Chapter 2 Practice Exercises 2.1

Chapter 2
Practice Exercises
2.1
The first sample has a ratio of
1.25 g Cd
0.357 g S
Therefore, the second sample must have the same ratio of Cd to S:
1.25 g Cd
x
=
0.357 g S 3.50 g S
Cross-multiplication gives,
(1.25 g Cd)(3.50 g S) = x(0.357 g S)
(1.25g Cd)(3.50 g S)
= x
0.357 g S
x = 12.3 g Cd
2.2 Compare the ratios of the mass of the compound before heating and the mass of the iron after heating, if they
are the same, the compounds are the same.
Sample
A
Ratio
25.36 g
= 1.574
16.11 g
B
15.42 g
= 1.86
8.28 g
C
7.85 g
= 1.86
4.22 g
D
11.87 g
= 1.57
7.54 g
Compounds A and D are the same, as are compounds B and C.
2.3
240
94 Pu
The bottom number is the atomic number, found on the periodic table (number of protons). The top
number is the mass number (sum of the number of protons and the number of neutrons). Since it is a neutral
atom, it has 94 electrons.
2.4
35
17 Cl
2.5
We can discard the 17 since the 17 tells the number of protons which is information that the symbol "Cl"
also provides. In addition, the number of protons equals the number of electrons in a neutral atom, so the
symbol "Cl" also indicates the number of electrons. The 35 is necessary to state which isotope of chlorine
is in question and therefore the number of neutrons in the atom.
2.6
2.24845 × 12 u = 26.9814 u
2.7
Copper is 63.546 u ÷12 u = 5.2955 times as heavy as carbon
2.8
(0.198 × 10.0129 u) + (0.802 × 11.0093 u) = 10.8 u
contains 17 protons, 17 electrons, and 18 neutrons.
15
Chapter 2
2.9
(a)
(b)
(c)
(d)
1 Ni, 2 Cl
1 Fe, 1 S, 4 O
3 Ca, 2 P, 8 O
1 Co, 2 N, 12 O, 12 H
2.10
(a)
(b)
(c)
(d)
2 N nitrogen,
4 H hydrogen, 3 O oxygen
1 Fe iron,
1 N nitrogen,
4 H hydrogen, 2 S sulfur,
8 O oxygen
1 Mo molybdenum,
2 N nitrogen,
11 O oxygen,
10 H hydrogen
6 C carbon,
4 H hydrogen, 1 Cl chlorine,
1 N nitrogen,
2 O oxygen
2.11
This is a balanced chemical equation, and the number of each atom that appears on the left is the same as
that on the right: 1 Mg, 2 O, 4 H, and 2 Cl.
2.12
Mg(OH)2(s) + 2HCl(aq) J MgCl2(aq) + 2H2O
2.13
6 N, 42 H, 2 P, 20 O, 3 Ba, and 12 C are on both the products and the reactants sides of the equation,
therefore the reaction is balanced.
