Chemistry, Third edition

130 8 · THE MOLE

Percentage yield

The previous calculations can be used to find out how much product you would

expect, if you carried out a reaction with a measured amount of reactants in the

laboratory. In practice it is often extremely difficult to obtain the calculated yield,

especially in organic chemistry, because of many reasons including:

1.Loss of reactants and/or product during experimental manipulations.

2.Competing side reactions, where other products are formed.

3.As you will see later, in some reactions (reversibleorequilibrium reactions), the

reactants do not completely change into the products (go to completion).

The yield of the product obtained by experiment, therefore, may be lower than

expected; sometimes it is a great deal lower. It is usual to report your percentage

yieldwhen you are doing preparative chemistry:

percentage yield 

experimental yield

100%

theoretical yield

where the experimental yield is the mass of product obtained and the theoretical yield

is the mass of product you calculated you would get from the balanced equation.

8.7

Calculating gas volumes from equations

All gas volumes are measured at room temperature and pressure

(i) What is the volume occupied by 0.25 mol of chlorine gas?

(ii)What is the mass of 600 cm^3 of sulfur dioxide gas?

(iii)What volume of hydrogen is obtained from the reaction of 4.8 g of magnesium metal with

excess hydrochloric acid?

Mg(s)2HCl(aq)MgCl 2 (aq)H 2 (g)

(iv)What volume of oxygen is needed for complete combustion of 2 dm^3 of propane (C 3 H 8 ).

C 3 H 8 (g)5O 2 (g)3CO 2 (g)4H 2 O(I)

(v)What volume of hydrogen chloride gas is produced by the reaction of 40 cm^3 of chlorine

with hydrogen? (First write a balanced equation.)

(vi)What mass of water is made when 10 dm^3 of hydrogen gas burns completely in excess air?

(First write a balanced equation.)

Exercise 8J

Example 8.8 (continued)

Convert moles to grams in the case of calcium carbonate and moles to cubic

decimetres in the case of carbon dioxide, because the question asks about the

volume of the gas:

100 g CaCO 3 24 dm^3 CO 2

Therefore,

50 g CaCO 3 12 dm^3 CO 2

So, 12 dm^3 CO 2 are produced.