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Lecture Presentation
Chapter 3
Chemical Reactions
and Reaction
Stoichiometry
© 2015 Pearson Education, Inc.
James F. Kirby
Quinnipiac University
Hamden, CT
Stoichiometry
• The study of the mass relationships in
chemistry
• Based on the Law of Conservation of
Mass (Antoine Lavoisier, 1789)
“We may lay it down as an
incontestable axiom that, in all the
operations of art and nature, nothing
is created; an equal amount of matter
exists both before and after the
experiment. Upon this principle, the
whole art of performing chemical
experiments depends.”
—Antoine Lavoisier
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Chemical Equations
Chemical equations are concise
representations of chemical reactions.
Stoichiometry
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What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Reactants appear on the left
side of the equation.
Stoichiometry
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What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Products appear on the right
side of the equation.
Stoichiometry
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What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
The states of the reactants and products are written
in parentheses to the right of each compound.
(g) = gas; (l) = liquid; (s) = solid;
(aq) = in aqueous solution
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Stoichiometry
What Is in a Chemical Equation?
CH4(g) + 2O2(g)
CO2(g) + 2H2O(g)
Coefficients are inserted to balance the equation
to follow the law of conservation of mass.
Stoichiometry
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Why Do We Add Coefficients Instead of
Changing Subscripts to Balance?
• Hydrogen and oxygen can make water
OR hydrogen peroxide:
 2 H2(g) + O2(g) → 2 H2O(l)
 H2(g) + O2(g) → H2O2(l)
Stoichiometry
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Three Types of Reactions
• Combination reactions
• Decomposition reactions
• Combustion reactions
Stoichiometry
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Combination Reactions
• In combination
reactions two or
more substances
react to form one
product.
• Examples:
– 2 Mg(s) + O2(g)
– N2(g) + 3 H2(g)
– C3H6(g) + Br2(l)
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2 MgO(s)
2 NH3(g)
C3H6Br2(l)
Stoichiometry
Decomposition Reactions
• In a decomposition
reaction one
substance breaks
down into two or
more substances.
• Examples:
– CaCO3(s)
– 2 KClO3(s)
– 2 NaN3(s)
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CaO(s) + CO2(g)
2 KCl(s) + O2(g)
2 Na(s) + 3 N2(g)
Stoichiometry
Combustion Reactions
• Combustion reactions
are generally rapid
reactions that produce
a flame.
• Combustion reactions
most often involve
oxygen in the air as a
reactant.
• Examples:
– CH4(g) + 2 O2(g)
– C3H8(g) + 5 O2(g)
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CO2(g) + 2 H2O(g)
3 CO2(g) + 4 H2O(g)
Stoichiometry
Formula Weight (FW)
• A formula weight is the sum of the atomic
weights for the atoms in a chemical formula.
• This is the quantitative significance of a
formula.
• The formula weight of calcium chloride,
CaCl2, would be
Ca: 1(40.08 amu)
+ Cl: 2(35.453 amu)
110.99 amu
Stoichiometry
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Molecular Weight (MW)
• A molecular weight is the sum of the atomic
weights of the atoms in a molecule.
• For the molecule ethane, C2H6, the molecular
weight would be
C: 2(12.011 amu)
+ H: 6(1.00794 amu)
30.070 amu
Stoichiometry
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Ionic Compounds and Formulas
• Remember, ionic compounds exist with
a three-dimensional order of ions.
There is no simple group of atoms to
call a molecule.
• As such, ionic compounds use empirical
formulas and formula weights (not
molecular weights).
Stoichiometry
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Percent Composition
One can find the percentage of the
mass of a compound that comes from
each of the elements in the compound
by using this equation:
(number of atoms)(atomic weight)
% Element =
(FW of the compound)
× 100
Stoichiometry
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Percent Composition
So the percentage of carbon in ethane is
(2)(12.011 amu)
%C =
=
(30.070 amu)
24.022 amu
30.070 amu
× 100
= 79.887%
Stoichiometry
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Counting numbers and the mole
Stoichiometry
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Avogadro’s Number
• In a lab, we cannot
work with individual
molecules. They are
too small.
