1.
False. Determining the proper dosage of medicine for a patient is a matter of quantitative analysis for a doctor.
2.
c
3.
Table 1 Qualitative and Quantitative Questions
Questions Qualitative Quantitative
Is the solution an acid or a base?
What is the pH of the solution?
Does the patient have sugar in his blood?
At what temperature does sugar melt?
What concentration of sugar does the patient have in his blood?
Does the lake contain pollutants called PCBs?
Is the baseball player using performance-enhancing substances?
X
X
X
X
X
X
X
What dose of anabolic steroids does an athlete take to improve performance?
Is sugar more soluble in hot water or cold water?
4. Figure 1
X
X
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-1

5.
c
6.
Descriptions may vary. Sample answer:
(a) The description provides qualitative information. Quantitative description: The alligator was very long. provides qualitative information.
(b) The description provides qualitative information. Quantitative description: The reaction of zinc and hydrochloric acid yielded 2.26 L of hydrogen gas.
(c) The description provides qualitative information. Quantitative description: Scientists have known for decades that consumption of over 50 g of carbohydrates each day promotes insulin release and the storage of fat in fat cells.
(d) The description provides quantitative information. Qualitative description: A small quantity of fat contains quite a few calories but consuming it does not promote insulin release so it does not contribute to diabetes
1.
False. The amount of salt the average Canadian consumes is more than twice Health Canada’s recommended daily intake.
2. (3) Sodium ion levels in the blood rise to high levels.
(1) A person consumes too much sodium chloride in his or her diet.
(5) Extra fluid retention swells blood vessels and increases blood pressure.
(6) Elevated blood pressure makes the heart work harder.
(7) Strain may damage the heart and cause cardiovascular disease.
(2) Sodium chloride ionizes in the blood after consumption.
(4) High sodium levels cause the body to retain excess water.
3.
In order to get to the recommended daily intake of sodium, the average Canadian would need to cut his or her daily salt consumption from 3400 mg to 1500 mg, a 56 % decrease.
4.
(a) The 2500 mg of sodium Randy consumes each day is 900 mg less than the sodium consumption of the typical Canadian. But, his diet is still not healthy because he consumes 1000 mg more salt each day than
Health Canada’s recommended daily intake of 1500 mg.
(b) Table salt accounts for 10 % of sodium intake. If Randy stops using table salt, his daily intake of 2500 mg will be reduced by 250 mg, putting him at 2250 mg, just below the maximum tolerable limit of 2300 mg. This is still well above the recommended level of 1500 mg, so Randy’s diet will not be healthy.
5.
Answers may vary. Sample answer:
Table 1 High-Sodium Foods and Low-Sodium Alternatives
Type of food High-sodium food Low-sodium alternative meat cheese fish vegetable snack bologna processed cheese slices canned tuna or salmon canned green beans tortilla chips sliced roast turkey natural (unprocessed) cheese fresh salmon fresh green beans fresh carrots
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-2

