Free Study Guide for
Cracolice • Peters
Introductory Chemistry: An Active Learning Approach
Second Edition
www.brookscole.com/chemistry
Chapter 7
Chemical Formula Relationships
Chapter 7–Assignment A: Chemical Formulas, Formula Masses
A chemical formula tells us the number of each atom that make up a given compound.
These atoms have a mass given in atomic mass units, abbreviated amu.
In Chapter 2 you learned that pure substances are either elements or compounds. In Chapter
5 you learned that relative masses of elements are given in amu. In this assignment you will
learn to find the relative masses of compounds, again using the atomic mass unit.
Look for these big ideas:
1)
A chemical formula tells how many atoms of each element are present in the
formula unit of a substance. The number of each atom is given by a subscript
following the symbol of that atom.
2)
Like atomic mass of elements, molecular mass or formula mass of compounds is
measured in atomic mass units using carbon-12 as the reference standard.
3)
To find the formula mass of a compound, add up the atomic masses of its atoms.
Learning Procedures
Study
Sections 7.1–7.2. Focus on Goals 1–3 as you study.
Strategy
Distinguishing among atomic, molecular, and formula mass will probably
be the most difficult part of this assignment. Study these definitions
carefully.
Answer
Questions, Exercises, and Problems 1–4. Check your answers with those at
the end of the chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 1–4.
39
Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.
No part of this work may be reproduced without the written permission of the publisher.
Study Guide for Introductory Chemistry: An Active Learning Approach
Chapter 7–Assignment B: The Mole, Avogadro's Number, and Molar Masses
The atomic mass unit, 1.66 ¥ 10– 2 4 gram, is far too small to detect with even the most
sensitive balance. To chemists, the term that suggests a weighable amount of something is
the mole, abbreviated mol. A mole of any substance contains 6.02 ¥ 102 3 units of that
substance. The experimentally derived number 6.02 ¥ 102 3 is called Avogadro's number.
Understanding the mole and being able to use it effectively is certainly one of the most
important skills you can acquire in this course.
Look for the big ideas in this assignment:
1)
One mole of anything contains the same number of objects as the number of atoms
in exactly 12 grams of carbon-12. This experimentally determined value is
Avogadro's number, 6.02 ¥ 102 3.
2)
The mass in grams of one mole of a substance is called the molar mass of that
substance.
3)
The molar mass of a compound is the sum of the molar masses of the atoms that
make up a formula unit of that compound.
4)
There are conversion factors between grams of a substance, number of particles of
that substance, and number of moles of that substance.
Learning Procedures
Study
Sections 7.3–7.5. Focus on Goals 4–8 as you study.
Strategy
There are two important equations used in this assignment.
1 mole of any substance = 6.02 ¥ 102 3 units of that substance
is one. This dimensional analysis conversion factor enables you to convert
between moles and number of units, exactly as you convert between dozens
of eggs and number of eggs. The other important equation is
molar mass = grams/mole
It enables you to convert between moles and grams of a substance. This
conversion is extremely important because it is the link between the
unseeable particulate world and the easily seen macroscopic world of the
laboratory or production plant. Learn these conversions well. You will use
them often.
Answer
Questions, Exercises, and Problems 5–20. Check your answers with those at
the end of the chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 5–20.
40
Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.
No part of this work may be reproduced without the written permission of the publisher.
Chapter 7
Chapter 7–Assignment C:
Formulas
Chemical Formula Relationships
Percentage Composition, Empirical and Molecular
In Assignment 7–B you learned to calculate the molar mass of a compound. Because
compounds conform to the Law of Definite Composition, the molar mass and the numbers
used to calculate it may be used to find the percentage composition of the elements in that
compound.
After learning percentage composition, you will use percentage composition calculations to
determine the empirical formula of a compound. Empirical formula? Let's try a formula
problem of a different sort, to explain.
I have some cats. Of these cats, 50% are male and 50% are female. How many cats do I
have?
You can't answer that question without more information. All you know is that the ratio of
female to male cats is 1:1. This ratio is the empirical formula for my cats. I might have a
total of 2 cats, 1 female and 1 male. I might have 6 cats, 3 female and 3 male. I might have
400 cats, 200 female and 200 male.