• 6.02 × 1023 atoms
or molecules is an
amount that brings
us to lab size. It is
ONE MOLE.
• One mole of 12C has
a mass of 12.000 g.
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Stoichiometry
Molar Mass
• A molar mass is the mass of
1 mol of a substance (i.e., g/mol).
• The molar mass of an
element is the atomic
weight for the element
from the periodic table.
If it is diatomic, it is twice
that atomic weight.
• The formula weight (in
amu’s) will be the same
number as the molar mass
(in g/mol).
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Stoichiometry
Using Moles
Moles provide a bridge from the molecular
scale to the real-world scale.
Stoichiometry
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Chemical arithmetic……
….…….how many conversions do I need to perform?
……....which conversions do I need to perform?
Stoichiometry
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Mole Relationships
• One mole of atoms, ions, or molecules contains
Avogadro’s number of those particles.
• One mole of molecules or formula units contains
Avogadro’s number times the number of atoms or
ions of each element in the compound.
Stoichiometry
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*Which calculations do or do not involve a single conversion?
Answer YES or NO (1 step). consider 15.02 g H2O as an example
1. Moles of water
Y
2. Moles of oxygen
N
3. Number of Atoms of hydrogen
N
4. Mass of hydrogen
Y
5. Number of atoms of oxygen
N
Stoichiometry
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Conversions (cont’d)
Consider 10.02 g Na3PO4
Mass (Na3PO4)
Mol (Na3PO4)
Mass (PO43-)
1 step
Mass (Sodium)
Mol (Sodium)
Mol (phosphate)
multistep
Ions (sodium)
Stoichiometry
Atoms (oxygen)
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Calculate the missing quantities in the table below
Compound
Mass of
Compound
Moles of…
Atoms of…
Mg3N2
Mg (1.52 mol) N
CuSO4.5H2O
H
O (3.75 X 1019 atoms)
H2S2O7
S (0.125 mol)
O
Iron(III) chloride 13.75 g
Fe
Cl
Stoichiometry
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Empirical Formula (Simplest Formula)
The formula of a substance written with the
smallest integer subscripts.
For example:
The empirical formula for N2O4 is NO2.
The empirical formula for H2O2 is HO.
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Stoichiometry
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Determining Empirical Formulas
*Finding the moles is the critical step
One can determine the empirical formula
from the percent composition by following
these three steps.
Stoichiometry
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Determining the empirical formula
1. Given % masses
Assume a fixed
total mass
2.Given element or decomposition
product amount
Calculate grams
of each element
Grams of
each element
Calculate moles
of each element
Convert decomposition products
into grams/moles of respective elements
Express as
Empirical formula
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Calculate the
mole ratio
Stoichiometry
Determining the Empirical Formula
Beginning with percent composition:
• Assume exactly 100 g so percentages
convert directly to grams.
• Convert grams to moles for each
element.
• Manipulate the resulting mole ratios to
Stoichiometry
obtain whole numbers.
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Manipulating the ratios:
Divide each mole amount by the smallest
mole amount.
If the result is not a whole number:
Multiply each mole amount by a factor to
make whole numbers.
For example:
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Stoichiometry
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Benzene is composed of 92.3% carbon
and 7.7% hydrogen. What is the empirical
formula of benzene?
1 mol C
92.3 g C
 7.685 mol C
12.01 g C
1 mol H
7.7 g H
 7.64 mol H
1.008 g H
7.685
1
7.64
7.64
1
7.64
Empirical formula: CH
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Molecular Formula
A formula for a molecule in which the
subscripts are whole-number multiples of
the subscripts in the empirical formula.
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To determine the molecular formula:
• Compute the empirical formula weight.
• Find the ratio of the molecular weight to
the empirical formula weight.
molecular weight
n
empirical formula weight
• Multiply each subscript of the empirical
formula by n.
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Benzene has the empirical formula CH.
Its molecular weight is 78.1 amu. What is
its molecular formula?