1.
You can calculate the quantity of entities in a large group by dividing the mass of the group by the mass of a single entity .
2.
True
3.
c
4.
Table 1 Accuracy of Estimation Using Mass
What you are measuring What you know
Accuracy of estimate Explanation bolts good nuts and bolts
• mass of each bolt
• total mass
• mass of each bolt and nut
• total mass poor small- and largesized bolts equal quantities of nuts and bolts
• mass of each bolt
• total mass
• mass of each bolt and nut
• total mass poor good bolts • mass of each bolt poor
5.
Given: m nail
= 14.1 g m bag
= 7.13 kg = 7.13
!
10 3 g
Required: number of nails in bag, n
Analysis:
Solution: number of nails in bag = n = m bag m nail mass of nails in bag mass of 1 nail
=
7.13
!
10 3
14.1 g g
Dividing total mass by mass of bolts will be accurate because bolts are identical.
You do not know the ratio of nuts to bolts in the bag, so the estimate is likely to be off.
You do not know the ratio of large bolts to small bolts, so the estimate will be off.
Knowing the ratio of nuts to bolts allows you to make an accurate calculation.
You do not know the total mass so you cannot make a calculation.
= 5.06
!
10 2
Statement: There are 506 nails in the bag.
6.
(a) A bolt has a greater mass than a nut. You know this because the box of 50 bolts and 50 nuts has a greater mass, 1.5 kg, than the box of 100 nuts, 1.25 kg.
(b) The mass of Box 3 is greater than the mass of Box 2. You know that a bolt has a greater mass than a nut, so a number of bolts will have a greater mass than the same number of nuts.
(c) The mass of Box 3 is greater than the mass of Box 1. You know that a bolt has a greater mass than a nut, so a box of 50 bolts and 50 nuts will have a smaller mass than a box of 100 bolts.
7.
(a) A mole of atoms, a mole of golf balls, and a mole of bowling balls all contain the same number of items.
(b) There are 6.02 × 10
23
atoms, 6.02 × 10
23
golf balls, and 6.02 × 10
23
bowling balls in a mole of each item.
8.
False. A mole is the number of carbon atoms in exactly 12 g of carbon-12.
9.
(a) 6.02 × 10
24
is greater than Avogadro’s number.
(b) 9 000 000 000 000 000 000 000 is less than Avogadro’s number.
(c) 6.02 × 10
22
is less than Avogadro’s number.
(d) 602 000 000 000 000 000 000 000 is equal to Avogadro’s number.
(e) 10
22
is less than Avogadro’s number.
(f) 7 × 10
22
is less than Avogadro’s number.
10.
d
11.
(a) vi; (b) vii; (c) v; (d) ii; (e) iii; (f) iv; (g) i
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-3

12.
(a) 4.2
!
10 –4 !
6 !
10 7 = 25.2
!
10 –4 + 7
= 25.2
!
10 3
(b) 5.756
!
10 1
(c) 1.8
!
10 –6 ÷
+ 4.200
!
10
= 2.5
!
10 4
–1 = 5.756
!
10 1
3 !
10 –5
= 5.798
!
10 1
= 0.6
!
10 –6 – (–5)
= 0.6
!
10 –1
+ 0.04200
!
10 1
= 6.0
!
10 –2
1.
False; For all non-molecular elements, molar mass and atomic mass have the same value.
2.
The mass of 1 molecule of a compound with a molar mass of 48.55 g/mol is 48.55 u.
3.
a
4.
(a) Molar mass of sulfur dioxide, SO
2
:
Given: sulfur dioxide, SO
2
Required: molar mass of SO
2
, M
Solution:
SO
2
M
S
M
SO
2
= 32.07 g/mol
= M
S
+ 2 M
O
; M
O
=
"#
!
32.07
g mol %&
$
+
"#
= 16.00 g/mol g mol %&
$
M
SO
2
= 64.07 g/mol
Statement: The molar mass of sulfur dioxide is 64.07 g/mol.
(b) Molar mass of ethanol, C
Given: ethanol, C
2
H
5
OH
2
H
5
OH:
Required:
Solution:
molar mass of C
2
H
5
OH, M
C
2
H
5
OH
M
C
= 12.01 g/mol;
M
C
2
H
5
OH
= 2 M
C
+ 6
M
M
H
H
= 1.01 g/mol; M
O
+ M
O
= 16.00 g/mol
=
"# g mol %&
$
+
"# g mol %&
$
+
"#
!
16.00
g mol %&
$
M
C
2
H
5
OH
= 46.08 g/mol
Statement: The molar mass of ethanol is 46.08 g/mol.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-4

