Knowledge
1.
(a)
2.
(b)
3.
(d)
4.
(b)
5.
(c)
6.
(c)
7.
(a)
8.
(b)
9.
(a)
10. (a) (ii), (iv)
(b) (i), (iii), (iv)
(c) (i)
11.
False. Quantitative analysis is used to measure the amount of a substance present.
12.
True
13.
True
14.
True
15.
False. The unit used for molar mass is g/mol .
16.
False. The number of entities in a sample can be found by multiplying the amount of the sample by Avogadro’s constant.
17.
False. Percentage composition is the percentage, by mass , of each element in a compound.
18.
True
19.
False. The formula C
4
H
8
O
2
is a molecular formula.
20. (a) Noting that a leaf is green is qualitative analysis.
(b) Measuring the length of a board to be 4.6 cm long is quantitative analysis.
21. (a) A graduated cylinder is a tool used in quantitative analysis.
(b) Your eyes are used in qualitative analysis.
22. (a) You can divide the molar mass of iron by Avogadro’s constant to find the mass of a single iron atom.
(b) The mass of a single iron atom is:
55.85
6.02
!
10 23 g mol atoms mol
=
"
#$
55.85 g mol
%
&'
"
# 6.02
!
1 mol
10 23 atoms
%
&
= 9.28
!
10 ( 23 g atom
23. (a) To multiply numbers in scientific notation, multiply the coefficients, then add the exponents. If necessary, express the product of the coefficients as a number between 1 and 10 and change the exponent accordingly.
(b) (2.4
!
10 14 )(4.6
!
10 4 ) = 11 !
10 18
= 1.1
!
10 19
24.
5.34
!
10
3.56
!
10
5
7
= 1.50
!
10 " 2
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-2

Understanding
25. (a) The mass of 1 mol of mercury is greater than the mass of 1 mol of bromine.
(b) The volume of 1 mol of mercury and 1 mol of bromine will be different, due to different types of particles, and the forces between these particles.
26. (a) The mass of 1 mol of zinc is different from the mass of 1 mol of sulfur because the mass of a zinc atom is different from the mass of a sulfur atom.
(b) The mass of 1 mol of zinc is greater than the mass of 1 mol of sulfur.
27.
The molar mass of aluminum is 26.98 g/mol.
28. (a) The molar mass of an oxygen atom is 16.00 g/mol.
(b) The molar mass of oxygen gas, O
2
, is 32.00 g/mol.
(c) The molar mass of ozone, O
3
, is 48.00 g/mol.
29. (a) The molar mass of a hydrate is calculated by adding the atomic masses of all atoms in the compound, including the atoms in water.
(b) Given: cobalt chloride hexahydrate, CoCl
2
•6H
2
O
Required: molar mass of CoCl
2
•6H
2
O
Solution:
Step 1.
Look up the molar masses of the elements.
M
Co
= 58.93
g mol
; M
Cl
= 35.45
g mol
; M
H
= 1.01
g mol
; M
O
= 16.00
g mol
Step 2.
Add the molar masses of the elements, multiplying the molar mass of each element by the number of atoms of that element in the compound.
M
CoCl
2
• 6H
2
O
= M
Co 2 +
+ 2 M
Cl -
+ 6(2 M
H
+ M
O
)
=
!
"#
58.93 g mol
$
%&
+
"# g mol
$
%&
+
"#
' 1.01 g mol
+ 16.00
g $ mol %&
M
CoCl
2
• 6H
2
O
= 237.95 g mol
Statement: The molar mass of cobalt chloride hexahydrate is 237.95 g/mol.
30.
The molar mass of a molecular compound, such as CH
4
or CO
2
, is the sum of the molar masses of the atoms in one molecule. The molar mass of an ionic compound, such as NaCl or
KNO
3,
is the sum of the molar masses of the ions in one formula unit of the compound.
31. (a) Given: potassium permanganate, KMnO
4
Required: molar mass of KMnO
4
Solution:
Step 1.
Look up the molar masses of the elements.
M
K
= 39.10 g mol
; M
Mn
= 54.94 g mol
; M
O
= 16.00 g mol
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-3

Step 2.
Add the molar masses of the elements, multiplying the molar mass of each element by the number of atoms of that element in the compound.
M
KMnO
4
= M
K
+
+ M
MnO
4
!
= M
K
+
+
=
"
#$
39.10
M
Mn g mol
%
&'
+ 4 M
O
+
"
#$
54.94 g % mol &'
+
"
#$
4 ( 16.00 g % mol &'
M
KMnO
4
= 158.04 g mol
Statement: The molar mass of potassium permanganate is 158.04 g/mol.
(b) Given:
Required: amount of potassium permanganate,
Solution: m
KMnO
4
= 258 g ; M
KMnO
4
= 158.04 g/mol n
KMnO
4
Use the mass of potassium permanganate and its molar mass to calculate the amount. n
KMnO
4
=
(
258 g
) !
"#
1 mol
158.04 g
$
%& n
KMnO
4
= 1.63 mol
Statement: There are 1.63 mol of potassium permanganate in 258 g of the compound.
32.
Given: n
Fe
= 2.7 mol
Required: mass of iron, m
Fe
Solution:
Step 1.
Look up the molar mass of iron.
M
Fe
= 55.85
g mol
Step 2.
Use the amount of iron and its molar mass to calculate the mass of iron. m
Fe
= (2.7 mol )
!
"#
55.85 g
1 mol
$
%& m
Fe
= 150 g
Statement: There are 150 g of iron in 2.7 mol of iron.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-4

