 X X X

 

I2 X X X

 

 

 

 

 

 

 

 

 

 

Write an ionic equation for the reaction where there is (V)

F2  (g)+                2Cl (aq)   -> 2F(aq)                     +        Cl2(g)

 

F2  (g)+                2Br (aq)  ->  2F(aq)                     +        Br2(aq)

 

F2  (g)+                2I (aq)     -> 2F(aq)                     +        I2(aq)

 

Cl2  (g)   +           2Br (aq)   ->           2Cl(aq)               +      Br2(aq)

 

Cl2  (g)+              2I (aq)      -> 2Cl(aq)           +       I2(aq)

 

Br2  (aq)+            2I (aq)-       ->  2Br(aq)             +      I2(aq)

(h) State one uses of:

  • Fluorine

         Manufacture of P.T.F.E (Poly tetra fluoroethene) synthetic fiber.

              Reduce tooth decay when added in small amounts/equations in tooth paste.

Note: large small quantities of fluorine /fluoride ions in water cause browning of teeth/flourosis.

              Hydrogen fluoride is used to engrave word pictures in glass.

 

  • Bromine

Silver bromide is used to make light sensitive photographic paper/films.

 

  • Iodide

Iodine dissolved in alcohol is used as medicine to kill bacteria in skin cuts. It is called tincture of iodine.

 

  • The table below to show some compounds of halogens.

 

                 Element

Halogen

H Na Mg Al Si C P
F HF NaF MgH2 AlF3 SiF4 CF4 PF3

 

Cl HCl NaCl MgCl AlCl3 SiCl3 CCl4 PCl3

 

Br HBr NaBr MgBr2 AlBo3 SiBr4 CBr4 PBr3

 

I Hl Nal Mgl2 All3 SiL4 Cl2 Pb3

 

 

 

 

 

 

 

 

 

(j) (i) Using dot (.) and Cross (x) to represent electrons, show the bonding in chlorine molecule.

  • Name the type of bond formed.

Covalent.

 

  • Below is the table showing the bond energy of four halogens.

Bond                           Bond energy k J mole-1

Cl-Cl                                     242

Br-Br                                    193

I-I                                       151

 

  1. What do you understand by the term “bond energy”

Bond energy is the energy required to break/ form one mole of chemical bond

 

  1. Explain the trend in bond Energy of the halogens above:

Decrease down the group from chlorine to Iodine

-Atomic radius increase down the group decreasing the energy required to break the covalent bonds between the larger atom with reduced effective nuclear  charge an outer energy level that take part in bonding.

 

(k) Some compounds of chlorine are in the table below the oxidation state of chlorine in each compound.

Compound              Oxidation state                 Name of compound

NaClO3                             +5                                    Sodium chlorate (V)

ClO2                        +4                                    Chloric (IV) oxide

KClO2                     +3                                    Potassium chlorate (III)

NaClO                    +1                                    Sodium Chlorite (I)

Cl2                                        0                                                         Chlorine Molecule

NaCl                       -1                                     Sodium Chloride (I)

MgCl2                     -1                                     Magnesium Chloride (I)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. HYDROGEN CHLORIDE

 

  1. Occurrence

Hydrogen Chloride does not occur free in the atmosphere or in nature

 

  1. Preparation

Hydrogen chloride may be prepared in the school laboratory by reacting solid sodium/potassium chloride crystals with concentrated sulphuric (Vi) acid as in the set up below.

 

 

  1. Properties of hydrogen chloride gas(questions)

 

  1. What precautions should be taken when handling concentrated sulphuric acid? Explain.

   -Wear protective clothing/gloves to avoid accidental contact with skin.

-Concentrated sulphuric (VI) acid is highly corrosive-it causes painful wounds when in contact with skin.

 

  1. What method of gas collection is used? Explain.

-Downward delivery// upward displacement of water

        -Hydrogen chloride is denser than air.

 

  1. a) Write the equation for the reaction that takes place.

NaCl(s) + H2SO4(l)   -> NaHSO4(aq)  +  HCl(g)

        KCl(s)   + H2SO4(l)   -> KHSO4(aq)    +  HCl(g)

    NaCl is commonly used because it is cheaper than KCl

 

  1. b) What property of concentrated sulphuric (VI) acid is used during the above reaction

-is the least volatile mineral acid, thus displace the more volatile hydrogen chloride from its salt (KCl/NaCl)

 

  1. i)What is the purpose of concentrated sulphuric (VI) acid.