5.
(a) Molar mass of calcium chloride, CaCl
2
:
Given: calcium chloride, CaCl
2
Required:
Solution:
molar mass of CaCl
2
, M
CaCl
2
M
Ca
= 40.08 g/mol; M
Cl
= 35.45 g/mol
Calcium chloride is an ionic compound containing one Ca
2+
ion and two Cl
–
ions per formula unit. Since the mass of an electron is very small, you can assume that the masses of the Ca
2+
ion and Cl
–
ions are the same as the masses of the Ca and Cl atoms, respectively.
M
CaCl
2
= M
Ca
+ 2 M
Cl
=
"#
!
40.08
g mol %&
$
+
"# g mol %&
$
M
CaCl
2
= 110.98 g/mol
Statement: The molar mass of calcium chloride is 110.98 g/mol.
(b) Molar mass of iron(III) nitrate, Fe(NO
3
Given: iron(III) nitrate, Fe(NO
3
)
3
)
3
:
Required: molar mass of Fe(NO
3
)
3
, M
Solution:
( )
3
M
Fe
= 55.85 g/mol; M
N
= 14.01 g/mol; M
O
= 16.00 g/mol
Iron(III) nitrate is an ionic compound containing one Fe
3+ unit. Since the mass of an electron is very small, you can assume that the masses of the Fe
O
2–
ion, three N
3–
ions and nine O
2–
ions per formula
3+
ion, N
3–
ion, and
ions are the same as the masses of the Fe, N, and O atoms, respectively.
M
( )
3
= M
Fe
+ 3 M
N
+ 9 M
O
=
"#
!
55.85
g mol %&
$
+
"# g mol %&
$
+
"# g mol %&
$
M
( )
3
= 241.88 g/mol
Statement: The molar mass of iron(III) nitrate is 241.88 g/mol.
6.
The equation n = m/M will tell you how the mass of a sample in grams is related to the amount of the substance in moles. If you know m, the mass in grams, and M, the molar mass of the substance, you can find n, the amount in moles.
7.
b
8.
Given: m
NaClO
= 60.0 g
Required: amount in moles of NaClO, n
NaClO
Analysis: Find the molar mass of NaClO, and then use n = m/M.
Solution:
M
NaClO
= M
Na
+
=
"#
!
22.99
3 M g
Cl mol
$
%&
+ 9 M
O
+
"#
!
35.45
g mol %&
$
+
"#
!
16.00
g mol %&
$
M
NaClO
= 74.44 g/mol n
NaClO
= m
NaClO
M
NaClO
=
(
60.0 g ##
!
"
) 1 mol
74.44 g
&&
$
% n
NaClO
= 0.806 mol
Statement: A 60.0 g sample of bleach is 0.806 mol of bleach.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-5

9.
Given: 900 g of CO
2
in 1 day
Required: number of moles of CO
Analysis: Determine mass of CO
Solution:
2
2
, n
CO
2
in 1 h,
, in 1 h m
CO
2
. Calculate the molar mass of CO
2
, M
CO
2
. Then, use n = m/M . m
CO
2
M
CO
2
= 900
= M
C
M n
CO
CO
2
2
=
=
"#
!
12.01
g mol %&
= 44.01 g/mol
$ m
CO
2
M
CO
2 g day
+ 2
!
M
O
1 day
24 h
+
"#
= 37.5 g in 1 h g mol %&
$
=
(
37.5 g ##
!
"
) 1 mol
44.01 g
&&
$
% n
CO
2
= 0.852 mol
Statement: In one hour, the average person exhales 0.852 mol of carbon dioxide .
1.
The number of entities in one mole of a substance is equal to Avogadro’s constant, N
A
.
2.
(a) No, a 10 g sample of gold and a 10 g sample of silver will not have the same number of atoms; atoms of gold and silver are different in mass so samples that are equal in mass will have a different number of particles.
(b) The sample of the substance whose atoms have the smaller mass will have more atoms, since it will take more of these atoms to equal the mass of the other sample. Since silver has a lower atomic mass than gold, the sample of silver will have more atoms.
3.
d
4.
Table 1 Properties of Elements
Entity name Cupcakes Days Minutes Atoms
Entity variable
Group name
Group variable
Group amount cups dozen doz
12 day week wk
7 min hours h
60
N mole
Number of groups n n n
Number of items: equation cups = n × doz days = n × wk min = nh
Number of items for n = 5 cups = (5)(12)
= 60 days = (5)(7)
= 35 min = (5)(60)
= 300
N
A
6.02 × 10
23 n
N = nN
A
N = (5)(6.02 × 10
23
)
= 3.01 × 10
24
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-6

