Chapter 8
Quantities
In Reactions
Homework
• Assigned Problems (odd numbers only)
• “Problems” 17 to 73
• “Cumulative Problems” 75-95
• “Highlight Problems” 97-99
Calculations Using Balanced Equations:
Stoichiometry
• Stoichiometry is the study of the quantitative
relationships among reactants and products in a
chemical reaction
• These chemical calculations can be used to
determine the amount of one reactant needed to
completely react with another
• Or, to determine the amount of reactant needed
to produce a desired amount of product
• To calculate chemical quantities in reactions
involves knowing how to interpret a balanced
chemical equation
Calculations Using Balanced Equations:
Law of Conservation of Mass
• As a chemical reaction proceeds:
–Reactants are consumed and new
materials with new chemical properties
are produced
–Bonds are broken, formed, or atoms
are rearranged which produces new
substances
–No material is lost or gained as original
substances change to new substances
Calculations Using Balanced Equations:
Law of Conservation of Mass
• Law of Conservation of Mass
–Quantity of matter does not change
during a chemical reaction
–The sum of the masses of products is
equal to the sum of masses of reactants
–Atoms are neither created nor destroyed
in chemical reactions
–Only a balanced equation obeys this law
Mole Relationships in Chemical Equations:
Conservation of Mass
• A balanced equation has the same
number of atoms on each side of the
arrow
CH 4 (g)  2 O 2 (g)  CO2 (g)  2 H 2 O (g)
Information Available from a
Balanced Chemical Equation
• “1 mole of methane gas reacts with 2 moles
of oxygen gas to produce 1 mole of carbon
dioxide and 2 moles of water vapor.”
• Multiplying each of the molar masses by the
coefficient will give the total mass of
reactants and products
Making Molecules:
Mole-Mole Conversions
A balanced chemical equations tell us:
– The formulas and symbols of the reactants and
products
– The physical state of each substance
– If special conditions such as heat are required
– The number of molecules, formula units, or atoms
of each type of molecule involved in the reaction
• Number can be in terms of single atoms, or
moles of atoms
– The relative number of moles of each reactant and
product
Making Molecules:
Mole-Mole Conversions
• In a balanced equation, conversion from moles of
one substance to another will be determined by
the values of the coefficients
• Balancing an equation will generate the
coefficients that equal the number of moles of
each reactant and product
CO(g)  2 H 2 (g) 
CH 3 OH()
• To determine how many moles of methanol would
be produced if 0.295 moles of hydrogen gas is
consumed:
• Requires a mole to mole relationship between
methanol and hydrogen gas
Making Molecules:
Mole-Mole Factors
CO (g)  2 H 2 (g)  CH3OH ()
• Converting the given mole amount of hydrogen
gas enables you to find the number of moles
methanol produced
Given: 0.295 mol H2
Find: mol CH3OH
• Coefficients of the balanced equation can be
used to make mole to mole relationships between
the different reactants and products
2 mol H2 = 1 mol CH3OH
Making Molecules:
Mole-Mole Factors
CO (g)  2 H 2 (g)  CH3OH ()
• The unit path begins with moles of H2 and ends
with moles of CH3OH
mol H2
Mole-mole
factor
mol CH3OH
1 mol CH 3OH
2 mol H 2
• From any mole-mole relationship, two mole-mole
conversion factors can be made: 2 mol H2 = 1 mol CH3OH
1 mol CH 3OH
2 mol H 2
or
1 mol CH 3OH
2 mol H 2
Making Molecules:
Mole-Mole Factors
• Write all of the possible mole-mole factors
for the following chemical equation
2 H2 (g)  O2 (g)  2 H2O ()
2 mol H2 = 1 mol O2
2 mol H2 = 2 mol H2O
1 mol O2 = 2 mol H2O
2 mol H 2
1 mol O 2
2 mol H 2
2 mol H 2 O
1 mol O 2
2 mol H 2 O
1 mol O 2
2 mol H 2
2 mol H 2 O
2 mol H 2
2 mol H 2 O
1 mol O 2
Using Mole-Mole Factors in Calculations:
Calculating Moles of a Product
• Calculations based on balanced
equations require the use of mole to
mole (conversion) factors
– Equation must be balanced
– Identify the known (given) and needed
(find) substances
– Make the conversion factor based on:
KNOWN ( given) number of moles 
coefficient of what is SOUGHT
