1. Percentage Yield

2. What is Yield?
It’s the amount of product that is produced in a chemical reaction  actual yield.
What you get in a chemical reaction may not be what you were supposed to get. The theoretical yield tells you how much product you should get according to the balanced equation.

3. Factors affecting Yield
Splattering when heating a solution
Filtering may not get all of the product
Using impure/contaminated reactants
Not letting wet solids dry properly
Use of equipment (old or faulty)
Side chemical reactions (eg. When you burn Mg in the air. It reacts with O2. However, air also has nitrogen, so Mg may react with N2 too!)

4. Percentage Yield
The comparison between the actual yield and theoretical yield of a product in a reaction.
Percentage yield = Actual Yield x 100%
Theoretical Yield

5. Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)
Example 1
Zinc reacts with hydrochloric acid to produce zinc chloride and hydrogen gas, as seen below:
Zn (s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g)
Calculate the theoretical yield and percentage yield of ZnCl2 if g of ZnCl2 (aq) is produced when g of HCl (aq) is used.

6. 2. Find # moles of product, using mole ratio
Balanced Eq’n
Zn (s) HCl (aq) → ZnCl2 (aq) H2 (g)
Given Mass (g)
Molar Mass (g/mol)
0.999 g
1.541 g (actual)
36.46 g/mol
g/mol
1. Find # moles reactant
nHCl = g = mol
36.46 g/mol
2. Find # moles of product, using mole ratio
nZnCl2 = mol HCl x __________ =
1 mol ZnCl2
2 mol HCl
mol

7. 3. Find mass of product
mZnCl2 = n x M
= mol x g/mol
= 1.86 g  theoretical yield
4. Calculate % yield
% yield = actual yield x 100%
theoretical yield
= g x 100%
1.86 g
= 82.8%

8. Example 2
In an experiment, 16.1 g of iron sulfide, FeS (s), is added to oxygen g of iron(III) oxide, Fe2O3 (s) is produced. Calculate the theoretical yield and percentage yield of Fe2O3 (s).
4FeS (s) + 7O2 (g) → 2Fe2O3 (s) + 4SO2 (g)

9. 2. Find # moles of product, using mole ratio
Balanced Eq’n
4FeS (s) O2 (g) → Fe2O3 (s) SO2 (g)
Given Mass (g)
Molar Mass (g/mol)
16.1 g
14.1 g (actual)
87.92 g/mol
159.7 g/mol
1. Find # moles reactant
nFeS = g = mol
87.92 g/mol
2. Find # moles of product, using mole ratio
nFe2O3 = mol FeS x __________ =
2 mol Fe2O3
4 mol FeS
mol

10. 3. Find mass of product
mFe2O3 = n x M
= mol x g/mol
= g  theoretical yield
4. Calculate % yield
% yield = actual yield x 100%
theoretical yield
= 14.1 g x 100%
14.61 g
= 96.5%