33.
Given :
Required: n
Cu(NO
3
)
2
= 4.5 mol mass of copper(II) nitrate, m
Cu(NO
3
Solution:
Step 1.
Determine the molar mass of Cu(NO
)
2
3
)
2
.
M
Cu(NO
3
)
2
= M
Cu 2 +
+ 2 M
NO
3
!
= M
Cu
2 +
+
=
"
#$
63.55
2( g
M mol
%
&'
N
+ 3
+
#$
M
O
) g mol
+ 3 ( 16.00 g % mol &'
M
Cu(NO
3
)
2
= 187.57 g mol
Step 2.
Use the amount of copper(II) nitrate and its molar mass to calculate the mass of the compound. m
Cu(NO
3
)
2
= (4.5 mol )
!
"#
187.57 g
1 mol
$
%& m
Cu(NO
3
)
2
= 840 g '
1 kg
1000 g m
Cu(NO
3
)
2
= 0.84 kg
Statement: The mass of 4.5 mol of copper(II) nitrate is 0.84 kg.
34.
Give n: n
Si
= 0.68 mol
Require d: number of atoms of silicon, N
Si
Solution: Calculate the number of atoms using an appropriate conversion factor.
N
Si
= (0.68 mol )
"
#$
6.02
!
10 23
1 mol atoms
%
&'
N
Si
= 4.1
!
10 23 atoms
Statement: There are 4.1 × 10 23 atoms in 0.68 mol of silicon.
35.
(a) Given: n
C
2
H
6
= 1.4 mol
Required: number of molecules of C
2
H
6
,
Solution:
N
C
2
H
6
Calculate the number of molecules using an appropriate conversion factor.
N
C
2
H
6
= (1.4 mol )
"
#$
6.02
!
10 23 molecules
1 mol
%
&'
N
C
2
H
6
= 8.428
!
10 23 molecules [2 extra digits carried]
Statement: There are 8.4 × 10 23 molecules in 1.4 mol of ethane.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-5

(b) Given: N
C
2
H
6
= 8.428
!
10 23 molecules
Required: number of atoms in C
2
Solution:
H
6
One molecule of C
2
H
6
contains 2 C atoms and 6 H atoms. Therefore, there are 8 atoms in one molecule of C
2
H
6
. Determine the number of atoms by multiplying the number of molecules by the number of atoms per molecule.
N atoms
= (8.428
!
10 23 molecules )
"
#$
8 atoms
1 molecule
%
&'
N atoms
= 6.7
!
10 24 atoms
Statement: There are 6.7 × 10 24 atoms in 1.4 mol of ethane.
36. Given:
Solution: m
H
2
O
= 4.56 g
Required: amount of water, n
H
2
O
Step 1.
Calculate the molar mass of water,
=
"#
+
!
"#
16.00
M
H
2
O
.
M
H
2
H
2
O g mol
$
%& g mol
$
%&
M
O
= 18.02 g mol
Step 2.
Use the mass of water and its molar mass to calculate the amount of water. n
H
2
O
= (4.56 g)
!
"#
1 mol
18.02 g
$
%& n
H
2
O
= 0.253 mol
Statement: The amount of water in a 4.56 g sample is 0.253 mol.
37. Percentage composition is the percentage, by mass, of each element in a compound. For example, for any sample of Na
2
O, 74.2 % of the mass of the sample is made up by Na. For any sample, 25.8 % of the mass of the sample is made up by oxygen.
38. Given: m sample
= 10.0 g ; m
C
= 7.50 g
Required: percentage of carbon, % C
Solution:
Use the percentage formula to calculate the percentage of carbon in the sample.
% C = m
C m sample
!
100 %
=
7.50 g
10.0 g
!
100 %
% C = 75.0 %
Statement: Carbon atoms make up 75.0 % by mass of methane.
.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-6

39.
Given: magnesium chloride, MgCl
2
Required: percentage composition of MgCl
2
Solution:
Step 1.
Calculate the molecular mass of the compound.
M
MgCl
2
= (24.31 u) + (2 !
35.45 u)
M
MgCl
2
= 95.21 u
Step 2.
Calculate the percentage of each of the elements Mg and Cl.
% Mg =
24.31 u
95.21 u
!
100 % % Cl =
70.90 u
95.21 u
!
100 %
% Mg = 25.53 % % Cl = 74.47 %
Statement: The percentage composition of magnesium chloride is 25.53 % magnesium and
74.47 % chlorine.
40.
The law of definite proportions states that a compound always has the same proportion of elements by mass.
41. Given:
m K
= 78.2 g ; m
N
= 28.0 g ; m
O
= 96.0 g
Required: percentage of each element: % K; % N; % O
Solution:
Step 1. Calculate the mass of the sample. m sample
= 78.2 g + 28.0 g + 96.0 g = 202.2 g
Step 2.
Calculate the percentage of each element.
% K = m
K m sample
!
100 % % N = m
N m sample
!
100 % % O = m
N m sample
!
100 %
=
78.2 g
202.2 g
!
100 % =
28.0 g
202.2 g
!
100 % =
96.0 g
202.2 g
!
100 %
% K = 38.7 % % N = 13.8 % % O = 47.5 %
Statement: The percentage composition of potassium nitrate is 38.7 % potassium,
13.8 % nitrogen, and 47.5 % oxygen.
42. (a) Given: m sample
= 24.9 g ;
m
Al
= 2.52 g
Required: mass of bromine,
m
Br
Solution:
Determine the mass of bromine by subtracting the mass of aluminum from mass of the sample.
m
Br
= 24.9 g !
2.52 g = 22.38 g (1 extra digit carried)
Statement: The mass of bromine in this sample is 22.4 g.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-7