coefficient of what is KNOWN
Using Mole-Mole Factors in Calculations:
Calculating Moles of a Product
• Using mole-mole factors from a
BALANCED chemical equation
–You can convert moles of one
compound to moles of another
compound using the correct mole-mole
factor
grams B
grams A
MM of A
MM of B
Stoichiometry
moles A
moles B
Mole-mole
moles
factorA
moles B
Using Mole-Mole Factors in Calculations:
Calculating Moles of a Product
• Calculate the moles of CO2 formed when 4.30
moles of C3H8 reacts with (the required) 21.5
moles of O2
C 3 H 8 (g)  5 O 2 (g) 3 CO 2 (g) 4 H 2 O(g)
• Balance the equation
• Plan to convert the given amount of moles to the
needed amount of moles
• Use coefficients to state the relationships and
mole-mole factors
• Set up the problem using the mole-mole factor
and canceling units
Using Mole-Mole Factors in Calculations:
Calculating Moles of a Product
C3 H8 (g)  5 O 2 (g)  3CO 2 (g)  4 H 2 O(g)
grams B
grams A
MWofofA
A
MM
moles A
Given:
MW
B
MMofof
Stoichiometry
4.30 mol C 3 H 8
B
moles B
moles A
Mole-mole
moles
factorB
Find:
mol CO 2
Mole-mole relation
Mole-mole factor
1 mol C3H8 = 3 mol CO2
1 mol C3 H 8
3 mol CO 2
and
3 mol CO 2
1 mol C3 H 8
Using Mole-Mole Factors in Calculations:
Calculating Moles of a Product
• Set up the problem using the mole-mole factor
that cancels given moles and provides needed
moles
C3 H8 (g)  5 O 2 (g)  3 CO 2 (g)  4 H 2 O(g)
given : 4.30 mol C 3 H 8
Mole-mole factor
find : mol CO 2
3 mol CO 2
1 mol C3 H 8
4.30 mol C3 H8 3 mol CO 2
 12.9 mol CO2
1 mol C3 H8
Making Molecules: Mass-to-Mass Conversions
• From the balanced equation
–It is also possible to start with a known
mass of one substance
–Then convert to moles of another
substance
–Start with a given amount (of grams) of
a substance
–Then use mole-mole factor to find the
sought-after mass of another substance
Making Molecules: Mass-to-Mass Conversions
• The most common type of stoichiometric
calculation is the mass-to-mass calculation
• In this type of problem:
• The mass of one substance involved in a
chemical reaction is given
• Find the mass of another substance involved in
the reaction
• If a chemist only has so many grams of a certain
chemical
– Can calculate how many grams of another substance can
be produced
– Can calculate how many grams of another reactant are
required to react with it
Mass-to-Mass Conversions
• To convert the grams of one substance to grams
of another substance:
• Find the mole-mole factor using the coefficients
in the balanced equation
– You can only relate (moles-moles) of two compounds, not
grams-to-grams
– Ratios ONLY apply to moles, NOT grams
– Must convert grams to moles, then use mole-mole factor
grams B
grams A
MW of
ofAA
MM
MMofofBB
MW
Stoichiometry
moles A
moles BA
moles
Mole-mole
factor
moles
moles AB
moles B
Mass-to-Mass Conversions
• Mass-to-mass conversions begin with a given
mass of substance A
• By use of the balanced equation, find the mass of
another (substance B)
1. Convert grams of A to moles of A
2. Convert the moles of A to moles of B by use of mole-mole
ratio
3. Convert moles of B to mass of B
grams B
grams A
MM of A
MM of B
Stoichiometry
moles A
moles B
moles B
moles A
Mass-to-Mass Conversions
(Mass of Product from Mass of Reactant)
• Calculate the mass of carbon dioxide produced when 96.1
g of propane react with sufficient oxygen.
C3 H8 (g)  5 O 2 (g)  3CO 2 (g)  4 H 2 O(g)
• Balance the equation
• Plan to convert the given mass to given moles
• Convert the given moles to sought-after moles by the use of molemole factor
• Convert the needed moles to needed mass
grams CO2
grams C3H8
MM of
C3H8
MM of CO2
Stoichiometry
moles C3H8
moles CO2
moles CO2
Mole-mole
moles
C3H8
factor
Mass-to-mass Conversions: Example 1
•
•
•
•
Write the equalities
1 mol C3H8 = 44.09 g C3H8
1 mol CO2 = 44.01 g CO2
1 mol C3H8 = 3 mol CO2 to create mole-mole factor
C3 H8 (g)  5 O 2 (g)  3 CO 2 (g)  4 H 2 O(g)
Find: g of CO2
Given: 96.1 g C3H8
grams CO2
grams C3H8
MM of A
MM of B
Stoichiometry
moles C3H8
3moles
mol CO
B2
1moles
mol C3A
H8
moles CO2
Mass-to-Mass Conversions: Example 1
C3 H8 (g)  5 O 2 (g)  3 CO 2 (g)  4 H 2 O(g)
96.1 g C 3 H8
96.1 g C3 H 8 1 mol C3 H 8