Empirical formula weight  13.02 amu
78.1
6
13.02
Molecular formula
C6H6
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Stoichiometry
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Sodium pyrophosphate is used in
detergent preparations. It is composed of
34.5% Na, 23.3% P, and 42.1% O. What
is its empirical formula?
1 mol Na
34.5 g Na
 1.501 mol Na
22.99 g Na
1 mol P
23.3 g P
 0.7523 mol P
30.97 g P
1 mol O
42.1 g O
 2.631 mol O
16.00 g O
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Stoichiometry
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1.501 mol Na
1.501
 2.00  2 = 4
0.7523
0.7523
0.7523 mol P
 1.00  2 = 2
0.7523
2.631 mol O
2.631
 3.50  2 = 7
0.7523
Empirical formula
Stoichiometry
Na4Copyright
P2O7 © Cengage
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Hexamethylene is one of the materials
used to produce a type of nylon. It is
composed of 62.1% C, 13.8% H, and
24.1% N. Its molecular weight is 116
amu. What is its molecular formula?
1 mol C
62.1 g C
 5.171 mol C
12.01 g C
1 mol H
13.8 g H
 13.69 mol H
1.008 g H
1 mol N
24.1 g N
 1.720 mol H
14.01 g N
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5.171 mol C
13.69 mol H
1.720 mol H
5.171
3
1.720
13.69
8
1.720
1.720
1
1.720
Empirical formula
C3H8N
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The empirical formula is C3H8N.
Find the empirical formula weight:
3(12.01) + 8(1.008) + 1(14.01) = 58.104
amu
116
n
58.10
2
Molecular formula: C6H16N2
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Stoichiometry
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Stoichiometry
The calculation of the quantities of
reactants and products involved in a
chemical reaction.
Interpreting a Chemical Equation
The coefficients of the balanced chemical
equation may be interpreted in terms of
either (1) numbers of molecules (or ions
or formula units) or (2) numbers of moles,Stoichiometry
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depending on your needs.
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To find the amount of B (one reactant or product)
given the amount of A (another reactant or
product):
1. Convert grams of A to moles of A
 Using the molar mass of A
2. Convert moles of A to moles of B
 Using the coefficients of the balanced
chemical equation
3. Convert moles of B to grams of B
 Using the molar mass of B
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Stoichiometry
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Propane, C3H8, is normally a gas, but it is
sold as a fuel compressed as a liquid in
steel cylinders. The gas burns according
to the following equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
How many grams of CO2 are produced
when 20.0 g of propane is burned?
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Molar masses
C3H8: 3(12.01) + 8(1.008) = 44.094 g
CO2: 1(12.01) + 2(16.00) = 44.01 g
1 mol C 3H8 3 mol CO2 44.01 g CO2
20.0 g C 3H8
44.094 g C 3H8 1 mol C 3H8 1 mol CO2
 59.8856987 3 g CO 2
59.9 g CO2
(3 significant figures)
Stoichiometry
3 | 44
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Propane, C3H8, is normally a gas, but it is
sold as a fuel compressed as a liquid in
steel cylinders. The gas burns according
to the following equation:
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
How many grams of O2 are required to
burn 20.0 g of propane?
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Stoichiometry
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Molar masses:
O2
2(16.00) = 32.00 g
C3H8 3(12.01) + 8(1.008) = 44.094 g
1 mol C 3H8
5 mol O 2 32.00 g O 2
20.0 g C 3H8
44.094 g C 3H8 1 mol C 3H8 1 mol O 2
 72.5722320 5 g O 2
72.6 g O2
(3 significant figures)
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Stoichiometry
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Limiting Reactant
The reactant that is entirely consumed
when a reaction goes to completion.
Once one reactant has been completely
consumed, the reaction stops.
Any problem giving the starting amount
for more than one reactant is a limiting
reactant problem.
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All amounts produced and reacted are
determined by the limiting reactant.
How can we determine the limiting
reactant?
1.Use each given amount to calculate the
amount of product produced.