5. Given: m
Al
= 80.00 g
Required: number of atoms of aluminum, N
Al
Analysis: Use molar mass, M
Al
Solution:
, to calculate amount in moles, n
Al
. Then use N = nN
A
.
M
Al n
Al
= 26.98 g/mol
=
(
80.00 g ##
!
"
) 1 mol
26.98 g
&&
$
% n
Al
N
Al
= 2.96515 mol [2 extra digits carried]
= n
Al
N
A
=
(
2.96515 mol $
"
#
) 6.02
!
10 23
1 mol atoms
'
%
&
N
Al
= 1.785
!
10 24 atoms
Statement: An 80.00 g sample of aluminum contains 1.785 × 10
24
atoms of aluminum.
6. Given: m
CCl
4
= 1.50 kg m
CCl
4
= 1.50
!
10 3 g
Required: number of molecules of carbon tetrachloride,
Analysis:
Solution:
Use molar mass, M
CCl
4
N
CCl
4
, to calculate amount in moles, n
CCl
4
M
CCl
4
= M
C
+ 4 M
Cl
=
"#
!
12.01
g mol %&
$
+
"# g mol %&
$
. Then use N = nN
A
.
M
CCl
4 n
CCl
4
= 153.81 g/mol
=
(
1.50
!
10 3 g
"
#
) 1 mol
153.81 g
%
& n
CCl
4
N
CCl
4
= 9.7522 mol [2 extra digits carried]
= n
CCl
4
N
A
=
(
9.7522 mol
)
$
"
#
6.02
!
10 23 molecules
1 mol
'
%
&
N
CCl
4
= 5.87
!
10 24 molecules
Statement: In this sample of 1.50 g of carbon tetrachloride, 5.87 × 10
24
molecules were released.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-7

7.
(a) Given: m
H
2
SO
4
= 5.0 g
Required: number of molecules of sulfuric acid,
Analysis: Use molar mass, M
CCl
4
N
H
2
SO
4
, to calculate amount in moles,
Solution: n
CCl
4
M
H
2
SO
4
= 2 M
H
+ M
S
+ 4 M
O
=
"#
!
1.01
g mol %&
$
+
"#
!
32.07
g mol %&
$
+
"# g mol %&
$
M
H
2
SO
4 n
H
2
SO
4
= 97.08 g/mol
=
( ) !
"
1 mol
97.08 g
$
% n
N
H
2
SO
4
H
2
SO
4
= 0.05150 mol [2 extra digits carried]
= n
H
2
SO
4
N
A
=
(
0.05150 mol $
"
#
) 6.02
!
10 23 molecules
1 mol
'
%
&
. Then use N = nN
A
.
N
H
2
SO
4
= 3.1
!
10 22 molecules
Statement: The 5.0 g sample of sulfuric acid contained 3.1 × 10
22
molecules.
(b) Given:
Required:
5.0 g of sulfuric acid, H
2
SO
4
, or 3.1 × 10
22
molecules
number of atoms of hydrogen, sulfur, and oxygen, N
H
, N
S
, and N
O
Analysis: Multiply the number of molecules in the sample by the number of each type of atom per molecule.
Solution:
N
H
=
(
3.1
!
10 22 molecules of H
2
SO
4
)
$
"
#
2 atoms of H
1 molecule of H
2
SO
4
'
%
&
N
H
N
S
= 6.2
!
10 22 atoms
=
(
3.1
!
10 22 molecules of H
2
SO
4
$
"
#
) 1 atoms of S
1 molecule of H
2
SO
4
'
%
&
N
S
N
O
= 3.1
!
10 22 atoms
=
(
3.1
!
10 22 molecules of H
2
SO
4
$
"
#
) 4 atoms of O
1 molecule of H
2
SO
4
'
%
&
N
O
= 1.2
!
10 23 atoms
Statement: A 5.0 g sample of sulfuric acid contains 6.2 × 10 and 1.2 × 10
23
atoms of oxygen.
22
atoms of hydrogen, 3.1 × 10
22
atoms of sulfur,
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-8

