Introductory Chemistry Essentials
Tro
Fourth Edition
ISBN 978-1-29202-072-3
9 781292 020723
Introductory Chemistry Essentials
Nivaldo J. Tro
Fourth Edition
Pearson Education Limited
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ISBN 10: 1-292-02072-5
ISBN 13: 978-1-292-02072-3
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A catalogue record for this book is available from the British Library
Printed in the United States of America
CHAPTER IN REVIEW
CHEMICAL PRINCIPLES
RELEVANCE
The Mole Concept: The mole is a specific number
(6.022 * 1023 ) that allows us to easily count atoms or
molecules by weighing them. One mole of any element
has a mass equivalent to its atomic mass in grams, and
a mole of any compound has a mass equivalent to its
formula mass in grams. The mass of 1 mol of an
element or compound is its molar mass.
The Mole Concept: The mole concept allows us to determine the number of atoms or molecules in a sample from
its mass. Just as a hardware store customer wants to know
the number of nails in a certain weight of nails, so we want
to know the number of atoms in a certain mass of atoms.
Since atoms are too small to count, we use their mass.
Chemical Formulas and Chemical Composition:
Chemical formulas indicate the relative number of each
kind of element in a compound. These numbers are
based on atoms or moles. By using molar masses, we
can use the information in a chemical formula to determine the relative masses of each kind of element in a
compound. We can then relate the mass of a sample of a
compound to the masses of the elements contained in
the compound.
Chemical Formulas and Chemical Composition: The
chemical composition of compounds is important because
it lets us determine how much of a particular element is
contained within a particular compound. For example, an
assessment of the threat to the Earth’s ozone layer from
chlorofluorocarbons (CFCs) requires knowing how much
chlorine is in a particular CFC.
Empirical and Molecular Formulas from Laboratory
Data: We can refer to the relative masses of the elements within a compound to determine the empirical
formula of the compound. If the chemist also knows
the molar mass of the compound, he or she can also
determine its molecular formula.
Empirical and Molecular Formulas from Laboratory
Data: The first thing a chemist wants to know about an
unknown compound is its chemical formula, because the
formula reveals the compound’s composition. Chemists
often arrive at formulas by analyzing compounds in the
laboratory—either by decomposing them or by synthesizing them—to determine the relative masses of the elements
they contain.
CHEMICAL SKILLS
EXAMPLES
Converting between Moles and Number of Atoms
(Section 3)
EXAMPLE 13
Converting between Moles
and Number of Atoms
Calculate the number of atoms in 4.8 mol of copper.
SORT
You are given moles of copper and asked to find the
number of copper atoms.
STRATEGIZE
To convert between moles and number of atoms, use
Avogadro’s number, 6.022 * 1023 atoms = 1 mol, as a
conversion factor.
GIVEN:
FIND:
4.8 mol Cu
Cu atoms
SOLUTION MAP
mol Cu
Cu atoms
6.022 1023 Cu atoms
1 mol Cu
RELATIONSHIPS USED
1 mol Cu = 6.022 * 1023 Cu atoms (Avogadro’s number,
from inside back cover)
218
Chemical Composition
SOLVE
SOLUTION
Follow the solution map to solve the problem.
4.8 mol Cu *
6.022 * 1023 Cu atoms
=
1 mol Cu
2.9 * 1024 Cu atoms
Check your answer. Are the units correct? Does the
answer make physical sense?
The units, Cu atoms, are correct. The answer makes physical sense because the number is very large, as you would
expect for nearly 5 moles of atoms.
Converting between Grams and Moles (Section 3)
EXAMPLE 14
CHECK
Converting between Grams
and Moles
Calculate the mass of aluminum (in grams) of 6.73
moles of aluminum.
SORT
You are given the number of moles of aluminum and
asked to find the mass of aluminum in grams.
STRATEGIZE
Use the molar mass of aluminum to convert between
moles and grams.
GIVEN:
FIND:
6.73 mol Al
g Al
SOLUTION MAP
g Al
mol Al
26.98 g Al
1 mol Al
RELATIONSHIPS USED
26.98 g Al = 1 mol Al (molar mass of Al from
periodic table)
SOLVE
Follow the solution map to solve the problem.
CHECK
Check your answer. Are the units correct? Does the
answer make physical sense?
SOLUTION
6.73 mol Al *
26.98 g Al
1 mol Al
= 182 g Al
The units, g Al, are correct. The answer makes physical
sense because each mole has a mass of about 27 g; therefore, nearly 7 moles should have a mass of nearly 190 g.
219
Chemical Composition
Converting between Grams and Number of Atoms or
Molecules (Section 3)
Converting between Grams and
Number of Atoms or Molecules
EXAMPLE 15
Determine the number of atoms in a 48.3-g sample of zinc.
48.3 g Zn
SORT
GIVEN:
You are given the mass of a zinc sample and asked to find
the number of Zn atoms that it contains.
FIND:
STRATEGIZE
SOLUTION MAP
Zn atoms
First use the molar mass of the element to convert from
grams to moles, and then use Avogadro’s number to
convert moles to number of atoms.
mol Zn
g Zn
1 mol Zn
65.39 g Zn
number of
Zn atoms
6.022 1023 Zn atoms
1 mol Zn
RELATIONSHIPS USED
65.39 g Zn = 1 mol Zn (molar mass of
Zn from periodic table)
1 mol = 6.022 * 1023 atoms (Avogadro’s number,
from inside back cover)
SOLVE
SOLUTION
Follow the solution map to solve the problem.