(b) Given: m sample
= 24.9 g ;
m Al
= 2.52 g ;
m Br
= 22.38 g
Required: percentage of each element: % Al; % Br
Solution: Calculate the percentage of each element.
% Al = m
Al m sample
!
100 % % Br = m m
Br sample
!
100 %
=
2.52 g
24.9 g
!
100 % =
22.38 g
24.9 g
!
100 %
% Al = 10.1 % % Br = 89.9 %
Statement: The percentage composition of the aluminum bromide is 10.1 % aluminum and
89.9 % bromine.
43.
Given: hydrogen sulfide, H
2
S
Required: percentage composition of H
2
Solution:
S
Step 1.
Calculate the molecular mass of the compound.
M
H
2
S
= (2 !
1.01 u) + (32.07 u)
M
H
2
S
= 34.09 u
Step 2.
Calculate the percentage of each element.
% H =
2.02 u
34.09 u
!
100 % % S =
32.07 u
34.09 u
!
100 %
% H = 5.93 % % S = 94.07 %
Statement: The percentage composition of hydrogen sulfide is 5.95 % hydrogen and
94.07 % sulfur.
44. The percentage composition of a compound is always the same. The percentage composition of a mixture can vary. For example, a mixture such as a salt solution could contain 5 g of salt
(sodium chloride) in 100 g of solution. The percentage composition of this mixture would be 5 % sodium chloride. The percentage composition of sodium in the compound sodium chloride is always 39.3 % Na.
45. No, K
2
F
2
could not be an empirical formula for potassium fluoride. The subscripts in an empirical formula must have no common factor, other than 1. The subscripts in K
2
F
2
have a common factor of 2.
46. (a) Given: % K = 23.6 %; % I = 76.4 %
Required: empirical formula of compound, K x
I y
Solution: A 100.0 g sample of this compound contains 23.6 g of potassium and 76.4 g of iodine.
Step 1.
Calculate the amount of each element in the 100.0 g sample. n
K
= (23.6 g)
!
"#
1 mol
39.10 g
$
%& n
I
= (76.4 g)
!
"#
1 mol
126.90 g
$
%& n
K
= 0.603 58 mol [2 extra digits carried] n
I
= 0.602 05 mol [2 extra digits carried]
Step 2.
Divide the amount of each element by the smallest amount. n
K n
I
=
0.603 58 mol
0.60205 mol
= 1.00
n n
I
I =
0.602 05 mol
0.602 05 mol
= 1.00
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-8

Step 3.
The ratio of potassium to iodine is 1:1. This ratio suggests an empirical formula of KI.
Statement: The empirical formula of the compound is KI.
47. (a) Given : % Na = 74.2 %; remainder is O
Required: percentage of oxygen, % O
Solution:
Determine the percentage of oxygen by subtracting the percentage of sodium from 100 %.
% O
= 100.0 % !
74.2 % = 25.8 %
Statement: In this compound, the percentage of oxygen is 25.8 %.
(b) Given : % Na = 74.2 %; % O = 74.2 %
Required: empirical formula of compound, Na x
O y
Solution: A 100.0 g sample of this compound contains 74.2 g of sodium and 25.8 g of oxygen.
Step 1.
Calculate the amount of each element in the 100.0 g sample. n
Na
= (74.2 g)
!
"#
1 mol
22.99 g
$
%& n
O
= (25.8 g)
!
"#
1 mol
16.00 g
$
%& n
K
= 3.2275 mol [2 extra digits carried] n
O
= 1.6125 mol [2 extra digits carried]
Step 2.
Divide the amount of each element by the smallest amount. n n
Na
O
=
3.2275 mol
1.6125 mol
= 2.00
n n
O
O
=
1.6125 mol
1.6125 mol
= 1.00
Step 3.
The ratio of sodium to oxygen is 2:1. This ratio suggests an empirical formula of Na
2
O.
Statement: The empirical formula of the compound is Na
2
O.
48. Given: % P = 20.2 %; % O = 10.4 %; remainder is Cl; mass of sample, m = 100.0 g
Required: empirical formula of compound, P x
O y
Cl z
Solution: Determine the percentage of chlorine by subtracting the sum of the percentages of the other elements from 100 %.
% Cl = 100.0 % !
(20.2 % + 10.4 %)
% Cl = 69.4 %
A 100.0 g sample of this compound contains 20.2 g of phosphorus, 10.4 g of oxygen, and
69.4 g of chlorine.
Step 1.
Calculate the amount of each element in the 100.0 g sample. n
P
= (20.2 g)
!
"#
1 mol
30.97 g
$
%& n
P
= 0.652 24 mol [2 extra digits carried] n n
O
O
= (10.4 g)
!
"#
1 mol
16.00 g
$
%&
= 0.65000 mol [2 extra digits carried] n
Cl
= (69.4 g)
!
"#
1 mol
$
35.45 g %& n
Cl
= 1.9577 mol [2 extra digits carried]
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-9

Step 2.
Divide the amount of each element by the smallest amount. n n
P
O
=
0.652 24 mol
0.650 00 mol
= 1.00
n n
O
O
=
0.650 00 mol
0.650 00 mol
= 1.00
n n
Cl
O
=
1.9577 mol
0.650 00 mol
= 3.01
Step 3.
The ratio of phosphorus to oxygen to chlorine is 1:1:3. This ratio suggests an empirical formula of POCl
3
.
Statement: The empirical formula of the compound is POCl
3
.
49. Given: empirical formula, NO
2
; molar mass of compound = 92.0 g/mol
Required: molecular formula of compound, N x
O y
Solution:
Step 1.
Calculate the empirical molar mass of the compound.
M
NO
2
=
!
"#
14.01 g mol
$
%&
+
!
"#
2 ' 16.00 g mol
$
%&
M
NO
2
= 46.01 g mol
Step 2.
Solve for x , the mass multiple.
92.0 g mol x =
46.01 g mol x = 2.00
Step 3.
The molar mass of the compound is 2 times the molar mass of the empirical formula.
Multiply each of the subscripts by 2 to give N
2
O
4
.
Statement: The molecular formula of the compound is N
2
O
4
.
50. Given: % P = 14.9 %; % Cl = 85.1 %; molar mass of compound = 208.5 g/mol
Required: molecular formula of the compound
Solution: A 100.0 g sample of this compound contains 14.9 g of phosphorus and 85.1 g of chlorine.
Step 1.
Calculate the amount of each element in the 100.0 g sample. n
P
= (14.9 g)
!
"#
1 mol
30.97 g
$
%& n
Cl
= (85.1 g)
!
"#
1 mol
35.45 g
$
%& n
P
= 0.481 11 mol [2 extra digits carried] n
Cl
= 2.4006 mol [2 extra digits carried]
Step 2.
Divide the amount of each element by the smallest amount. n n
P
P
=
0.481 11 mol
0.481 11 mol
= 1.00
n n
Cl
P
=
2.4006 mol
0.481 11 mol
= 4.99
The ratio of phosphorus to chlorine is 1:5. This ratio suggests an empirical formula of PCl
5
.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-10