1.
In combustion of hydrocarbons, a hydrocarbon with the general formula C x and carbon dioxide.
H y
O z
produces oxygen, water,
2.
c
3.
True
4.
(a) Table 1 Combustion Analysis of C x
H y
O z
Sample number
Mass of compound (g)
Mass of carbon (g)
Mass of hydrogen (g)
Mass of oxygen (g)
1
2
20.00
30.00
10.44
15.66
2.60
3.90
6.96
10.44
3 40.00 20.88 5.20 13.92
4 50.00 26.10 6.50 17.40
(b) The law of definite proportions states that a compound always contains the same proportion of elements by mass. For example, if you double the mass of the compound, you double the mass of carbon, hydrogen, and oxygen in the sample. The values in this table show that kind of proportionality, so they are consistent with the law of definite proportions.
5. (a) If Sample 1 and Sample 2 are consistent with the law of definite proportions, then they are different compounds with different compositions.
(b) Sample 4 is the same compound as Sample 1 because they have the same percentage composition: they both contain 52.2 % C, 13 % H, and 34.8 % O.
(c) Sample 3 is the same compound as Sample 2 because they have the same percentage composition: they both contain 38.7 % C, 9.7 % H, and 51.6 % O.
6. Given: m
C
6
H
6
O
2
= 82.59 g; m
O
= 24.00 g; m
H
= 4.55 g
Required: percentage composition of hydroquinone: % O; % H; % C
Analysis: Use the solution:
% O =
= m
O m
C
6
H
6
O
2
24.00 g
82.59 g
!
100 %
!
100 %
% H
= 29.06 %
= m
H m
C
6
H
6
O
2
!
100 %
=
4.55 g
82.59 g
!
100 %
= 5.51 %
% C = 100 % !
(29.06 % + 5.51 %)
= 100 % !
34.57 %
= 65.43 %
Statement: The percentage composition of hydroquinone is 29.06 % oxygen, 5.51 % hydrogen, and 65.43 % carbon.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-9

7. Given: potassium sulfate, K
2
SO
4
Required: percentage composition of K
2
SO
4
Analysis: Calculate M
K
2
SO
4 and then calculate the percentage of each element by dividing the total mass of the element by the molecular mass and multiplying by 100 %.
Solution:
M
K
2
SO
4
= 2 M
(
K
+ M
S
= 2 39.10 u
)
+ 4 M
+
(
O
32.07 u
)
+ 4 16.00 u
)
M
K
2
SO
4
= 174.27 u
% K =
(
174.27 u
= 44.87 %
)
!
100 %
% S =
32.07 u
174.27 u
% O =
= 18.40 %
( )
174.27 u
!
100 %
!
100 %
= 36.72 %
Check: 44.87 % + 18.40 % + 36.72 % = 99.99 %
Statement: The percentage composition of potassium sulfate is 44.87 % potassium, 18.40 % sulfur, and
36.72 % oxygen.
8. Given: glucose, C
6
H
12
O
6
; hexane, C
6
H
14
.
Required: percentage composition of carbon in each molecule
Solution:
M
C
6
H
12
O
6
= 6 M
C
+ 12 M
H
+ 6 M
O g $ g $ g $
=
"# mol %&
+
"# mol %&
+
"# mol %&
M
C
6
H
12
O
= 180.18 g/mol
M
C
6
H
14
= 6 M
C
+ 14 M
H
=
"# g mol %&
$
+
"# g mol %&
$
M
C
6
H
14
% C
C
6
H
12
O
6
% C
C
6
H
14
= 86.20 g/mol
=
6(12.01 g)
180.18 g
!
100 %
= 39.99 %
=
6(12.01 g)
86.20 g
!
100 %
= 83.59 %
Statement: The percentage of carbon in glucose is 39.99 %. In hexane, the percentage of carbon is 83.59 %.
So, the percentage of carbon in hexane makes up a greater proportion of the total composition.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-10

