Chapter 3
The Mole and
Stoichiometry
Chapter 3
Table of Contents
3.1
3.2
3.3
3.4
3.5
3.6
3.9
3.10
Counting by Weighing
Atomic Masses
The Mole
Molar Mass
Percent Composition by Mass of Compounds
Determining the Formula of a Compound
Stoichiometric Calculations: Amounts of Reactants
and Products
Calculations Involving a Limiting Reactant
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2
Section 3.1
Counting by Weighing
Objectives 3.1 – 3.3
1. To understand the concept of average mass
2. To learn how counting can be done by
massing
3. To understand atomic mass and learn how it
is determined
4. To understand the mole concept and
Avogadro’s number
5. To learn to convert among moles and mass.
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3
Section 3.1
Counting by Weighing
•
•
Need average mass of the object.
Objects behave as though they were all
identical.
Find the average mass of one bean.
Find the mass of 1000 beans.
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4
Section 3.1
Counting by Weighing
Exercise:
A pile of marbles weigh 394.80 g. 10 marbles weigh
37.60 g. How many marbles are in the pile?
Average mass of 1 marble = 37.60 g = 3.76g/marble
10 marbles
394.80 g
1 marble
3.76 g
= 105 marbles
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5
Section 3.2
Atomic Masses
Counting
by Weighing
•
•
Elements occur in nature as mixtures of isotopes.
Carbon = 98.89% 12C
1.11% 13C
< 0.01% 14C
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6
Section 3.2
Atomic Masses
Counting
by Weighing
Average Atomic Mass for Carbon
98.89% of 12 amu + 1.11% of 13.0034 amu =
exact number
(0.9889)(12 amu) + (0.0111)(13.0034 amu) =
12.01 amu
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7
Section 3.2
Atomic Masses
Counting
by Weighing
Average Atomic Mass for Carbon
•
•
Even though natural carbon does not
contain a single atom with mass 12.01, for
stoichiometric purposes, we can consider
carbon to be composed of only one type of
atom with a mass of 12.01.
This average enables us to count atoms of
natural carbon by weighing a sample of
carbon.
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Section 3.2
Atomic Masses
Counting
by Weighing
Schematic Diagram of a Mass Spectrometer
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Section 3.2
Atomic Masses
Counting
by Weighing
Exercise
An element consists of 62.60% of an isotope
with mass 186.956 amu and 37.40% of an
isotope with mass 184.953 amu.
• Calculate the average atomic mass and
identify the element.
186.2 amu
Rhenium (Re)
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Section 3.3
The Mole by Weighing
Counting
• The number equal to the number of carbon
atoms in exactly 12 grams of pure 12C.
• 1 mole of anything = 6.022 x 1023 units of that
thing (Avogadro’s number).
• 1 mole C = 6.022 x 1023 C atoms = 12.01 g C
• Other units? Dozen? Fathom?? Stone??
Hand?
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Section 3.3
The Mole by Weighing
Counting
The Mole !!!!!
3
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Section 3.3
The Mole by Weighing
Counting
The Mole !!!!!!
• Let’s use the Periodic Table to determine the
Atomic Mass of the following: (Rounding?)
• H
• Co
• Rn
• Fr
• W
• Ne
• How many atoms do each of the above atomic
mass values contain? What are the units?
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13
Section 3.3
The Mole by Weighing
Counting
How many is a mole?
• 6.022 x 1023 Casio EXP below 3, TI30 XA EE above 7,
XIIS 2ndEE above 7, XZ multiview x10n 3 above #8
• Al Foil
• 1 mole of marbles would cover the earth
• To a depth of 50 miles
• Would you accept $1 million ($1 x 106) to count to a
mole?
• If you counted 1 per second, it would take you
2 x 1016 years to finish
• Your hourly wage would be $5 x 10-15 per hour
• It would take hundreds of millions of years to earn
$0.01
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Section 3.3
The Mole by Weighing
Counting
How to convert using unit analysis.
• First, let’s learn to convert from moles to grams.
• What is the atomic mass (molar mass) of Fe?
• What are the units of atomic mass (molar mass)?
• How many grams are contained in 1 mol of Fe?
• How many grams are contained in 2 mol of Fe?
• How many grams are contained in 0.5 mol of Fe?
• How many grams are contained in 2.4 mol of Fe?
• 2.4 mol Fe
55.85 g Fe =
•
1 mol Fe
• Know to Go approach, Unit Analysis!!
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Section 3.3
The Mole by Weighing
Counting
How to convert using unit analysis.
• Use Unit Analysis to convert from moles to grams:
• 0.250 mol Al
• 1.28 mol Ca
• 0.371 mol P
• 10.24 mol Au
• To convert from moles to grams, multiply by the
molar mass.
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16
Section 3.3
The Mole by Weighing
Counting
How to convert using unit analysis.
• Second, let’s learn to convert from grams to moles
• What is the atomic mass (molar mass) of Al?
• What are the units of atomic mass (molar mass)?
• How many moles of atoms do 27 g of Al contain?
• How many moles of atoms do 54 g of Al contain?
• How many moles of atoms do 108 g of Al contain?