-Drying agent / to dry the gas.

 

  1. ii) What property of concentrated sulphuric (VI) acid is used during the above use.

-Is hygroscopic – absorbs water but do not form solution.

 

           iii) Name another substance which can be used to dry chlorine gas.

-anhydrous Calcium chloride

                        – silica gel

 

iv)Using a chemical equation, explain why anhydrous calcium oxide cannot be used in flask B

-Calcium oxide reacts with water /moisture to form calcium hydroxide. The calcium hydroxide formed reacts with chlorine to form calcium hypochlorite.

        Chemical equations:

                   CaO(s)  +  2H2O(l)  ->   Ca(OH)2(aq)  +  H2O(l)

                   Ca(OH)2(aq)  +  Cl2 (g)  ->  CaOCl2(aq)  +  H2O(l)

 

This reduces the amount of Chlorine produced.

 

d)Blue and red litmus papers were dipped in the hydrogen chloride prepared above. The Procedure was repeated with damp/wet/moist litmus papers. Explain the differences in observations made.

Dry blue litmus papers remain blue

          -Dry red litmus papers remain red

          -Damp/moist/wet blue litmus papers turn red

          -Damp/moist/wet red litmus paper turns red.

    -Dry hydrogen chloride is a molecular compound that is joined by covalent bonds between the atoms. The gas is polar thus dissolves in water and ionize completely to free H+ that are responsible to turning blue litmus paper red.

 

  1. Dry hydrogen chloride gas was bubbled in two separately beakers containing water and in methylbenzene.
  • Classify the two solvents as either “polar” or “non-polar”

Water – polar

        Methylbenzene – non-polar

 

(ii) State and explain the observations made in the beaker containing:

(i)Methylbenzene

Colour of litmus solution remain.

 Hydrogen chloride is a molecular substance. When dissolved in non-polar solvent, it does not dissociate / ionize to release Hions that changes the colour of litmus solution.

 

(ii)Water

Colour of litmus solution change to red.

Hydrogen chloride is a molecular substance. When dissolved in polar solvent like water, it dissociate/ionize to release H+ ions that changes litmus solution to red.

 

(iii)Why should an inverted filter funnel be used to dissolve hydrogen chloride.

– The filter funnel is dipped just below the water surface to increase the surface area of dissolving the gas and prevent suck back.

 

(iv)Name the solution formed when hydrogen chloride dissolves in water.

Hydrochloric acid

 

(f) Describe the test for presence of hydrogen chloride gas.

-Dip a glass rod in ammonia. Bring it to the mouth of a gas jar containing a gas suspected to be hydrogen chloride

          -White fumes of ammonia chloride are formed.

 

(g) Place 5cm3 of dilute hydrochloric acid into a four separate test tubes. To separate test tube add zinc, magnesium iron and copper metals. State and explain the observations made.

Observation

          – Effervescence/bubbles/fizzing in all cases except copper

  • Colourless solution formed with zinc and magnesium.
  • Green solution formed with ion.
  • Gas produced that extinguishes splint with explosion.

Explanation

 Metals above hydrogen in reactivity series react with hydrochloric and liberating hydrogen gas.

Chemical Equation:

Concentrated hydrochloric acid is a weak oxidizing agent than other concentrated acids i.eSulphuric (VI) acid and nitric (V) acid that react with all metals even those lower in the reactivity series.

 

(h) Place 5cm3 of dilute hydrochloric acid into five separate test tubes. To separate test tubes, add calcium carbonate, silver carbonate, copper carbonate, iron (II) carbonate and Sodium hydrogen carbonate. Explain the observations made.

Observation

         Effervescence/bubbles/fizzing vigorously except in silver carbonate and lead (II) carbonate that stop later.

  • Colourless solution formed except with iron (II) carbonate and copper (II) carbonate
  • Green solution formed with iron (II) carbonate
  • Blue solution formed with copper (II) carbonate

Explanation.

      Carbonates and hydrogen carbonate react with dilute hydrochloric acid to produce carbon (IV) oxide, water and form chlorides.

     All chlorides formed are soluble Except Lead (II) Chloride (soluble on heating/warming) and silver chloride.