1.
a
2.
False. Two compounds with the same empirical formula can have very different properties, but they must share the same percentage composition.
3.
d
4.
(a) empirical formula: CH
(b) empirical formula: C
(c)
2
empirical formula: CH
H
4
(d) empirical formula: CH
2
4
3
O
O
2
O; molecular formula: C
2
H
O
2
3
; molecular formula: C
; molecular formula: C
5. Given: % Cu = 75.5 %; % P = 24.5 %
Required: empirical formula of compound
2
2
; molecular formula: CH
4
H
O
H
4
4
6
O
O
3
2
2
O
4
Solution: Assume a 100.0 g sample of the compound. m
Cu
M
Cu
= 75.5 g; m
P
= 24.5 g
= 63.55 g/mol; M
P
= 30.97 g/mol n
Cu
=
(
75.5 g ##
!
"
) 1 mol
63.55 g
&&
$
% n
Cu
= 1.188 mol [extra digits carried] n
P n
P
=
(
24.5 g ##
!
"
) 1 mol
30.97 g
&&
$
%
= 0.7911 mol [extra digits carried] n
Cu n
P
=
1.188 mol
0.7911 mol
= 1.50
n
P n
P
=
0.7911 mol
0.7911 mol
= 1
Ratio of copper to phosphorus is 1.5 : 1, or 3 : 2, so the formula is Cu
Statement: The empirical formula of the compound is Cu
3
P
2
.
3
6.
Given: % H = 2.6 %; % Si = 36.0 %; % O = 61.5%
Required: empirical formula of silicic acid.
Solution: Assume a 100.0 g sample of the compound.
P
2
.
m
H
M
H
= 2.6 g; m
Si
= 36.0 g; m
O
= 61.5 g
= 1.01 g/mol; M
Si
= 28.09 g/mol; M
O
= 16.00 g/mol n
H
=
( )
##
!
"
1 mol
1.01 g
&&
$
% n
H n
Si n
Si
= 2.574 mol [extra digits carried]
=
(
36.0 g ##
!
"
) 1 mol
28.09 g
&&
$
%
= 1.2815 mol [extra digits carried] n
O n
O
=
(
61.5 g ##
!
"
) 1 mol
16.00 g
&&
$
%
= 3.8437 mol [extra digits carried]
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-11

n
H n
Si
=
2.574 mol
1.2815 mol
= 2 n
Si n
Si
=
1.2815 mol
1.2815 mol
= 1 n
O n
Si
=
3.8437 mol
1.2815 mol
= 3
Ratio of hydrogen to silicon to oxygen is 2 : 1 : 3, so the formula is H
Statement: The empirical formula of the compound is H
2
SiO
3.
2
SiO
3.
1.
The experiment described in this Chemistry Journal is a controlled experiment, because all variables were controlled except for the counting machines (manipulated variable) and responding variable (amount of cocaine).
2.
False. The data collected in this experiment shows that money-counting machines are partially responsible for the contamination of British currency with cocaine.
3.
If money-counting machines are a source of contamination of banknotes with cocaine, then you should see evidence of cocaine contamination on clean paper that passes through the machines.
4.
a
5.
Table 1 Desorption and Mass Spectrometer Analysis
Step What happens Purpose
Thermal desorption Test strip is heated to 285 ºC for 1 s.
Mass spectrometer ionization
Molecules are broken into charged ions.
Mass spectrometer deflection
Molecules are deflected by a magnetic field.
Organic compounds are vaporized and can be collected for analysis.
Ions can now be deflected by a magnetic field.
The amount of deflection identifies molecular mass.
Mass spectrometer spectrum
Data is displayed as a mass spectrum.
The spectrum can be compared to known substances.
6.
Answers may vary. Sample answer: From the data collected in this experiment, I conclude that the counting machines are only partly responsible for the contamination. The fact that all banknotes seem to be contaminated, but only half of the strips got contaminated, indicates that there is at least one other source of contamination.
1.
Volatile organic compounds, VOCs, can be synthetic or natural and consist primarily of carbon and hydrogen.
2.
False. If you know the chemical formula for a compound, you can always figure out the compound’s empirical formula.
3.
The most likely molecular mass of nitromethane is around 61 u. It is the largest fragment, so it is most likely to be the compound. Other smaller samples are likely to be fragments that broke apart as a result of ionization.
4.
b
5.
The largest fragment with a molecular mass of 30 is likely to be NO
CH
3
NO
2
that has a molecular mass of exactly 30. because that is the only piece of
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-12