Step 3.
Calculate the empirical molar mass of PCl
5
.
M
PCl
5
=
!
"#
30.97 g mol
$
%&
+
!
"#
5 ' 35.45 g mol
$
%&
M
PCl
5
= 208.22 g mol
Step 4.
Solve for x , the mass multiple.
208.5
g mol x =
208.22
g mol x = 1.001
Step 5.
The mass multiple is 1, and so the molecular formula is the same as the empirical formula.
Statement: The molecular formula of the compound is PCl
5
.
Analysis and Application
51. (a) The student uses qualitative analysis when he uses his eyes to observe which plants are the tallest.
(b) The student uses quantitative analysis when he measures the amount of fertilizer added.
Quantitative analysis is important, since exact measurements of the amount of fertilizer used must be taken, and measurements of plant height must be taken. This quantitative data allows the results to be mathematically analyzed.
52. (a) The student might use qualitative analysis when she observes that more bubbles form on one metal than form on the other.
(b) The student might use quantitative analysis when she collects the hydrogen and measures the volumes produced by each metal.
53. (a) A metric ruler is useful in quantitative analysis.
(b) A balance is useful in quantitative analysis.
(c) An eye is useful in qualitative analysis.
(d) The skin on a finger is useful in qualitative analysis.
(e) A measuring cup is useful in quantitative analysis.
54. Given : number of cells = 3.00 × 10 13 cells
Required : 1 mol of cells
Solution : Divide the number of red blood cells by Avogadro’s constant. amount cells
=
=
3.00
!
10 13 cells
6.02
!
10 23 cells
(3.00
!
10 13 mol cells )
"
#$
1 mol
6.02
!
10 23
% cells &' amount cells
= 4.98
!
10 ( 11 mol
Statement: The number of 3.00 × 10 13 red blood cells represents 4.98 × 10 -11 mol of red blood cells.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-11

55. Given : distance to the Large Magellanic Cloud = 1.63 × 10 5 light years;
1 light year = 9.5 × 10 12 km
Required : distance of the large Magellanic Cloud from Earth in km
Solution : Multiply the distance in light years by the number of kilometers in one light year. distance
Cloud
= (1.63
!
10 5
" light years ) 9.5
#
!
10 12 km light year
%
'
& distance
Cloud
= 1.5
!
10 18 km
Statement: The Large Magellanic Cloud is 1.5 × 10 18 km from Earth.
56. The mass of chlorine added would be less than the mass of bromine added because 1 mol contains the same number of molecules and bromine molecules are heavier than chlorine molecules.
57. (a) The molar mass of bromine atoms is 79.9 g/mol whereas the molar mass of bromine molecules is 159.8 g/mol.
(b) You need to know whether you are working with the bromine atom, Br, or the bromine molecule, Br
2
.
58. (a) One chemical formula can represent two or more compounds if the atoms in the compounds are arranged differently. That is, these compounds have different structural formulas.
(b) The molar masses of these two compounds are the same because the compounds contain the same number and type of atoms.
59. Given: molar mass of hydrate #1 = 246.4 g/mol; molar mass of hydrate #2 = 210.4 g/mol
Required: chemical formulas of two MgSO
4
hydrates
Solution:
Step 1.
Calculate the molar mass of MgSO
4
and the molar mass of H
2
O.
M
MgSO
4
=
!
"#
24.31 g mol
$
%&
+
!
"#
32.07 g mol
$
%&
+
!
"#
4 ' 16.00 g mol
$
%&
M
MgSO
4
= 120.38 g mol
M
H
2
H
2
O
=
"
#$
2 !
1.01 g mol
%
&'
+
"
#$
16.00 g mol
%
&'
M
O
= 18.02 g mol
Step 2.
Calculate the mass of water in 1 mol of each hydrate.
M
M hydrate #1 hydrate# 1
= 246.4
= 126.02 g mol
!
120.38 g mol g mol
[1 extra digit carried]
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-12

M hydrate# 2
= 210.4 g mol
!
120.38 g mol
M hydrate# 2
= 90.02 g mol
[1 extra digit carried]
Step 3.
Determine the number of water molecules in each hydrate by dividing the mass of water in 1 mol by the molar mass of water.
Hydrate #1: number of molecules =
126.02 g mol
18.02 g mol number of molecules = 6.993
Hydrate #2: number of molecules =
90.02 g mol
18.02 g mol number of molecules = 4.996
Step 4.
The number of water molecules per 1 mol of each hydrate is 7 in hydrate #1 and
5 in hydrate #2.
Statement: The chemical formulas for the two hydrates are MgSO
4
•7H
2
O and MgSO
4
•5H
2
O.
60. (a) Given:
Solution: n
Al
2
O
3
= 2.70 mol
Required: mass of aluminum oxide, m
Al
2
O
3
Step 1.
Calculate the molar mass of aluminum oxide,
M
Al
2
O
3
=
"
#$
2 !
26.98 g mol
%
&'
+
"
#$
3 !
16.00 g mol
%
&'
M
Al
2
O
3
.
M
Al
2
O
3
= 101.96 g mol
Step 2.
Use the molar mass to calculate the mass of aluminum oxide. m
Al
2
O
3
=
"
(2.70 mol ) 3
#$
!
16.00 g
% mol &' m
Al
2
O
3
= 275 g
Statement: The mass of aluminum oxide in 2.70 mol of the compound is 275 g.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-13

(b) Given: m
Al
2
O
3
= 67.5 g ; M
Al
2
O
3
= 101.96
Required: amount of aluminum oxide,
Solution: n
Al
2
O
3 g mol
Use the mass of aluminum oxide and its molar mass to calculate the amount of aluminum oxide. n
Al
2
O
3
= (67.5 g)
!
"#
1 mol
101.96 g
$
%& n
Al
2
O
3
= 0.662 mol
Statement: There are 0.622 mol of aluminum oxide in 67.5 g of the compound.
61. (a) Given: m
C
2
H
2
= 526 g
Required: amount of ethyne,
Solution: n
C
2
H
2
Step 1.
Calculate the molar mass of ethyne,
M
C
2
H
2
=
"
#$
2 !
12.01 g mol
%
&'
+
"
#$
2 !
1.01 g mol
%
&'
M
C
2
H
2
.
M
C
2
H
2
= 26.04 g mol
Step 2.
Use the mass of ethyne and its molar mass to calculate the amount of ethyne. n
C
2
H
2
= (526 g)
!
"#
1 mol
26.04 g
$
%& n
C
2
H
2
= 20.2 mol
Statement: The amount of ethyne in a tank that contains 526 g of the gas is 20.2 mol.
(b) Given:
Solution: n
C
2
H
2
= 25.7 mol ; M
Required: mass of ethyne, m
C
2
C
2
H
2
H
2
= 26.04 g g mol
Use the amount of ethyne and its molar mass to calculate the mass of ethyne. m
C
2
H
2
= (25.7 mol)
!
"#
26.04 g
$
1 mol %& m
C
2
H
2
= 669 g
Statement: The mass of ethyne in a tank that is filled with 25.7 mol of the gas is 669 g.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-14

