# Chapter 9 Stoichiometry

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```					 Chapter 9:
Stoichiometry
Coach Kelsoe
Chemistry
Pages 298–325
Section 9-1:
Introduction to Stoichiometry
Coach Kelsoe
Chemistry
Pages 299–301
Section 9-1 Objectives
n   Define stoichiometry
n   Describe the importance of the mole ratio in
stoichiometric calculations.
n   Write a mole ratio relating two substances in
a chemical equation
Stoichi-what?
n   Stoichiometry (pronounced “sto-key-ometry”)
comes from two Greek words:
n Stoicheion means “element”
n Metron means “measure”
Introduction to Stoichiometry
n   A lot of our knowledge of chemistry is based
on the careful quantitative analysis of
substances involved in chemical reactions.
n   Composition stoichiometry deals with the
mass relationships of elements in compounds.
n   Reaction stoichiometry involves the mass
relationships between reactants and products
in a chemical reaction.
Introduction to Stoichiometry
n   The good news: this is not entirely new
composition stoichiometry in Section 3-2. We
just never called it that!
n   Reaction stoichiometry is based on chemical
equations and the law of conservation of
matter.
n   ALL reaction stoichiometry calculations start
with a balanced equation.
Reaction-Stoichiometry Problems
n   The problems in this chapter can be classified
according to the information given in the problem
and the information you are expected to find, called
the unknown.
n   The given and unknown may both be reactants,
they may both be products, or one may be a
reactant and the other a product.
n   Masses should be expressed in grams.
n   Stoichiometric problems are solved by using ratios
from the balanced equation to convert the given
quantity using the following methods.
Reaction-Stoichiometry Problems
n   Problem Type 1:
Given and unknown quantities are in moles.
n When you are given the amount of substance in
moles and asked to calculate the amount in moles
of another substance in the chemical reaction, the
general plan is
n Amount of given               Amount of unknown
substance (in mol)      substance (in mol)
Reaction-Stoichiometry Problems
n   Problem Type 2:
Given is an amount in moles and the
unknown is a mass that is often expressed in
grams.
n   When you are given the amount in moles of one
substance and asked to calculate the mass of
another substance in the chemical reaction, the
general plan is
n   Amount of            Amount of        Mass of unknown
given substance   unknown substance     substance
(in mol)              (in mol)          (in grams)
Reaction-Stoichiometry Problems
n   Problem Type 3:
Given is a mass in grams and the unknown is
an amount in moles.
n   When you are given the mass of one substance
and asked to calculate the amount in moles of
another substance in the chemical reaction, the
general plan is
n   Mass of              Amount of      Amount of unknown
given substance   given substance       substance
(in grams)            (in mol)           (in mol)
Reaction-Stoichiometry Problems
n   Problem Type 4:
Given is a mass in grams and the unknown is
a mass in grams.
n   When you are given the mass of one substance
and asked to calculate the mass of another
substance in a chemical reaction, the general plan
is
n   Mass of      Amount of   Amount of    Mass of
given          given     unknown     unknown
substance    subtance    substance   substance
(in grams)   (in mol)     (in mol)   (in grams)
Mole Ratio
n   Solving any reaction-stoichiometry problem
requires the use of a mole ratio to convert
from moles or grams of one substance in a
reaction to moles or grams of another
substance.
n   A mole ratio is a conversion factor that
relates the amounts in moles of any two
substances involved in a chemical reaction. It
is obtained directly from the balanced
chemical equation.
Mole Ratio
n   Consider the following equation:
2Al2O3(l) à 4Al(s) + 3O2(g)
n   Remember that the coefficients in a chemical
equations satisfy the law of conservation of
matter and represent the relative amounts of
moles of reactants and products.
n   Therefore, 2 mol of aluminum oxide
decompose to produce 4 mol of aluminum
and 3 mol of oxygen gas. See board for the
six ratios.
Mole Ratio
n   For the decomposition of aluminum oxide, the
appropriate mole ratio would be used as a
conversion factor to convert a given amount in
moles of one substance to the corresponding
amount in moles of another substance.
n   For example, if given 13 mol of aluminum oxide
and need to find out how much oxygen it
decomposes into:

13.0 mol Al2O3 x 3 mol O2 = 19.5 mol O2
2 mol Al2O3
Mole Ratio
sandwich:
4 slices of bread + 1 cup PB + 1 cup J à 2 PB&J sandwiches
n   Let’s assume this equation is balanced.
n   If we have 16 slices of bread, how much PB
will we need?
n   16 bread x 1 cup PB = 4 cups PB
n   We can do this for any ingredient!
Mole Ratio
n   Mole ratios are exact, so they do not limit the
number of significant figures in a calculation.
n   The number of significant figures in the
answer is therefore determined only by the
number of significant figures of any measured
quantities in a particular problem.
n   In layman’s terms, you should have the same
number of sig figs when you multiply by your
Molar Mass
n   Remember that the molar mass is the mass,
in grams, of one mole of a substance.
n   The molar mass is the conversion factor that
relates the mass of a substance to the
amount in moles of that substance.
n   To solve reaction-stoichiometry problems, you
will need to determine the molar masses
using the periodic table.
Molar Mass
n   Returning to the aluminum oxide problem:
2Al2O3(l) à 4Al(s) + 3O2(g)
n   The molar masses can be expressed as so:
Aluminum      101.96 g          1 mol Al2O3
n
1 mol Al2O3   OR    101.96 g
oxide

n   Aluminum      26.98 g      OR   1 mol Al
1 mol Al          26.98 g

32.00 g      OR   1 mol O2
n   Oxygen
1 mol O2           32.00 g
Molar Mass
n   To find the number of grams of aluminum
equivalent to 26.0 mol of aluminum, the
calculation would be as follows:
n    26.0 mol Al x 26.98 g Al = 701 g Al
1 mol Al
Vocabulary
n   Composition stoichiometry
n   Mole ratio
n   Reaction stoichiometry

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