Percentage yield in organic preparations.
Reactions involving organic compounds are unlikely
to give a 100% yield of product. In practice many organic preparations give
very low yields. There may be many reasons for less than a 100% yield
including;
¨
the reactants may not be pure
¨
the reaction may involve an equilibrium, so that
it does not go to completion
¨
the possibility of side-reactions occurring, which
lead to the formation of other products
¨
some of the products may be left behind in the
reaction vessel
¨
the product may be difficult to purify
From the balanced
equation for a reaction it is possible to find the theoretical yield of the product (i.e. the maximum possible yield
assuming that all reactants are converted to products).
The actual yield obtained can be compared with the
yield calculated from the molar masses of the reactants, to obtain the percentage yield in the reaction.
Percentage yield
= actual yield of product
x 100
theoretical yield of product
e.g.
C4H9OH
(l) +
HBr (g) C4H9Br
(l) + H2O (l)
1 mol 1
mol
74 g 137
g
Therefore, 74 g of butan-1-ol reacting with excess
hydrogen bromide would give a theoretical yield of 137 g of 1-bromobutane.
If this reaction was carried out using 10 g of
butan-1-ol the theoretical yield would be:
10
g x 137g = 18.5 g
74
g
If you actually carry out this reaction in the
laboratory you will find that you obtain less than 18.5 g of 1-bromobutane for
the reasons given above.
Example
In a preparation of 1-bromobutane using 10 g of
butan-1-ol, 15 g of 1-bromobutane were obtained after purification. Calculate
the percentage yield.
Percentage yield
= actual yield of product
x 100
theoretical yield of product
= 15 g x 100 =
81.1%
18.5 g
Exercise 1
PROBLEMS ON PERCENTAGE YIELD
1.
Phenol, C6H5OH, is converted
into trichlorophenol, C6H2Cl3OH. If 488 g of
product are obtained from 250 g of phenol, calculate the percentage yield.
2.
29.5 g of ethanoic acid, CH3COOH, are
obtained from the oxidation of 25.0 g of ethanol, C2H5OH.
What percentage yield does this represent?
3.
100 cm3 of barium chloride
solution of concentration 0.0500 mol dm-3
were treated with an excess of sulphate ions in solution. The precipitate of
barium sulphate formed was dried and weighed. A mass of 1.1558 g was recorded.
What percentage yield does this represent?
LIMITING REACTANT
In chemical reactions
the reactants are often mixed in proportions
which are different than those given in the balanced chemical equation.
One of the reactants is often in excess and is not completely used up in the
reaction. The amount of product is then determined by the amount of reactant
which is not in excess and is therefore used up completely in the reaction.
This is called the limiting reactant. You first have to decide
which is the limiting reactant before you can calculate the amount of product
formed.
Exercise 2
1.
In the blast furnace, the overall reaction is
2Fe2O3
(s) +
3C (s) 3CO2
(g) +
4Fe (s)
what is the maximum mass
of iron that can be obtained from 700 tonnes of iron (III) oxide and 70 tonnes
of coke?
[1 tonne = 1000 kg]
2.
in the manufacture of the fertiliser ammonium
sulphate
H2SO4
(aq) +
2NH3 (g) (NH4)2SO4
(aq)
What is the maximum mass
of ammonium sulphate that can be obtained from 2.0 kg of sulphuric acid and 1.0
kg of ammonia?
3.
In the thermit reaction
2Al
(s) +
Cr2O3 (s) 2Cr
(s) +
Al2O3 (s)
Calculate the percentage
yield when 180 g of chromium are obtained from a reaction between 100 g of
aluminium and 400 g of chromium (III) oxide.