62. Given: n
HNO
3
=
1 .
25 mol
Required: number of atoms in nitric acid
Solution:
Step 1.
Use Avogadro’s number to calculate the number of molecules of nitric acid.
N
HNO
3
= (1.25 mol )
"
#$
6.02
!
10 23 molecules
1 mol
%
&'
N
HNO
3
= 7.525
!
10 23 molecules [1 extra digit carried]
Step 2.
One molecule of HNO
3
contains 1 H atom, 1 N atom, and 3 O atoms. Therefore, there are 5 atoms in one molecule of HNO
3
.
Determine the number of atoms by multiplying the number of molecules by the number of atoms per molecule.
N atoms
= (7.525
!
10 23 molecules )
"
#$
5 atoms
1 molecule
%
&'
N atoms
= 3.76
!
10 24 atoms
Statement: There are 3.76 × 10 24 atoms in 1.25 mol of nitric acid.
63.
(a) Given: m
SO
3
= 3.20 g
Required: number of oxygen atoms in sulfur trioxide, N
Solution:
O
. Step 1.
Calculate the molar mass of sulfur trioxide,
M
SO
3
=
!
"#
32.07 g mol
$
%&
+
!
"#
3 ' 16.00 g mol
$
%&
M
M
SO
3
= 80.07 g mol
Step 2.
Calculate the amount of sulfur trioxide.
SO
3 n
SO
3
= (3.20 g)
!
"#
1 mol
80.07 g
$
%& n
SO
3
= 0.039 965 mol [2 extra digits carried]
Step 3.
Use Avogadro’s number to calculate the number of molecules of sulfur trioxide.
N
SO
3
=
(
0.039 965 mol
) "
#$
6.02
!
10 23 molecules
1 mol
%
&'
N
SO
3
= 2.4059
!
10 22 molecules [2 extra digits carried]
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-15

Step 4.
One molecule of SO
3
contains 3 O atoms. Determine the number of O atoms by multiplying the number of molecules by the number of O atoms per molecule.
N
O
= (2.4059
!
10 22 molecules)
"
#$
3 atoms
%
1 molecule &'
N
O
= 7.22
!
10 22 atoms
Statement: In 3.20 g of sulfur trioxide, there are 7.22 × 10 22 oxygen atoms.
(b) Given: m
NO
2
= 2.50 g
Required: number of oxygen atoms in nitrogen dioxide, N
Solution:
O
Step 1.
Calculate the molar mass of nitrogen dioxide, M
M
NO
2
NO
2
=
!
"#
14.01 g mol
$
%&
+
!
"#
2 ' 16.00 g mol
$
%&
M = 46.01
g mol
Step 2.
Calculate the amount of nitrogen dioxide. n
NO
2
= (2.50 g)
!
"#
1 mol
46.01 g
$
%&
NO
2
. n
NO
2
= 0.054 336 mol [2 extra digits carried]
Step 3.
Use Avogadro’s number to calculate the number of molecules of nitrogen dioxide.
N
NO
2
= (0.054 336 mol )
"
#$
6.02
!
10 23 molecules
1 mol
%
&'
N
NO
2
= 3.2710
!
10 22 molecules [2 extra digits carried]
Step 4.
One molecule of NO
2
contains 2 O atoms. Determine the number of O atoms by multiplying the number of molecules by the number of O atoms per molecule.
N
O
= (3.2710
!
10 22 molecules )
"
#$
2 atoms
1 molecule
%
&'
N
O
= 6.54
!
10 22 atoms
Statement: In 2.50 g of nitrogen dioxide, there are 6.54 × 10 22 oxygen atoms.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-16

64. (a) Given: m
SO
3
= 5.2 g
Required : number of molecules of SO
3
Solution :
Step 1.
Calculate the molar mass of sulfur trioxide,
M
SO
3
=
!
"#
32.07 g mol
$
%&
+
!
"#
3 ' 16.00 g mol
$
%&
M
M
SO
3
= 80.07 g mol
Step 2.
Calculate the amount of sulfur trioxide.
SO
3 n
SO
3
= (5.2 g)
!
"#
1 mol
80.07 g
$
%&
. n
SO
3
= 0.064 94 mol [2 extra digits carried]
Step 3.
Use Avogadro’s number to calculate the number of molecules of sulfur trioxide.
N
SO
3
= (0.064 94 mol )
"
#$
6.02
!
10 23 molecules
1 mol
%
&'
N
SO
3
= 3.9
!
10 22 molecules [2 extra digits carried]
Statement : There are 3.9 × 10 22 sulfur trioxide molecules in 5.2 g of the gas.
(b) Given: m
H
2
SO
4
= 254 g
Required : number of atoms in sulfuric acid
Solution :
Step 1.
Calculate the molar mass of sulfuric acid,
M
H
2
SO
4
=
"
#$
2 !
1.01 g mol
%
&'
+
"
#$
32.07 g mol
%
&'
+
"
#$
4 !
M
H
2
16.00
SO
4 g
. mol
%
&'
M
H
2
SO
4
= 98.09 g mol
Step 2.
Calculate the amount of sulfuric acid. n
H
2
SO
4
= (254 g)
!
"#
1 mol
98.09 g
$
%& n
H
2
SO
4
= 2.5895 mol (2 extra digits carried)
Step 3.
Use Avogadro’s number to calculate the number of molecules of sulfuric acid.
N
H
2
SO
4
= (2.5895 mol )
"
#$
6.02
!
10 23 molecules
1 mol
%
&'
N
H
2
SO
4
= 1.5589
!
10 24 molecules [2 extra digits carried]
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-17

