Calculations from Chemical Equations
Accurate measurement and dosage calculations are critical
in dispensing medicine to patients all over the world.
Foundations of College Chemistry, 14th Ed.
Morris Hein and Susan Arena
Copyright © 2014 John Wiley & Sons, Inc. All rights reserved.
Chapter Outline
9.1 Introduction to Stoichiometry
9.2 Mole-Mole Calculations
9.3 Mole-Mass Calculations
9.4 Mass-Mass Calculations
9.5 Limiting Reactant and Yield Calculations
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole/Molar Mass Review
Molar Mass (MM): sum of the atomic mass of the atoms
in an element, compound, or formula unit.
Mole: Avogadro’s number (6.022 x 1023) of units
(atoms, molecules, ions etc.)
Useful Conversion Factors
grams of a substance
Molar mass =
moles of the substance
MM allows conversion between g and mol of a substance.
number of units of substance
Moles of a substance =
6.022 x 1023 units of substance
© 2014 John Wiley & Sons, Inc. All rights reserved.
Introduction to Stoichiometry
Equations must always be balanced before calculation of
any mass, moles, or volume of a reactant or product!
Stoichiometry: area of chemistry that deals with
quantitative relationships between products and
reactants in chemical equations.
Example
aA+bB
cC+dD
Using X.X g of A, how much C will be formed?
Solving stoichiometry problems always requires the use of:
1. A balanced chemical equation (coefficients must be known!)
2. Conversion factors in units of moles (i.e. mole ratios)
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Mole Ratios
Mole ratio: ratio (conversion factor) between any two
species in a chemical reaction.
Example
2 H2 (g) + O2 (g)
2 H2O (l)
The coefficients of a balanced chemical equation are
used to generate mole ratios.
6 possible mole ratios exist:
2 mol H2
1 mol O2
2 mol H2
2 mol H2O
1 mol O2
2 mol H2O
1 mol O2
2 mol H2
2 mol H2O
2 mol H2
2 mol H2O
1 mol O2
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole Ratios
The mole ratio can be used as a conversion factor to
convert between moles of one substance and another.
Example
2 H2 (g) + O2 (g)
2 H2O (l)
If 4.0 mol of oxygen are present,
how many moles of H2O could be formed?
4.0 mol O2
2 mol H2O
×
1 mol O2
= 8.0 mol H2O
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole Ratios Practice
Given the following balanced chemical equation, write
the mole ratio need to calculate:
a. The moles of H2O produced from 3 moles of CO2
b. The moles of H2 needed to produce 3 moles of H2O.
CO2 (g) + 4 H2 (g)
a.
3.0 mol CO2 ×
CH4 (g) + 2 H2O (l)
2 mol H2O
1 mol CO2
= 6.0 mol H2O
Mole ratio
Desired quantity in the numerator of the mole ratio:
known quantity in the denominator
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole Ratios Practice
Given the following balanced chemical equation, write
the mole ratio need to calculate:
a. The moles of H2O produced from 3 moles of CO2
b. The moles of H2 needed to produce 3 moles of H2O.
CO2 (g) + 4 H2 (g)
b.
3.0 mol H2O ×
CH4 (g) + 2 H2O (l)
4 mol H2
2 mol H2O
= 6.0 mol H2
Mole ratio
Desired quantity in the numerator of the mole ratio:
known quantity in the denominator
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole Ratios Practice
Given the following balanced chemical equation,
what is the mole ratio needed to calculate the following:
the moles of KCl produced when 4.5 moles of
O2 are formed?
2 KClO3 (s)
2 mol KCl
3 mol O2
2 KCl (s) + 3 O2 (g)
3 mol O2
2 mol KCl
3 mol KCl
2 mol O2
2 mol O2
3 mol KCl
Calculate
4.5 mol O2 ×
2 mol KCl
3 mol O2
= 3.0 mol KCl
Mole ratio
© 2014 John Wiley & Sons, Inc. All rights reserved.
Problem Solving for
Stoichiometry Problems
1. Make sure the equation is balanced!
2. If needed, convert the quantity of known substance to
moles.
1 mol substance
Moles = (grams) x
molar mass substance
3. Convert the moles of known substance to desired
substance using a mole ratio.
Mole ratio =
moles of desired substance
moles of known substance
Moles desired substance = Moles of known substance x Mole ratio
From Step 2
© 2014 John Wiley & Sons, Inc. All rights reserved.
Problem Solving for
Stoichiometry Problems
4. Convert moles of desired substance to the desired units
from the problem.
If answer is in moles, you are finished.
If answer is in grams, multiply by the compound’s molar mass.
grams = (moles) x
Molar mass (in g)
1 mole
If answer is in atoms/molecules, multiply by Avogadro’s number.
