Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction based on the law of conservation of mass. Balanced chemical equations show the mole ratios between substances reacting and forming. These mole ratios allow for conversions between amounts in moles and masses. Stoichiometric calculations can determine limiting reactants, theoretical and actual product yields, and masses of substances involved in a reaction. Percent yield compares the actual product amount to the theoretical maximum.

1. Stoichiometry Notes
Chemistry

2. Stoichiometry
■ Stoichiometry is the quantitative relationship
between the amount of reactants used and the
amount of products formed.
■ It is based on the law of conservation of mass.
– The law states that matter is neither created nor
destroyed in a chemical reaction.
– Meaning that the mass of the reactants MUST equal
the mass of the products.

3. Using Balanced Equations
■ Remember that the coefficients of
reactants and products can represent # of
atoms as well as the # of moles that
reacted.
■ Moles do not directly tell us the mass of
the reactants; however, remember that
the number of moles can be converted
into grams.

4. Relationship from balanced equations
IRON + OXYGEN → Iron (III) Oxide
4Fe + 3O2 → 2Fe2O3
4 atoms Fe + 3 molecules O2 → 2 formula units of
Fe2O3
4 moles Fe + 3 moles O2 → 2 moles Fe2O3
223.4 g Fe + 96.0 g O2 → 319.4 g Fe2O3
319.4 g reactants 319.4 g products

5. Mole Ratios
■ You can use the coefficients in a
balanced chemical equation to write
mole ratios.
■ A mole ratio is a ratio between the
number of moles of any 2 substances in
a chemical equation.
2H2O2 → O2 + 2H2O
Ex: 2 mol H2O2
1 mol O2
Ex: 1 mol O2
2 mol H2O

6. Stoichiometric Conversions
■ Mole to Mole
■ Mole to Mass
■ Mass to Mass

7. Stoichiometric Calculations
Mole-to-Mole conversions
■ When determining the number of moles
produced from a reaction, you must first
begin with a balanced equation!
■ After writing your balanced equation, you
then need to identify the substance you
know and the one you need to determine.
■ Use mole ratios to determine the
substance that you need to find.

8. Stoichiometric Calculations cont.
Mole-to-Mole conversions
■ Example: A vigorous reaction between
potassium and water produces hydrogen gas
and potassium hydroxide.
■ Steps:
1. Write a balanced equation:
2K + 2H2O → 2KOH + H2
2. How many moles of hydrogen are produced if
0.0400 moles of potassium are used?
Moles of known x (mole ratio) = moles of unknown
0.0400 mol K x (1 mol H2/2 mol K) = 0.0200 mol H2
Hint: the mole ratio should be the moles of the
unknown over the moles of the known

9. Stoichiometric Calculations
Mole-to-Mass conversions
■ Now suppose you know the number of
moles of a reactant or product and want
to calculate the mass of another product
or reactant.
■ This is done the same way using mole
ratios from the equation to get the
number of moles of the unknown. Once
you have the number of moles, use the
molar mass to determine the number of
grams that were produced.

10. Stoichiometric Calculations cont.
Mole-to-Mass conversions
How many grams of glucose (C6H12O6) are produced
when 24.0 mol of carbon dioxide reacts in excess
water?
1. Determine number of moles of glucose produced
by the given amount of carbon dioxide.
24.0 mol CO2 X 1 mol C6H12O6 = 4.00 mol C6H12O6
6 mol CO2

11. Stoichiometric Calculations cont.
Mole-to-Mass conversions
2. Convert the number of moles of glucose to
grams of glucose by multiplying # moles by
molar mass.
4.00 mol C6H12O6 X 180.18 g C6H12O6 = 721 g C6H12O6
1 mol C6H12O6
721 grams of glucose is produced from 24.0 moles of carbon dioxide.

12. Stoichiometric Calculations
Mass-to-Mass conversions
■ Using the mole ratios and mole/mass
calculations, you can determine the needed
masses to react and the masses that will be
produced.
■ First – write the balanced chemical equation for
the reaction.
■ Then convert the mass of the reactant to moles.
■ Use the mole ratios to determine the moles of
the unknown you are looking for.
■ Then use molar mass to convert moles of the
unknown to mass of the unknown.