     Chemical equation:    

             CaCO3 (s) + 2HCl(aq) → CaCl2(aq) + H2O(l) + CO2(g)

                                                (Colourless solution)   

 

     Chemical equation:  

           Ag2CO3 (s) + 2HCl(aq) → AgCl(s)   +       H2O(l)         + CO2(g)

                                               (Coats/Cover Ag2CO3)

 

Chemical equation:   

          CuCO3 (s) + 2HCl(aq) → CuCl2(aq) + H2O(l) + CO2(g)  

                               (Blue Solution)

 

     Chemical equation:   

           FeCO3 (s) + 2HCl(aq) → FeCl2(aq) + H2O(l) + CO2(g)

     

Chemical equation:     

          NaHCO3 (s) + HCl(aq) → NaCl(aq) + H2O(l) + CO2(g)

 

  • Place 5cm3 of dilute sodium hydroxide, Potassium hydroxide and aqueous ammonia solution into three separate test tubes. Add one drop of phenolphthalein indicator drop wise, add dilute hydrochloric acid. Explain the observations made.

Observation

Colour of Phenophthalein indicator change from pink to colourless.

Explanation

Hydrochloric acid neutralizes alkalis to salt and water

     When all the alkali has reacted with the acid, An extra slight excess acid turns the indicator used to colourless.

     Chemical equation:    

          NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) 

     Chemical equation:    

          KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)

     Chemical equation:    

NH4OH(aq) + HCl(aq) → NHaCl(aq) + H2O(l)

 

(j) Place 5cm3 of hydrochloric acid into four separate test tube tubes Separately add about 1g of each of copper (II) Oxide, Zinc (II) Oxide, Lead (II) Oxide< Calcium (II) Oxide. What happens to each test tube? Explain.

 

    Observation:

All Solid dissolves except Lead (II) Oxide

Colourless solution formed with zinc Oxide and calcium (II) Oxide blue solution        formed with copper (II) Oxide.

 

   Explanation:

Metal oxides dissolves in dilute hydrochloric acid to form water and chloride salt Insoluble Lead (II) chloride and silver chloride once formed cover/coat unreacted oxides stopping further reaction.

   Chemical equation:      CuO(s) + HCl (aq) → CaCl2(aq) + H2O(l)

    

   Chemical equation: CaO(s) + HCl (aq) → CaCl2 (aq) + H2O (l)

    

   Chemical equation: PbO(s) + 2HCl (aq) → PbCl2 (aq) + H2O (l)

    

    Chemical equation: ZnO(s) + HCl (aq) → ZCl2 (aq) + H2O (l)

 

(k) Manufacture of Hydrochloric acid.

 

         (i) Raw Materials

  1. Hydrogen

(i) During electrolysis of Brine from the flowing mercury-cathode cell during the manufacture of sodium hydroxide solution.

(ii)From water gas  by passing steam in heated charcoal.

C(s)   +  H2O   →   CO(g)  +  H2(g)

                      (iii)From partial oxidation of natural gas/methane

CH4(g)  +  O2(g) → CO(g)  +  3H2(g)

 

2.Chlorine

(i)From electrolysis of fused/solid sodium chloride in the downs process during extraction of sodium

(ii)From electrolysis of brine/concentrated sodium chloride solution in the flowing mercury-cathode during the manufacture of sodium hydroxide solution.

 

(ii)Chemical processes.

  • Hydrogen and chlorine gases are separately passed through concentrated sulphuric(VI) acid to act as a drying
  • Small amount of pure hydrogen is continuously ignited in a chamber with continous supply of pure dry chlorine.

Large amount of hydrogen explodes.

  • Hydrogen burns in chlorine to form hydrogen chloride gas.

Chemical Equation

H2(g)   +   Cl(g)   →   2HCl(g)

 

  • The hydrogen chloride produced is then passed up to meet a downward flow of water in the absorbtion Hydrogen chloride is very soluble in water and dissolves to form 35% concentrated hydrochloric acid.

Chemical Equation

HCl(g)      +      (aq)       →      HCl(aq)

 

The absorption chamber is shelved and packed with broken glass beads to

(i)Slow down the downward flow of water.

(ii)Increase surface area over which the water dissolves

The hydrochloric acid is then transported in steel tanks lined with rubber for market

 

         (iii)Uses of Hydrochloric Acid

  • To standardize the pH of (alcohol and wines)
  • Regenerating ion-exchange resin during removal of hardness of water.
  • Pickling of metals to remove oside layers on their surfaces.
  • In the manufacture of dyes and drugs.
  • Making zinc chloride for making dry cells.

 

Cold water

(iv)Diagram Showing Industrial manufacture.

Dry Hydrogen gas
Dry chlorine gas
Burning Hydrogen gas
Absorbtion

chamber

35% Conc. HCl

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

(ii)Environmental effects of manufacturing HCl.