• How many moles of atoms do 56 g of Al contain?
• 56 g Al
1 mole Al =
•
27 g Al
• Know to Go approach, Unit Analysis!!
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Section 3.3
The Mole by Weighing
Counting
How to convert using unit analysis.
• Use Unit Analysis to convert from grams to moles:
• 220 g Al
• 568 g Si
• 3.61 mg He
• 26.7 g Li
• To convert from grams to moles, divide by the molar
mass.
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18
Section 3.3
The Mole by Weighing
Counting
Objectives Review 3.1 – 3.3
1. To understand the concept of average mass
2. To learn how counting can be done by
massing
3. To understand atomic mass and learn how it
is determined
4. To understand the mole concept and
Avogadro’s number
5. To learn to convert among moles and mass.
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19
Section 3.3
The Mole by Weighing
Counting
Objectives 3.3 part 2
1. To learn to convert among moles, mass, and
# of atoms.
2. Unit Analysis! Multi Step!
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20
Section 3.3
The Mole by Weighing
Counting
• Conversions Help Page!
• To convert from moles to grams, multiply by the molar mass.
• To convert from grams to moles, divide by the molar mass.
• 1 mole = 6.02 X 1023
• Grams  moles  # atoms
• Xg
• 1 mol
or
• 6.02 X 1023 atoms
•
1 mol
or
1 mol
Xg
1 mol
.
6.02 X 1023 atoms
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Section 3.3
The Mole by Weighing
Counting
• Know to Go approach, Unit Analysis!! Let it Guide
you!
• Grams  moles  # atoms
• How many atoms in 79 g of Se? (Know to go?)
• 79 g Se 1 mol Se
6.02 X 1023 atoms Se =
•
78.96 g Se
1 mol Se
• How many atoms in 2.35 mol of #%^$?
• 2.35 mol #%^$
6.02 X 1023 atoms #%^$ =
•
1 mol #%^$
• Remember that a mole measures quantity!!
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Section 3.3
The Mole by Weighing
Counting
• Know to Go approach, Unit Analysis!! Let it Guide
you!
• Grams  moles  # atoms
• How many atoms are contained in 4.17 mol of Al?
• How many atoms are contained in 4.17 mol of Si?
• How many atoms are contained in 1.67 mol of Al?
• How many atoms are contained in 2 X 10 -4 mol of Al?
• How many moles are present in 2.13 X 1024 C atoms?
• What is the mass in g of the above moles of C?
• What is the mass in g of 9.4 X 1025 atoms of Mg?
• How many atoms are in 365 g of V?
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Section 3.3
The Mole by Weighing
Counting
Concept Check
Calculate the number of copper atoms in a
63.55 g sample of copper.
6.022×1023 Cu atoms
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Section 3.3
The Mole by Weighing
Counting
Concept Check
Calculate the number of grams contained in a
sample of 1.21 X 1023 Al atoms.
5.42 g Al
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Section 3.3
The Mole by Weighing
Counting
Exercise
Rank the following according to number of
moles (greatest to least):
a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
b)
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a)
c)
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26
Section 3.3
The Mole by Weighing
Counting
Exercise
Rank the following according to number of
atoms (greatest to least):
a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
b)
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a)
c)
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27
Section 3.3
The Mole by Weighing
Counting
Objectives Review 3.3 part 2
1. To learn to convert among moles, mass, and # of
atoms.
2. Unit Analysis! Multi Step!
3. Work Session: Page 117 # 21,27(amu = g),34,35
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Section 3.4
Molar Mass
Objectives
1. To understand the definition of
molar mass for atoms and compounds
2. To learn to convert between
moles and mass for compounds
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Section 3.4
Molar Mass
1. Molar Mass
• A compound is a collection of atoms bound together.
• The molar mass of a compound is obtained by summing
the masses of the component atoms.
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Section 3.4
Molar Mass
1. Molar Mass
• For compounds containing ions the molar mass is
obtained by summing the masses of the component
ions.
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Section 3.4
Molar Mass
• Conversions Help Page!
• 1 mole = 6.02 X 1023
• Grams  moles  # atoms or molecules
• For molecules…
• 6.02 X 1023 molecules
•
1 mol
or
1 mol
.
6.02 X 1023 molecules
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Section 3.4
Molar Mass
2. Calculations Using Molar Mass (~Atomic Mass)
• What is the Molar Mass of methane gas CH4?
• Use Unit Analysis to convert:
• How many molecules are contained within 1 mole of
methane gas CH4?
• How many C atoms are contained within 1 mole of
methane gas CH4?
• How many H atoms are contained within 1 mole of
methane gas CH4?
• How many g of C are contained within 1 mole of
methane gas?
• How many g of H are contained within 1 mole of
methane gas?
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33
Section 3.4
Molar Mass
2. Calculations Using Molar Mass (~Atomic Mass)
• Find the Molar Mass of SO2
• Find the Molar Mass of NaCl
• Find the Molar Mass of CaCO3
• What would the mass of 4.86 mol CaCO3 be?
• Find the Molar Mass of Na2SO4
• How many moles in 300 g of Na2SO4? (know-go?)