6. Given: empirical formula for ribose, CH
Required: molecular formula of ribose, C
2 x
H
O; molar mass of ribose = 150.13 g/mol
Solution: y
O z
M
CH
3
O
= M
C
+ 2 M
H
+ M
O
=
"#
!
12.01
g mol %&
$
+
"# g mol %&
$
+
"#
!
16.00
g mol %&
$
M
CH
3
O x =
= 30.03 g/mol molar mass of compound molar mass of empirical formula g
150.13
mol
= g
30.03
mol x = 4.999
The molar mass of the compound is 5 times the molar mass of the empirical formula.
Multiplying each of the subscripts of CH
2
Statement:
O by 5 gives C
5
H
10
The molecular formula for ribose is C
O
5
.
5
H
H
10
O y
O z
5
.
7. Given: 53.3 % C; 11.2 % H; 35.5 % O; molar mass = 90.14 g/mol
Required: molecular formula of the compound, C x
Solution: A 100.0 g sample of this compound contains 53.3 g C, 11.2 g H, and 35.5 g O.
M
C
= 12.01 g/mol; M
H
= 1.01 g/mol; M
O
= 16.00 g/mol n
C
=
(
53.3 g ##
!
"
) 1 mol
12.01 g
&&
$
% n
C n
H
= 4.4379 mol [extra digits carried]
=
(
11.2 g ##
!
"
) 1 mol
1.01 g
&&
$
% n
H
= 11.089 mol [extra digits carried] n
O n
O n
C n
O
=
(
35.5 g ##
!
"
) 1 mol
16.00 g
&&
$
%
= 2.2187 mol [extra digits carried]
=
4.4379 mol
2.2187 mol
= 2 n
H n
O
=
11.089 mol
2.2187 mol
= 4.99
n n
O n
O
C
=
: n
M
C
2
H
H
5
O
2.2187 mol
2.2187 mol
: n
O
= 1
= 2 : 5 : 1, so the empirical formula is C
= 2 M
C
+ 5 M
H
+ M
O
2
H
5
O
=
"# g mol %&
$
+
"# g mol %&
$
+
"#
!
16.00
g mol %&
$
M
C
2
H
5
O
= 45.07 g/mol
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-13

x = molar mass of compound molar mass of empirical formula
90.14
g mol
=
45.07
g mol
= 2.00
The molar mass of the compound is 2 times the molar mass of the empirical formula.
Multiplying each of the subscripts of C
Statement:
2
H
5
O by 2 gives C
4
The molecular formula for the compound is C
H
4
H
10
10
O
2
O
2
.
.
1.
Descriptions may vary. Sample answer:
(a) The description provides quantitative information. Qualitative description: The patient’s blood sugar was high.
(b) The description provides qualitative information. Quantitative description: A 50 g sample of glucose will dissolve in water that is 70 °C.
2.
False. Most of the excess salt in the average Canadian’s diet comes from eating processed foods.
3. Figure 1
4. Figure 2
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-14

5.
(a) The pure silver nugget has more atoms than the pure gold nugget.
(b) Given: m
Ag
= 60.2 g; m
Au
= 75.4 g
Required: Difference in number of atoms in nugget of gold and silver
Solution:
M
Ag
= 107.87 g/mol n
Ag n
Ag
N
Ag
=
(
60.2 g ##
!
"
) 1 mol
107.87 g
&&
$
%
= 0.55807 mol [2 extra digits carried]
= n
Ag
N
A
=
(
0.55807 mol #
!
"
) 6.02
' 10 23
1 mol atoms
&
$
%
N
M
Ag
Au n
Au n
Au
N
Au
= 3.36
' 10 23 atoms
= 196.97 g/mol
=
(
75.4 g ##
!
"
) 1 mol
196.97 g
&&
$
%
= 0.38279 mol [2 extra digits carried]
= n
Au
N
A
=
(
0.38279 mol #
!
"
) 6.02
' 10 23
1 mol atoms
&
$
%
N
Au
= 2.30
' 10 23 atoms
N
Ag
– N
Au
= 3.36
!
10 23 atoms – 2.30
!
10 23 atoms
=1.06
!
10 23 atoms
Statement: The pure silver nugget has 1.06 × 10
23
more atoms than the pure gold nugget.
6.
(a) Sample 4 contains the same compound as Sample 1. Both samples have the percentage composition
62 % C, 10% H, and 28% O, so they are the same compound, by the law of definite proportions.
(b) Samples 2 and 6 contain the same compound, and samples 3 and 5 contain a third compound, by the law of definite proportions. This is true because samples 2 and 6 have the same percentage composition, 55 % C,
9% H, and 36% O, so they are the same compound; and, samples 3 and 5 have the same percentage composition, 39 % C, 9% H, and 52% O, so they are the same compound
7.
(a) Given: m
C
= 18.6 g; m
H
= 3.1 g; m
O
= 8.3 g
Required: empirical formula of compound
Solution:
M
C
= 12.01 g/mol; M
H
= 1.01 g/mol; M
O
= 16.00 g/mol n
C
=
(
18.6 g ##
!
"
) 1 mol
12.01 g
&&
$
% n
C
= 1.5487 mol [extra digits carried] n
H n
H
=
( )
##
!
"
1 mol
1.01 g
&&
$
%
= 3.069 mol [extra digits carried] n
O
=
( )
##
!
"
1 mol
16.00 g
&&
$
% n
O
= 2.1593 mol [extra digits carried]
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-15