2.14
The term "octa' means eight, therefore there are 8 carbon atoms in octane. The formula for an alkane is
CnH2n+2, so octane has 8 carbons and ((2 × 8) + 2) = 18 H. The condensed formula is
CH3CH2CH2CH2CH2CH2CH2CH3, and the structural format is:
H H H H H H H H
H C C C C C C C C H
H H H H H H H H
2.15
The condensed formula is CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
The structural formula is:
H H H H H H H H H H
H C C C C C C C C C C H
H H H H H H H H H H
2.16
(a)
Propanol: CH3CH2CH2OH
H H H
H C C C O H
H H H
(b)
Butanol: CH3CH2CH2CH2OH
H H H H
H C C C C O H
H H H H
2.17
(a)
(b)
(c)
(d)
Fe: 26 protons and 26 electrons
Fe3+: 26 protons and 23 electrons
N3–: 7 protons and 10 electrons
N: 7 protons and 7 electrons
2.18
(a)
(b)
O: 8 protons and 8 electrons
O2–: 8 protons and 10 electrons
16
Chapter 2
(c)
(d)
Al3+: 13 protons and 10 electrons
Al: 13 protons and 13 electrons
2.19
(a) NaF
(b) Na2O
(c) MgF2
(d) Al4C3
2.20
(a) Ca3N2
(b) AlBr3
(c) Na3P
d) CsCl
2.21
(a)
(b)
CrCl3 and CrCl2, Cr2O3 and CrO
CuCl, CuCl2, Cu2O and CuO
2.22
(a)
(b)
Au2S and Au2S3, Au3N and AuN
SnS and SnS2, Sn3N2 and Sn3N4
2.23
(a) KC2H3O2
(b) Sr(NO3)2
2.24
(a) Na2CO3
(b) (NH4)2SO4
2.25
(a)
(b)
(c)
2.26
(a) AsCl5
(b) SCl6
(c) S2Cl2
2.27
(a) K2O
(b) BaBr2
(c) Na3N
2.28
(a)
(c)
aluminum chloride
sodium bromide
(b)
(d)
barium sulfide
calcium fluoride
2.29
(a)
(c)
postassium sulfide
nickel(II) chloride
(b)
(d)
magnesium phosphide
iron(III) oxide
2.30
(a) Al2S3
2.31
(a)
(b)
lithium carbonate
iron(III) hydroxide
2.32
(a)
KClO3
2.33
diiodine pentaoxide
2.34
chromium(III) acetate
(c) Fe(C2H3O2)3
phosphorous trichloride
sulfur dioxide
dichlorine heptaoxide
(b) SrF2
(d) Al2S3
(c) TiO2
d) Au2O3
(b)
Ni3(PO4)2
Review Questions
2.1
The first law of chemical combination is the law of conservation of mass: no detectable gain or loss of mass
occurs in chemical reactions. The other law is the law of definite proportions: in a given chemical
compound, the elements are always combined in the same proportions by mass.
2.2
Isotopes of a particular element have nearly identical chemical properties and the average mass of an atom
is independent, almost, of the source of the atom.
2.3
Conservation of mass derives from the postulate that atoms are not destroyed in normal chemical reactions.
The Law of Definite Proportions derives from the notion that compound substances are always composed
of the same types and numbers of atoms of the various elements in the compound.
17
Chapter 2
2.4
This is the Law of Definite Proportions, which guarantees that a single pure substance is always composed
of the same ratio of masses of the elements that compose it.
2.5
(a)
To test the law of conservation of mass, a reaction would have to be carried out in which the mass
of the reactants and the mass of the products are weighed and shown to be the same.
The law of definite proportions could be shown by demonstrating that no matter how a compound
is made, the same proportions by mass are used. This could be done by decomposing a compound
and showing that the masses of the elements are always in the same ratio.
To test the law of multiple proportions, two different compounds made up of the same elements
would have to be decomposed. The amount used would have to keep mass of one of the elements
constant, and then the masses of other the element from the different samples would have to be in
a ratio of small whole numbers.
(b)
(c)
2.6
A nucleon is a subatomic particle found in the atomic nucleus. We have studied neutrons and protons.
2.7
Protons, 11 p + , +1 charge
Electron, 00 e− , –1 charge
Neutron, 01 n , no charge
2.8
The atomic number is equal to the number of protons in the nucleus of the atom, and the mass number is
the sum of the number of neutrons and the number of protons. The atomic number (symbol Z) is
designated by a subscript preceding the chemical symbol and the mass number (symbol A) is a superscript
preceding the chemical symbol.
2.9
Nearly all of the mass is located in the nucleus, because this is the portion of the atom where the proton and
the neutron are located.
2.10
(a)
2.11
(a)
2.12
Strontium and calcium are in the same Group of the periodic table, so they are expected to have similar
chemical properties. Strontium should therefore form compounds that are similar to those of calcium,
including the sorts of compounds found in bone.