•
•
•
•
SO2: 64 g/mol
NaCl: 58.5 g/mol
CaCO3: 100 g/mol, 486 g
Na2SO4: 142 g/mol, 2.11 mol
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34
Section 3.4
Molar Mass
2. Calculations Using Molar Mass (~Atomic Mass)
• How many moles in 1.56 g of Juglone C10H6O3?
• (Dye and herbicide made from black walnuts)
• 8.96 X 10 -3 mol C10H6O3
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Section 3.4
Molar Mass
2. Calculations Using Molar Mass (~Atomic Mass)
• Bees release 1 X 10-6 g of isopentyl acetate (C7H14O2)
during a sting. (Also a banana scent!)
• How many moles of isopentyl acetate are released?
• 8 X 10-9 mol C7H14O2
• How many molecules of isopentyl acetate are
released?
• 5 X 1015 molecules C7H14O2
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Section 3.4
Molar Mass
2. Calculations Using Molar Mass (~Atomic Mass)
• How many molecules are contained in 135 g of
Teflon (C2F4)?
• 8.127 X 1023 molecules of C2F4
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Section 3.4
Molar Mass
Objectives Review
1. To understand the definition of molar mass
for atoms and compounds
2. To learn to convert between moles and
mass for compounds
3. Work Session: 118 # 37,39,41,43,45,47,49
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Section 3.5
Percent Composition of Compounds
Objectives 3.5 and 3.6
1. To learn to calculate the mass percent of
an element in a compound
2. To learn to calculate empirical formulas
3. To learn to calculate the molecular
formula of a compound
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Section 3.5
Percent Composition of Compounds
•
Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound
•
For iron in iron(III) oxide, (Fe2O3):
2( 55.85 g)
111.70 g
mass % Fe =
=
× 100% = 69.94%
2( 55.85 g)+ 3( 16.00 g) 159.70 g
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Section 3.5
Percent Composition of Compounds
1. Percent Composition of Compounds
• Rules to determine the percent
composition by mass of each element:
• 1. Determine the molar mass of the compound
• 2. Divide the mass of each element type by the
total mass of the compound (#1)
• 3. Multiply by 100 for percent
Mass percent = mass of a given element in 1 mol of compound 100%
mass of 1 mol of compound
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Section 3.5
Percent Composition of Compounds
1. Percent Composition of Compounds
• Determine the percent composition by mass of
each element in methane gas CH4
• Molar Mass
• Each Element
• X 100
• 75% C, 25% H
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Section 3.5
Percent Composition of Compounds
1. Percent Composition of Compounds
• Determine the percent composition by
mass of each element in SO2
• Molar Mass
• Each Element
• X 100
• 50/50
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Section 3.5
Percent Composition of Compounds
1. Percent Composition of Compounds
• Determine the percent composition by mass of
each element in NaCl
• Molar Mass
• Each Element
• X 100
• 39% Na, 61% Cl
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Section 3.5
Percent Composition of Compounds
1. Percent Composition of Compounds
• Determine the percent composition by mass of
each element in CaCO3
• Molar Mass
• Each Element
• X 100
• 40% Ca, 12% C, 48% O
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Section 3.6
Determining the Formula of a Compound
Analyzing for Carbon and Hydrogen
•
Device used to determine the mass
percent of each element in a compound.
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Section 3.6
Determining the Formula of a Compound
Formulas
•
Empirical formula = CH
 Simplest whole-number ratio
• Molecular formula = (empirical formula)n
[n = integer]
• Molecular formula = C6H6 = (CH)6
 Actual formula of the compound
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Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
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Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
• Patch’s Interpretation:
• Calculate moles if you don’t already have
them.
• Divide all by the smallest # of moles
• Get to Whole Number Multiples.
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Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
• Determine the empirical formula if an analysis of
a C, H, and O compound resulted in the
following :
• 0.0806 g C
• 0.01353 g H
• 0.1074 g O
• 0.00671 mol C, 0.01353 mol H, 0.00671 mol O
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50
Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
• Now, let’s find the empirical formula:
• 0.00671 mol C
• 0.01353 mol H
• 0.00671 mol O
• CH2O(simplest) C2H4O2 ; C4H8O4 ;C5H10O5
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Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
• Determine the empirical formula if an
analysis of a Ni and O compound resulted
in the following :
• 0.2636 g Ni
• 0.0718 g O
• NiO
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52
Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
• AlxOy
• 4.151 g Al
• 3.692 g O
• Al2O3
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53
Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
• It is observed that 0.664 g of lead
combined with 0.2356 g of chlorine.
Determine the empirical formula.
• PbCl2
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Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
• Determine the empirical formula of cisplatin, a cancer
tumor treatment drug that has the following
% composition by mass:
• 65.02% Pt
• 9.34% N
• 2.02% H
• 23.63% Cl
• Assume some mass (100g)
• PtN2H6Cl2
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Section 3.6
Determining the Formula of a Compound
2. Calculation of Empirical Formulas
• The most common form of nylon (Nylon-6)
is 63.68% C, 12.38% N, 9.80% H, and
14.4% O by mass. Calculate the empirical
formula for Nylon-6.
