Basic Stoichiometry

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Basic Stoichiometry
Part One

• Pisgah High School
• M. Jones
• Revision history 5/16/03, 02/04/12, 04/27/12

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The word stoichiometry comes from the Greek words
stoicheion which means element and metron which
means measure.
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Stoichiometry deals with the amounts of reactants
and products in a chemical reaction.
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Stoichiometry deals with moles.
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Recall that
1 mole the molar mass
1 mole 22.4 L of any gas at STP
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The word mole is one that represents a very large
number.
Much like dozen means 12,
mole means 6.022 x 1023
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The key to doing stoichiometry is the balanced
chemical equation.
2 H2 O2 ? 2 H2O
2
2
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The coefficients give the relative number of
atoms or molecules of each reactant or product
as well as the number of moles.
2 H2 O2 ? 2 H2O
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2 H2 O2 ? 2 H2O
2 molecules of hydrogen
1 molecule of oxygen
2 molecules of water
Two molecules of hydrogen combine with one
molecule of oxygen to make two molecules of water.
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2 H2 O2 ? 2 H2O
2 molecules of hydrogen
1 molecule of oxygen
2 molecules of water
The balanced chemical equation also gives the
smallest integer number of moles.
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2 H2 O2 ? 2 H2O
2 moles of hydrogen
1 mole of oxygen
2 moles of water
The balanced chemical equation also gives the
smallest integer number of moles.
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2 H2 O2 ? 2 H2O
2 moles of hydrogen
1 mole of oxygen
2 moles of water
Two moles of hydrogen combine with one mole
of oxygen to make two moles of water.
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How can we show that this is really true?
Consider the combustion of hydrogen in oxygen
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2 H2 O2 ? 2 H2O
2 moles of hydrogen
1 mole of oxygen
2 moles of water
What do each of the reactants and product weigh?
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2 H2 O2 ? 2 H2O
2 moles of hydrogen
1 mole of oxygen
2 moles of water
2 x 2.0 g
1 x 32.0 g
2 x 18.0 g


4.0 g
32.0 g
36.0 g
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2 H2 O2 ? 2 H2O
2 moles of hydrogen
1 mole of oxygen
2 moles of water
The Law of Conservation of Matter


4.0 g
32.0 g
36.0 g
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2 H2 O2 ? 2 H2O
2 moles of hydrogen
1 mole of oxygen
2 moles of water
Matter can neither be created nor destroyed, only
changed in form.


4.0 g
32.0 g
36.0 g
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The oxidation of iron
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Consider the oxidation of iron
Fe(s) O2(g) ? Fe2O3(s)
4
3
2
4 moles
3 moles
2 moles
The coefficients give the ratio of moles.
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Consider the oxidation of iron
Fe(s) O2(g) ? Fe2O3(s)
4
3
2
If these react
then we have the following
4 moles
3 moles
2 moles
8 moles
6 moles
4 moles
2 moles
1.5 moles
1 mole
0.50 mol
0.375 mol
0.25 mol
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Consider the oxidation of iron
Fe(s) O2(g) ? Fe2O3(s)
4
3
2
The 0.375 moles was not as easy to predict.
0.375 mole O2
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Consider the oxidation of iron
Fe(s) O2(g) ? Fe2O3(s)
4
3
2
The 0.375 moles was not as easy to predict.
0.375 mole O2
Use a conversion factor to determine the number
of moles of an unknown.
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Consider the oxidation of iron
Fe(s) O2(g) ? Fe2O3(s)
4
3
2
0.375 mole O2
The conversion factor comes from the coefficients
in the balanced equation.
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Back to the oxidation of iron
Fe(s) O2(g) ? Fe2O3(s)
4
3
2
4 moles
3 moles
2 moles
Calculate the masses of these moles.
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Back to the oxidation of iron
Fe(s) O2(g) ? Fe2O3(s)
4
3
2
4 moles
3 moles
2 moles
2 x 159.6 g
4 x 55.8 g
3 x 32.0 g