2.13
For all group IA elements (the alkali metals), the formula is MX, that is one Cl per atom of metal. For all
group IIA elements (the alkaline earth metals), the formula is MX2, that is two Cl atoms per atom of metal.
The correspondence in formula and the similarities in chemical behavior allowed Mendeleev to locate
theses two series into their separate groups on the periodic table.
2.14
Mendeleev constructed his periodic table by arranging the elements in order of increasing atomic weight,
and grouping the elements by their recurring properties. The modern periodic table is arranged in order of
increasing number of protons.
2.15
Cadmium is in the same periodic table group as zinc, but silver is not. Therefore cadmium would be
expected to have properties similar to those of zinc, whereas silver would not.
2.16
Silver and gold are in the same periodic table group as copper, so they might well be expected to occur
together in nature, because of their similar properties and tendencies to form similar compounds.
mass number
131
53 I
(b)
90
38 Sr
(b)
(c)
atomic number
137
55 Ce
(d)
18
18
9F
Chapter 2
2.17
(a)
(b)
(c)
(d)
(e)
(f)
(g)
Li
I
W
Xe
Sm
Pu
Mg
2.18
The superscript before the symbol indicates the mass number; the superscript after the symbol indicates the
charge on the atom; the subscript before the symbol indicates the atomic number; and the subscript after the
symbol indicates the number of atoms in the compound.
mass number charge
14
For example: atomic
number N number of atoms
7 N2
Nitrogen has 7 protons and 7 electrons for a neutral atom. The molecule has two atoms in it, and the
isotope with 7 neutrons gives it a mass number of 14
2.19
See Figure in the margin of page 50.
2.20
Mercury is used in thermometers because it is a liquid, and tungsten is used in light bulbs because is has
such a high melting point.
2.21
They are semiconductors.
2.22
Mercury and bromine
2.23
The noble gases: He, Ne, Ar, Kr, Xe, and Rn
2.24
See figure 2.7, page 51.
2.25
Luster, electrical conductivity, thermal conductivity, ductility, and malleability are the characteristic
properties of metals.
2.26
(a)
(b)
(c)
In general, melting points decrease from left to right across the periodic table and increases from
top to bottom.
In general, boiling points decrease from left to right across the periodic table and increases from
top to bottom.
In general, density has a maximum in the middle of the periodic table and falls off to the right and
left. Also, the density increases moving down a group.
2.27
Luster, malleability, color, and brittleness are some trends that are mentioned in terms of moving from the
metals to the nonmetals across the periodic table, or moving down a group from nonmetals to metals.
2.28
Metals which are used to make jewelry are those that do not corrode, silver, gold, and platinum. Iron would
be useless for jewelry because it is susceptible to rusting. Potassium reacts violently with water to form
hydrogen and potassium hydroxide.
2.29
The heavy line separates the metals from the nonmetals, and the metalloids border the line.
2.30
This may stand for the name of an element or for the name of one atom of an element.
2.31
The smallest particle that is representative of a particular element is the atom of that element. A molecule is
a representative unit that is made up of two or more atoms linked together.
2.32
H2, hydrogen
Cl2, chlorine
N2, nitrogen
Br2. bromine
O2, oxygen
I2, iodine
F2, fluorine
19
Chapter 2
2.33
Reactants are the substances to the left of the arrow in a reaction that are present before the reaction begins.
Products are the substances to the right of the arrow in a reaction and they are formed during the reaction
and are present when the reaction is over.
2.34
A chemical reaction is balanced when there is the same number of each kind of atom on both the reactant
and product side of the equation; and the total charges on both the reactant and product sides of the
equation are the same. These conditions must be met due to the law of conservation of matter.
2.35
2C8H18(l) + 25O2(g) → 18CO2(g) + 18H2O(l)
2.36
(a)
(b)
(c)
(d)
2.37
S8, P4
2.38
HAt
2.39
The noble gases: He, Ne, Ar, Kr, Xe, and Rn
2.40
PH3
2.41
SnH4
2.42
Nonmetals
2.43
(a) CH4
2.44
(a)
2.45
C10H22 or CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
2.46
(a) CH4, component of natural gas
(c) CH3CH2CH3, gas-fired barbecues
2.47
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3
or CH3(CH2)21CH3 or C23H48
2.48
All of the elements are nonmetals, and the formula is not in the smallest whole number ratio.