• C6NH11O
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56
Section 3.6
Determining the Formula of a Compound
3. Calculation of Molecular Formulas
• The molecular formula is the exact formula of
the molecules present in a substance.
• The molecular formula is a whole number
multiple (n) of the empirical formula.
n = molar mass molecular formula
molar mass empirical formula
1. Find n
2. (empirical formula) n = molecular formula
**You have to be given the molar mass of the
molecular formula!
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Section 3.6
Determining the Formula of a Compound
3. Calculation of Molecular Formulas
• Calculate the molecular formula for a
compound if the empirical formula is P2O5
and the molecular molar mass is 283.88
g/mol.
• P4O10 Check the mm=(4(31)+10(16))=284
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Section 3.6
Determining the Formula of a Compound
3. Calculation of Molecular Formulas
• Calculate the empirical and molecular
formulas for a compound that gives the
following mass percentages upon analysis:
• 71.65% Cl
• 24.27% C
• 4.07% H
• The molar mass is known to be 98.96 g/mol
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Section 3.6
Determining the Formula of a Compound
3. Calculation of Molecular Formulas
• 71.65% Cl
• 24.27% C
• 4.07% H
• ClCH2 empirical
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Section 3.6
Determining the Formula of a Compound
3. Calculation of Molecular Formulas
• ClCH2 (empirical)
• The molar mass is known to be 98.96 g/mol
• n = molar mass molecular formula
molar mass empirical formula
• Cl2C2H4
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Section 3.6
Determining the Formula of a Compound
3. Calculation of Molecular Formulas
• Caffeine (MM = 194.2 g/mol) contains by
mass:
• 49.48% C
• 5.15% H
• 28.87% N
• 16.49% O
• Determine the molecular formula.
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Section 3.6
Determining the Formula of a Compound
3. Calculation of Molecular Formulas
• 49.48% C
• 5.15% H
• 28.87% N
• 16.49% O
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Section 3.6
Determining the Formula of a Compound
3. Calculation of Molecular Formulas
• Caffeine (MM = 194.2 g/mol)
• n = molar mass molecular formula
molar mass empirical formula
C8H10N4O2
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Section 3.6
Determining the Formula of a Compound
Hydrate Lab
Calculate the molar mass
MgSO4 * 7H2O
Mg + S + 4(O) + 7(H2O)
Calculate the % mass of waters…
7(H2O)
. X 100
Mg + S + 4(O) + 7(H2O)
MM = 246g/mol : % H2O = 51.22%
Silica Gel Packaging…
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65
Section 3.6
Determining the Formula of a Compound
Objectives Review 3.5 and 3.6
1. To understand the meaning of empirical
formula
2. To learn to calculate empirical formulas
3. To learn to calculate the molecular
formula of a compound
4. Work Session: pg 119 # 60, 67, 69, 68, 74
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Objectives 3.9 Part 1
1. To understand the information given in a
balanced equation
2. To use a balanced equation to determine
relationships between moles of reactant and
products (mole ratio) (Stoichiometry)
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Sandwich Shoppe!
1. Let’s create a sandwich recipe…PHeT
2. Slice, dozen, 100, 6.02 X 1023, one mole
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Quinoa Side Dish
www.allrecipes.com
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
1. Information Given by Chemical Equations
• The coefficients of a balanced equation give
the relative numbers of molecules.
• A balanced chemical equation gives relative
numbers (or moles) of reactant and product
molecules that participate in a chemical
reaction.
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• A balanced equation can predict the moles of
product that a given number of moles of reactants
will yield. How many moles 02 from __ moles H2O?
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• The mole ratio allows us to convert from
moles of one substance in a balanced
equation to moles of a second substance in
the equation.
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
•
•
•
•
•
The mole ratio is determined by:
Mole ratio = new mole substance A
balanced mole substance A
How many mole ___ from ___ moles ___?
2H2O  2H2 + O2
• mole
mole ratio
mole
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction.
2.
3. Use the balanced equation to set up the
appropriate mole ratios.
4. Use the appropriate mole ratios to
calculate the number of moles of desired
reactant or product.
5.
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Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• Must first have a BALANCED equation
• __C3H8(g) + __O2  __CO2(g) + __H2O(g) + heat
• 4.30 C3H8 +__O2  __ CO2 + __ H2O + heat
• Calculate the number of moles of CO2 formed
when 4.30 mol C3H8 combust.
• mole
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mole ratio
mole
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77
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• From this new mole value, we can calculate the
new recipe!
•
C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat
• 4.30 C3H8 +__O212.9 CO2 + __ H2O + heat
• Calculate the number of moles of H2O formed
when 4.30 mol C3H8 combust.
• mole
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mole ratio
mole
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78
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• For all reactants AND products!
•
C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat
• 4.30 C3H8 +__O212.9 CO2 + 17.2 H2O + heat
• Calculate the number of moles of O2 consumed
when 4.30 mol C3H8 combust.