2.The limiting reactant will produce the lesser
or least amount of product.
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Magnesium metal is used to prepare
zirconium metal, which is used to make
the container for nuclear fuel (the nuclear
fuel rods):
ZrCl4(g) + 2Mg(s)  2MgCl2(s) + Zr(s)
How many moles of zirconium metal can
be produced from a reaction mixture
containing 0.20 mol ZrCl4 and 0.50 mol
Mg?
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Stoichiometry
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1 mol Zr
0.20 mol ZrCl 4
 0.20 mol Zr
1 mol ZrCl 4
1 mol Zr
0.50 mol Mg
 0.25 mol Zr
2 mol Mg
Since ZrCl4 gives the lesser amount of Zr,
ZrCl4 is the limiting reactant.
0.20 mol Zr will be produced.
Stoichiometry
3 | 50
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Urea, CH4N2O, is used as a nitrogen
fertilizer. It is manufactured from
ammonia and carbon dioxide at high
pressure and high temperature:
2NH3 + CO2(g)  CH4N2O + H2O
In a laboratory experiment, 10.0 g NH3
and 10.0 g CO2 were added to a reaction
vessel. What is the maximum quantity (in
grams) of urea that can be obtained?
How many grams of the excess reactant
Stoichiometry
are left at the end of the reactions?
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Molar masses
NH3
1(14.01) + 3(1.008) = 17.02 g
CO2
1(12.01) + 2(16.00) = 44.01 g
CH4N2O 1(12.01) + 4(1.008) + 2(14.01) +
1(16.00) = 60.06 g
1 mol NH 3 1 mol CH4 N 2 O 60.06 g CH4 N 2 O
10.0 g NH 3
17.024 g NH 3 2 mol NH 3
1 mol CH4 N 2 O
 17.6 g CH4 N 2 O
1 mol CO 2 1 mol CH4 N2 O 60.06 g CH4 N 2 O
10.0 g CO 2
44.01 g CO 2 1 mol CO 2
1 mol CH4 N 2 O
 13.6 g CH4 N 2 O
CO2 is the limiting reactant since it gives the lesser
amount of CH4N2O.
Stoichiometry
3 | 52
13.6 g CH4N2O will be produced.
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To find the excess NH3, we need to find how much
NH3 reacted. We use the limiting reactant as our
starting point.
1 mol CO2 2 mol NH 3 17.02 g NH 3
10.0 g CO2
44.01 g CO2 1 mol CO2 1 mol NH 3
 7.73460577 1 g NH 3
 7.73 g NH 3 reacted
Now subtract the amount reacted from the
at start
starting amount: 10.0
−7.73 reacted
2.27 g remains
2.3 g NH3 is left unreacted.
(1 decimal place)
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Stoichiometry
3 | 53
Theoretical/Expected Yield
Theorerical yield is the maximum amount
of product that can be obtained by a
reaction from given amounts of reactants.
This is a calculated amount based on a
balanced chemical equation.
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Actual/Experimental Yield
The amount of product that is actually
obtained.
This is a measured amount.
Percentage Yield
actual yield
percentage yield 
 100%
theoretical yield
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2NH3 + CO2(g)  CH4N2O + H2O
When 10.0 g NH3 and 10.0 g CO2 are
added to a reaction vessel, the limiting
reactant is CO2. The theoretical yield is
13.6 of urea. When this reaction was
carried out, 9.3 g of urea was obtained.
What is the percent yield?
3 | 56
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Stoichiometry
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Theoretical yield = 13.6 g
Actual yield = 9.3 g
9.3 g
Percent yield =
 100%
13.6 g
= 68% yield
(2 significant figures)
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Stoichiometry
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Silver chloride can be made according to the following chemical
equation
NaCl(aq)
+
AgNO3(aq)
→
NaNO3(aq)
+
AgCl(s)
When solutions containing 0.351 g AgNO3 and 0.250 g NaCl
were mixed, the amount of AgCl precipitated was 0.250 g.
Calculate the percent yield of AgCl.
Stoichiometry
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