Stoichiometry Mole Relationships
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1
2SCH 3U
Name: _________________________
Stoichiometry
Mole Relationships in Chemical Reactions
The quantitative relationship among reactants and products is called stoichiometry. The term stoichiometry is
derived from two Greek words: stoicheion (meaning "element") and metron (meaning "measure"). Stoichiometry
calculations are based on the fact that atoms are conserved. They cannot be destroyed or created. Numbers and kinds
of atoms before and after the reactions are always the same. This is the Law of Conservation of mass and is why
chemical reactions must be balanced. The molar coefficients from a balanced chemical equation, represent the ratio in
which reactants come together and make products in terms of moles.
2 H
For example, in the reaction 2 H
+ O
O
one mole of oxygen gas reacts with two moles of hydrogen
2(g)
2(g)
2
(g)
gas to make 2 moles of water.
In most stoichiometry problems, amounts will be given in something other than moles and therefore you must
always convert to moles before working with the molar ratio given by the balanced chemical equation. The following
flow chart outlines the basic process to any stoichiometry calculation.
mass of unknown
mass of material
use the molar ratio in
the balanced chemical
concentration and
moles of
concentration and
moles of
equation to determine
volume of unknown
unknown
known
volume of material in
the number of moles of
material
material in solution
material
solution
the unknown material
P V T of a gaseous
P V T of a gaseous
unknown material
material
Steps to Solving a Stoichiometry Problem
1. Write a balanced chemical equation for the reaction.
2. Identify the known and unknown materials in the problem.
3. Calculate the number of moles of the known or given material.
4. Use the molar ratio from the balanced chemical equation to determine the number of moles of the unknown.
5. Convert the number of moles of the unknown material to whatever is asked for in the question.
Sample Problem: Calculate the mass of oxygen gas produced by the decomposition of 12.26 g of potassium chlorate into
potassium chloride and oxygen.
Solution:
2 KCl + 3 O
2 KClO
3
2
12.26 g
?
(
)
Therefore 4.800 g of oxygen gas are produced.
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