 

  • Hydrochloric acid is acidic. Any leakage from a manufacturing plant to nearby rivers/lake causes exess acidity that lowers pH of water killing marine life.
  • Hydrogen chloride leakage into atmosphere dissolves to form “acidic rain” that accelerate corrosion in buildings, Breathing problems to human beings and kill fauna and flora around the paint.
  • Chlorine leakage causes breathing and sight problems to human being. It accelerates bleaching of dyed metals.
  • Hydrogen leakage can cause an explosion because impure hydrogen explodes on ignition.

 

  • Factors considered in setting hydrochloric acid manufacturing plant.

 

  1. Nearness to the manufacturing of sodium hydroxide because the by products of electrolysis of brine are the raw materials for hydrochloric acid plant.
  2. Availability of natural gas for extraction of hydrogen.
  3. Nearness/Availability of water to dissolve the hydrogen chloride gas.
  4. Availability of labour, market, capital and good means of transport.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

                           D: CHLORIDE (Cl) SALTS

 

 

  1. Chlorides are salts derived from hydrochloric acid. Hydrochloric acid is a monobasic (HX) salt with only one ionazable/replaceable “H” in its molecule. All chlorides are therefore normal salts.
  2. All metals exist as chloride salt except platinum and gold as below

 

 

Metal K Na Li Mg Ca Al Zn Fe Pb H. Cu Ag Hg
Formula of chloride KCl NaCl LiCl MgCl2 CaCl2 AlCl3 ZnCl2 FeCl2

FeCl3

PbCl

PbCl4

HCl CuCl

CuCl2

AgCl Hg2Cl2

HgCl2

 

(i)Both FeCl2 and FeCl3 exists but FeCl2 is readily oxidized to FeCl3 because it is more stable.

(ii)PbCl2 and PbCl4 exist but PbCl4 is only oxidized to form PbCl2 by using excess chlorine. It is less stable.

(iii)CuCl and CuCl2 exists but CuCl­2 is (thermodynamically) more stable    than CuCl. CuCl disproportionate to Cu and CuCl2..

(iv)HgCl and HgCl2exists as molecular compounds.

 

  1. All chlorides are soluble/dissolves in water except silver chloride(AgCl), Copper (I) chloride CuCl, Mercury (I) Chloride Hg2Cl2 and Lead (II) Chloride PbCl2 that dissolves in warm
  2. Most chlorides are very stable compounds. They do not decompose on gentle or strong bunsen burner heating in a school laboratory except Ammonium Chloride.

 

  1. Heating ammonium chloride

 

Place about 2g of solid ammonium chloride crystals in a clean dry boiling tube.Heat gently then strongly.

 

Observation

         –red litmus paper turn blue

         -blue litmus paper remains blue

Then later:

         -both blue litmus papers turn red

 

   Explanation:

Ammonium chloride on heating decomposes through chemical sublimation to ammonia and hydrogen chloride gas. Ammonia gas is less dense than hydrogen chloride. It is a basic gas and diffuses out faster to turn red litmus paper to blue. Hydrogen chloride is an acidic gas .It is denser than ammonia gas and thus diffuses slower than ammonia gas to turn the already both blue litmus paper to red.

Chemical equation

NH4Cl(s)     ->       HCl(g)         +                  NH3 (g)

(acidic gas)                      (basic/alkaline gas)

 

  • Test for Cl ions
  1. The following experiment shows the test for the presence of Clions in solids chloride salts.

 

  • Procedure:

Place about 1g of sodium chloride, Zinc chloride and copper (II) chloride in separate boiling tubes. Place moist blue and red litmus papers on the mouth of the test tube. Carefully, add three drops of concentrated sulphuric (VI) acid.

Dip a glass rod in aqueous ammonia solution then bring it to the mouth of the boiling tube.

 

observation inference
-red litmus paper remain red

-blue litmus paper turn red

 

-vigorous effervescence/fizzing

/bubbling

 

-white fumes produced on                     

H+ ions

 

 

Cl ions

 

 

HCl gas suspected

 

          (b)Explanation:

Concentrated sulphuric (VI) acid is the less volatile mineral acid.

It vigorously displaces chlorine in metal chlorides to evolve acidic hydrogen chloride gas fumes.

