Chapter

1
Chapter 3
Stoichiometry
2
Mole-Mass Relationships in Chemical Systems
3.1 The Mole
3.2 Determining the Formula of an Unknown Compound
3.3 Writing and Balancing Chemical Equations
3.4 Calculating the Amounts of Reactant and
Product
3.5 Fundamentals of Solution Stoichiometry
3
Definition of the Mole
mole - the amount of a substance that contains
the same number of particles as there are atoms
in exactly 12 g of carbon-12.
This amount contains 6.022 x 1023 particles. The
number, 6.022 x 1023, is called Avogadros number
and is abbreviated N.
One mole (1 mol) contains 6.022 x 1023 particles
(to four significant figures)
4
Counting Objects of Fixed Relative Mass
Figure 3.1
12 red marbles at 7 g each 84g 12 yellow
marbles at 4 g each 48g
55.85 g Fe 6.022 x 1023 atoms Fe 32.07 g S
6.022 x 1023 atoms S
5
Oxygen 32.00 g
One Mole of Common Substances
Water 18.02 g
CaCO3 100.09 g
Figure 3.2
Copper 63.55 g
6
Table 3.1 Summary of Mass Terminology
Term
Definition
Unit
isotopic mass
mass of an isotope of an element
amu
atomic mass
amu
average of the masses of the naturally occurring
isotopes of an element weighted according to
their abundance
(also called atomic weight)
molecular (or formula) mass
amu
sum of the atomic masses of the atoms (or ions)
in a molecule (or formula unit)
(also called molecular weight)
g/mol
mass of 1 mole of chemical particles (atoms,
ions, molecules, formula units)
molar mass (M)
(also called gram-molecular weight)
7
Information Contained in the Chemical Formula of
Glucose C6H12O6 ( M 180.16 g/mol)
Table 3.2
Oxygen (O)
Carbon (C)
Hydrogen (H)
Atoms/molecule of compound
6 atoms
6 atoms
12 atoms
Moles of atoms/ mole of compound
6 moles of atoms
12 moles of atoms
6 moles of atoms
Atoms/mole of compound
6 (6.022 x 1023) atoms
12 (6.022 x 1023) atoms
6 (6.022 x 1023) atoms
Mass/moleculeof compound
6 (12.01 amu) 72.06 amu
12 (1.008 amu) 12.10 amu
6 (16.00 amu) 96.00 amu
Mass/mole of compound
96.00 g
72.06 g
12.10 g
8
Interconverting Moles, Mass, and Number of
Chemical Entities
9
Calculating the Mass and the Number of Atoms in a
Given Number of Moles of an Element
Sample Problem 3.1
(a) Silver (Ag) is used in jewelry and tableware
but no longer in US coins. How many grams of Ag
are in 0.0342 mol of Ag?
(b) Iron (Fe), the main component of steel, is
the most important metal in industrial society.
How many Fe atoms are in 95.8 g of Fe?
PLAN
(a) To convert mol of Ag to g, we use the number
of g Ag per mol Ag, the molar mass M.
SOLUTION
0.0342 mol Ag x
3.69 g Ag
(b) To convert g of Fe to atoms we first find the
number of moles of Fe and then convert moles to
atoms.
PLAN
SOLUTION
95.8 g Fe x
1.72 mol Fe
1.04 x1024 atoms Fe
1.72 mol Fe x
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Sample Problem 3.2
Calculating the Moles and Number of Formula Units
in a Given Mass of a Compound
PROBLEM
Ammonium carbonate is white solid that decomposes
with warming. Among its many uses, it is a
component of baking powder, fire extinguishers,
and smelling salts. How many formula units are
in 41.6 g of ammonium carbonate?
PLAN
After writing the formula for the compound, we
find M by adding the masses of the elements,
convert the given mass (41.6 g) to moles using M,
and then convert moles to formula units using
Avogadros number.
SOLUTION
The formula is (NH4)2CO3.