2.49
methane:
Magnesium reacts with oxygen to give (yield) magnesium oxide.
The reactants are Mg and O2.
The product is MgO.
2Mg(s) + O2(g) → 2MgO(s)
(b) NH3
(c) TeH2
CH3OH
(d) HI
(b)
CH3CH2OH
(b) CH3CH3, component of natural gas
(d) CH3CH2CH2CH3, cigarette lighters
ethane:
propane:
20
Chapter 2
butane:
methanol:
ethanol:
decane (10 carbons):
23 carbon hydrocarbon
21
Chapter 2
2.50
Nonmetals react with metals, nonmetals, and metalloids.
2.51
(a)
(b)
An ionic compound is formed by the transfer of electrons, and it is accompanied by the formation
of ions of opposite charge.
Molecular compounds arise from the sharing of electrons between atoms, rather than from the
complete transfer of electrons as in (a).
2.52
Metals react with nonmetals.
2.53
Nonmetals are more frequently found in compounds because of the large variety of ways they may
combine. A particularly illustrative example is the combination of carbon, a nonmetal, with other elements.
So many compounds are possible that there is one entire area of chemistry devoted to the study of carbon
compounds, organic chemistry.
2.54
In ionic substances, no molecules exist. Rather we have a continuous array of cations and anions, which are
present in a constant ratio. The ratio is given by the formula unit.
2.55
(a)
(b)
(c)
(d)
(e)
2.56
A cation is a positively charged ion with one or fewer electrons than its neutral atom. An anion is a
negatively charged ion with one or more electrons than its neutral atom. A polyatomic ion is made up of
more than one atom; the whole unit is the ion.
2.57
Al2Cl6 is molecular because the smallest whole number ratio of elements is not used in the formula.
2.58
An ion is a charged particle. It can be monatomic or polyatomic, and it can have either a positive or a
negative charge. It is derived from an atom or a molecule by gain or loss of electrons. Atoms and molecules
are neutral.
2.59
Nitrogen gained 3 electrons to form N3–; it has 7 protons and 10 electrons.
2.60
Titanium lost four electrons to form Ti4+; it has 22 protons and 18 electrons.
2.61
The incorrect ones are a, d, and e.
2.62
The formula should have the smallest whole numbers possible. The formula should be TiO2.
2.63
Negative
2.64
(a)
(b)
(c)
(d)
(e)
(f)
2.65
Rb forms a +1 cation (Rb+) and Cl forms a –1 anion (Cl–), so the formula should be RbCl. The cation is
first in the formula; therefore the formula should be Na2S.
2.66
(a)
(d)
Na, Na+
These particles have the same number of nuclei.
These particles have the same number of protons.
These particles could have different numbers of neutrons, if they are different isotopes.
These particles do not have the same number of electrons; Na+ has one less electron.
Fe2+, Fe3+
Co2+, Co3+
Hg2+, Hg22+
Cr2+, Cr3+
Sn2+, Sn4+
Cu+, Cu2+
dichromate ion
carbonate ion
(b)
(e)
hydroxide ion
cyanide ion
22
(c)
(f)
acetate ion
perchlorate ion
Chapter 2
(a)
(b)
(c)
(d)
Ca(s) + Cl2(g) J CaCl2(s)
2Mg(s) + O2(g) J 2MgO(s)
4Al(s) + 3O2(g) J 2Al2O3(s)
S(s) + 2Na(s) J Na2S(s)
or
2.68
(a)
(d)
CN–
SO32–
(b)
(e)
NH4+
ClO3–
(c)
f)
NO3–
SO42–
2.69
(a)
(d)
OCl–
H2PO4–
(b)
(e)
HSO4–
MnO4–
(c)
(f)
PO43–
C2O42–
2.70
(a)
(b)
C3H8(g) + 5O2(g) J 3CO2(g) + 4H2O(g)
2Na(s) + 2H2O J 2NaOH(aq) + H2(g)
2.71
(a)
(b)
Fe(OH)3(s) + 3HCl(g) J H2O + FeCl3(aq)
2AgNO3(aq) + BaCl2(aq) J 2AgCl(s) + Ba(NO3)2(aq)
2.72
For naming ionic compounds of the transition elements it is essential to know the charge on the anion since
that will help determine the charge on the transition element. Transition elements can have more than one
charge.