Step 4.
One molecule of H
2
SO
4
contains 2 H atoms, 1 s atom, and 4 O atoms. Therefore, there are 7 atoms in one molecule of H
2
SO
4
. Determine the number of atoms by multiplying the number of molecules by the number of atoms per molecule.
N atoms
= (1.5589
!
10 24 molecules )
"
#$
7 atoms
1 molecule
%
&'
N atoms
= 1.09
!
10 25 atoms
Statement : There are 1.09 × 10 25 atoms in 254 g of sulfuric acid.
65. Given: m sample
= 24.3 g ; m
Fe
= 11.3 g
Required: percentage of iron in pyrite, % Fe
Solution: Use the % element formula to calculate the percentage of iron.
% Fe = m
Fe m sample
!
100 %
=
11.3 g
24.3 g
!
100 %
% Fe = 46.5 %
Statement: The percentage by mass of iron in pyrite is 46.5 %.
66. Given: ethanol, C
2
H
5
OH
Required: percentage composition of ethanol
Solution:
Step 1. Calculate the molecular mass of the compound.
M
C
2
H
5
OH
= (2 !
12.01 u) + (5 !
1.01 u) + (16.00 u) + (1.01 u)
M
C
2
H
5
OH
= 46.08 u
Step 2.
Calculate the percentage of each of element.
% C =
24.02 u
46.08 u
!
100 % % H =
6.06 u
46.08 u
!
100 % % O =
16.00 u
46.08 u
!
100 %
% C = 52.13 % % H = 13.2 % % O = 34.72 %
Statement: The percentage composition of ethanol, to 0.1 %, is 52.1 % carbon,
13.2 % hydrogen, and 34.7 % oxygen.
67. The law of definite proportions says that the elements in a compound are always in the same proportion by mass. All samples of Mg(OH)
2
contain 41.7 % Mg.
68.
Given: m
Fe
= 5.58 g ; m
O
= 2.40 g
Required: percentage composition of iron(III) oxide
Solution:
Step 1.
Determine the mass of the sample. m sample
= 5.58 g + 2.40 g = 7.98 g
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-18

Step 2.
Calculate the percentage of each element.
% Fe = m
Fe !
100 % % O = m sample m
Fe m sample
!
100 %
=
5.58 g
7.98 g
!
100 % =
2.40 g
7.98 g
!
100 %
% Fe = 69.9 % % O = 30.1 %
Statement: The percent composition of iron(III) oxide is 69.9 % iron and 30.1 % oxygen.
69.
Given: m sample
= 245 kg ; m
Ca
= 98.1 kg ; m
C
= 29.4 kg
Required: percentage composition of calcium carbonate
Solution:
Step 1.
Determine the mass of oxygen by subtracting the sum of the masses of the other elements from 245 kg. m
O
= 245 kg !
(98.1 kg + 29.4 kg) m
O
= 117.5 kg [1 extra digit carried]
Step 2.
Calculate the percentage of each element.
% Ca = m
Ca m sample
!
100 % % C = m
C m sample
!
100 % % O = m
O m sample
!
100 %
=
98.1 kg
245 kg
!
100 % =
29.4 kg
245 kg
!
100 % =
117.5 kg
245 kg
!
100 %
% Ca = 40.0 % % C = 12.0 % % O = 48.0 %
Statement: The percentage composition of calcium carbonate is 40.0 % calcium, 12 % carbon, and 48.0 % oxygen.
70. Given: tetraphosphorus trisulfide, P
4
S
3
Required: percentage composition of tetraphosphorus trisulfide
Solution:
Step 1.
Calculate the molecular mass of the compound.
M
P
4
S
3
= (4 !
30.97 u) + (3 !
32.07 u)
M
P
4
S
3
= 220.09 u
Step 2.
Calculate the percentage of each of the element.
% P =
123.88 u
220.09 u
!
100 % % S =
96.21 u
220.09 u
!
100 %
% P = 56.29 % % S = 43.71 %
Statement: The percentage composition of tetraphosphorus trisulfide is 56.29 % phosphorus and
43.71 % sulfur.
71. Answers may vary. Sample answers:
(a) Examples of formulas that are both molecular and empirical include CO
2
and CH
4
.
(b) Examples of formulas that are molecular but not empirical include C
6
H
12
O
6
and H
2
O
2
.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-19

72. (a) Mark was incorrect.
(b) The sample size of 100.0 g is usually used because it is easy to use. For example, if the percentage composition of a compound is 57.1 % P and 42.9 % S, using a sample size of 100.0 g allows you to easily calculate masses of 57.1 g P and 42.9 g S in the sample. However, any sample size could be used.
73. (a) Given: % Mg = 43.1 %; remainder is S
Required: percentage of sulfur
Solution:
Determine the percentage of sulfur by subtracting the percentage of magnesium from 100 %.
% S
= 100.0 % !
43.1 % = 56.9 %
Statement: The percentage of sulfur in magnesium sulfide is 56.9 %.
(b) Given: % Mg = 43.1 %; % S = 56.9 %
Required: empirical formula of magnesium sulfide
Solution: A 100.0 g sample of this compound contains 43.1 g of magnesium and 56.9 g of sulfur.
Step 1.
Calculate the amount of each element in the 100.0 g sample. n
Mg
= (43.1 g)
!
"#
1 mol
24.31 g
$
%& n
S
= (56.9 g)
!
"#
1 mol
32.07 g
$
%& n
Mg
= 1.77 mol n
S
= 1.77 mol
Step 2.
The ratio of magnesium to sulfur is 1:1. This ratio suggests an empirical formula of MgS.
Statement: The empirical formula of magnesium sulfide is MgS.
74.
Given: % Ca = 36.1 %; remainder is Cl
Required: empirical formula of calcium chloride
Solution:
Step 1.
Determine the percentage of chlorine by subtracting the percentage of calcium from 100 %.
% Cl
= 100.0 % !
36.1 % = 63.9 %
A 100.0 g sample of this compound contains 36.1 g of calcium and 63.9 g of chlorine.
Step 2.
Calculate the amount of each element in the 100.0 g sample. n
Ca
= (36.1 g)
!
"#
1 mol
40.08 g
$
%& n
Cl
= (63.9 g)
!
"#
1 mol
35.45 g
$
%& n
Ca
= 0.900 70 mol [2 extra digits carried] n
Cl
= 1.8025 mol [2 extra digits carried]
Step 3.
Divide the amount of each element by the smallest amount. n
Ca n
Ca
=
0.900 70 mol
0.900 70 mol
= 1.00
n
Cl n
Ca
=
1.8025 mol
0.900 70 mol
= 2.00
The ratio of calcium to chlorine is 1:2. This ratio suggests an empirical formula of CaCl
2
.
Statement: The empirical formula of calcium chloride is CaCl
2
.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-20