• mole
mole ratio
mole
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79
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• Or in other words…(X) There is some factor X
that distributes to all of the coefficients…
•
C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat
• 1(X)C3H8 + 5(X)O23(X)CO2 + 4(X) H2O+heat
• That X is the mole ratio =
New
Balanced
• 4.30 C3H8 +__O2__ CO2 + __ H2O + heat
• __ C3H8 +__O2__ CO2 + __ H2O + heat
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80
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• C3H8(g) + 5O2  3CO2(g) + 4H2O(g) + heat
• Calculate the number of moles of all reactants
and products when 2.70 mol C3H8 combust.
• 2.7 C3H8 +__O2 __ CO2 + __ H2O + heat
• 2.7 C3H8 +13.5 O2 8.1 CO2 + 10.8 H2O + heat
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81
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• Balance the following equation and determine
how much NH3 can be made from 1.3 mol H2.
• __N2 + __H2  __NH3
•
N2 + 3H2  2NH3
• How much N2 will be needed to completely
react the 1.3 mol H2?
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82
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mole-mole Relationships
• Using the balanced equation, determine the
new mole recipe given the new mole value.
•
2 K + 2 H2O 
H2 + 2 KOH
• ___K + 1.7 H2O  ___ H2 + ___KOH
• 1.7 K + 1.7 H2O  0.85 H2 + 1.7 KOH
•
4 NH3 + 5 O2  4 NO + 6 H2O
• ___ NH3 + ___ O2  0.6 NO + ___ H2O
• 0.6 NH3 + 0.75 O2  0.6 NO + 0.9 H2O
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83
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction.
2.
3. Use the balanced equation to set up the
appropriate mole ratios.
4. Use the appropriate mole ratios to
calculate the number of moles of desired
reactant or product.
5.
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84
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Objectives Review 3.9 Part 1
1. To understand the information given in a
balanced equation
2. To use a balanced equation to determine
relationships between moles of reactant and
products (mole ratio) (Stoichiometry)
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85
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Objectives 3.9 Part 2
1. To learn to use Stoichiometry to relate
masses of reactants and products in a
chemical reaction
2. To perform mass calculations that involve
scientific notation
3. Calculate the theoretical predications for a
reaction that will be performed in the lab
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86
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
1. Mass Calculations
• Stoichiometry is the process of using a
balanced chemical equation to determine the
relative masses (quantities) of reactants and
products involved in a reaction.
– Scientific notation can be used for the
masses of any substance in a chemical
equation.
• Mole ratio = new mole substance A
•
balanced mole substance A
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87
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
1. Mass Calculations
•
N2 + 3H2  2NH3
• How much NH3 will be produced when 2.6 g H2
completely reacts with N2?
• **Can’t do a mass ratio!!**
• Must first convert from grams to moles.
• Don’t forget the mole!
• grams
• mole
grams
mole ratio
mole
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88
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
1. Mass Calculations
•
N2 + 3H2  2NH3
• How much NH3 will be produced when 2.6 g H2
completely reacts with N2?
• 2.6 grams H2
• mole
grams
mole ratio
• 0.867 mol NH3: 14.71 g NH3
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mole
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89
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
1. Mass Calculations
•
N2 +
3H2 
2NH3
•
__ N2 + 1.3 H2  0.867 NH3 (new mol recipe)
• __ g N2 2.6 g H2
14.71 g NH3 (new g recipe)
• How much N2 will be consumed when 2.6 g H2
completely reacts?
• Use the same mole ratio—
• 0.433 mol N2: 12.12 g N2
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90
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
1. Mass Calculations
•
N2 +
3H2 
2NH3
• 0.433 N2 + 1.3 H2  0.867 NH3 (new mol recipe)
• 12.12 g N2 2.6 g H2
14.71 g NH3 (new g recipe)
• How does this relate to the Law of Conservation of
Mass?
• Does the mass reactants = mass products?
• 12.12 g + 2.6 g = 14.72 g
• Compared to 14.71 g Why the 0.01 g difference?
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91
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the
reactant or product to moles of that
substance.
3. Use the balanced equation to set up the
appropriate mole ratios.
4. Use the appropriate mole ratios to
calculate the number of moles of desired
reactant or product.
5. Convert from moles back to grams if
required by the problem.
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92
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
1. Mass Calculations
•
__Al(S) + __I2(g)  __AlI3(s)
• Balance the equation and calculate the mass of
I2 needed to completely react with 35.0 g Al.
Also calculate the mass of AlI3 that will be
formed.
• grams
grams
• mole
mole ratio
mole (next)
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93
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
•
•
•
2Al(S) + 3I2(g)
__Al(S) + __I2(g)
35.0 g
 2AlI3(s)
 __AlI3(s) (new mole)
• Balance the equation and calculate the mass of I2
needed to completely react with 35.0 g Al. Also
calculate the mass of AlI3 that will be formed.
• 35.0 g Al
(all new moles)
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94
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
•
•
•
2 Al(S) +
3 I2(g)
1.3 Al(S) + 1.95 I2(g)
35.0 g
 2 AlI3(s)
 1.3 AlI3(s) (new mole)
• Balance the equation and calculate the mass of I2
needed to completely react with 35.0 g Al. Also
calculate the mass of AlI3 that will be formed.