Chemical equation

 

NaCl(s)  +  H2SO4(l)        ->  NaHSO4(aq)  +  HCl(g)

KCl(s)    +  H2SO4(l)       ->  KHSO4(aq)    +  HCl(g)

                    CuCl2(s) +  H2SO4(l)       -> CuSO4(aq)     +  2HCl(g)

ZnCl2(s) +  H2SO4(l)       ->  ZnSO4(aq)      +  2HCl(g)

 

Hydrogen chloride and ammonia gases react and form white fumes of ammonium chloride that confirms presence of  Cl ions in the solid substance.

Chemical equation

NH3(g)     +     HCl(g)     ->    NH4Cl(s)

 

 

  1. The following experiment shows the test for the presence of Clions in solution /aqueous chloride salts.

 

(i)Using aqueous Lead (II) nitrate(V)

 

(a)Procedure:

I.Place about 5cm3 of sodium chloride, Iron (III) chloride and copper (II) chloride in separate boiling tubes. Add four drops of Lead (II) nitrate(V) solution to each. Preserve.

 

Observation                 Inference
White precipitate/ppt     SO42-, SO32-, Cl,CO32-

 

 

II.To the preserved sample, add six drops of nitric (V) acid. Preserve.

Observation                 Inference
White precipitate/ppt persist               SO42-,  Cl

 

 

III. To the preserved sample, heat the mixture to boil

Observation                 Inference
White precipitate/ppt dissolves on boiling/warming                     Cl

 

 

Explanation:

 

I.When Lead(II) nitrate(V) solution is added to an unknown salt , a white precipitate/ppt of Lead(II) sulphate(VI) Lead(II) carbonate(IV) Lead(II) sulphate(IV) Lead(II) chloride(I) are formed.

Ionic equation:

Pb2+ (aq)     +        SO42-(aq)        ->      PbSO4(s)

Pb2+ (aq)     +        SO32-(aq)        ->      PbSO3(s)

Pb2+ (aq)     +        CO32-(aq)        ->      PbCO3(s)

Pb2+ (aq)     +         Cl(aq)            ->      PbCl2(s)

 

II.When the white precipitate/ppt formed is acidified with dilute nitric(V) acid, the white precipitate of Lead(II) sulphate(VI)  and  Lead(II) chloride(I) persist/remain while that of Lead(II) carbonate(IV) and Lead(II) sulphate(IV)  dissolves.

 

III.On heating /warming Lead (II) chloride (I) dissolves but on cooling it recrystallizes.This shows the presence of Clions in aqueous solutions

.

     (ii)Using aqueous silver (I) nitrate(V)

 

Procedure

  1. Place about 5cm3 of sodium chloride, Iron (III) chloride and copper (II) chloride in separate boiling tubes. Add four drops of silver(I) nitrate(V) solution to each. Preserve.

 

Observation              Inference
White precipitate/ppt               Cl,  CO32-

 

 

  1. To the preserved sample, add six drops of nitric (V) acid. Preserve.
Observation              Inference
White precipitate/ppt persist                   Cl

 

 

Explanation:

I.When silver(I) nitrate(V) solution is added to an unknown salt , a white precipitate /ppt of  silver(I) carbonate(IV) and silver(I) chloride(I) are formed.

Ionic equation:

2Ag+ (aq)     +        CO32-(aq)        ->      Ag2CO3(s)

Ag+ (aq)       +         Cl(aq)            ->       AgCl(s)

 

  1. When the white precipitate/ppt formed is acidified with dilute nitric (V) acid, the white precipitate of silver (I) chloride (I) persist/remain. This shows the presence of Clions in aqueous solutions.

Silver (I) carbonate (IV) dissolves when reacted with nitric (V) acid.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

COMPREHENSIVE REVISION QUESTIONS

 

  1. In an experiment ,dry hydrogen chloride gas was passed through heated zinc turnings as in the set up below.The gas produced was the passed through copper(II) oxide

 

 

 

  1. Write the equation for the reaction :

(i)For the school laboratory preparation of hydrogen chloride gas.

NaCl(s) + H2SO4(l) -> NaHSO4(aq)  +  HCl(g)

 

(ii)in tube S

Zn(s)  + 2HCl(aq)   ->  ZnCl2(aq)  +   H2(g)

 

    b)State and explain the observation made  in tube V.

Observations-colour of solid changes from black to brown

                              -colourless liquid forms   on the cooler parts of tube V

Explanation-Hydrogen produced in tube S reduces black copper(II) oxide to brown copper metal and the gas oxidized to water vapour that condense on cooler parts..

Chemical equation.

