1. Percentage Atom Economy
Yr 12 IB Chemistry
Percentage Yield and
Percentage Atom Economy
During a Reaction

2. Theoretical Yield (as predicted by reaction equation) in Grams
Percentage Yield During a Reaction
Actual Yield in Grams Produced During an Experiment
% Yield =
x 100
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

3. 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 :
Actual Yield in Grams
% Yield =
x 100
Theoretical Yield in Grams
N.B. This assumes all other reactants are present in excess, or at least in stoichiometric amounts

4. e. g1 12. 0g ethanol (C2H5OH) is reacted to produce 5
e.g g 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
= 7.3g
5.3 4.
% Yield =
x 100
= %
7.3

5. e. g2 8. 85 g of sodium hydroxide (NaOH) is reacted to produce 9
e.g 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 / ( ) =
2.      2NaOH + CO2    Na2CO3  +  H2O. 
Theoretical number of moles of Na2CO3 formed
= 0.221/2 (since 2:1 reaction ) =
3.      Theoretical mass of Na2CO3 formed
= x (2(23)+12+3(16))
= 11.73g
9.45 4.
% Yield =
x 100
= %
11.73

6. % Atom Economy 4. massproduct x 100
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 :
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
massproduct x 100
massreactants
% Atom Economy =
Note : reactions involving only the production of a SINGLE PRODUCT will have 100% atom economy
e.g. H2 + Cl2  2HCl C2H4 + H2  C2H6
N H2  2NH3

7. 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 = (16.0) = g
Mproduct = = g
% Atom Economy = x
100.1
= 56.0%

8. Example 2
The equations for the extraction of titanium from titanium(IV) oxide are:
TiO2 + C + 2Cl2  TiCl4 + CO2
TiCl Mg  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 = (16.0) (2 x 35.5) + 2(24.3) = g
Mproduct = g
% Atom Economy = x
282.5
= 17.0%
Such a low % atom economy is typical of a MULTISTAGE process.

9. Mreactants = (63.5 + 32.1) + (2 x 16) = 127.6 g
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
CuS O  Cu SO2
Mreactants = ( ) + (2 x 16) = g
Mproduct = g
% Atom Economy = x
127.6
= 49.8%
2. Theoretically, in excess oxygen, one mole CuS  one mole Cu
i.e. ( ) g CuS  g Cu
i.e g CuS  g Cu
i.e. 1.0 g CuS  / = g Cu
i.e. 1.0 tonnes CuS  tonnes Cu
i.e tonnes CuS  x = tonnes Cu
= 75.2%
% Yield Cu =
4.80 x 100
6.38

10. The End