• Now convert mole  grams for each
• 1.95 mol I2
=
• 1.3 mol AlI3
=
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95
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
•
•
•
2 Al(S) +
3 I2(g)
1.3 Al(S) + 1.95 I2(g)
35.0 g
495.3 g
 2 AlI3(s)
 1.3 AlI3(s) (new mole)
530.4 g
• Balance the equation and calculate the mass of I2
needed to completely react with 35.0 g Al. Also
calculate the mass of AlI3 that will be formed.
• Check the Law of Conservation of Mass!
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96
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Scientific Notation (CO2 scrubber on space vehicles)
•
•
•
•
•
•
__LiOH(s) + __CO2(g)  __Li2CO3(s) + __H2O(l)
What mass of CO2 can 1 X 103 g LiOH absorb?
What mass of H2O will be available for use?
What mass of Li2CO3 will be generated?
First Balance! Then, below!!
grams
grams
• mole
mole ratio
mole
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97
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Scientific Notation (CO2 scrubber on space vehicles)
•
•
•
•
2 LiOH(s) + CO2(g)  Li2CO3(s) +
H2O(l)
__LiOH(s) + __CO2(g)  __Li2CO3(s) + __H2O(l)
What mass of CO2 can 1 X 103 g LiOH absorb?
Do new mole recipe next…
• grams
• mole
grams
mole ratio
mole
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98
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Scientific Notation (CO2 scrubber on space vehicles)
• 2 LiOH(s) + CO2(g)  Li2CO3(s) +
H2O(l)
• 41.67LiOH + 20.8CO2 20.8Li2CO3 + 20.8H2O
• Now convert all moles to grams.
• 20.8 mol CO2
=
• 20.8 mol Li2CO3
=
• 20.8 mol H2O
=
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99
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Scientific Notation (CO2 scrubber on space vehicles)
•
•
•
•
•
•
•
•
•
2 LiOH(s) + CO2(g)  Li2CO3(s) +
H2O(l)
41.67LiOH + 20.8CO2 20.8Li2CO3 + 20.8H2O
1 X 103 g
915.2 g
1539.2 g
374.4 g
What mass of CO2 can 1 X 103 g LiOH absorb?
What mass of H2O will be available for use?
What are some possible uses for the H2O?
What mass of Li2CO3 will be generated?
What will be done with the Li2CO3?
Conserved?
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100
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations Using Scientific Notation
• Hydrofluoric acid, an aqueous solution
containing dissolved hydrogen fluoride, is used
to etch glass by reacting with silica, SiO2, in the
glass to produce gaseous silicon tetrafluoride
and liquid water. The unbalanced reaction is:
• __HF(aq) + __SiO2(s)  __SiF4(g) + __H2O(l)
• Calculate the mass of HF that is needed and
the products that are produced to react
completely with 5.68 g SiO2.
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101
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations Using Scientific Notation
• 4 HF(aq) +
SiO2(s) 
SiF4(g) + 2 H2O(l)
• __ HF(aq) + __ SiO2(s)  __ SiF4(g) + __ H2O(l)
• 5.68 g SiO2
=
• Mole ratio
• grams
•
• mole
grams
mole ratio
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mole
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102
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations Using Scientific Notation
• 4 HF(aq) +
SiO2(s) 
SiF4(g) + 2 H2O(l)
• 0.378
9.45 X10-2  9.45 X10-2 + 0.189
(moles)
• 7.56 g
5.68 g
 9.828 g
3.41 g
(grams)
• Was mass Conserved?
• grams
• mole
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grams
mole ratio
mole
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103
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations Comparisons
• Which is the better antacid? Baking Soda,NaHCO3,or
MOM, Mg(OH)2?
• How many moles of HCl will react with 1.00 g of each
antacid?
• NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g) Demo
• Mg(OH)2(s) + 2HCl(aq)  2H2O(l) + MgCl2(aq)
• grams
• mole
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grams
mole ratio
mole
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104
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations Comparisons
• Which is the better antacid? Baking Soda,NaHCO3,or
MOM, Mg(OH)2?
• How many moles of HCl will react with 1.00 g of each
antacid?
• NaHCO3(s) + HCl(aq)  NaCl(aq) + H2O(l) + CO2(g)
• 1.0 g/1.19 X 10-2 mol NaHCO3
• 1.19 X 10-2 mol HCl
• Mg(OH)2(s) + 2HCl(aq)  2H2O(l) + MgCl2(aq)
• 1.0 g/1.71 X 10-2 mol Mg(OH)2
• 3.42 X 10-2 mol HCl
• MOM is better!
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105
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations for Lab- don’t round too much!
• Predict the charged balanced products and
their solubility. Then balance the equation, and
calculate the number of grams of all reactants
and products needed to completely react with
0.01 mol SrCl2.
• ___SrCl2 + ___Na2CO3
•
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106
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations for lab
•
SrCl2 +
Na2CO3
SrCO3 + 2NaCl
• 0.01 SrCl2 + ___Na2CO3___SrCO3+___NaCl
• 0.01SrCl2 + 0.01Na2CO3  0.01SrCO3 + 0.02NaCl
•
1.59 g +
1.06 g

1.48 g
+
1.17 g
• Conservation of mass?