A calculation from the percentage composition can give only the empirical formula, or ratio
between the elements present, just as percentage composition of cats can give only the ratio
of cats present. You need more information to determine the actual molecular formula. You
will be given that information when it is needed in this assignment.
The main ideas in this assignment are:
1)
The molar mass and the chemical formula of a compound are used to determine the
percentage by mass of each element in that compound.
2)
The empirical formula gives the ratio in lowest terms of the elements in a
compound.
3)
Empirical formulas are calculated from percentage composition data. They are also
found from the mass of each element in a sample of a compound.
4)
Empirical formulas may or may not be the actual molecular formulas of compounds.
The molar mass of the compound is needed to determine molecular formulas from
simplest formulas.
Learning Procedures
Study
Sections 7.6–7.9. Focus on Goals 9–13 as you study.
Strategy
The proficiencies you need to acquire in this assignment are primarily
quantitative problem-solving skills. Practice by solving lots of problems.
Answer
Questions, Exercises, and Problems 21–32. Check your answers with those
at the end of the chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 21–32.
41
Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.
No part of this work may be reproduced without the written permission of the publisher.
Study Guide for Introductory Chemistry: An Active Learning Approach
Chapter 7–Assignment D: Summary and Review
A review of Chapter 7 is actually a review of the entire course so far. Chapter 7 ties together
what you learned in Chapter 3 about exponential notation, measurements, and calculations,
in Chapter 5 about the atom and atomic mass, and in Chapter 6 about chemical formulas.
The key to understanding Chapter 7 is understanding the idea of the mole. The mole is the
“chemist's dozen,” a handy bulk measure of a substance. Because you can use the mole as
a conversion relationship between the huge number of particles in a sample, or the
particulate world, and the mass of the sample, or the macroscopic world, the mole is our
handle on laboratory-sized chemical mass relationships.
The empirical formula of a compound comes from mass ratio measurements. Like a fraction
in lowest terms, these ratios cannot be reduced further. The empirical formula may not be
the molecular formula; make sure you can tell them apart. The word empirical means
“derived from experiment or observations,” and these formulas come from mass data
obtained by laboratory analysis.
Be sure that you work the example question in Section 7.7; there are a lot of useful facts
there. If you understand that question, you won't get atomic mass, molecular mass, and
molar mass confused.
Learning Procedures
Review
your lecture and textbook notes.
the Chapter in Review and the Key Terms and Concepts, and read the Study
Hints and Pitfalls to Avoid.
Answer
Concept-Linking Exercises 1–2. Check your answers with those at the end
of the chapter.
Questions, Exercises, and Problems 33–35. Include Questions 36–37 if
assigned by your instructor. Check your answers with those at the end of the
chapter.
Workbook
If your instructor recommends the Active Learning Workbook, do
Questions, Exercises, and Problems 33–34. Include Questions 35–36 if
assigned by your instructor.
Take
the chapter summary test that follows. Check your answers with those at the
end of this assignment.
Chapter 7 Sample Test
Instructions: Pick the letter of the best answer choice for Questions 3 and 4. Answer all
other questions with a complete setup. You may use a “clean” periodic table.
1)
How many nitrogen atoms and how many oxygen atoms are in one molecule of
dinitrogen pentoxide?
42
Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.
No part of this work may be reproduced without the written permission of the publisher.
Chapter 7
Chemical Formula Relationships
2)
Write the formula of the compound composed of one phosphorus atom and five
fluorine atoms. Name this compound.
3)
Identify the incorrect statement among the following:
a) A mole is that quantity of a substance that contains 6.02 ¥ 102 3 particles of that
substance.
b) One mole of any substance contains the same number of particles as one mole
of any other substance.
c) One mole of any substance contains the same number of particles as the number
of atoms in exactly 12 grams of carbon-12.
d) A mole is a quantity of a substance that has a mass of exactly 12 grams.
4)
Identify the correct statement among the following:
a) The molar mass of atoms of an element is numerically equal to its atomic mass.
b) The molar mass is the mass in amu of one mole of any substance.
c) Molar mass and atomic mass are always equal both numerically and in units.
d) Molar masses of compounds are always equal to atomic masses of those
compounds.
