Chapter 3: Stoichiometry
• Stoichiometry - The study of
quantities of materials
consumed and produced in
chemical reactions.
• In order to understand
stoichiometry, we must first
understand atomic masses
Atomic Masses
• Elements occur in nature as
mixtures of isotopes
• Carbon = 98.89% 12C
1.11% 13C
<0.01% 14C Carbon
atomic mass = 12.01 amu
Try this…
Naturally occurring lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
• Find the average atomic mass
of lithium.
7.42(6.015 )  92.58  7.016 
100
 6 .9 4 1a m u
How about this one?
Fill in the blanks.
Isotope
28
Si
29
Si
32
Si
Mass (amu)
Abundance
27.98
29.97
4.70%
3.09%
The mole
• A unit to count the number of
molecules
Dozen = 12
Pair = 2
• 1 mol = 6.022 x 1023
Samples of 1 mole of
Cu, Al, Fe, S, I, and Hg
eggs
Molar mass is the mass of 1 mole of shoes in grams
marbles
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
1 mole 12C atoms = 12.00 g 12C
1 mole lithium atoms = 6.941 g of Li
For any element
atomic mass (amu) = molar mass (grams)
Try this!
Which of the following 100.0 g
samples contains the greatest
number of atoms?
a) Magnesium
b) Zinc
c) Silver
Rank the following according to number of
atoms (greatest to least):
a) 100.0 g of silver
b) 62.0 g of zinc
c) 21.0 g of magnesium
Molar mass
• My expectation is that you do
NOT need a refresher in molar
mass! So try this!!!
• What is the molar mass of
isopentyl acetate (C7H14O2)?
If 1 g of C7H14O2 is released in a bee sting,
how many molecules are released?
• HINT:
– g = 1x10-6g
– g  mol  molecules
• How many atoms of carbon are
present?
How many C atoms in bee sting?
• HINT: each molecule of
isopentyl acetate has 7 atoms
of carbon
Percent Composition or Mass %
• There are two ways to describe
the composition of a
compound.
– The number of constituents
atoms
– The percent of each of its
elements (by mass)
isopentyl acetate (C7H14O2)
• Let’s look at mass percent of
an element
m ass% 
n  m olar m ass of elem ent
 100%
m olar m ass of com poun d
• Find the mass % of C
m a ss% 
7  1 2 .0 1
 100%
1 3 0 .1 8
m a ss% o f C  6 4 .5 8 %
Try this one.
Consider separate 100.0 gram
samples of each of the
following:
H2O, N2O, C3H6O2, CO2
• Rank them from highest to
lowest percent oxygen by
mass.
Formulas
• molecular formula = (empirical
formula)n
[n = integer]
• molecular formula = C6H6 = (CH)6
• empirical formula = CH
– The lowest ratio of atoms in a
molecule (mole ratio)
Working backwards
• From the percent
composition, you can
determine the empirical
formula.
Determine the empirical formula of a compound
that has the following percent composition by
mass: K 24.75%, Mn 34.77% , O 40.51%
Assume 100 g sample of the compound
n K  24.75g 
n M n  34.77g 
1m ol K
 0.6330m olK
39.10g K
1m ol M n
 0.6329m olM n
54.94g M n
n O  40.51g 
1m ol O
16.00g O
 2.532m ol O
nK = 0.6330, nMn = 0.6329, nO = 2.532
K 
0 .6 3 3 0
 1 .0
0 .6 3 2 9
Mn 
0 .6 3 2 9
 1 .0
0 .6 3 2 9
O 
2 .5 3 2
 4 .0
0 .6 3 2 9
KMnO4
This one has a twist!
• A compound is made of only
sulfur and nitrogen. It is 69.6%
S by mass. Its molar mass is
184 g/mol. What is its
empirical and molecular
formula?
How to “Read” Chemical
Equations
2 Mg + O2  2 MgO
2 atoms Mg + 1 molecule O2 makes 2
formula units MgO
2 moles Mg + 1 mole O2 makes 2 moles
MgO
48.6 grams Mg + 32.0 grams O2 makes
80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
Things to know about equations
• They are sentences
• They describe what happens in
a chemical reaction
• Reactants  Products
• They should be balanced
• They have the same number of
each kind of atoms on both
sides
Try this
Chromium compounds exhibit a variety of bright
colors. When solid ammonium dichromate
(NH4)2Cr2O7, a vivid orange compound is ignited
a spectacular reaction occurs. Assume that the
products are chromium (III) oxide, nitrogen gas,
and water vapor. Write and balance the
equation.
And this one
• At 1000oC, ammonia gas, (NH3)
reacts with oxygen gas to form
gaseous nitric oxide, and water
vapor. This reaction is the first
step in the commercial production
of nitric acid by the Ostwald
process. Write and balance the
equation for this reaction.
Stoichiometry
How much do you remember?
• Methanol burns in air according to the
equation
2CH3OH + 3O2  2CO2 + 4H2O
If 209 g of methanol are used up in the
combustion, what mass of water is produced?
A good FRQ for class.
• Methane (CH4) reacts with the oxygen in
the air to produce carbon dioxide and
water.
• Ammonia (NH3) reacts with the oxygen
in the air to produce nitrogen monoxide
and water.
• What mass of ammonia would produce
the same amount of water as 1.00 g of
methane reacting with excess oxygen?
1.4g ammonia
Limiting Reactant
• Reactant that determines the
amount of product formed.
• The one you run out of first.
• Makes the least product.
• Three methods to solve:
– Ratio method shown in text
– Have versus need
– Do 2 stoichiometry problems to
find the amount of product. The
reactant that produces the LEAST
amount is the limiting reagent.
Let’s try together!
• Nitrogen gas can be prepared by passing
gaseous ammonia over solid copper(II) oxide
at high temperatures. The other products of
the reaction are solid copper and water vapor.
If a sample containing 18.1g of NH3 is reacted
with 90.4g of CuO, which is the limiting
reactant? How many grams of nitrogen gas will
be formed?
• Have versus need method:
g NH3mol NH3  mol CuO neededg CuO needed
2NH3 + 3CuO  N2 +3Cu + 3H2O
1 8 .1g N H 3 
1m o lN H 3
1 7 .0 4 g N H 3