• Theoretical vs Real…
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107
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations for lab
• Calculate the amount of grams contained in
0.01 mol the following hydrated crystals:
• SrCl2 x 6H2O
• Na2CO3 X H2O
• 2.67 g SrCl2 x 6H2O
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1.24 g Na2CO3 X H2O
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108
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
2. Mass Calculations for lab
• 0.01SrCl2 x 6H2O + 0.01Na2CO3 X H2O  0.01SrCO3 + 0.02NaCl
•
2.67 g
+
1.24 g
 1.48 g + 1.17 g
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109
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
1. Balance the equation for the reaction.
2. Convert the known mass of the reactant
or product to moles of that substance.
3. Use the balanced equation to set up the
appropriate mole ratios.
4. Use the appropriate mole ratios to
calculate the number of moles of desired
reactant or product.
5. Convert from moles back to grams if
required by the problem.
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110
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Objectives Review 3.9 Part 2
1. To understand the information given in a
balanced equation
2. To use a balanced equation to determine
relationships between moles of reactant and
products (mole ratio) (Stoichiometry)
3. Calculate the theoretical predications for a
reaction that will be performed in the lab
4. Work Session: pg 121 # 89, 91, 94, 95 and
Challenge Question next 3 slides.
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111
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Work Session Challenge Question
Methane (CH4) reacts with the oxygen in the
air to produce carbon dioxide and water.
Ammonia (NH3) reacts with the oxygen in the
air to produce nitrogen monoxide and water.
 Write balanced equations for each of
these reactions.
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112
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Work Session Challenge Question
Methane (CH4) reacts with the oxygen in the
air to produce carbon dioxide and water.
Ammonia (NH3) reacts with the oxygen in the
air to produce nitrogen monoxide and water.
 What mass of ammonia would produce
the same amount of water as 1.00 g of
methane reacting with excess oxygen?
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113
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Let’s Think About It
•
Where are we going?

•
To find the mass of ammonia that would
produce the same amount of water as 1.00 g of
methane reacting with excess oxygen.
How do we get there?

We need to know:
• How much water is produced from 1.00 g of
methane and excess oxygen.
• How much ammonia is needed to produce
the amount of water calculated above.
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114
Section 3.10
The Concept of Limiting Reagent
Objectives
1. To understand the concept of limiting
reactants
2. To learn to recognize the limiting reactant in
a reaction
3. To learn to use the limiting reactant to do
stoichiometric calculations
4. To learn to calculate percent yield
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115
Section 3.10
The Concept of Limiting Reagent
Limiting Reactants
•
•
Limiting reactant – the reactant that is
consumed first and therefore limits the
amounts of products that can be formed.
Determine which reactant is limiting to
calculate correctly the amounts of
products that will be formed.
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116
Section 3.10
The Concept of Limiting Reagent
1. The Concept of Limiting Reactants
• Stoichiometric mixture (balanced)
– N2(g) + 3H2(g)  2NH3(g)
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117
Section 3.10
The Concept of Limiting Reagent
1. The Concept of Limiting Reactants
• Limiting reactant mixture (runs out first)
– N2(g) + 3H2(g)  2NH3(g)
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118
Section 3.10
The Concept of Limiting Reagent
Mixture of CH4 and H2O Molecules Reacting
CH4 + H2O  3H2 + CO
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119
Section 3.10
The Concept of Limiting Reagent
CH4 + H2O  3H2 + CO
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120
Section 3.10
The Concept of Limiting Reagent
Limiting Reactants
•
•
•
•
The amount of products that can form is
limited by the methane.
Methane is the limiting reactant.
Water is in excess.
Limiting and Excess Reactants depend
on the actual amounts of reactants
present and must be calculated for each
different reaction. (Water won’t always
be in excess!)
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121
Section 3.10
The Concept of Limiting Reagent
Limiting Reactants
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122
Section 3.10
The Concept of Limiting Reagent
2. Calculations Involving a Limiting Reactant
• What mass of water is needed to react with
249 g methane (CH4)?
• CH4 + H2O  3H2 + CO
• 249 g CH4
• How many g of H2 and CO are produced?
• 279 g H2O, 93 g H2, 434 g CO
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123
Section 3.10
The Concept of Limiting Reagent
2. Calculations Involving a Limiting Reactant
• How many g of H2 and CO are produced
when 249 g methane (CH4) reacts with
300 g H2O?
• CH4 + H2O  3H2 + CO
• 249 g + 279 g  93 g + 434 g (last slide)
• 249 g + 300 g  93 g + 434 g
• The 249 g CH4 will react with only 279 g
H2O, thereby leaving 21 g H2O in excess.
CH4 is the reactant that will run out first and
is the LIMITING REACTANT.
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124
Section 3.10
The Concept of Limiting Reagent
2. Calculations Involving a Limiting Reactant
• So, how do we figure this out using Stoich?
• 1) Balance Reaction
• 2) Do Stoich to relate EACH reactant to one
product
• 3) Whichever reactant produces the LEAST
amount of product is the LIMITING
REACTANT
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125
Section 3.10
The Concept of Limiting Reagent
2. Calculations Involving a Limiting Reactant
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126
Section 3.10
The Concept of Limiting Reagent
2. Calculations Involving a Limiting Reactant
• Suppose 2.50 X 104 g N2 reacts with 5 X 103 g H2.