M (2 x 14.01 g/mol N) (8 x 1.008 g/mol H
(12.01 g/mol C) (3 x 16.00 g/mol O)
96.09 g/mol
41.6 g (NH4)2CO3 x
x

2.61 x 1023 formula units (NH4)2CO3
12
mass of element X
atoms of X in formula x atomic mass of X (amu)
x 100
molecular (or formula) mass of compound(amu)
13
Flow Chart of Mass Percentage Calculation
Multiply by M (g/mol) of X
Divide by mass (g) of one mole
of compound
Multiply by 100
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16
Empirical and Molecular Formulas
Empirical Formula
The simplest formula for a compound that agrees
with the elemental analysis and gives rise to the
smallest set of whole numbers of atoms.
Molecular Formula
The formula of the compound as it exists, it may
be a multiple of the empirical formula.
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Sample Problem 3.4
Determining the Empirical Formula from Masses of
Elements
PROBLEM
Elemental analysis of a sample of an ionic
compound gave the following results 2.82 g of
Na, 4.35 g of Cl, and 7.83 g of O. What are the
empirical formula and name of the compound?
PLAN
Find the relative number of moles of each
element divide by the lowest mole amount to find
the relative mole ratios (empirical formula).
SOLUTION
2.82 g Na x
0.123 mol Na
4.35 g Cl x
0.123 mol Cl
7.83 g O x
0.489 mol O
Na1 Cl1 O3.98
NaClO4
NaClO4 is sodium perchlorate.
18
Sample Problem 3.5
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM
During physical activity, lactic acid (M 90.08
g/mol) forms in muscle tissue and is responsible
for muscle soreness. Elemental analysis shows
that it contains 40.0 mass C, 6.71 mass H, and
53.3 mass O.
(a) Determine the empirical formula of lactic
acid.
(b) Determine the molecular formula.
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Sample Problem 3.5 (continued)
Assuming 100 g of lactic acid, the constituents
are
SOLUTION
40.0 g C x
6.71 g H x
53.3 g O x
3.33 mol C
6.66 mol H
3.33 mol O
C3.33
H6.66
O3.33
CH2O
empirical formula
C3H6O3 is the molecular formula
3
20
Combustion Apparatus for the Determination
of Formulas of Organic Compounds
Figure 3.4
21
Sample Problem 3.6
Determining a Molecular Formula from Combustion
Analysis
difference (after-before) mass of oxidized
element
PLAN
find the mass of each element in its combustion
product
molecular formula
preliminary formula
empirical formula
find the moles
22
Sample Problem 3.6
(continued)
SOLUTION
There are 12.01 g C per mol CO2
1.50 g CO2 x
0.409 g C
There are 2.016 g H per mol H2O
0.41 g H2O x
0.046 g H
O must be the difference 1.000 g - (0.409
0.046) 0.545
0.0341 mol C
0.0461 mol H
0.0341 mol O
C1H1.3O1
C3H4O3
2.000
C6H8O6
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Table 3.4 Two Compounds with Molecular Formula
C2H6O
Property
Ethanol
Dimethyl Ether
46.07
M (g/mol)
46.07
colorless
color
colorless
-138.5 0C
melting point
-117 0C
boiling point
78.5 0C
-25 0C
density (20 0C)
0.789 g/mL (liquid)
0.00195 g/mL (gas)
use
intoxicant in alcoholic beverages
refrigeration
structural formulas and space-filling models
25
Formation of HF gas macroscopic and molecular
levels
Figure 3.6
26
A three-level view of the chemical reaction in a
flashbulb
Figure 3.7
27
Sample Problem 3.7
Balancing Chemical Equations
PROBLEM
Within the cylinders of a cars engine, the
hydrocarbon, octane (C8H18), which is one of the
many components of gasoline, mixes with oxygen
from air and burns to form carbon dioxide and
water vapor. Write a balanced equation for this
reaction.