5)
Calculate the molar mass of potassium chloride.
6)
What is the mass of 1.06 ¥ 102 4 molecules of carbon monoxide?
7)
What mass of sodium nitrate contains 1.62 g of nitrogen?
8)
How many moles of potassium sulfate are in 37.5 grams?
43
Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.
No part of this work may be reproduced without the written permission of the publisher.
Study Guide for Introductory Chemistry: An Active Learning Approach
9)
Calculate the percentage composition of barium hydroxide.
10)
From the following, pick those that are empirical formulas. C1 0H6 ; C2 H6 O2 ; CH3 ;
C2 H2 ; C3 0H5 0O
11)
A compound has the percentage composition 92.25% C and 7.75% H. Calculate the
empirical formula for this compound. The molar mass of this compound is
78!g/mol. Calculate the true molecular formula of this compound.
Answers to Chapter 7 Sample Test
1)
Dinitrogen pentoxide, N2 O5 : 2 nitrogen atoms and 5 oxygen atoms
2)
PF5 , phosphorus pentafluoride
3)
d
5)
KCl: 39.10 + 35.45 = 74.55 g/mol
6)
CO: 12.01 + 16.00 = 28.01 g/mol
GIVEN: 1.06 ¥ 102 4 molecules CO
4) a
WANTED: mass CO (assume g)
¥ 10 23 molecules CO/mol CO
PER/PATH: molecules CO æ6.02
æ æ æ æ æ æ æ æ æ æ æ ææÆ mol CO
28.01 g CO/mol CO
ææææææææ
æÆ g CO
† CO ¥
1.06 ¥ 102 4 molecules
1 mol CO
28.01 g CO
¥
=
23
mol CO
6.02 ¥ 10 molecules CO
†
49.3 g CO
†
44
†
Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.
No part of this work may be reproduced without the written permission of the publisher.
Chapter 7
7)
Chemical Formula Relationships
NaNO3 : 22.99 + 14.01 + 3(16.00) = 85.00 g/mol
GIVEN: 1.62 g N
WANTED: mass NaNO3 (assume g)
g N/85.00 g NaNO 3
PER/PATH: g N æ14.01
æææææææææ
æÆ g NaNO3
85.00 g NaNO 3
1.62 g N ¥
= 9.83 g NaNO3
14.01 g N
†
K2 SO4 : 2(39.10) + 32.07 + 4(16.00) = 174.27 g/mol
8)
G†
IVEN: 37.5 g K2 SO4
WANTED: mol K2 SO4
g K SO 4 /mol K 2SO 4
PER/PATH: g K2 SO4 æ174.27
æ æ æ æ2æ æ
ææææ
æÆ mol K2 SO4
1 mol K 2SO 4
37.5 g K2 SO4 ¥
= 0.215 mol K2 SO4
174.27 g K 2SO 4
†
Ba(OH)2 : 137.3 + 2(16.00) + 2(1.008) = 171.32 g/mol
9)
†
% Ba =
137.3 g/mol Ba
¥ 100 = 80.14% Ba
171.32 g/mol Ba(OH)2
%O=
2(16.00) g / mol O
¥ 100 = 18.68% O
171.32 g / mol Ba(OH) 2
% H =!
2(1.008) g/mol H
¥ 100 = 1.177% H
171.32 g/mol Ba(OH)2
†
†
80.14 + 18.68 + 1.177 = 100.00%
10) † Both CH3 and C3 0H5 0O could be empirical formulas; the others are not empirical
formulas because they are not in “lowest terms.”
11)
Element Grams
C
H
92.25
7.75
78 g/mol
=6
13 g/mol
Moles
7.681
7.69
Mole
Ratio
1.00
1.00
Formula
Ratio
1
1
Empirical
Formula
CH
(CH)6 = C6 H6
†
45
Copyright © 2004 Brooks/Cole, a division of Thomson Learning, Inc.
No part of this work may be reproduced without the written permission of the publisher.
Molecular
Formula
C6 H6