2.73
Binary compounds, such as CCl4 contain two elements only. A diatomic substance is composed of
molecules having two atoms, such as HCl or N2. In the latter, the two atoms may or may not be the same.
2.74
When naming the compound, molecular compounds need the prefixes to specify the number of atoms in the
molecule. Ionic compounds made with transition metals or post-transition metals need to have the charge
of the metal specified.
2.75
(a)
2.67
(b)
S8(s) + 16Na(s) → 8Na2S(s)
Greek prefixes are used to specify the number of atoms of each element in a molecular compound
(PCl5 is phosphorous pentachloride); specify the number of water molecules in a hydrate
(CuSO4·5H2O is copper sulfate pentahydrate); specify the number of atoms in polyatomic ions;
and organic compounds use Greek prefixes to specify the number of carbon atoms in the
compound (pentane has five carbons).
Roman numerals are used in the name of transition metal compounds to specify the charge of the
metal.
Review Problems
2.76
5.54 g ammonia. From the ratio, Problem 2.80, we see that for every 4.56 g nitrogen there needs to be 0.98
g hydrogen. According to the Law of Conservation of Mass 5.54 g ammonia will be produced.
9.33 g nitrogen 4.56 g nitrogen
=
x g hydrogen
2.00 g hydrogen
x = 0.977 g hydrogen
mass of NH3 = 4.56 g nitrogen + 0.977 g hydrogen = 5.54 g NH3
2.77
55.0 g of the phosphorus chloride. From the ratio, Problem 2.81, we see that for every 12.5 g phosphorus it
is necessary to have 42.9 g chlorine. According to the Law of Conservation of Mass, 55.0 g of the
phosphorus chloride compound was formed.
1.20 g phosphorous 12.5 g phosphorous
=
x g chlorine
4.12 g chlorine
x = 42.9 g chlorine
mass of PCl3 = 12.5 g phosphorous + 42.9 g chlorine = 55.4 g PCl3
23
Chapter 2
2.78
2.286 g of O. The first nitrogen-oxygen compound has an atom ratio of 1 atom of N for 1 atom of O, and a
mass ratio of 1.000 g of N for 1.143 g of O. The second compound has an atom ratio of 1 atom of N for 2
atoms of O; therefore, the second compound has twice as many grams of O for each gram of N, or 2.286 g
of O.
2.79
This ratio should be 4/2 = 2/1, as required by the formulas of the two compounds.
Twice 0.597 g Cl, or 1.19 g Cl.
2.80
From the first ratio we see that there is a ratio of 9.33 g of nitrogen to 2.00 g of hydrogen. Multiplying the
mass of nitrogen by 3.14 we see that for every 6.28 g hydrogen there will be 29.3 g nitrogen.
9.33 g nitrogen
x g nitrogen
=
2.00 g hydrogen 6.28 g hydrogen
x = 29.3 g nitrogen
2.81
From the ratio of phosphorous to chlorine, we see that for every 1.20 g phosphorus, there are 4.12 g
chlorine. Dividing the mass of chlorine by 3.43, we find that for every 6.22 g chlorine there will be 1.81 g
phosphorus.
1.20 g phosphorous x g phosphorous
=
4.12 g chlorine
6.22 g chlorine
x =1.81 g phosphorous
2.82
Regardless of the definition, the ratio of the mass of hydrogen to that of carbon would be the same. If C-12
were assigned a mass of 24 (twice its accepted value), then hydrogen would also have a mass twice its
current value, or 2.01588 u.