translate the statement
8 CO2 9 H2O
C8H18 O2
8
25/2
9
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Sample Problem 3.8
Calculating Amounts of Reactants and Products
PROBLEM
In a lifetime, the average American uses 1750 lb
(794 kg) of copper in coins, plumbing, and
wiring. Copper is obtained from sulfide ores,
such as chalcocite, or copper(I) sulfide, by a
multistage process. After an initial grinding
step, the first stage is to roast the ore (heat
it strongly with oxygen gas) to form powdered
copper(I) oxide and gaseous sulfur dioxide. (a)
How many moles of oxygen are required to roast
10.0 mol of copper(I) sulfide? (b) How many grams
of sulfur dioxide are formed when 10.0 mol of
copper(I) sulfide are roasted? (c) How many
kilograms of oxygen are required to form 2.86 kg
of copper(I) oxide?
PLAN
write and balance equation
find mols O2
find mols SO2
find mols Cu2O
find g SO2
find mols O2
find kg O2
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Sample Problem 3.8
(continued)
SOLUTION
10.0 mol Cu2S x
(a)
15.0 mol O2
10.0 mol Cu2S x
(b)
641 g SO2
x
2.86 kg Cu2O x
(c)
20.0 mol Cu2O
x
20.0 mol Cu2O x
x
x
0.960 kg O2
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Summary of the Mass-Mole-Number Relationships in
a Chemical Reaction
Figure 3.8
32
Sample Problem 3.9
Calculating Amounts of Reactants and Products in
a Reaction Sequence
PLAN
SOLUTION
write balanced equations for each step
cancel reactants and products common to both
sides of the equations
or
sum the equations
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An Ice Cream Sundae Analogy for Limiting Reactions
Figure 3.9
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Sample Problem 3.10
Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
PROBLEM
A fuel mixture used in the early days of rocketry
was composed of two liquids, hydrazine (N2H4) and
dinitrogen tetraoxide (N2O4), which ignite on
contact to form nitrogen gas and water vapor.
How many grams of nitrogen gas (N2) form when
1.00 x 102 g of N2H4 and 2.00 x 102 g of N2O4 are
mixed?
PLAN
Start with a balanced chemical equation and find
the number of moles of reactants identify
limiting reagent determine grams of N2 formed.
divide by M
molar ratio
mol of N2
mol of N2
35
Sample Problem 3.10
(continued)
SOLUTION
2
4
3
1.00 x 102g N2H4 x
3.12 mol N2H4
N2H4 is the limiting reactant because it produces
less product, N2, than does N2O4.
3.12 mol N2H4 x
4.68 mol N2
2.00 x 102g N2O4 x
2.17 mol N2O4
4.68 mol N2 x
2.17 mol N2O4 x
6.51 mol N2
131g N2
36
Formation of side-products during chemical
reactions
37
Sample Problem 3.11
Calculating Percent Yield
PROBLEM
Silicon carbide (SiC) is an important ceramic
material that is made by allowing sand (silicon
dioxide, SiO2) to react with powdered carbon at
high temperature. Carbon monoxide is also
formed. When 100.0 kg of sand are processed,
51.4 kg of SiC are recovered. What is the
percent yield of SiC in this process?
PLAN
SOLUTION
write balanced equation
1664 mol SiO2
x
100.0 kg SiO2 x
find mol reactant product
mol SiO2 mol SiC 1664 mol
find g product predicted
66.73 kg
1664 mol SiC x
x
percent yield
x100
77.0
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Sample Problem 3.12
Calculating the Molarity of a Solution
PROBLEM
Hydrobromic acid (HBr) is a solution of hydrogen
bromide gas in water. Calculate the molarity of
a hydrobromic acid solution if 455 mL contains
1.80 moles of hydrogen bromide.