44.09 g C3 H 8
X g CO2
2.1796 mol C3 H8
2.1796 mol C3 H8 3 mol CO 2
 6.539 mol CO2
1 mol C 3H 8
6.539 mol CO2 44.01 g CO 2

1 mol CO 2
288 g CO2
Mass-to-Mass Conversions: Example 2
• What mass of carbon monoxide
and what mass of hydrogen are
required to form 6.0 kg of methanol
by the following reaction:
CO(g)  2 H 2 (g)  CH3OH()
? g CO ? g H2
6.0 kg CH 3OH
Mass-to-Mass Conversions: Example 2
CO (g)  2 H 2 (g)  CH3OH (l )
6.0 kg CH 3OH 1000 g
 6000 g CH3OH
1 kg
6000 g CH3OH 1 mol CH 3 OH
 187.27 mol CH3OH
32.04 g CH 3OH
187.27 mol CH3OH
1 mol CO
 187.27 mol CO
1 mol CH 3OH
187.27 mol CH3OH
2 mol H 2
 374.53 mol H2
1 mol CH 3OH
Mass Calculations Example 2
187.27 mol CO 28.01 g CO
 5245 g CO
1 mol CO
374.53 mol H 2 2.0156 g H 2
 754.9 g H2
1 mol H2
Limiting Reactant, Theoretical Yield,
and Percent Yield
• The chemical reactants are usually not present in
the exact mole-mole ratios as stated in the
balanced chemical equation
• Often, one of the reactants is purposely added in
an excess amount
• Reasons include:
• Increase the rate of reaction
• To ensure that one reactant is completely used
up (reacted)
• Reactants are not completely converted to
products as stated on paper (theory)
Limiting Reactant, Theoretical Yield,
and Percent Yield
• Chemical reactions with two or more reactants
will continue until one of the reactants is used up
(consumed)
• If one of the reactants is used up, the reaction will
stop because there is not enough of the other
reactant to react with it
• The reactant used up is called the limiting
reactant (reagent)
• This reactant limits the amount of product that
can be made
Limiting Reactant
• When you make peanut
butter sandwiches
• Required: 2 slices of
bread and 1 tbsp. peanut
butter per sandwich
2 slices of br. +1 tbsp p.b. =1sndw.
• The reaction of nitrogen
gas and hydrogen gas
forming ammonia gas
• Required: 1 molec. N2 gas
and 3 molec. H2 gas
N2 (g) + 3H2 (g) = 2 NH3 (g)
Limiting Reactant
To determine the limiting reactant
between two reactants:
1. Balance the equation
2. State the mole-mole relationships to
make conversion factors
3. Convert the initial masses (reactants)
to moles of each reactant
4. Calculate how many moles of product
can be produced by each reactant
Limiting Reactant
5. Convert the moles of product to
number of grams of product that each
of the reactants would produce
6. Compare the numbers: The reactant
producing the least amount of product
(grams) is the limiting reactant
Limiting Reactant Problem
• Lithium nitride, an ionic compound
containing Li+ and N3- ions, is
prepared by the reaction of lithium
metal and nitrogen gas. Calculate the
mass of lithium nitride formed from
56.0 g of nitrogen gas and 56.0 g of
lithium metal.
6 Li (s)
56.0 g Li

N 2 (g)  2 Li 3 N (s)
56.0 g N2
X g Li 3 N
Limiting Reactant Problem
6 Li(s)  N 2 (g)  2Li 3N(s)
Given: 56.0 g Li
Given: 56.0 g N2
Find: g of Li3N
Equalities and Conversion Factors6 mol Li = 2 mol Li3N
1 mol N2 = 2 mol Li3N
1 mol Li = 6.941g Li
1 mol N2 = 28.00 g N2
grams N
Li2
Li2
MM of N
Li2
moles N
1 mol Li3N= 34.83 g Li3N
Solution Map:
grams Li3N
Stoichiometry
MM of Li3N
moles Li3N
moles Li3N
moles
moles Li
N2
Limiting Reactant Problem
6 Li (s) 
Limiting reactant
56.0 g Li
N 2 (g)  2Li 3 N(s)
56.0 g N 2 (x) g Li 3 N
56.0 g Li 1 mol Li
 8.07 mol Li
6.941 g Li
2 mol Li3 N 34.82 g Li3 N
8.07 mol Li 

 93.67 g Li3N
6 mol Li
1 mol Li3 N
56.0 g N 2 1 mol N 2
 2.00 mol N2
28.02 g N 2
2.00 mol N 2 
2 mol Li3 N 34.82 g Li3 N