Determine the limiting reactant.
• ___N2 + ___H2  ___NH3
• 2.50 X 104 g N2
• 5 X 103 g H2
• From N2  1785.72 mol NH3
• From H2  1666.67 mol NH3 ; 28333.3 g NH3
• Not just because original mass of H2 was less.
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127
Section 3.10
The Concept of Limiting Reagent
2. Calculations Involving a Limiting Reactant
• How much N2 will be produced when 18.1 g NH3
and 90.4 g CuO react? Determine the limiting
reactant.
• __NH3 + __CuO  __N2 + __Cu + __H2O
• 18.1 g NH3
• 90.4 g CuO
• NH3 0.53 mol N2
• CuO 0.38 mol N2 *limiting reactant* 10.64 g N2
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128
Section 3.10
The Concept of Limiting Reagent
Concept Check
• You know that chemical A reacts with
chemical B. You react 10.0 g of A with
10.0 g of B.
 What information do you need to
know in order to determine the mass
of product that will be produced?
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129
Section 3.10
The Concept of Limiting Reagent
Exercise
You react 10.0 g of A with 10.0 g of B. What
mass of product will be produced given that
the molar mass of A is 10.0 g/mol, B is
20.0 g/mol, and C is 25.0 g/mol? They react
according to the equation:
A + 3B  2C
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130
Section 3.10
The Concept of Limiting Reagent
3. Percent Yield
• Theoretical Yield
– The maximum amount of a given product that can be
formed when the limiting reactant is completely
consumed. Just what we’ve been calculating!
• The actual yield (amount produced is usually given) of a
reaction is usually less than the max theoretical yield
because of side or incomplete reactions.
• Percent Yield The actual amount of a given product as the
percentage of the theoretical yield.
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Section 3.10
The Concept of Limiting Reagent
3. Percent Yield
• From the previous set of calculations, we
determined that 10.64 g of N2 would theoretically
be produced from the reaction:
• 2NH3 + 3CuO  N2 + 3Cu + 3H2O
• What would the Percent Yield be if 7.25 g of
N2 were actually produced?
•
•
7.25 g N2
10.64 g N2
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X 100 = 68.14 %
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132
Section 3.10
The Concept of Limiting Reagent
3. Percent Yield
• Consider the following reaction:
• __CO(g) +
__H2(g)  __CH3OH(l)
• If 6.85 X 104 g CO reacts with 8.6 X 103 g H2 to
produce an actual yield of 3.57 X 104 g CH3OH,
determine the limiting reactant, the theoretical yield,
and the percent yield.
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133
Section 3.10
The Concept of Limiting Reagent
3. Percent Yield
• CO(g)
+
• 2446.4 mol CO
• 2446.4 mol CH3OH
2H2(g) 
CH3OH(l)
4300 mol H2
2150 mol CH3OH *Limiting
• 2150 mol CH3OH *32g/mol = 68800 g CH3OH
• % yield = 3.57 X 104 g/ 6.88 X 104 g = 51.89 %
• 28 g/mol CO: 2 g/mol H2: 32 g/mol CH3OH
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134
Section 3.10
The Concept of Limiting Reagent
3. Percent Yield
• Consider the following reaction:
• __TiCl4
+
__O2  __TiO2 + __Cl2
• If 6.71 X 103 g TiCl4 reacts with 2.45 X 103 g O2
with a percent yield of 75%, determine the limiting
reactant, the theoretical yield, and the actual yield
of TiO2.
• TiCl4 is limiting
• 2119.6 g TiO2 is actually produced in this reaction.
• 189.68 g/mol TiCl4: 32 g/mol O2: 79.88 g/mol TiO2
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135
Section 3.10
The Concept of Limiting Reagent
Objectives Review
1. To understand the concept of limiting
reactants
2. To learn to recognize the limiting reactant in
a reaction
3. To learn to use the limiting reactant to do
stoichiometric calculations
4. To learn to calculate percent yield
5. Work Session: pg 121 # 97, 101, 104, 105
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136
Section 3.10
The Concept of Limiting Reagent
Notice
•
We cannot simply add the total moles of
all the reactants to decide which reactant
mixture makes the most product. We
must always think about how much
product can be formed by using what we
are given, and the ratio in the balanced
equation.
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137
Section 3.9
Stoichiometric Calculations: Amounts of Reactants and Products
Calculating Masses of Reactants and Products in Reactions
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138
Section 3.10
The Concept of Limiting Reagent
Let’s Think About It
•
Where are we going?

•
To determine the mass of product that will be
produced when you react 10.0 g of A with 10.0 g
of B.
How do we get there?

We need to know:
• The mole ratio between A, B, and the product
they form. In other words, we need to know
the balanced reaction equation.
• The molar masses of A, B, and the product
they form.
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139
Section 3.10
The Concept of Limiting Reagent
Percent Yield
•
An important indicator of the efficiency of
a particular laboratory or industrial
reaction.
Actual yield
 100%  percent yield
Theoretica l yield
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140