75. (a) Given: % C = 40.0 %; % H = 6.67 %
Required: percentage of oxygen in glucose
Solution: Determine the percentage of oxygen by subtracting the sum of the percentages of carbon and hydrogen from 100 %.
% O = 100.0 % !
(40.0 % + 6.67 %)
% O = 53.33 % [1 extra digit carried]
Statement: Glucose contains 53.3 % of oxygen.
(b) Given: % C = 40.0 %; % H = 6.67 %; % O = 53.33 %
Required: empirical formula of glucose
Solution: A 100.0 g sample of glucose contains 40.0 g of carbon, 6.67 g of hydrogen, and 53.33 g of oxygen.
Step 1.
Calculate the amount of each element. n
C
= (40.0 g)
!
"#
1 mol
12.01 g
$
%& n
C
= 3.3306 mol [2 extra digits carried] n
H
= (6.67 g)
!
"#
1 mol
1.01 g
$
%& n
H
= 6.6040 mol [2 extra digits carried] n
O
= (53.33 g)
!
"#
1 mol
16.00 g
$
%& n
O
= 3.3331 mol [2 extra digits carried]
Step 2.
Divide the amount of each element by the smallest amount. n n
C
C
=
3.3306 mol
3.3306 mol
= 1.00
n n
H
C
=
6.6040 mol
3.3306 mol
= 1.98
n n
O
C
=
3.3331 mol
3.3306 mol
= 1.00
The ratio of carbon to hydrogen to oxygen is 1:2:1. This ratio suggests an empirical formula of CH
2
O.
Statement: The empirical formula of glucose is CH
2
O.
76. (a) Given: % H = 2.20 %; % C = 26.7 %; remainder is O
Required: percentage of oxygen in oxalic acid
Solution: Determine the percentage of oxygen by subtracting the sum of the percentages of hydrogen and carbon from 100 %.
% O = 100.0 % !
(2.20 % + 26.7 %)
% O = 71.1 %
Statement: In oxalic acid, the percentage of oxygen is 71.1 %.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-21

(b) Given: % H = 2.20 %; % C = 26.7 %; % O = 71.1 %
Required: empirical formula of oxalic acid
Solution: A 100.0 g sample of this compound contains 2.2 g of hydrogen, 26.7 g of carbon, and 71.1 g of oxygen.
Step 1.
Calculate the amount of each element. n
H
= (2.20 g)
!
"#
1 mol
1.01 g
$
%& n
H
= 2.1782 mol [2 extra digits carried] n
C
= (26.7 g)
!
"#
1 mol
12.01 g
$
%&
= 2.2231 mol [2 extra digits carried] n
C n
O
= (71.1 g)
!
"#
1 mol
16.00 g
$
%& n
O
= 4.4438 mol [2 extra digits carried]
Step 2.
Divide the amount of each element by the smallest amount. n n
H
H
=
2.1782 mol
2.1782 mol
= 1.00
n n
C
H
=
2.2231 mol
2.1782 mol
= 1.02
n n
O
H
=
4.4438 mol
2.1782 mol
= 2.04
The ratio hydrogen to carbon to oxygen is 1:1:2. This ratio suggests an empirical formula of HCO
2
.
Statement: The empirical formula of oxalic acid is HCO
2
.
(c) Given: empirical formula, HCO
2
; molar mass of compound is 90.0 g/mol
Required: molecular formula of oxalic acid, H x
C y
O z
Solution:
Step 1.
Calculate the empirical molar mass by adding the molar masses of the elements.
M
HCO
2
=
!
"#
1.01 g mol
$
%&
+
!
"#
12.01 g mol
$
%&
+
!
"#
2 ' 16.00 g mol
$
%&
M
HCO
2
= 45.02 g mol
Step 2.
Solve for x , the mass multiple.
90.0 g x = mol
45.02 g mol x = 2.00
Step 3.
The molar mass of the compound is 2 times the molar mass of the empirical formula.
Multiply each of the subscripts by 2 to give H
2
C
2
O
4
.
Statement: The molecular formula of oxalic acid is H
2
C
2
O
4
.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-22

77. Given: empirical formula, P
2
O
5
; molar mass of compound = 285 g/mol.
Required: molecular formula of compound, P x
Solution:
O y
Step 1.
Calculate the empirical molar mass by adding the molar masses of the elements.
M
P
2
O
5
=
"
#$
2 !
30.97 g mol
%
&'
+
"
#$
5 !
16.00 g mol
%
&'
M
P
2
O
5
= 141.94 g mol
Step 2.
Solve for x , the mass multiple.
285 g mol x =
141.94 g mol x = 2.01
Step 3.
The molar mass of the compound is 2 times the molar mass of the empirical formula.
Multiply each of the subscripts by 2 to give P
4
O
10
.
Statement: The molecular formula of the compound is P
4
O
10
.
Evaluation
78.
Answers may vary. Students’ answers should show strong justification for their sides. They might mention that quantitative analysis allows them to draw inferences from measurements or that qualitative analysis allows them to observe changes that are not measurable.
79. (a) You need to order the amounts needed so that there is not a large excess of one reactant.
(b) To make ammonia, the amount of hydrogen is 3 times the amount of nitrogen needed. Use the amount of hydrogen to find the amount of nitrogen. n
N
2
= n
H
2
!
1
3
= 270 mol !
1
3 n
N
2
= 9.0
!
10 mol
You would order 9.0 × 10 mol of nitrogen to react with 270 mol of hydrogen.
(c) Given: m
N
= 840 kg ; 1 mol N
2
reacts with 2 mol H
Required: mass of hydrogen, in kg
2
Solution: To make ammonia, the amount of hydrogen is 3 times the amount of nitrogen needed.
Step 1.
Calculate the amount of 840 kg of nitrogen. n
N
2
= 840 kg !
1000 g
1 kg
!
1 mol
14.01 g n
N
2
= 5.996
!
10 4 mol [1 extra digit carried]
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-23

