Chemistry Stoichiometry

1. Chemistry State Standards in Chapter 9
• C5.2A Balance simple chemical equations
applying the conservation of matter.
• C5.2B Distinguish between chemical and
physical changes in terms of the properties of
the reactants and products.
• C5.2d Calculate the mass of a particular
compound formed from the masses of starting
materials.
• C5.2e Identify the limiting reagent when given
the masses of more than one reactant.

2. Section 9.1: Using Chemical Equations
Because atoms are just rearranged (not created
or destroyed) in a chemical reaction, we must
always balance a chemical equation by
choosing the correct coefficients.
It is important to recognize that the coefficients in
a balanced chemical equation give the relative
numbers of molecules, moles, or dozens of
molecules, etc.
Chapter 9: Chemical Quantities

3. Mole Ratio – the ratio of moles of one
substance to moles of another substance in
the same balanced chemical equation.
OR you could say it is the number of moles of
one substance divided by the number of
moles of another substance in the same
balanced chemical equation.

4. For instance, the following eqn:
2 H2 + O2 → 2 H2O
means 2 moles H2 + 1 mole O2 → 2 moles H2O
or
2 molecules H2 + 1 molecule O2 → 2 molecules H2O
There are many different mole ratios possible
from each eqn, some include:
2 moles H2 1 mole O2 2 moles H2O etc.
________________________ _______________________ _______________________
1 mole O2 2 moles H2 2 moles H2

5. A balanced chemical eqn verifies the law of conservation of
matter.
To prove this we can look an example:
2 H2 + O2 → 2 H2O
To compare the masses of the reactant to the mass of the products, we
could write:
2 moles H2 2.0158 g H2 1 mole O2 31.998 g O2
______________ ________________ ______________ _________________
1 1 mole H2 1 1 mole O2
2 moles H2O 18.0148 g H2O
_________________ _____________________
1 1 mole H2O
4.0316 g + 31.998 g = 36.0296 g
36.0296 g = 36.0296 g
Therefore matter (or mass) is conserved!
+ →x x
x
( () )
)(

6. Mole–mole problems – converting moles of one
substance to moles of another (this is what
we will do in 9-1 Practice Problems).
The major types of stoichiometry problems are
mass–mass, mass–volume, and
volume–volume. (Note: There are other
types, but these are the major ones).

7. Section 9.2: Solving Stoichiometry
Problems
Stoichiometry – is the process of using a balanced
chemical equation to determine the relative masses of
reactants and products involved in a reaction.
Mass-mass problems –
you are given the mass of one substance and
are asked to find the mass of another substance
involved in the same chemical reaction.

8. Calculating Masses of Reactants and Products
for mass–mass problems
1.1. Balance the chemical equation.Balance the chemical equation.
2.2. Convert mass of given substance to moles (theConvert mass of given substance to moles (the
given could be a reactant or a product).given could be a reactant or a product).
3.3. Use proper mole ratio to convert fromUse proper mole ratio to convert from
moles of “given” substance to moles of “wanted”moles of “given” substance to moles of “wanted”
substance.substance.
4.4. Then convert moles of “wanted” substance toThen convert moles of “wanted” substance to
grams of that substance.grams of that substance.

9. After you balance the chemical reaction and have
identified the “given” and the “wanted” substances,
you can use the following template to complete the
calculation.
g “given” 1 mole “given” # mol “wanted” # g “wanted” =
_____________________ ___________________________________ ______________________________________ _________________________________
# g “given” # mol “given” 1 mol “wanted"
g “given” ÷ molar mass x molar x molar mass = final
of “given” ratio of “wanted” answer

10. Working a Stoichiometry ProblemWorking a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
1. Identify reactants and products and write
the balanced equation.
Al + O2 Al2O3
b. What are the reactants?
a. Every reaction needs an arrow.
c. What are the products?
d. What are the balanced coefficients?
4 3 2

11. Working a Stoichiometry ProblemWorking a Stoichiometry Problem
6.50 grams of aluminum reacts with an excess of
oxygen. How many grams of aluminum oxide are
formed?
4 Al + 3 O2 → 2Al2O3
=
6.50 g Al
? g Al2O3
1 mol Al
26.98 g Al 4 mol Al
2 mol Al2O3
1 mol Al2O3
101.96 g Al2O3
6.50 x 2 x 101.96 ÷ 26.98 ÷ 4 = 12.3 g Al2O3