SOLUTION
mol of HBr
divide by volume
x
3.96 M
concentration (mol/mL) HBr
103 mL 1L
molarity (mol/L) HBr
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Sample Problem 3.13
Calculating Mass of Solute in a Given Volume of
Solution
PROBLEM
How many grams of solute are found in 1.75 L of a
0.460 M solution of sodium monohydrogen phosphate?
Molarity is the number of moles of solute per
liter of solution. Knowing the molarity and
volume allows us to find the number of moles and
then the number of grams of solute. The
molecular formula of the solute is Na2HPO4.
PLAN
volume of soln
SOLUTION
multiply by M
1.75 L
x
0.805 mol Na2HPO4
moles of solute
0.805 mol Na2HPO4
multiply by M
x
grams of solute
114 g Na2HPO4
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Sample Problem 3.14
Preparing a Dilute Solution from a Concentrated
Solution
PROBLEM
Isotonic saline is a 0.15 M aqueous solution of
NaCl that simulates the total concentration of
ions found in many cellular fluids. Its uses
range from a cleaning rinse for contact lenses to
a washing medium for red blood cells
(erythrocytes). How would you prepare 0.80 L of
isotonic saline from a 6.0 M stock solution?
PLAN
The number of moles of solute does not change
during the dilution but volume does. The new
volume will be the sum of the two volumes, that
is, the total final volume.
Mdil x Vdil mol solute Mconc x Vconc
SOLUTION
0.80 L soln
0.12 mol NaCl
x
0.020 L soln
0.12 mol NaCl
x
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Converting a Concentrated Solution to a Dilute
Solution
Figure 3.13
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Figure 3.11
43
Sample Problem 3.15
Calculating Amounts of Reactants and Products for
a Reaction in Solution
PROBLEM
Specialized cells in the stomach release HCl to
aid digestion. If they release too much, the
excess can be neutralized with antacids. A
common antacid contains magnesium hydroxide,
which reacts with the HCl to form water and
magnesium chloride solution. As a government
chemist testing commercial antacids, you use 0.10
M HCl to simulate the acid concentration in the
stomach. How many liters of stomach acid react
with a tablet containing 0.10 g of magnesium
hydroxide?
PLAN
Write a balanced equation for the reaction find
the moles of Mg(OH)2 determine the mole ratio of
reactants and products use moles to convert to
molarity.
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Sample Problem 3.15
(continued)
SOLUTION
0.10 g Mg(OH)2 x
1.7 x 10-3 mol Mg(OH)2
1.7 x 10-3 mol Mg(OH)2 x
3.4 x 10-3 mol HCl
3.4 x 10-3 mol HCl x
3.4 x 10-2 L HCl
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Sample Problem 3.16
Solving Limiting Reactant Problems for Reactions
in Solution
PROBLEM
Mercury and its compounds have many uses, from
filling teeth (as an alloy with silver, copper,
and tin) to the industrial production of
chlorine. Because of their toxicity, however,
soluble mercury compounds, such mercury(II)
nitrate, must be removed from industrial
wastewater. One removal method reacts the
wastewater with sodium sulfide solution to
produce solid mercury(II) sulfide and sodium
nitrate solution. In a laboratory simulation,
0.050 L of 0.010 M mercury(II) nitrate reacts
with 0.020 L of 0.10 M sodium sulfide. How many
grams of mercury(II) sulfide form?
PLAN
Write a balanced chemical equation for the
reaction. Since this is a problem concerning a
limiting reactant, we find the amount of product
that would be made from each reactant. We then
choose the reactant that gives the lesser amount
of product.
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Sample Problem 3.16
(continued)
SOLUTION
0.050 L Hg(NO3)2
x 0.010 mol/L x
5.0 x 10-4 mol HgS
0.020 L Na2S
x 0. 10 mol/L x
2.0 x 10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
5.0 x 10-4 mol HgS x
0.12 g HgS
47
Laboratory Preparation of Molar Solutions
Figure 3.12
48
Key Mass/Mole Relationships
Figure 3.14