Step 2.
Calculate the amount of hydrogen needed. n
H
2
= n
H
2
!
3
= 5.996
!
10 4 mol !
3 n
H
2
= 1.799
!
10 5 mol [1 extra digit carried]
Step 3.
Calculate the amount of hydrogen in kilograms. m
H
2
= (1.799
!
10 5 mol )
"
#$
1.01 g
1 mol
%
&'
"
#$
1 kg
1000 g
%
&' m
H
2
= 180 kg
Statement: To make ammonia, you need 180 kg of hydrogen if you are ordering 840 kg of nitrogen.
80. (a) You need to know Avogadro’s constant.
(b) Divide the number of carbon dioxide molecules by Avogadro’s constant to find the amount.
(c) Each carbon dioxide molecule contains 2 oxygen atoms, so the amount of oxygen atoms in a sample of CO
2 is twice the number of molecules of CO
2
. You would need to divide the number of oxygen atoms by 2 and by Avogadro’s constant to find the amount of carbon dioxide.
81.
A chemical formula might be both a molecular formula and an empirical formula (CH
4
,
NO
2
), or it might be both an empirical formula and show a formula unit (NaCl, MgO). Molecular formulas apply only to molecular compounds, and formula units apply only to ionic compounds, so these two types of formulas will never apply to the same compound.
82.
Any formula of an ionic compound shows the simplest ratio of ions in the compound, and so it is always an empirical formula. It cannot be a molecular formula because ionic compounds do not form molecules. For example, a crystal of sodium chloride really contains a very large number of each ions. We could express this as Na
1000
Cl
1000
, but since we always reduce when writing ionic compounds, the formula would simplify to NaCl. Similarly, a crystal of magnesium bromide could be Mg
2000
Br
4000
, which would simplify to MgBr
2
.
Reflect on Your Learning
83.
Answers may vary. Students’ answers will likely include a product like shampoo.
(a) In the manufacturing of shampoo, qualitative analysis might show whether the colour is appealing and that it lathers well.
(b) For the shampoo to be the same every time, exact amounts of ingredients must be used.
These ingredients must be measured quantitatively.
84. Answers may vary. Sample answers:
(a) Examples of expressing an amount in millimoles include the amount of enzymes in the human body or the amount of a certain pollutant in relatively pure water.
(b) Examples of expressing an amount in kilomoles include the amount of material used in a manufacturing process and the amount of metal present in a large ore deposit.
85. (a)
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-24

(b)
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-25

86. (a) Both 500 sheets of paper and Avogadro’s constant express the number of entities in a whole unit (ream or mole). Dividing the number of sheets by the sheets in a ream to find the number of reams is similar to dividing the number of entities present by Avogadro's constant to find the number of moles.
(b) Both 144 and Avogadro’s constant express the number of entities in a whole unit (gross or mole). Multiplying 144 pencils/gross by 4.5 gross to find the number of pencils is similar to multiplying Avogadro's constant by the number of moles to find the number of entities.
Research
87.
Students should explain that both sugars have the same molecular formula, C
6
H the structures are different.
12
O
6
, but that
88. (a) The molecular formula for ethylene is C
2
H
4
. The empirical formula for ethylene is CH
2
.
(b) Other compounds with the same empirical formula as ethylene include propylene (C
3
H
6
), butylene (C
4
H
8
), and cyclohexane (C
6
H
12
).
(c) Students’ answers should include different uses of the three compounds.
Sample answer: Propylene is used in the manufacture of plastics (injection moulding) and fibers
(carpets). Butylene is used in the manufacture of synthetic rubber. Cyclohexane is used in the production of nylon.
89. (a) Students’ answers may vary but could include the production of ethane in the petrochemical industry through a process known as steam cracking. In steam cracking, complex hydrocarbons, are broken down into simpler hydrocarbons, by breaking carbon-carbon bonds.
High temperatures and a catalyst are required in the process.
(b) Use the conversion factor
200.0 t !
1000 kg
1 t
= 2.000
!
1000 kg
10 5
1 t
kg
to convert mass in t to mass in kg.
The mass of 200.0 t of ethene is equivalent to 2.000 × 10 5 kg.
(c) Calculate the molar mass of ethene.
M
M
C
2
H
4
C
2
H
4
=
=
"
#$
2 !
12.01
28.06 g mol g mol
%
&'
+
"
#$
4 !
1.01 g mol
%
&'
Convert mass of ethene to amount in moles. n
C
2
H
4
= 2.000
!
10 5 kg !
1000 g
1 kg
!
1 mol
28.06 g n
C
2
H
4
= 7.128
!
10 6 mol
Since the number of moles is so large, kilomoles may be more convenient to use.
90.
(a) Students should determine that empirical relates to information gained by experimentation or observation.
(b) This definition fits with an empirical formula being determined by data related to percentage composition.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-26

91. (a) Sodium and chlorine are in salt. It is the amount of sodium present that reflects the amount of salt in a product.
(b) Answers may vary.
Students could check fruits, vegetables, or meat to find different forms of the same food. Sample answer: A 213 g can of sockeye salmon has 880 mg of salt, some of which is naturally occurring and some which is added. The equivalent mass of fresh sockeye salmon contains approximately 95 mg of natural occurring salt. The canned salmon has more than twice the amount of sodium than the fresh salmon.
(c) The recommended daily intake of sodium increases from childhood to adulthood, then decreases slightly for older adults.
(d) A low-sodium food contains a maximum of 140 mg of sodium per serving, and a no-sodium food contains 5 mg of sodium per serving, or less.
92. (a) Australia developed plastic currency.
(b) In the mass spectrum of heroin, most of the peaks in the spectrum are at different locations and show ions with different masses than in the mass spectrum of cocaine. The molecular mass of heroin is 370 g/mol, the largest molecular mass on the spectrum.
Copyright © 2011 Nelson Education Ltd. Chapter 6: Quantities in Chemical Formulas 6-27