CuO(s)  +H2(g)   ->Cu(s)  +   H2O(l)

 

   (c)How would the total mass of tube S and tube V and their contents compare before and after the experiment.

Tube S- Mass increase/rise because Zinc combine with chlorine to form heavier Zinc Chloride.

Tube V- Mass decrease/falls/lowers because copper (II) oxide is reduced to lighter copper and oxygen combine with hydrogen to form water vapour that escape.

  1. Chlorine is prepared by using solid sodium chloride, concentrated sulphuric(VI) acid and potassium manganate(VII)

a)What is the role of the following in the reaction;

(i) concentrated sulphuric(VI)

To produce hydrogen chloride gas by reacting with the solid sodium chloride.

 

(ii)potassium manganate(VII)

To oxidize hydrogen chloride gas to chlorine

 

3.Use the flow chart below to answer the questions that follow.

 

 

 

 

a)(i) Name:

gas X                               Hydrogen chloride

          solution W                       hydrochloric acid

gas Q                               chlorine

          bleaching agent Z  sodium chlorate(V)

b)Write the chemical equation for the formation of :

(i) gas X

          NaCl(s) + H2SO4(l) -> NaHSO4(aq)  +  HCl(g)

 

(ii)solution W

HCl(g)         +        (aq)             ->       HCl(aq)

 

(iii)gas Q

2KMnO4  +  16HCl(aq) ->  2KCl(aq)  +  2MnCl2(aq)  +  8H2O(l)  +  5Cl2(g)

 

(iv)bleaching agent Z

6NaOH(aq)     +        3Cl2(g)  ->NaCl(aq) + NaClO3(aq) +  3H2O(l)

 

c)State and explain the following observations;

(i) a glass rod dipped in aqueous ammonia is brought near gas X

Observation: Dense white fumes

     Explanation:Ammonia gas reacts with hydrogen chloride gas to form dense white fumes of ammonium chloride.

     Chemical equation: NH3(g)  +HCl(g)  ->  NH4Cl(s)

(ii)Wet blue and red litmus papers were dipped into gas Q

Observations: Blue litmus paper turned red the both are bleached

/decolorized.

Explanations: chlorine reacts with water to form both acidic hydrochloric and chloric (I) acids that turn blue litmus paper red. Unstable chloric (I) acid oxidizes the dye in the papers to colourless.

Chemical equations

Cl2(g)  + HCl(aq)   ->HCl(aq)  +   HClO(aq)

Coloured dye  +HClO(aq)   ->HCl(aq)  +   (Colourless dye +O)//

(Coloured dye-O)  + HClO(aq)   ->HCl(aq)  +   Colourless dye

 

4.Use the flow chart below to answer the questions that follow

 

 

 

  1. Name

Liquid A                          Concentrated sulphuric(VI) acid

Process Z                         Neutralization

White solid X                 Ammonium chloride     

 

b)Write the equation for the formation of:

(i) Hydrogen chloride

NaCl(s) + H2SO4(l) -> NaHSO4(aq)  +  HCl(g)

 

          (ii) B

HCl(g)         +        (aq)             ->       HCl(aq)

          (iii)process Z (using ionic equation)

                    H+ (aq)        +        OH(aq)      ->       H2O(l)

 

(iv)C (using ionic equation)

                    Ag+ (aq)      +        Cl(aq)         ->       AgCl(s)

 

c)Describe how solution B is obtained.

Bubbling hydrogen chloride gas through inverted funnel into distilled water until no more dissolve.

5 The results obtained when halogens are bubbled into test tubes containing solutions of halide A,B and C is as in the table below. Tick(v) means a reaction took place.Cross(x) means no reaction took place.

 

 

Halogens

Halide ions in solution
A B C
I2 x x
Br2 x v
Cl2 v v

a)Identify the halide ions represented by letter

A                 Cl

B                       I

C                  Br

b)Write the ionic equation for the reaction that take place with halide:

(i)  C

Cl2(g)          +        2Br(aq)      ->       2Cl(aq)       +        Br2(aq)

 

(ii) B

Cl2(g)          +        2Br(aq)      ->       2Cl(aq)       +        Br2(aq)

Cl2(g)          +        2I(aq)         ->       2Cl(aq)       +        I2(aq)

 

6.The diagram below shows a set up of apparatus for  the school laboratory collection of dry chlorine gas.

 

 

 

 

 

a)Name:

(i) substance Q

Concentrated hydrochloric acid

          (ii)suitable drying agent L

                    -Concentrated sulphuric(VI) acid

                    -anhydrous calcium chloride

                    -silica gel

  1. b) State a missing condition for the reaction to take place faster.