 139.3 g Li3N
1 mol N 2
1 mol Li3 N
Limiting Reactant Problem
• Lithium is the limiting reactant. We calculated
the number of grams of lithium nitride which is
formed in the reaction based on the limiting
reactant
• This is the calculated amount of lithium nitride
formed if the reaction proceeds completely as
described by its balanced chemical equation
34.83 g Li3 N
2.69 mol Li3 N 
 93.7 g Li N
3
1 mol Li3 N
Theoretical yield
Percent Yield
• The calculated amount of product that
should be obtained is called the theoretical
yield
• Assumes all reactants are converted to
product based on the mole-mole ratios of
reactant to product
• Rarely do you get the maximum amount of
product
– Side reactions
– Loss during transfer
– Accidental spills
Percent Yield
• Theoretical Yield
– The calculated amount of product
• Actual Yield
– The actual amount of product
– Something less than the theoretical
• Percent Yield
– The fraction of the theoretical yield actually
obtained is expressed as a percent
Actual Yield
% Yield 
100%
Theoretical Yield
Percent Yield Example
• Suppose, in the previous limiting
reactant problem, you actually
produced 90.8 g of Li3N. What is the
percent yield of this reaction?
90.8 g
actual
100 %  96.9 % yield
100 % 
93.7 g
theoretical
Enthalpy
• Chemical reactions are associated with an
absorption or evolution of heat
– A change in energy occurs as bonds are
broken (reactants) and new ones form
(products)
– Nearly all chemical reactions absorb or
produce heat
– Measured by the heat of reaction or enthalpy
• Enthalpy change is the amount of heat
produced or consumed in a process (∆H )
Sign of ∆Hrxn
• Endothermic reactions absorb heat as they
occur
– If (∆H ) is positive, then heat is added to the
reaction
– If heat supply is removed, the reaction stops
• Exothermic reactions produce heat as they
occur
– If (∆H ) is negative, then heat is evolved by
the reaction
Sign of ∆Hrxn
Enthalpy of Reaction
• Photosynthesis reaction
– Carbon dioxide reacts with water to produce glucose
and oxygen
6 CO2 (g)  6 H2 O ()  C6H12 O 6 (s)  6 O 2 (g)
∆H = +2801 kJ
• Cell metabolism
– Glucose reacts with oxygen to produce carbon dioxide
and water
C6H12 O 6 (s)  6 O 2 (g)  6 CO2 (g)  6 H2 O ()
∆H = -2801 kJ
Stoichiometry of ∆Hrxn
• The coefficients in a given chemical reaction
represent the number of moles of reactants and
products that produce the given heat of reaction
(enthalpy change)
• The combustion of methane gas:
CH4 (g)  2 O2 (g)  CO2 (g)  2 H2O(g) ∆Hrxn = -890 kJ
• This information gives a quantitative relationship
between the heat evolved per mole of methane
and oxygen gas
1 mol CH4 = -890 kJ
2 mol O2 = -890 kJ
Stoichiometry of ∆Hrxn
• If the combustion of 1 mol of CH4 with 2 mol O2
releases 890 kJ of heat, the combustion of 2 mol
of CH4 with 4 mol of O2 produces twice as much
heat
2 CH4 (g)  4 O2 (g)  2 CO2 (g)  4 H2O(g) ∆Hrxn = -1780 kJ
• The equivalence statements can be used to make
conversion factors between the amounts of
reactants or products and the amount of heat
absorbed or emitted in a given reaction
• Calculate the amount of heat emitted when a
certain amount in grams undergoes combustion
Stoichiometry Involving ∆H
• Calculate the heat associated with the complete
combustion of 4.50 g of methane gas
CH4 (g)  2 O2 (g)  CO2 (g)  2 H2O(g) ∆H rxn= -890 kJ
Given: 4.50 g CH4
Conversion Factors:
Solution Map:
Find: kJ
1 mol CH4= -890 kJ
g CH4
mol CH4
1 mol CH 4
16.00 g CH 4
Solution: 4.50 g CH4 
1 mol CH4 = 16.00 g
 890 kJ
1 mol CH 4
kJ CH4
1 mol CH 4
 890 kJ

 -2.50 × 102 kJ
16.00 g CH 4
1 mol CH 4
Stoichiometry Involving ∆H
• The combustion of sulfur dioxide
• It reacts with oxygen to produce sulfur
trioxide
2 SO2 (g)  O2 (g)  2 SO3 (g) ∆H rxn= -99.1 kJ
• Calculate the heat produced when 75.2 g of
sulfur trioxide is produced
Given 75.2 g SO3
Find:
Heat in kJ produced
when SO3 is formed
Stoichiometry Involving ∆H
Solution Map: Relation between g of SO3 and heat released
Grams
of SO3
Molar
mass
Moles
of SO3
Heat
of rxn
kj
Write the necessary conversion factors:
1 mol SO3  80.07 g SO3
1 mol SO3
80.07 g SO3
and
80.07 g SO3
1 mol SO3
2 mol SO3  99.1 kJ
2 mol SO3
99.1 kJ
and
99.1 kJ
2 mol SO3
Set up the problem:
1 mol SO3
99.1kJ
75.2 g SO3 

 46.5 kJ
80.07 g SO3 2 mol SO3
• end