319.2 g
96.0 g
223.2 g
Mass is conserved.
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The decomposition of ammonium carbonate
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Now consider the decomposition of solid ammonium
carbonate.
(NH4)2CO3 ? 2 NH3 CO2 H2O
Suppose 96.0 grams of ammonium carbonate
decomposes. How many grams of each of the gases
will be produced?
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Now consider the decomposition of solid ammonium
carbonate.
(NH4)2CO3 ? 2 NH3 CO2 H2O
96.0 grams
? g
? g
? g
The coefficients tell moles, not grams.
Convert 96.0 g (NH4)2CO3 to moles.
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Now consider the decomposition of solid ammonium
carbonate.
(NH4)2CO3 ? 2 NH3 CO2 H2O
96.0 grams
? g
? g
? g
Find the molar mass of ammonium carbonate.
2x14 8 12 3x16
96.0 g/mol
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Now consider the decomposition of solid ammonium
carbonate.
(NH4)2CO3 ? 2 NH3 CO2 H2O
96.0 grams
? g
? g
? g
Isnt that convenient! We have one mole of
ammonium carbonate.
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Now consider the decomposition of solid ammonium
carbonate.
(NH4)2CO3 ? 2 NH3 CO2 H2O
96.0 grams
? g
? g
? g
1 mol
1 mol
1 mol
2 mol
2 moles of ammonia are produced, along with 1
mole of carbon dioxide and 1 mole of water vapor.
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Now consider the decomposition of solid ammonium
carbonate.
(NH4)2CO3 ? 2 NH3 CO2 H2O
96.0 grams
? g
? g
? g
1 mol
1 mol
1 mol
2 mol
How many grams of each product are formed?
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Now consider the decomposition of solid ammonium
carbonate.
(NH4)2CO3 ? 2 NH3 CO2 H2O
96.0 grams
? g
? g
? g
1 mol
1 mol
1 mol
2 mol
34.0 g 44.0 g 18.0 g
96.0 g
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The reaction between dinitrogen pentoxide and
water
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Consider the reaction between dinitrogen
pentoxide and water.
N2O5 H2O ? 2 HNO3
There is only one product.
What kind of reaction is it?
It must be a synthesis reaction.
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Consider the reaction between dinitrogen
pentoxide and water.
N2O5 H2O ? 2 HNO3
There are several kinds of synthesis reactions.
Check the reference tables.

1. Hydrogen nonmetal binary acid
2. Metal nonmetal salt
3. Metal water base
4. Nonmetal oxide water ternary acid

Nitric acid is a ternary acid.
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N2O5 H2O ? 2 HNO3
Suppose we needed to make 100.0 grams of nitric
acid.
How many grams of dinitrogen pentoxide would we
need to react with excess water?
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N2O5 H2O ? 2 HNO3
100.0 g
??? g
1.59 mol/2
1.59 mole
0.794 mole
1.59 mole HNO3
Divide 1.59 mole HNO3 by 2 to get moles of N2O5
85.7 g N2O5
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N2O5 H2O ? 2 HNO3
100.0 g
85.7 g
1.59 mol/2
1.59 mole
0.794 mole
But we dont have 85.7 grams of N2O5. There are
only 60.0 grams available.
How many grams of nitric acid could we actually
make?
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N2O5 H2O ? 2 HNO3
??? g
60.0 g
0.556 mol x 2
1.11 mole
0.556 mole
0.556 mol N2O5
Multiply 0.556 mol by 2 to get moles of HNO3.
70.0 g HNO3
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Description of stoichiometry problems
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Stoichiometry problems will usually take one of
the following forms

1. Mole-mole problem where you might be given moles
   and asked to find moles of another substance.
2. Mole-mass problem where you might be given moles
   and asked find the mass of another substance.

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1. Mass-mass problem where you might be given a mass
   and asked to find the mass of another substance.

1. Mass-volume problem where you might be given a
   mass and asked to find the volume of a gas.
2. Volume-volume problem where you might be given a
   volume and asked to find another volume.