-Heat/Heating

 

c)Red and blue litmus papers were dipped into the chlorine gas from the above set up .State and explain the observations made.

Observation: Blue litmus paper remain blue. Red litmus paper remain red.

 Explanation: Dry chlorine has no effect on dry litmus papers.

 

d)Write the equation for the reaction taking place in the conical flask

MnO4 (s) +  4HCl(aq) ->   MnCl2(aq)  +  2H2O(l)  +  Cl2(g)

 

e)Name two other substances that can be used in place of MnO2

          Lead(IV) oxide (PbO2)

          Potassium manganate(VI)(KMnO4)

Potassium dichromate(K2Cr2O4)

Bleaching powder(CaOCl2)

7.The set up below shows the apparatus used to prepare and collect anhydrous iron(III) chloride.

 

 

a)Name salt K

          Iron(III)cchloride

 

  1. Write the equation for the reaction for the formation of salt K

2Fe(s)          +  3Cl2 (g)   ->       2FeCl3 (s/g)

 

  1. State and explain the following

(i)Small amount of water is added to iron (II) chloride in a test tube then shaken

Solid dissolves to form a green solution. Iron(II) chloride is soluble in water

 

(ii)I.Three drops of aqueous sodium hydroxide is added to aqueous iron(II) chloride and then added excess of the alkali.

Observation:

Green precipitate is formed that persist/remain /insoluble in excess akali.

          Explanation:

Iron(II) chloride reacts with aqueous sodium hydroxide to form a green precipitate of iron(II) hydroxide.

          Ionic equation:

Fe2+(aq)      +        OH(aq)      ->       Fe(OH)2(s)

 

II.Six drops of hydrogen peroxide is added to the mixture in d(ii) above.

Observation:

Effervescence/bubbling/fizzing take place and the green precipitate dissolve to form a yellow/brown solution.

Explanation:

hydrogen peroxide oxidizes green Fe2+to yellow/ brown Fe3+solution.

 

 

 

 

 

 

 

9.Use the flow chart below to answer the questions that follow.

 

 

 

 

a)Write the chemical equation for the formation of gas A

          NaCl(s) + H2SO4(l) -> NaHSO4(aq)  +  HCl(g)

 

b)Identify:

(i) four possible ions that can produce white precipitate B

SO42-,SO32-, CO32-,  Cl

              

               (ii)two possible ions that can produce;

I.White precipitate C

SO42-,Cl

II.colourless solution D

SO32-, CO32-

(iii)possible ions present in

I.White precipitate E

 SO42-

II.colourless solution F

          Cl

 

  1. Below is a set up in the preparation of a particular salt. Study it and answer the questions that follow.

 

State the observation made when aluminium wool is heated.

Glows red hot.

 

b)(i) Identify salt A

aluminium(III) chloride// AlCl3

 

               (ii)Write the equation for the formation of salt A

2Al(s)  + 3Cl2(g)   ->  2AlCl3(s/g)

 

          (iii)What property of salt A is exhibited as shown in the experiment.

It sublimes//sublimation.

 

(iv)Calculate the minimum volume of chlorine required to form 700kg of  iron(III) chloride at room temperature.(Fe= 56.0, Cl=35.5, 1 mole of a gas =24000cm3, 1000g = 1kg)

Mole ratio   Fe  : Cl2      =  2: 3  molar mass FeCl3 = 162.5g

Method 1

           2 x 162.5 g  FeCl3  -> 3x 22400 cm3 Cl2

                              700 x1000 gFeCl3  ->  (700 x1000 x3 x22400)/(2 x 162.5)         

                                                                =1.4474 x 10-8 cm3

Method 2

Moles of FeCl3= mass/ molar mass

 =>  (700 x 1000) / 162.5 = 4307.6923 moles

                    Moles of Cl2= 3/2  moles of FeCl3

                                             =>3/2 x 4307.6923  = 6461.5385 moles

                    Volume of chlorine= moles x molar gas volume

                                        =>6461.5385 x 24000 = 1.5508 x 10-8 cm3

 

  1. c) Name another metal that can produce similar results as salt K.

Iron

 

d)(i) What is the purpose of anhydrous calcium chloride.

 

-ensure the apparatus are water free.

-prevent water from the atmosphere from entering and altering//hydrolysing salt A

 

(ii) Write the equation for the reaction that take place if anhydrous calcium chloride is not used in the above set up.

