% Yield
Definitions
1. Theoretical Yield
The maximum amount of product that
can be formed from a given amount of
reactants
In other words, the calculated amount
predicted through stoichiometry
% Yield
Definitions
Actual Yield
The amount that is actually formed
when the reaction is carried out in the
laboratory.
% Yield =
actual yield X 100
theoretical yield
% yield will never be over 100%
Most likely it will never even be 100%
Why will % yield never be 100%
 Advantageous to add an excess of an inexpensive reagent
to ensure that all of the more expensive reagents react
 Reactant may not be 100% pure
 Materials are lost during the reaction
If a reactions takes place in a solution it may be
impossible to get all of the reactants or products out of
the solution
 If the reactions takes place at a high temperature,
materials may be vaporized and escape into the air
 Side reactions may occur
Example Mg burned in air. Some of Mg reacts with
nitrogen reducing the amount of MgO produced.
 Loss of product when filtering or transferring
 If reactants are not carefully measured
% yield example problem
In a reaction between barium chloride
and potassium sulfate, 3.89 g of barium
sulfate is produced from 3.75 g of
barium chloride. What is the percent
yield?
% yield example problem
answer
BaCl2 + K2SO4 --> BaSO4 + 2 KCl
3. 75 g BaCl2 1 mol BaCl2
1 mol BaSO4 233.4 g BaSO4
208.3 g BaCl2 1 mol BaCl2
1 mol BaSO4
= 4.20 g BaSO4
3.89 g BaSO4 x 100 = 92.6 %
4.20 g BaSO4
% yield example problem
your turn
13.35 grams of magnesium hydroxide is
produced when 42.50 grams of
magnesium nitrate reacts with an
excess of aluminum hydroxide. What is
the percent yield?
% yield example problem
answer
3 Mg(NO3)2 + 2 Al(OH)3 --> 3 Mg(OH)2 + 2 Al(NO3)3
42.50g Mg(NO3)2 1mol Mg(NO3)2
3mol Mg(OH)2 58.3g Mg(OH)2
148.3g Mg(NO3)2 3mol Mg(NO3)2 1mol Mg(OH)2
= 16.71 g Mg(OH)2
% yield = 13.35 g Mg(OH)2 x 100 = 79.89 % Mg(OH)2
16.71 g Mg(OH)2