13. Stoichiometric Calculations cont’d
Mass-to-Mass conversions
How many grams of NaOH are needed to
completely react with 50.0 g of H2SO4 to form
Na2SO4 and H2O?
1. Write the balanced equation.
2NaOH + H2SO4 → Na2SO4 + 2H20

14. Stoichiometric Calculations cont’d
Mass-to-Mass conversions
2. Convert grams of sulfuric acid to moles of
NaOH
50 g H2SO4 X 1 mol H2SO4 = 0.510 mol H2SO4
98.09 g H2SO4
0.510 mol H2SO4 X 2 mol NaOH = 1.02 mol NaOH
1 mol H2SO4

15. Stoichiometric Calculations cont’d
Mass-to-Mass conversions
3. Calculate the mass of NaOH needed.
1.02 mol NaOH X 40.00 g NaOH = 40.8 g NaOH
1 mol NaOH
50 g of H2SO4 reacts completely with 40.8 g of NaOH

16. Limiting Reactants and
Percent Yield

17. Limiting Reactants
■ A chemical reaction will proceed until all of one
reactant is used up.
■ The reactant that is used up is the limiting
reactant.
■ The left over reactants are called excess
reactants.

18. Identifying the Limiting Reactant
■ You can determine the limiting reactant in
a reaction when given the mass of the
reactants.
Example: 40 g of sodium hydroxide reacts
with 60 g of sulfuric acid.
1. Write the balanced chemical equation:
2 NaOH + H2SO4 → Na2SO4 + 2 H2O

19. 2. Convert the mass of the reactants into actual
moles of reactants available.
40.0 g NaOH X 1 mol NaOH = 1.00 mol NaOH
40.0 g NaOH
60.0 g H2SO4 X 1 mol H2SO4 = .612 mol H2SO4
98.09 g H2SO4

20. 3. Compare the actual mole ratio calculated to
the mole ratio from the balanced chemical
equation.
Actual mole ratio: 1.00 mol NaOH : 0.612 mol H2SO4
Mole ratio from equation: 2 mol NaOH : 1 mol H2SO4
Divide ratio by 2: 1 mol NaOH : 0.5 mol H2SO4
All of the 1 mol of NaOH is used up (limiting reactant) and there
is an excess of H2SO4 (0.612 mol – 0.5 mol = .112 mol left
over)

21. Calculating mass of product
produced when given mass of
reactants.
Multiply the number of moles of the limiting
reactant by the mole ratio of the product to the
limiting reactant. Then multiply by the molar
mass of the product.
1.00 mol NaOH x 1 mol Na2SO4 x 142.04 g Na2SO4 = 71.0 g Na2SO4
2 mol NaOH 1 mol Na2SO4

22. Calculating Percent Yield
■ Balance the equation
■ Do a mass → mass conversion between
the known and unknown
■ Use the given “actual yield” to set up the
ratio:
– Actual yield is the amount of product that
was ACTUALLY produced (found in an
experiment)
– Theoretical yield is the MAXIMUM amount
of a reactant produced (from your
stoichiometric calculations)
■ Percent yield=(actual/theoretical) x 100

23. Calculating Percent Yield
Ex: Aspirin (C9H8O4) can be made from salicylic
acid (C7H6O3) and acetic anhydride (C4H6O3).
What is the percent yield of aspirin when you
mix 13.2 g salicylic acid with an excess of
acetic anhydride and obtain 5.9 g of aspirin
and some water?
Actual yield of aspirin = 5.9 g
1. Write the balanced equation.
2 C7H6O3 + C4H6O3 → 2 C9H8O4 + H20

24. 2. Calculate the theoretical yield. Salicylic acid is the
limiting reactant.
13.2 g C7H6O3 X 1 mol C7H6O3 = 0.0956 mol C7H6O3
138.1 g C7H6O3
0.0956 mol C7H6O3 X 2 mol C9H8O4 = 0.0956 mol C9H8O4
2 mol C7H6O3
0.0956 mol C9H8O4 X 180.2 g C9H8O4 = 17.2 g C9H8O4
1 mol C9H8O4

25. Calculate the Percent Yield
■ Percent yield = 5.9 g C9H8O4 X 100 = 34%
17.2 g C9H8O4