1. Calculations Based on Chemical Equations
STOICHIOMETRY
Calculations Based on
Chemical Equations

2. Iron (III) oxide reacts with carbon monoxide to form iron and carbon dioxide.
Fe2O3 + CO  Fe + CO2 3 2 3
How many CO molecules are required to react with 25 particles of Fe2O3?

3. Iron (III) Oxide reacts with carbon monoxide to form iron and carbon dioxide.
Fe2O3 + CO  Fe + CO2 3 2 3
How many iron atoms can be produced by the reaction of 2.5 x 105 particles of Fe2O3?

4. STOICHIOMETRIC RELATIONSHIPS
Formulas can also represent MOLES of substances involved in chemical reactions.
Equations define reaction ratios, i.e. the molar ratios of reactants and products

5. What mass of CO is required to react with 146 grams of iron (III) oxide?
Fe2O3 + 3CO  2Fe + 3CO2

6. What mass of iron (III) oxide is required to produce 8
What mass of iron (III) oxide is required to produce 8.65 grams of carbon dioxide?
Fe2O3 + 3CO  2Fe + 3CO2

7. Limiting Reactants (Reagents)
and Percent Yield

8. Calculations need to be based on the limiting reactant.
Example 1: Suppose a box contains 87 bolts, 110 washers and 99 nails. How many sets of 1 bolt, 2 washers and 1 nail can you use to create? What is the limiting factor?
55 sets; washers limit the amount

9. Calculations need to be based on the limiting reactant.
Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?
CS2 + O2  CO2 + SO2
Start by balancing the equation…

10. Calculations need to be based on the limiting reactant.
Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?
CS2 + 3O2  CO2 + 2SO2
Now solve the problem…

11. Which reactant is LIMITING?
Example 2: What is the maximum mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?
CS2 + 3O2  CO2 + 2SO2
Which reactant is LIMITING?

12. O2 limits the amount of SO2 that can be produced. CS2 is in excess.
Example 2: What is the mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?
CS2 + 3O2  CO2 + 2SO2
O2 limits the amount of SO2 that can be produced. CS2 is in excess.

13. 134 g of SO2 can be produced in this reaction.
Example 2: What is the mass of sulfur dioxide that can be produced by the reaction of 95.6 g carbon disulfide with 100. g oxygen?
CS2 + 3O2  CO2 + 2SO2
134 g of SO2 can be produced in this reaction.

14. Calculations need to be based on the limiting reactant.
Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2?
CH4 + O2  CO2 + H2O
Start by balancing the equation…

15. Calculations need to be based on the limiting reactant.
Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2  CO2 + 2H2O
Now solve the problem…

16. Which reactant is LIMITING?
Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2  CO2 + 2H2O
Which reactant is LIMITING?

17. CH4 limits the amount of CO2 that can be produced. O2 is in excess.
Example 3: What mass of CO2 could be formed by the reaction of 8.0 g CH4 with 48 g O2?
CH4 + 2O2  CO2 + 2H2O
CH4 limits the amount of CO2 that can be produced. O2 is in excess.

18. Many chemical reactions do not go to completion (reactants are not completely converted to products). Percent Yield: indicates what percentage of a desired product is obtained.

19. So far, the masses we have calculated from chemical equations were based on the assumption that each reaction occurred 100%.
The THEORETICAL YIELD is the yield calculated from the balance equation.
The ACTUAL YIELD is the amount “actually” obtained in an experiment.

20. Look back at Example 2. We found that 134 g of SO2 could be formed from the reactants.
In an experiment, you formed 130 g of SO2. What is your percent yield?

21. CH3COOH + C2H5OH  CH3COOC2H5 + H2O
Example: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What percent yield of ethyl acetate is this?
CH3COOH + C2H5OH  CH3COOC2H5 + H2O

22. Example: A 10.0 g sample of ethanol, C2H5OH, was boiled with excess acetic acid, CH3COOH, to produce 14.8 g of ethyl acetate, CH3COOC2H5. What percent yield of ethyl acetate is this?