6.022 x 1023 molecules
Atoms/molecules = (moles) x
1 mole
© 2014 John Wiley & Sons, Inc. All rights reserved.
Problem Solving for
Stoichiometry Problems
Flow Chart for Stoichiometry Problems
Grams of Known
Grams of Desired
Step 2
Step 4
Step 3
Moles of Desired
Moles of Known
Step 2
Using the
Mole Ratio
Step 4
Atoms/Molecules
of Desired
Atoms/Molecules
of Known
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole-Mole Calculations
Known substance is given in moles;
desired substance is requested in moles.
How many moles of CO2 will be produced by reaction of
2.0 mol of glucose, given the following balanced equation?
C6H12O6 + 6 O2
Solution Map
6 CO2 + 6 H2O
mol C6H12O6
mol CO2
The mole ratio needed relates mol C6H12O6 to mol CO2.
6 mol CO2
1 mol C6H12O6
Calculate
2.0 mol C6H12O6 ×
6 mol CO2
1 mol C6H12O6
= 12 mol CO2
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole-Mole Calculations Practice
How many H2O molecules are produced when
0.010 mol O2 react, given the following balanced equation?
2 H2 + O2
2 H2O
a. 8.3 x 10-27 molecules
b. 3.3 x 10-26 molecules mol O2
c. 3.0 x 1021 molecules
d. 1.2 x 1022 molecules
Calculate
Solution Map
mol H2O
molecules H2O
Mole ratio:
2 mol H2O
1 mol O2
2 mol H2O
6.022 x 1023 molecules H2O
0.010 mol O2 ×
×
1 mole H2O
1 mol O2
= 1.2 x 1022 molecules
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole-Mole Calculations Practice
How many moles of Al are produced when 0.5 mol of O2
react, given the following balanced equation?
4 Al + 3 O2
2 Al2O3
Solution Map
mol O2
mol Al
a. 0.38 moles
b. 0.67 moles
Mole ratio:
c. 1.0 moles
4 mol Al
3 mol O2
d. 0.25 moles
Calculate
0.5 mol O2 ×
4 mol Al
= 0.67 mol Al
3 mol O2
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mole-Mass Calculations
What mass of H2 can be produced when 6.0 mol of Al
reacts with HCl?
2 Al + 6 HCl
Solution Map
mol Al
2 AlCl3 + 3 H2
mol H2
g H2
The mole ratio and molar mass of H2 are needed:
3 mol H2
2 mol Al
2.016 g H2
1 mol H2
Calculate
2.016 g H2
3 mol H2
6.0 mol Al ×
×
1 mol H2
2 mol Al
© 2014 John Wiley & Sons, Inc. All rights reserved.
= 18 g H2
Mole-Mass Calculations Practice
How many moles of water are produced when 325 g of
octane (C8H18) are burned?
2 C8H18 + 25 O2
Solution Map
16 CO2 + 18 H2O
g C8H18
mol C8H18
moles H2O
The mole ratio and molar mass of C8H18 are needed:
18 mol H2O
2 mol C8H18
1 mol C8H18
114.2 g C8H18
1 mol C8H18
18 mol H2O
Calculate
325. g C8H18 ×
114.2 g C8H18
×
2 mol C8H18
© 2014 John Wiley & Sons, Inc. All rights reserved.
= 25.6 mol H2O
Mole-Mass Calculations Practice
How many grams of AgNO3 are needed to produce
0.25 mol of Ag2S?
2 AgNO3 + H2S
Ag2S + 2 HNO3
a. 42.5 g
b. 57.1 g
c. 2.19 x 10-3 g
d. 85.0 g
Calculate
mol Ag2S
Solution Map
mol AgNO3
g AgNO3
The mole ratio and
molar mass of AgNO3 are needed:
2 mol AgNO3
169.9 g AgNO3
1 mol Ag2S
1 mol AgNO3
2 mol AgNO3
169.9 g AgNO3
0.25 mol Ag2S ×
×
= 85.0 g AgNO3
1 mol Ag2S
1 mol AgNO3
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mass-Mass Calculations
How many grams of HNO3 are required to produce
8.75 g of N2O from the following reaction?
4 Zn + 10 HNO3
4 Zn(NO3)2 + N2O + 5 H2O
Solution Map g N2O
mol N2O
mol HNO3
g HNO3
The mole ratio and molar masses of N2O and HNO3 are needed:
10 mol HNO3
1 mol N2O
1 mol N2O
44.02 g N2O
63.02 g HNO3
1 mol HNO3
Calculate
1 mol N2O
10 mol HNO3
63.02 g HNO3
8.75 g N2O ×
×
×
1 mol N2O
44.02 g N2O
1 mol HNO3
= 125. g HNO3
© 2014 John Wiley & Sons, Inc. All rights reserved.