n
H n
O n
C n
O
=
1.5487 mol
0.51875 mol
=
3.069 mol
0.51875 mol
= 3.0
= 5.9
n
O n
C n
C n
O n
O
=
0.51875 mol
= 1
0.51875 mol
Ratio of carbon to hydrogen to oxygen is 3 : 6 : 1, so the formula is C
3
Statement: The empirical formula of Sample 1 is C
3
H
6
O.
(b) Samples 3 and 5 both are more than 50 % oxygen by mass.
H
6
O
.
Given: Sample 5: m
C
= 7.8 g; m
H
= 1.8 g; m
O
= 10.4 g
Required: empirical formula of samples 3 and 5
Solution:
M
C n
C
= 12.01 g/mol; M
=
( )
##
!
"
1 mol
12.01 g
&&
%
H
$
= 1.01 g/mol; M
O
= 16.00 g/mol n
C n
H n
H
= 0.6494 mol [extra digits carried]
=
( )
##
!
"
1 mol
1.01 g
&&
$
%
= 1.782 mol [extra digits carried] n
O
=
(
10.4 g ##
!
"
) 1 mol
16.00 g
&&
$
%
= 0.65000 mol [extra digits carried]
=
0.6494 mol
0.6494 mol
= 1 n
H n
C
=
1.782 mol
0.6494 mol
= 2.7
n
O n
C
=
0.65000 mol
= 1.0
0.6494 mol
Ratio of carbon to hydrogen to oxygen is 1 : 2.7 : 1, which is equivalent to 3 : 8 : 3, so the formula for
Sample 5 is C formula.
3
H
8
O
3
. Sample 3 has the same percentage composition as Sample 5, so it has the same empirical
Statement: The empirical formula of both Sample 3 and Sample 5 is C
3
H
8
O
3
.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-16

n
H n
O n
C n
O
8. Given: 60.9 % C; 4.4 % H; 34.8 % O; molar mass = 138.1 g/mol
Required: molecular formula of the compound, C x
H y
O z
Solution: A 100.0 g sample of this compound contains 60.9 g C, 4.4 g H, and 34.8 g O.
M
C
= 12.01 g/mol; M
H
= 1.01 g/mol; M
O
= 16.00 g/mol n
C
=
(
60.9 g ##
!
"
) 1 mol
12.01 g
&&
$
% n
C
= 5.0707 mol [extra digits carried] n
H n
H
=
( )
##
!
"
1 mol
1.01 g
&&
$
%
= 4.356 mol [extra digits carried] n
O n
O
=
(
34.8 g ##
!
"
) 1 mol
16.00 g
&&
$
%
= 2.1750 mol [extra digits carried]
=
5.0707 mol
2.1750 mol
=
4.356 mol
2.1750 mol
= 2.33
= 2 n n
O n
O
=
2.1750 mol
2.1750 mol
C
: n
H
: n
O
= 1
= 2.33 : 2 : 1, which is equivalent to 7 : 6 : 3, so the empirical formula is C
7
H
6
M
C
7
H
6
O
3
= 7 M
C
+ 6 M
H
+ 3 M
O
O
3
.
=
"# g mol %&
$
+
"# g mol %&
$
+
"# g mol %&
$
M
C
7
H
6
O
3
= 138.13 g/mol
The molar mass of the compound is the same as the molar mass of the empirical formula, so the molecular formula for the compound is the same as the empirical formula, C
Statement: The molecular formula for the compound is C
7
H
6
O
3
.
7
H
6
O
3
.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-17