Chapter 8
Chemical Quantities
In Reactions
Homework
• Assigned Problems (odd numbers only)
• “Questions and Problems” 8.1 to 8.31
(begins on page 230)
• “Additional Questions and Problems”
8.39 to 8.51 (begins on page 247)
• “Challenge Questions” 8.53 and 8.55,
page 248
• “Combining Ideas” CI 11 and CI 13,
page 250
Mole Relationships in Chemical Equations:
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
Mole Relationships in Chemical Equations:
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
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
Information Available from a
Balanced Chemical Equation
• Balanced chemical equations
tell us:
–Relative number of atoms reacting
and produced
–Number can be in terms of single
atoms, or moles of atoms
CO(g)  2H 2 (g) 
CH 3OH(l)
Mole-Mole Factors
• In a balanced equation, conversion from moles
of one substance to another will be determined
by the values of the coefficients
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
• Will require a mole to mole conversion
• Converting the given mole amount of hydrogen
gas to the needed mole amount of methanol
Mole-Mole Factors
CO(g)  2 H 2 (g) 
CH 3OH(l)
• Coefficients of the balanced
equation can be used to make
ratios (conversion factors)
between the different reactants
and products
1 mol CO
2 mol H 2
1 mol CO
1 mol CH 3OH
2 mol H2
1 mol CH 3OH
Mole-Mole Factors
• Write all of the possible mole ratios for
the following chemical equation
2 H 2 (g) 
O 2 (g)  2 H 2 O()
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
• Calculations based on balanced
equations require the use of mole to
mole conversions
– Equation must be balanced
– Identify the known and needed
substances
– Make the conversion factor based on
needed number of moles
given number of moles 
given number of moles
Stoichiometry
• Using mole ratios from a
BALANCED chemical equation
–Can convert moles of one compound to
moles of another compound using the
correct mole ratio
grams B
grams A
MM of A
MM of B
Stoichiometry
moles A
moles A
Mol-mol
moles B
factor
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)  3 CO 2 (g)  4 H 2 O(g)
grams B
grams A
MW of A
moles A
Given:
MW of B
Stoichiometry
4.30 mol C 3 H 8
moles B
moles A
Mol-mol
moles
factorB
Needed:
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
needed : mol CO 2
1 mol C3H 8
Mole-mole factor
3 mol CO 2
4.30 mol C3 H8 3 mol CO 2
 12.9 mol CO2
1 mol C3 H8
Mass Calculations for Reactions
• 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 known amount of moles of
a substance
–Then convert to the sought mass of
another substance
Mass Calculations
• Convert the moles of one substance to
moles of another substance
• Find the mole-mole factor using the
coefficients in the balanced equation
– Ratios ONLY apply to moles, NOT grams
– Convert grams to moles, then use mole ratio
grams B
grams A
MW of A
MW of B
Stoichiometry
moles A
moles B
moles
A
Mol-mol
moles
factorB
Mass Calculations
• You can only relate moles of two
compounds
–Ratios ONLY apply to moles, NOT
grams
–Convert grams to moles, then use mole
ratio
grams B
grams A
MW of A
MW of B
Stoichiometry
moles A
moles B
moles A
moles B
Mass Calculations For Reactions:
Mass of Product from Mass of Reactant
• What mass of carbon dioxide is produced when
96.1 g of propane react with sufficient oxygen?
C3 H8 (g)  5 O 2 (g)  3CO 2 (g)  4H 2 O(g)
• Balance the equation
• Plan to convert the given mass to given moles
• Convert the given moles to needed moles by the use of mole-mole
factor
• Convert the needed moles to needed mass
grams CO2
grams C3H8
MM of A
MM of B
Stoichiometry
moles C3H8
moles CO2
moles CO2
moles C3H8
Mass Calculations 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)
Mass needed: g of CO2
Given: 96.1 g C3H8
grams CO2
grams C3H8
MM of A
MM of B
Stoichiometry
moles C3H8
moles CO2
moles A
moles B
Mass Calculations Example 1
C3 H8 (g)  5 O 2 (g)  3 CO 2 (g)  4 H 2 O(g)
96.1 g C 3 H8
X g CO2
96.1 g C3 H8 1 mol C 3 H8
 2.1796 mol C3 H8
44.09 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
 287.76 g CO2
1 mol CO 2
Mass Calculations 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 Calculations Example 2
CO(g)  2 H 2 (g) 
CH 3OH(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 H2 2.0156 g H 2
 754.9 g H2
1 mol H2
Limiting Reactants
• Chemical reactions with two or
more reactants will continue until
one of the reactants is used up
• The reactant used up is called the
limiting reactant (reagent)
• This limits the amount of product
that can be made
Reactions
• In the lab, reactants do not always
combine in exact mole ratios
–Often reactions are run with more of
one reactant than is needed (excess)
• A reaction will continue until one
of the reactants runs out
–Reactant that runs out first is the
Limiting Reactant
Calculating Moles of Product from
a Limiting Reactant
• To determine the limiting reactant
between two reactants
–Balance the equation
–Determine the number of moles of each
reactant
–Calculate the number of moles of
product that each of the reactants
(moles) would produce
–The reactant producing the least
amount of product (moles) is the
limiting reactant
Calculating Moles of Product from
a Limiting Reactant
• Determine the number of moles
of product that each reactant
can make
–The smallest number is the
MAXIMUM product you can make
–The reactant that gives the smallest
amount of product is the limiting
reactant
Calculating Mass of Product from a
Limiting Reactant
1) Balance the equation
2) Determine the amount of
product that can be made by
each reactant
3) Compare numbers
 Reagent that gives the smaller
number is limiting and that is the
maximum amount of product that
can be made
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)  N 2 (g)  2 Li 3N(s)
56.0 g Li
56.0 g N2
? g Li 3N
Limiting Reactant Problem
6 Li(s) 
Given: 56.0 g Li
N 2 (g)  3 Li 3N(s)
Given: 56.0 g N2
Needed: g of Li3N
Equalities and Conversion Factors
1 mol Li = 6.941g Li
1 mol Li
6.941 g Li
and
6.941 g Li
1 mol
1 mol N2 = 28.00 g N2
1 mol N 2
6.941 g N 2
and
6.941 g N 2
1 mol N 2
1 mol Li 3 N
34.83 g Li 3 N
and
34.83 g Li 3 N
1 mol Li 3 N
Grams Li3N
grams Li
MM of Li
1 mol Li3N= 34.83 g Li3N
MM of Li3N
Stoichiometry
moles Li3N
moles Li
moles Li3N
moles Li
Calculate the Limiting Reactant
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
8.07 mol Li 
2 mol Li3 N
 2.69 mol Li3N
6 mol Li
56.0 g N 2 1 mol N 2
 2.00 mol N2
28.02 g N 2
2 mol Li3 N
2.00 mol N 2 
 3.997 mol Li3N
1 mol N 2
Limiting Reactant Problem
Lithium is the limiting reagent.
Calculate the number of grams of lithium nitride
formed in the reaction based on the limiting
reactant: 2.69 mol of lithium
34.83 g Li3 N
2.69 mol Li3 N 
 93.7 g Li N
3
1 mol Li3 N
Product 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
• 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