1. Chapter 9: Stoichiometry

2. Objective: To perform mole to mole conversion problems.

3. Stoichiometry
Stoichiometry is the branch of chemistry that deals with quantities of substances in chemical reactions.

4. Mole to Mole Problems: The Steps
1. Write chemical equation 2. Balance chemical equation using coefficients 3. Use the following setup to perform calculation: A is the known quantity B is the unknown quantity 4. Don’t forget units on your final answer!!
Mole Ratio

5. 9.2 Mole to Mass and Mass to Mole
Objective: To perform mole to mass and mass to mole conversion problems.

6. Mole to Mass Molar Mass Unknown Moles Given Unknown Moles (Eqn.)
Grams Unknown
Given Moles (Eqn.)
1 mole Unknown
Mole Ratio

7. Mass to Mole Unknown Moles (Eqn.) Grams Given 1 mol Given
Moles Unknown
Molar Mass Given
Given
Moles (Eqn.)
Mole Ratio

8. 9.3 Mass to Mass
Objective: To perform mass to mass conversion problems.

9. Stoichiometry Roadmap
Unknown Moles (Eqn.)
Molar Mass Unknown
Grams Given
1 mol Given
Molar Mass Given
Given
Moles (Eqn.)
1 mole Unknown
Grams Unknown
Mole Ratio

10. 9.4 Limiting Reactant Objectives:
To calculate the theoretical yield of a chemical reactions.
To determine the limiting reactant and excess reactant in a chemical reaction.

11. Limiting Reactant
Any reactant that is used up first in a chemical reaction.
It determines the amount of product that can be formed in the reaction.

12. Excess Reactant
The reactant that is not completely used up in a reaction.

13. Limiting Reactant Problems
Use the mass to mass conversions
Grams Given
1 mol Given
Unknown Moles
Molar Mass of Unknown
Molar Mass Given
Given Moles
1 mol of Unknown

14. Example
Copper reacts with sulfur to form copper(I) sulfide according to the following balanced equation:
2Cu + S  Cu2S
What is the limiting reactant when 80.0 grams of Cu reacts with 25.0 grams of S?

15. Example 2Cu + S  Cu2S The general equation is: Start with Copper:
Grams Given
1 mol Given
Unknown Moles
Molar Mass of Unknown
Molar Mass Given
Given Moles
1 mol of Unknown
Start with Copper:
80.0 g Cu
1 mol Cu
1 mol Cu2S
g Cu2S
63.55 g Cu
2 mol Cu
1 mol of Cu2S
= g Cu2S
Now use Sulfur:
25.0 g S
1 mol S
1 mol Cu2S
g Cu2S
32.07 g S
1 mol of Cu2S
= g Cu2S

16. Example The limiting reactant is copper.
The excess reactant is sulfur.
The amount of Cu2S that is produced is g Cu2S.

17. 9.5 Percent Yield
Objective: To calculate percent yield.

18. Introduction to Percent Yield…
If you get 15 out of 20 questions correct on a test, what percentage did you receive on the test? How did you figure this out?
15/20 x 100 = 75%
If Sammy Sosa gets 25 hits in the month of May and has 113 bats, what is his batting average for the month of May? Explain how you arrived at your answer.
25/113 = 0.221
As a percentage, this is written, 25/113 x 100 = 22.1%

19. · It is a measure of efficiency of a reaction.
Percent Yield
·Percent yield is the ratio of the actual yield to the theoretical yield for a chemical reaction expressed as a percentage.
· It is a measure of efficiency of a reaction.
Percent Yield = actual yield x 100%
theoretical yield

20. Actual Yield
The amount of product that forms when a reaction is carried out in the laboratory.

21. Theoretical Yield
The amount of product that could form during a reaction calculated from a balanced chemical equation.
It represents the maximum amount of product that could be formed from a given amount of reactant.

22. Example #1
The equation for the complete combustion of ethene (C2H2) is
2C2H O2  4CO H2O
If 0.10 g of C2H2 is reacted with g of O2, identify the limiting reactant.
What is the theoretical yield of H2O?
If the actual yield of H2O is 0.05 g, calculate the percent yield.

23. Example #1 2C2H2 + 5O2  4CO2 + 2H2O Start with C2H2:
0.10 g C2H2
1 mol C2H2
2 mol H2O
18.00 g H2O
26.02 g C2H2
2 mol C2H2
1 mol H2O
= 0.07 g H2O
Theoretical Yield
Limiting Reactant = C2H2
Now start with O2:
g O2
1 mol O2
2 mol H2O
18.00 g H2O
32.00 g O2
5 mol O2
1 mol H2O
= g H2O

24. Example #1 2C2H2 + 5O2  4CO2 + 2H2O Actual Yield = 0.05 g H2O
Theoretical Yield = 0.07 g H2O
Percent Yield = actual yield x 100%
theoretical yield
Percent Yield = g H2O x 100% = %
0.07 g H2O

25. Example #2
Determine the percent yield for the reaction between 3.74 g of Na and excess O2 if 5.34 g of Na2O2 is recovered?
First, write the chemical equation:
Na + O2  Na2O2
Second, balance the chemical equation:
2Na + O2  Na2O2

26. Example #2 2Na + O2  Na2O2 3.74 g Na 1 mol Na 1 mol Na2O2
Solve the mass-mass problem, starting with Na:
Actual Yield = 5.34 g Na2O2
Theoretical Yield = 6.34 g Na2O2
Percent Yield = actual yield x 100%
theoretical yield
Percent Yield = g x 100% = 84.23%
6.34 g
3.74 g Na
1 mol Na
1 mol Na2O2
77.98 g Na2O2
22.99 g Na
2 mol Na
= 6.34 g Na2O2