48.3 g Zn *
1 mol Zn
6.022 * 1023 Zn atoms
*
65.39 g Zn
1 mol Zn
= 4.45 * 1023 Zn atoms
CHECK
Check your answer. Are the units correct? Does the
answer make physical sense?
Converting between Moles of a Compound and
Moles of a Constituent Element (Section 5)
The units, Zn atoms, are correct. The answer makes
physical sense because the number of atoms in any
macroscopic-sized sample should be very large.
Converting between Moles of a
Compound and Moles of a
Constituent Element
EXAMPLE 16
Determine the number of moles of oxygen in 7.20 mol of
H2SO4.
SORT
You are given the number of moles of sulfuric acid and
asked to find the number of moles of oxygen.
STRATEGIZE
To convert between moles of a compound and moles of a
constituent element, use the chemical formula of the
compound to determine a ratio between the moles of the
element and the moles of the compound.
GIVEN:
FIND:
7.20 mol H 2SO4
mol O
SOLUTION MAP
mol O
mol H2SO4
4 mol O
1 mol H2SO4
RELATIONSHIPS USED
4 mol O : 1 mol H2SO4
SOLVE
SOLUTION
Follow the solution map to solve the problem.
7.20 mol H2SO4 *
CHECK
Check your answer. Are the units correct? Does the
answer make physical sense?
220
4 mol O
= 28.8 mol O
1 mol H2SO4
The units, mol O, are correct. The answer makes physical sense because the number of moles of an element in
a compound is equal to or greater than the number of
moles of the compound itself.
Chemical Composition
Converting between Grams of a Compound and
Grams of a Constituent Element (Section 5)
EXAMPLE 17
Converting between Grams of a
Compound and Grams of a
Constituent Element
Find the grams of iron in 79.2 g of Fe2O3.
79.2 g Fe2O3
SORT
GIVEN:
You are given the mass of iron (III) oxide and asked to
find the mass of iron contained within it.
FIND:
STRATEGIZE
SOLUTION MAP
Use the molar mass of the compound to convert from
grams of the compound to moles of the compound. Then
use the chemical formula to obtain a conversion factor to
convert from moles of the compound to moles of the
constituent element. Finally, use the molar mass of the
constituent element to convert from moles of the element
to grams of the element.
g Fe
g Fe2O3
mol Fe2O3
1 mol Fe2O3
159.70 g Fe2O3
mol Fe
2 mol Fe
1 mol Fe2O3
g Fe
55.85 g Fe
1 mol Fe
RELATIONSHIPS USED
Molar mass Fe2O3
= 2(55.85) + 3(16.00)
= 159.70 g/mol
2 mol Fe : 1 mol Fe2O3 (from given chemical formula)
SOLVE
SOLUTION
Follow the solution map to solve the problem.
79.2 g Fe2O3 *
1 mol Fe2O3
2 mol Fe
*
*
159.70 g Fe2O3
1 mol Fe2O3
55.85 g Fe
1 mol Fe
CHECK
Check your answer. Are the units correct? Does the
answer make physical sense?
= 55.4 g Fe
The units, g Fe, are correct. The answer makes physical
sense because the mass of a constituent element within
a compound should be less than the mass of the
compound itself.
221
Chemical Composition
Using Mass Percent Composition as a Conversion
Factor (Section 6)
EXAMPLE 18
Using Mass Percent Composition
as a Conversion Factor
Determine the mass of titanium in 57.2 g of titanium(IV)
oxide. The mass percent of titanium in titanium(IV)
oxide is 59.9%.
SORT
GIVEN:
You are given the mass of titanium(IV) oxide and the
mass percent titanium in the oxide. You are asked to find
the mass of titanium in the sample.
59 g Ti
100 g TiO2
FIND:
STRATEGIZE
57.2 g TiO2
g Ti
SOLUTION MAP
Use the percent composition as a conversion factor
between grams of titanium(IV) oxide and grams of
titanium.
g Ti
g TiO2
59.9 g Ti
100 g TiO2
RELATIONSHIPS USED
59.9 g Ti : 100 g TiO2
SOLVE
SOLUTION
Follow the solution map to solve the problem.
57.2 g TiO2 *
CHECK
Check your answer. Are the units correct? Does the
answer make physical sense?
Determining Mass Percent Composition from a
Chemical Formula (Section 7)
59.9 g Ti
100 g TiO2
= 34.3 g Ti
The units, g Ti, are correct. The answer makes physical
sense because the mass of an element within a compound
should be less than the mass of the compound itself.
EXAMPLE 19
Determining Mass Percent
Composition from a Chemical
Formula
Calculate the mass percent composition of potassium in
potassium oxide (K 2O).
K 2O
SORT
GIVEN:
You are given the formula of potassium oxide and asked
to determine the mass percent of potassium within it.
FIND:
STRATEGIZE
SOLUTION MAP
The solution map shows how the information derived
from the chemical formula can be substituted into the
mass percent equation to yield the mass percent of the
element.
Mass % K
Chemical
formula
Mass % K Mass % K
2 Molar mass K
Molar mass K2O
100%
RELATIONSHIPS USED
Mass percent of element X
Mass of element X in 1 mol of compound
=
Mass of 1 mol of compound
(mass percent equation, from Section 6)
222
* 100%