Answer Key
Chapter 6: Basic Review Worksheet
1. The average atomic mass of an element represents the weighted average mass, on the
relative atomic scale, of all the isotopes of an element. Average atomic masses are usually
given in terms of atomic mass units
2. The molar mass of a compound is the mass in grams of one mole of the compound (6.022 _
1023 molecules of the compound) and is calculated by summing the average atomic masses
of all the atoms present in a molecule (or empirical formula unit for an ionic substance) of
the compound. For example, a unit of the compound K2O contains two potassium atoms
and one oxygen atom. The molar mass is obtained by adding up the average atomic masses
of these atoms: Molar mass K2O = 2(39.10 g) + 1(16.00 g) = 94.20 g.
3. The percent composition (by mass) of a compound shows the relative amount of each
element present in the compound on a mass basis. For compounds whose formulas are
known (and whose molar masses therefore are known), the percentage of a given element
present in the compound is given by
Mass of the element present in 1 the
molcompoun
of
_
Mass of 1 mol of the compound
100
The percent composition of water, therefore, is 11.2% hydrogen and 88.8% oxygen.
4.
a. Cu (molar mass = 63.55 g)
Mol Cu =5.00 g _ 1 mol = 0.0787 mol
63.55
Atoms Cu = 0.0787 mol _
23
6.022 10atoms
1 mol
= 4.74 _ 1022 atoms
b. NH3 (molar mass = 17.03 g)
Mol NH3 = 5.00 g _ 1 mol = 0.294 mol
17.03 g
Molecules NH3 = 0.294 mol _
23
6.022 10molecules
1 mol
Atoms N = 1.77 _ 1023 molecules _
1 N atom
=
1 molecul
Atoms H = 1.77 _ 1023 molecules _
3 H atom
1 molecul
= 1.77 _ 1023 molecules
1.77 _ 1023 atoms N
= 5.31 _ 1023 atoms H
c. KClO3 (molar mass = 122.6 g)
Mol KClO3 = 5.00 g _ 1 mol = 0.0408 mol KClO3
122.6 g
6.022 1023formula units
= 2.46 1022 formula units KClO3
1 mol
K = 2.46 _ 1022 formula units _ 1 K atom = 2.46 _ 1022 atoms K
1 formula un
22
Cl = 2.46 _ 10 formula units T 1 Cl atoms = 2.46 _ 1022 atoms Cl
1 formula un
O = 2.46 _ 1022 formula units _ 3 O atoms = 7.38 _ 1022 atoms O
1 formula un
0.0408 mol KClO3 _
Atoms
Atoms
Atoms
d. Ca(OH)2 (molar mass = 74.096 g)
Mol Ca(OH)2 = 5.00 g _
1 mol
74.096
= 0.0675 mol Ca(OH)2
6.022 1023formula units
= 4.06 _ 1022 formula units Ca(OH)2
1 mol
Ca = 4.06 _ 1022 formula units 1 Ca atom = 4.06 _ 1022 atoms Ca
1 formula un
22
O = 4.06 _ 10 formula units 2 O atoms = 8.12 _ 1022 atoms O
1 formula un
22
H = 4.06 _ 10 formula units 2 H atoms = 8.12 _ 1022 atoms H
1 formula un
0.0675 mol Ca(OH)2 _
Atoms
Atoms
Atoms
5.
a. 100% Cu
b. NH3: %N =
%H =
14.01 g
_
17.03 g
3(1.008 g H
_
17.03 g
c. KClO3: %K =
100% = 82.27% N
100% = 17.76% H
39.10 g K
_
122.6 g
%Cl =
35.45 g C
_
122.6 g
%O =
3(16.00 g O
_
122.6 g
d. Ca(OH)2: %Ca =
100% = 31.89% K
100% = 28.92% Cl
100% = 39.15% O
40.08 g C
_
74.096 g
100% = 54.09% Ca
%O =
2(16.00 g O
_
74.096 g
100% = 43.19% O
%H =
2(1.008 g H
_
74.096 g
100% = 2.721% H
6. The empirical formula of a compound represents the smallest ratio of the relative number
of atoms of each type present in a molecule of the compound, whereas the molecular
formula represents the actual number of atoms of each type present in a real molecule of
the compound. For example, both acetylene (molecular formula C2H2) and benzene
(molecular formula C6H6) have the same relative number of carbon and hydrogen atoms
(one hydrogen for each carbon atom) and so have the same empirical formula (CH).
7. 7.99 g C _
1 mol
=
12.01 g
0.665 mol C
2.01 g H _
1 mol
=
1.008 g
1.99 mol H
1.99/0.665 = 2.99. Thus there are three hydrogen atoms for every one carbon atom. The
empirical formula is CH3.
8. The molar mass of the empirical formula CH3 is 15.034 [12.01 + 3(1.008)]. This is half the
molar mass of the compound; thus the compound must have the molecular formula C2H6.