waecE-LEARNING
Chemistry Paper 2 (Essay) ,May/June 2010  
Questions:   1 2 3 4 5 6 7 8   Main
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Question 4

(a) (i) Give the two reasons why soda lime is used instead of caustic soda in the preparation
of methane.
(ii) List two physical properties of methane.
(Hi) A hydrocarbon with a vapour of29 contains 82.76% carbon and 17.24% hydrogen.
Determine the:


I. empirical formula;
ii. molecular formula of the hydrocarbon.
[H = 1.00 C = 12.00] [ 12 marks]

(b) (i)What is meant by the term isomerism?
(ii) Draw the structures of the two isomers of the compound with the molecular formula C2H60.
(iii) Give the name of each of the isomers in4(b)(ii).
 (iv) State the major difference between the isomers.                                                  [ 7 marks]

(c) Give three deductions that could be made from the qualitative and quantitative analysis of a given organic compound.         [ 3 marks]

(d) Give one chemical test to distinguish between propene and propane. [3 marks]


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OBSERVATION

The question was popular among candidates and the performance was good.

In part(a)(i), candidates gave two reasons why soda lime was used instead of caustic soda in the preparation of methane thus:

- soda lime does not attack glass apparatus unlike caustic soda
- soda lime is not deliquescent unlike caustic soda

In (a)(ii), candidates listed two physical properties of methane from among the following:
- it is a gas at room temperature;
- colourless gas;
- odourless gas;
- slightly soluble in water;
- less dense than air.

In (a)(iii), candidates determined correctly both empirical and molecule formulae of the hydrocarbon thus:

 

 

hydrogen

Carbon

%

17.24

82.76

 


                       17.24
                 1.00

82.76
12

 

 

 

 


17.24

6.90

 


17.24
 6.90

6.90
6.90

 

2.5
5

1

 

 

2

 

 

 


1. Empirical formula C2H5
II. Molecular formula (C2 H5)n = V.D x 2


→(12 x 2 + 1 x 5)n = 58
                          29n    =    58
                       n = 2
 →(C2 H5)2            = C4 H10

In part (b), candidates stated what was meant by isomerism, drew the structures of the two isomers of the compound C2H60, gave the name of the two isomers in (i) - (v) respectively.

However, some of them lost marks because of spelling errors, omission of bonds etc.
The expected responses from candidates were as follows:

  (i)            The existence of two or more compounds with the same molecular formula but
different molecular structures/different arrangements of atoms.



  (ii)     H    H
             I     I
        H-   C-   C-   O  -H
             I      I
           H       H

           H              H
             I               I
        H-   C-   O-   C  -H
             I                I
           H                 H

 



[All bonds must be shown however, accept -OH was accepted]
  (iii)           Ethanol and methoxymethane respectively.
  (iv)         The two isomers belong to different homologous series/they have different functional groups

In part (c), most candidates could not give three deductions made from qualitative and quantitative analysis of a given organic compound hence lost marks. The required answers from candidates were:

- functional groups
- number of atoms of different elements
- types of elements
- types of bonds
- percentage composition of the elements in compound
- spatial arrangements of atoms in a molecule.

In parts( d), only very few candidates could correctly give one chemical test to distinguish between propene and propane. The expected answer from candidates was as written below:

- Pass each of the gases into (acidified) KMnO4J bromine water/bromine in CCl4/ bromine, propene decolourizes bromine water/KMnO4/bromine in CCl4/bromine whereas propane does not.

 

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