Mass-Mass Calculations Practice
How many grams of CrCl3 are required to produce
75.0 g of AgCl using the following reaction?
CrCl3 + 3 AgNO3
a. 204 g
b. 249 g
c. 22.6 g
d. 27.6 g
g AgCl
Cr(NO3)3 + 3 AgCl
Solution Map
mol AgCl
mol CrCl3
g CrCl3
Mole ratio/molar masses needed:
1 mol CrCl3
3 mol AgCl
1 mol AgCl
143.3 g AgCl
Calculate
158.4 g CrCl3
1 mol CrCl3
1 mol AgCl
1 mol CrCl3
158.4 g CrCl3
75.0 g AgCl ×
×
×
143.3 g AgCl
3 mol AgCl
1 mol CrCl3
= 27.6 g CrCl3
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactants
In many chemical reactions,
one reactant is used in excess.
The maximum amount of product formed depends on the
amount of reactant not in excess (the limiting reactant).
A Nonchemical Analogy
To put together a bicycle,
you need several parts.
The number of bicycles
is limited by the part you
have the least of.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactants
Chemical Example
If you started with 5 molecules of H2 and 3 molecules
of Cl2, how much HCl could you make?
H2 + Cl2
2 HCl
Because you need 1 molecule of H2 for each
molecule of Cl2, the Cl2 limits the reaction.
With 3 molecules of Cl2,, you can make a total of 6
molecules of HCl (because of the reaction coefficients).
2 molecules of H2 remain unused (are in excess).
When the coefficients of the balanced equation are more
complex, a general method should be used.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactants
Problem Solving Strategy for Limiting Reactant Problems
1. Calculate the amount of product formed from each
reactant present.
2. The reactant that gives the least amount of product is
limiting; the other reactant is in excess.
3. The amount of product is determined by the
calculation from Step 1 with the limiting reactant.
4. If the amount of excess reactant is desired, determine
the amount of excess reactant needed to consume the
limiting reactant and subtract from the initial quantity
present.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactant Problems
How many moles of HCl can be produced from 4.0 mol of
H2 and 3.5 mol of Cl2? What is the limiting reactant?
H2 + Cl2
2 HCl
Step 1 Calculate the moles of HCl formed from each
reactant using the appropriate mole ratios.
2 mol HCl
4.0 mol H2 ×
= 8.0 moles HCl
1 mol H2
3.5 mol Cl2 ×
2 mol HCl
1 mol Cl2
= 7.0 moles HCl
Step 2 Less HCl is formed with Cl2, so it is the limiting
reactant. The maximum amount of product is 7.0 moles.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactant Problems
How many moles of Fe3O4 can be produced from
16.8 g Fe and 10.0 g H2O? What is the limiting reactant?
3 Fe + 4 H2O
Fe3O4 + 4 H2
Step 1 Calculate the moles of Fe3O4 formed from each
reactant using the appropriate molar masses and
mole ratios.
1 mol Fe
1 mol Fe3O4
= 0.1001 mol Fe3O4
16.8 g Fe ×
×
3 mol Fe
55.45 g Fe
1 mol H2O
1 mol Fe3O4
10.0 g H2O ×
= 0.139 mol Fe3O4
×
4 mol H2O
18.02 g H2O
Step 2 Less Fe3O4 is formed with Fe, so it is the limiting
reactant. The maximum amount of product is 0.1001
moles.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactant Problems
How many grams of AgBr can be produced from
50.0 g MgBr2 and 100.0 g AgNO3?
How much excess reactant remains?
MgBr2 + 2 AgNO3
2 AgBr + Mg(NO3)2
Step 1 Calculate the mass of AgBr formed from each
reactant using the appropriate molar masses and
mole ratios.
1 mol MgBr2
187.4 g AgBr
2 mol AgBr
50.0 g MgBr2 ×
×
×
1 mol MgBr2
1 mol AgBr
184.1 g MgBr2
= 102 g AgBr
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactant Problems
How many grams of AgBr can be produced from
50.0 g MgBr2 and 100.0 g AgNO3?
How much excess reactant remains?
MgBr2 + 2 AgNO3
2 AgBr + Mg(NO3)2
Step 1 Calculate the mass of AgBr formed from each
reactant using the appropriate molar masses and
mole ratios.
1 mol AgNO3
2 mol AgBr
187.4 g AgBr
100.0 g AgNO3 ×
×
×
2 mol AgNO3
169.9 g AgNO3
1 mol AgBr
= 110.3 g AgBr
Step 2 Less AgBr is formed with MgBr2, so it is the limiting
reactant. The maximum amount of product is 102 g.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactant Problems
From the problem on the previous slide,
how much excess reactant remains?