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Volume-volume stoichiometry problems are easiest
when you use Gay-Lussacs Law.
The ratio between the volumes of the reactant
gases and the products can be expressed in simple
whole numbers.
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Simply put, Gay-Lussacs Law says this
The volumes of the gases are in the same ratio as
the coefficients in the balanced equation.
2 H2 O2 ? 2 H2O
2 moles
1 mole
2 moles
2 L
1 L
2 L
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Applications of Gay-Lussacs Law
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C3H8(g) 5 O2(g) ? 3 CO2(g) 4 H2O(g)
1 mol
5 mols
3 mols
4 mols
1 L
5 L
3 L
4 L
Suppose 3.5 L of propane gas at STP is burned in
oxygen, how many L at STP of oxygen will be
required?
17.5 L O2
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C3H8(g) 5 O2(g) ? 3 CO2(g) 4 H2O(g)
1 mol
5 mols
3 mols
4 mols
1 L
5 L
3 L
4 L
Suppose 3.5 L of propane gas at STP is burned in
oxygen, how many L at STP of oxygen will be
required?
17.5 L O2
How many liters of carbon dioxide gas and water
vapor at STP would be produced?
10.5 L CO2 and 14.0 L H2O
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CH4(g) 4 Cl2(g) ? CCl4(g) 4 HCl(g)
When methane gas is allowed to react with an
excess of chlorine gas, tetrachloromethane and
hydrogen chloride gas will be produced.
How many L of methane will react with 0.800 L of
chlorine gas at STP?
0.200 L Cl2
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Stoichiometry problems involving gases
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Find the mass of HOCl that can be produced when
2.80 L of chlorine gas at STP reacts with excess
hydrogen peroxide.
Cl2(g) H2O2 (l) ? 2 HOCl (l)
2.80 L
??? g
Convert 2.80 L of Cl2 gas at STP to moles
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Cl2(g) H2O2 (l) ? 2 HOCl (l)
2.80 L
??? g
.125 x 2
0.250 mol
0.125 mol
0.125 mol Cl2
13.1g HOCl
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The reaction between copper and nitric acid
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Will copper dissolve in acids?
Consider hydrochloric acid
Cu 2HCl ? CuCl2 2H2 (g)
No Reaction
Most metals react with HCl to produce a metal
chloride solution and H2 gas.
Not copper
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Will copper dissolve in acids?
Consider hydrochloric acid
Cu 2HCl ? CuCl2 2H2 (g)
No Reaction
Copper is below hydrogen in the activity series.
Copper metal will only replace elements that are
below it in the activity series.
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What about other acids?
Cu HBr ? NR Cu HI ? NR Cu HF ? NR Cu
H2SO4 ? NR Cu HC2H3O2 ? NR
The same is true for all acids except nitric acid
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Nitric acid is the only acid that will dissolve
copper.
3Cu 8HNO3? 3Cu(NO3)2 2NO 4H2O
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Nitric acid is the only acid that will dissolve
copper.
3Cu 8HNO3? 3Cu(NO3)2 2NO 4H2O
The penny begins to disappear and the solution
turns blue-green and a brown gas is given off.
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Nitric acid is the only acid that will dissolve
copper.
3Cu 8HNO3? 3Cu(NO3)2 2NO 4H2O
The gas produced in the reaction is NO, which is
colorless. Reddish brown NO2 forms when NO reacts
with the oxygen in the air.
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Nitric acid is the only acid that will dissolve
copper.
3Cu 8HNO3? 3Cu(NO3)2 2NO 4H2O
The penny is gone and the solution turns a dark
blue. The brown NO2 gas escapes from the open
beaker.
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Nitric acid is the only acid that will dissolve
copper.
3Cu 8HNO3? 3Cu(NO3)2 2NO 4H2O
Calculate the volume of NO gas at STP when 20.0
grams of copper dissolves.
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Nitric acid is the only acid that will dissolve
copper.
3Cu 8HNO3? 3Cu(NO3)2 2NO 4H2O
20.0 g
??? L
x 0.667
0.210 mol
0.315 mol
0.315 mol Cu
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Nitric acid is the only acid that will dissolve
copper.
3Cu 8HNO3? 3Cu(NO3)2 2NO 4H2O
20.0 g
??? L
x 0.667
0.210 mol
0.315 mol
4.70 L NO
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Part Two of Basic Stoichiometry will include
the gas laws.