3m o l C u O
2m o lN H 3

7 9 .5 5 g C u O
 127g CuO needed
1m o l C u O
• The problem says the we HAVE 90.4g of CuO.
• Using ALL of the NH3 given we NEED 127g of
CuO; therefore, the CuO is the limiting
reactant since we don’t have enough of CuO
• You can do this starting with the CuO as well.
Let’s double check our work.
2NH3 + 3CuO  N2 +3CuO + 3H2O
9 0 .4 g C u O 
1m o l C u O
7 9 .5 5 g C u O

2m o lN H 3
3m o l C u O

1 7 .0 4 g N H 3
1m o lN H 3
 1 2 .9 g N H 3 n e e d e d
• The problem says the we HAVE 18.1g of NH3.
• Using ALL of the CuO given we NEED 12.9g of
NH3; therefore, the CuO is the limiting reactant
since we have an excess of NH3.
• There is an excess of 5.2g of NH3
2NH3 + 3CuO  N2 +3CuO + 3H2O
• Do the stoichiometry problem twice. The
reactant that gives you the LEAST amount of
product is the limiting reactant.
9 0 .4 g C u O 
1 8 .1g N H 3 
1m o l C u O

1m o lN 2

2 8 .0 2g N 2
7 9 .5 5C u O
3m o l C u O
1m o lN 2
1m o lN H 3
1m o lN 2
2 8 .0 2g N 2
1 7 .0 4 g N H 3

2m o lN H 3

1m o lN 2
 1 0 .6 g N 2 m a d e
 1 4 .9 g N 2 m a d e
• The CuO is the limiting reactant. This confirms
our results from the have versus need method.
Try one on your own.
• When 124 grams of aluminum are
reacted with 601 grams of iron(III) oxide,
aluminum oxide and solid iron are
produced. Calculate the mass of
aluminum oxide formed.
Excess Reactant (reagent)
• The reactant you don’t run out of.
• The amount of stuff you make is the
yield.
• The theoretical yield is the amount you
would make if everything went perfect.
• The actual yield is what you make in the
lab.
Percent Yield
% yie ld 
a ctu a l
 100%
th e o re tica l
OR
% yie ld 
w h a t yo u g o t
w h a t yo u co u ld h a v e
 100%
Try this
Methanol is the simplest alcohol. It is used
as a fuel in race cars and is a potential
replacement for gasoline. Suppose
68.5kg CO(g) is reacted with 8.6kg H2(g).
Calculate the theoretical yield of
methanol.
If 3.57x104g CH3OH is actually produced,
what is the percent yield?