1. Percentage Yield

2. Theoretical Yield – amount of product that is predicted using stoichiometry
Actual Yield – amount of product that is obtained in an experiment
Percent Yield – compares the mass of product obtained by experiment with the mass of product determined by stoichiometric calculations

3. Reasons For Low Yield Sources of error. (Experimental procedures)
Impurities in reagents used (different grades of chemicals)
Side reactions (other products formed)
Reactions are reversible

4. Reasons for High Yield
1. Sources of error (experimental procedures)

5. Solving Percent Yield Problems
First, determine the theoretical yield (how much you should get) using stoichiometry
Second using the experimental/ actual yield, use the formula.
% yield = Experimental yield x 100%
theoretical yield

6. Example 1: In a particular experiment 10
Example 1: In a particular experiment 10.0 g of sugar should be produced but only g is produced. What is the percentage yield?
% yield = x 100
= x 100
= 6.64 %
Experimental yield
theoretical yield
0.664 g
10.0g

7. Step 1: Write out the balanced chemical equation
Example 2: Aluminium reacts with oxygen to form aluminium oxide. If 635g of Aluminium oxide is obtained from reacting 1150g of aluminium, what is the percentage yield?
Step 1: Write out the balanced chemical equation
Al + O2  Al2O3
4Al + 3O2  2Al2O3

8. Step 2: Fill in chart with information you know
Balanced
Equation
4Al
3O2
2Al2O3
Mole Ratio 4 3 2
Mass (m)
1150g
Molar Mass (M)
26.98g/mol
Moles (n)

9. Step 3: Convert given mass into moles (n=m/M)
Part 1: Find the theoretical yield through stoichmetry (how much you should have got)
Step 3: Convert given mass into moles (n=m/M)
Moles of Al = 1150g
27.0g/mol
= mol of Al

10. 4Al 3O2 2Al2O3 4 3 2 1150g Fill in chart with information you know
Balanced
Equation
4Al
3O2
2Al2O3
Mole Ratio 4 3 2
Mass (m)
1150g
Molar Mass (M)
26.98g/mol
Moles (n)
42.59 mol

11. Step 4: Use moles of given substance to find moles of required substance (mole to mole ratio)
4Al + 3O2  2Al2O3
x mols of Al2O = 2 mol of Al2O mol of Al mol of Al
= mols of Al2O3 or
42.59 mols of Al mol of Al2O3
4 mol Al X
= mols of Al2O3

12. 4Al 3O2 2Al2O3 4 3 2 1150g Fill in chart with information you know
Balanced
Equation
4Al
3O2
2Al2O3
Mole Ratio 4 3 2
Mass (m)
1150g
Molar Mass (M)
26.98g/mol
102g/mol
Moles (n)
42.59 mol
21.31 mol

13. Step 5: Convert moles of required substance to the mass (m=n x M)
mass of Al2O3 = mol of Al2O3 X 102g/mol
= 2173 g of Al2O3
This is your theoretical yield

14. Part 2: Calculate Percent Yield
Step 6: Using experimental and theoretical yield find percent yield
% yield = x 100
= x 100
= 29.2 %
Experimental yield
theoretical yield
635 g
2173g

15. Step 1: Write out the balanced chemical equation
Example 3:
Iron (III) oxide reacts with carbon monoxide to produce carbon dioxide and iron. If 300g of iron is produced when 425 of iron ore is used. What is the percentage yield?
Step 1: Write out the balanced chemical equation
Fe2O CO  2Fe + 3CO2
m= 425g
m= ?

16. Step 2: Fill in chart with information you know
Balanced
Equation
Fe2O3
3CO
2Fe
3CO2
Mole Ratio 1 3 2
Mass (m)
425g
? g
Molar Mass (M)
159.7 g/mol
Moles (n)

17. Step 3: Convert given mass into moles (n=m/M)
Part 1: Find the theoretical yield through stoichmetry (how much you should have got)
Step 3: Convert given mass into moles (n=m/M)
Moles of Fe2O3 = 425g
159.7g/mol
= 2.66 mol of Fe2O3

18. Fe2O3 3CO 2Fe 3CO2 425g Fill in chart with information Balanced
Equation
Fe2O3
3CO
2Fe
3CO2
Mole Ratio 1 3 2
Mass (m)
425g
? g
Molar Mass (M)
159.7 g/mol
Moles (n)
2.66mol

19. = 5.32 mols of Fe or 1 mols Fe2O3 X = 5.32 mols of Fe
Step 4: Calculate the number of mols of the required substance (mol to mole ratio)
Fe2O CO  2Fe + 3CO2
x mols of Fe = 2 mol of Fe mols of Fe2O mol of Fe2O3
= 5.32 mols of Fe or
2.66 mols of Fe2O mol of Fe
1 mols Fe2O3 X
= 5.32 mols of Fe

20. Fe2O3 3CO 2Fe 3CO2 425g Fill in chart with information Balanced
Equation
Fe2O3
3CO
2Fe
3CO2
Mole Ratio 1 3 2
Mass (m)
425g
? g
Molar Mass (M)
159.7 g/mol
55.85 g/mol
Moles (n)
2.66mol
5.32 mol

21. Step 5: Convert moles of required substance to the mass
mass of Fe = 5.32 mols of Fe X 55.85g/mol of Fe
= 297g of Fe

22. Part 2: use percent yield formula
Step 6: Using experimental and theoretical yield find percent yield
% yield = x 100
= x 100
= 101 %
Experimental yield
theoretical yield
300 g
297g