 

                    AlCl3(s)   + 3H2O(l)   ->  Al(OH)3(aq)  + 3HCl(g)

 

(iii) Write the equation for the reaction that take place when Iron metal is reacted with dry hydrogen chloride gas.

 

Fe(s)   + 2HCl(g)   -> FeCl2(s)  + H2(g)

 

(iv)Calculate the mass of Iron(II)chloride formed when 60cm3 of hydrogen chloride at r.t.p is completely reacted. (1 mole of a gas =24dm3 at r.t.p, Fe = 56.O, Cl= 35.5)

Chemical equation  Fe(s)   +    2HCl(g)   ->  FeCl2(s)   +  Cl2(g)

 

Mole ratio  HCl: FeCl2 = 1:1

Molar mass  FeCl2 = 127g

 

Moles of HCl used = 60cm3 /24000cm3 = 2.5 x 10 -3 moles

Moles of FeCl2 = Moles of HCl  => 2.5 x 10 -3 moles

 

          Mass of FeCl2 = moles x molar mass => 2.5 x 10 -3 x 127  =0.3175g

 

 

12.Study the flow chart below and use it to answer the questions that follow

 

a)Identify substance:

P        Iron(II) chloride//FeCl2

Q       Chlorine // Cl2

R        Iron(III) chloride//FeCl3

 

b)Write the equation for the reaction for the formation of:

(i) gas  Q

2KMnO4 (s) +  16HCl(aq) ->   2KCl(aq) + 2MnCl2(aq)  +  8H2O(l)  + 5Cl2(g)

 

(ii) the green precipitate (using ionic equation)

Ionic equation:

Fe2+(aq)      +        2OH(aq)    ->       Fe(OH)2(s)

 

(ii) the brown precipitate (using ionic equation)

Ionic equation:

Fe3+(aq)      +        3OH(aq)    ->       Fe(OH)3(s)

 

c)A glass rod was dipped in aqueous ammonia. The rod was then brought near hydrogen chloride. State and explain the observation made.

Observation:

White fumes

          Explanation:

 Ammonia gas reacts with hydrogen chloride gas to form white fumes of ammonium chloride.

          Chemical equation:

NH3(g)  +  HCl(g)  ->  NH4Cl(s)

  1. Using dot(.)and cross(x)to represent electrons,show the bonding in aluminium chloride in vapour phase.

 

 

 

 (b)How many electrons in :

(i)aluminium atoms are used in bonding.

          Six electrons(three valence electrons in each aluminium atom)

(ii)chlorine atoms atoms are used in dativebonding.

          four electrons(two lone pairs of valence electrons in two chlorine atoms)

(iii)the molecule are used in bonding.

          Sixteen electrons

-six valence electrons from aluminium atom through covalent bond

-six valence electrons from chlorine atoms through covalent bond.

– four valence electrons from chlorine atoms through dative bond

 

(c)How many lone pair of electrons do not take part in bonding within the molecule.

          Sixteen(16) lone pairs from six chlorine atoms(32 electrons)

 

(d)Aluminium chloride does not conduct electricity in molten state but Magnesium chloride conduct.

Aluminium chloride is a molecular compound that has no free mobile Al3+ and Cl ions which are responsible for conducting electricity. Magnesium chloride has free mobile Mg2+ and Cl ions because it is an ionic compound.

 

 

 

 

 

 

 

 

  1. Use the flow chart below to answer the questions that follow:

 

 

a)Write an equation for the school laboratory formation  of hydrogen chloride gas

NaCl(s) + H2SO4(l) -> NaHSO4(aq)  +  HCl(g)

KCl(s) + H2SO4(l) -> KHSO4(aq)  +  HCl(g)

b)Name:

  1. solid Y Iron (II) chloride (FeCl2)                         

          II green precipitateIron (II) hydroxide (Fe (OH)2

          III Gas Y                         Chlorine (Cl2)

  1. Bleaching agent A Sodium hypochlorite (NaOCl)

c)Blue and red litmus papers were dipped into bleaching agent A. Write the equation for the reaction that takes place.

               Coloured dye  +NaOCl(aq)   ->NaCl(aq)  +   (Colourless dye + O)//

          (Coloured dye-O)  + NaOCl(aq)   ->NaCl(aq)  +   Colourless dye

 

d)State four uses of gas Z

  1. Bleaching agent
  2. Manufacture of hydrochloric acid
  3. Chlorination of water to kill germs
  4. Manufacture of PVC pipes

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