MgBr2 + 2 AgNO3
2 AgBr + Mg(NO3)2
Step 3 Calculate how much AgNO3 reacts with the limiting
reactant, assuming all MgBr2 reacts.
50.0 g MgBr2 ×
1 mol MgBr2
2 mol AgNO3
169.9 g AgNO3
×
×
1 mol MgBr2
1 mol AgNO3
184.1 g MgBr2
= 92.3 g AgNO3
Unreacted AgNO3 = initial – amount reacted
= 100.0 g – 92.3 g = 7.7 g AgNO3 unused
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactant Problems
How many grams of BaSO4 can be produced from
200.0 g of Ba(NO3)2 and 100.0 g of Na2SO4?
Ba(NO3)2 + Na2SO4
BaSO4 + 2 NaNO3
Step 1 Calculate the mass of BaSO4 formed from each
reactant using the appropriate molar masses and
mole ratios.
200.0 g Ba(NO3)2 ×
1 mol Ba(NO3)2
×
1 mol BaSO4
261.4 g Ba(NO3)2 1 mol Ba(NO3)2
×
233.4 g BaSO4
1 mol BaSO4
= 178.6 g BaSO4
© 2014 John Wiley & Sons, Inc. All rights reserved.
Limiting Reactant Practice
How many grams of BaSO4 can be produced from
200.0 g of Ba(NO3)2 and 100.0 g of Na2SO4?
Ba(NO3)2 + Na2SO4
BaSO4 + 2 NaNO3
Step 1 Calculate the mass of BaSO4 formed from each
reactant using the appropriate molar masses and
mole ratios.
1 mol Na2SO4
1 mol BaSO4 233.4 g BaSO4
100.0 g Na2SO4 ×
×
×
1 mol BaSO4
142.0 g Na2SO4 1 mol Na2SO4
= 164.4 g BaSO4
Step 2 Less BaSO4 is formed with Na2SO4, so it is the
limiting reactant. The maximum amount of product is
the smaller amount, 164.4 g.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Reaction Yield
The amount of products formed calculated by
stoichiometry are the maximum yields possible (100%).
Yields are often lower due to side reactions, loss of
product while isolating/transferring the material, etc.
Theoretical yield: maximum possible yield for a reaction,
calculated based on the balanced chemical equation.
Actual yield: actual yield obtained from the reaction.
Percent yield: ratio of the actual and theoretical yield
Actual yield
× 100
% yield =
Theoretical yield
© 2014 John Wiley & Sons, Inc. All rights reserved.
Calculate % Yield
Calculate the percent yield of AgBr if 375.0 g of the
compound are prepared from 200.0 g of MgBr2.
MgBr2 + 2 AgNO3
Mg(NO3)2 + 2 AgBr
To calculate the % yield, calculate the theoretical yield.
Solution Map
g MgBr2
mol MgBr2
mol AgBr
g AgBr
200.0 g MgBr2 × 1 mol MgBr2 × 2 mol AgBr × 187.8 g AgBr
184.1 g MgBr2
1 mol MgBr2
1 mol AgBr
= 408.0 g AgBr
With theoretical yield, we can calculate % yield
Actual yield
375.0 g
×
× 100 = 91.91 %
100 =
% yield =
408.0 g
Theoretical yield
© 2014 John Wiley & Sons, Inc. All rights reserved.
% Yield Practice
Calculate the percent yield of Al2O3 if
125.0 g of Al give 100.0 g of Al2O3.
2 Al + 3 CrO
Al2O3 + 3 Cr
To calculate the % yield, calculate the theoretical yield.
Solution Map
g Al
mol Al
mol Al2O3
g Al2O3
1 mol Al
1 mol Al2O3
102.0 g Al2O3
×
×
125.0 g Al ×
2 mol Al
26.98 g Al
1 mol Al2O3
= 236.3 g Al2O3
With theoretical yield, we can calculate % yield
Actual yield
100.0 g
% yield =
× 100 =
× 100 = 42.32 %
Theoretical yield
236.3 g
© 2014 John Wiley & Sons, Inc. All rights reserved.
Learning Objectives
9.1 Introduction to Stoichiometry
Define stoichiometry and describe the strategy required to
solve problems based on chemical equations.
9.2 Mole-Mole Calculations
Solve problems where the reactants and products
are both in moles.
9.3 Mole-Mass Calculations
Solve problems where known mass is given and the answer
is to be determined in moles or the moles of known are
given and mass is determined.
© 2014 John Wiley & Sons, Inc. All rights reserved.
Learning Objectives
9.4 Mass-Mass Calculations
Solve problems where mass is given and the desired
unit to be determined is mass.
9.5 Limiting Reactant and Yield Calculations
Solve problems involving limiting reactants and yield.
© 2014 John Wiley & Sons, Inc. All rights reserved.