Yr 12 IB Chemistry
Percentage Yield
and
Percentage Atom
Economy
During a Reaction
Percentage Yield During a Reaction
Actual Yield in Grams Produced During an Experiment
% Yield =
Theoretical Yield (as predicted by reaction equation) in Grams
% yield will usually be less than 100% because :
1. The reaction may be incomplete
2. Other reactions may occur
3. Product is lost during separation and purification
x 100
Steps to Calculate a Percentage Yield
1. Convert the mass of main reactant to a number of moles
using mol = mass/molar mass
2. Calculate the theoretical number of moles of product using
the ratio from the balanced equation
3. Calculate the theoretical mass of product using mass =
mol x molar mass
4. Calculate the percentage yield using :
% Yield =
Actual Yield in Grams
x 100
Theoretical Yield in Grams
N.B. This assumes all other reactants are present in
excess, or at least in stoichiometric amounts
e.g1 12.0g ethanol (C2H5OH) is reacted to produce
5.3g of ethene (C2H4). Calculate the % yield.
1.
Number of moles of ethanol used
= 12.0 / 46 = 0.26
2.
C2H5OH  C2H4 + H2O.
Theoretical number of moles of ethene formed
= 0.26 since 1:1 reaction
3.
Theoretical mass of ethene formed
= 0.26 x 28
4.
% Yield =
= 7.3g
5.3
7.3
x 100
=
72.5%
e.g2 8.85 g of sodium hydroxide (NaOH) is reacted to
produce 9.45 g of sodium carbonate (Na2CO3).
Calculate the % yield.
1.
Number of moles of NaOH used
= 8.85 / (23.0 + 16.0 + 1.0) = 0.221
2.
2NaOH + CO2  Na2CO3 + H2O.
Theoretical number of moles of Na2CO3 formed
= 0.221/2 (since 2:1 reaction ) = 0.111
3.
Theoretical mass of Na2CO3 formed
= 0.111 x (2(23)+12+3(16)) = 11.73g
4.
% Yield =
9.45
11.73
x 100
=
80.6%
% Atom Economy
A measure of the percentage of the starting materials that actually ends up in the
useful products.
The greater the % atom economy of a reaction, the more 'efficient' or 'economic' it is
likely to be,
To calculate the % Atom Economy :
1. Write the balanced equation for the reaction
2. Calculate the total mass of the reactants for the numbers of moles in the reaction
equation
3. Calculate the total mass of the “desired product” for the numbers of moles in the
reaction equation
4.
massproduct x 100
% Atom Economy =
massreactants
Note : reactions involving only
the production of a
SINGLE PRODUCT will
have 100% atom
economy
e.g. H2 + Cl2  2HCl
C2H4 + H2  C2H6
N2 + 3H2  2NH3
Example 1
The equation for the production of lime (CaO) from limestone (CaCO3) is:
CaCO3  CaO + CO2
Calculate the % atom economy of this process.
Mreactants = 40.1 + 12.0 + 3(16.0) = 100.1 g
Mproduct
= 40.1 + 16.0 = 56.1 g
% Atom Economy = 56.1 x 100
100.1
=
56.0%
Example 2
The equations for the extraction of titanium from titanium(IV) oxide are:
TiO2 + C + 2Cl2  TiCl4 + CO2
TiCl4 + 2Mg  Ti + 2MgCl2
Calculate the % atom economy of the overall process.
Combining 2 equations for an overall equation gives:
TiO2 + C + 2Cl2 + TiCl4 + 2Mg  TiCl4 + CO2 + Ti + 2MgCl2
Mreactants = 47.9 + 2(16.0) + 12.0 + 2(2 x 35.5) + 2(24.3) = 282.5 g
Mproduct
= 47.9 g
% Atom Economy = 47.9 x 100
282.5
=
17.0%
Such a low % atom economy is typical of a MULTISTAGE process.
Example 3
The equation for the extraction of copper from copper sulphide is:
CuS + O2  Cu + SO2
1 Calculate the % atom economy of the reaction.
2 If 9.60 tonnes of copper sulphide produce 4.80 tonnes of copper, calculate the % yield
1
CuS
+
O2

Cu
+
SO2
Mreactants = (63.5 + 32.1) + (2 x 16) = 127.6 g
Mproduct
= 63.5 g
% Atom Economy = 63.5
x 100
=
127.6
49.8%
2. Theoretically, in excess oxygen, one mole CuS  one mole Cu
i.e. (63.5 + 32.1) g CuS  63.5 g Cu
i.e. 95.6 g CuS  63.5 g Cu
i.e. 1.0 g CuS  63.5 / 95.6 = 0.664 g Cu
i.e. 1.0 tonnes CuS  0.664 tonnes Cu
i.e. 9.60 tonnes CuS  9.60 x 0.664 = 6.38 tonnes Cu
% Yield Cu = 4.80 x 100
6.38
=
75.2%
The End