2.83
Taking the mass ratio of 109Ag to 12C and multiplying it by 12 we see that the mass of Ag-109 is 108.90 u.
12 × 9.0754 = 108.90 u
2.84
Since we know that the formula is CH4, we know that one fourth of the total mass due to the hydrogen
atom constitutes the mass that may be compared to the carbon. Hence we have 0.33597 g H ÷ 4 = 0.083993
g H and 1.00 g assigned to the amount of C-12 in the compound. Then it is necessary to realize that the
ratio 1.00 g C ÷ 12 for carbon is equal to the ratio 0.083993 g H ÷ X, where X equals the relative atomic
mass of hydrogen.
⎛ 1.000 g C ⎞
⎛ 0.083993 g H ⎞
⎜ 12 u C ⎟ = ⎜
⎟ = 1.008 u
X
⎝
⎠
⎝
⎠
2.85
Using the ratio of the number of atoms of O and X and the atomic mass of O, we can compare that to the
ratio of the masses of O and X to calculate the atomic mass of X:
⎞
1.125 g X ⎛ 2 atoms X ⎞ ⎛
uX
=
⎜
⎟
1.000 g O ⎜⎝ 3 atoms O ⎟⎠ ⎝ 15.9994 u O ⎠
u X = 27.00 u
The element is aluminum.
2.86
(a)
(b)
(c)
(d)
Radium-226
206
Pb
Carbon-14
23
Na
neutrons
138
124
8
12
protons
88
82
6
11
24
electrons
88
82
6
11
Chapter 2
2.87
(a)
(b)
(c)
(d)
electrons
55
92
53
79
Cesium-137
238
U
Iodine-131
197
Au
protons
55
92
53
79
neutrons
82
146
78
118
2.88
(0.6917 × 62.9396 u) + (0.3083 × 64.9278 u) = 63.55 u
2.89
(0.7899 × 23.9850 u) + (0.1000 × 24.9858 u) + (0.1101 × 25.9826 u) = 24.31 u
2.90
(a)
(b)
(c)
(d)
(e)
2 K, 2 C, 4 O
2 H, 1 S, 3 O
12 C, 26 H
4 H, 2 C, 2 O
9 H, 2 N, 1 P, 4 O
2.91
(a)
(b)
(c)
(d)
(e)
3 Na, 1 P, 4 O
1 Ca, 4 H, 2 P, 8 O
4 C, 10 H
3 Fe, 2 As, 8 O
3 C, 8 H, 3 O
2.92
1 Cr, 6 C, 9 H, 6 O
2.93
3 Ca, 5 Mg, 8 Si, 24 O, 2 H
2.94
(a)
(b)
(c)
(d)
(e)
1 Ni, 2 Cl, 8 O
1 Cu, 1 C, 3 O
2 K, 2 Cr, 7 O
2 C, 4 H, 2 O
2 N, 9 H, 1 P, 4 O
2.95
(a)
(b)
(c)
(d)
(e)
6 C, 12 H, 2 O
1 Mg, 1 S, 14 H, 11 O
1 K, 1 Al, 2 S, 20 O, 24 H
1 Cu, 2 N, 6 O
4 C, 10 H, 1 O
2.96
MgSO4
2.97
KNaC4H4O6·4H2O
2.98
(a)
(b)
(c)
6 N, 3 O
4 Na, 4 H, 4 C, 12 O
2 Cu, 2 S, 18 O, 20 H
2.99
(a)
(b)
(c)
14 C, 28 H, 14 O
4 N, 8 H, 2 C, 2 O
10 K, 10 Cr, 35 O
2.100
(a)
6
(b)
3
(c)
27
2.101
(a)
16
(b)
36
(c)
50
25
Chapter 2
2.102
(a)
(d)
K+
S2–
(b)
(e)
Br–
Al3+
(c)
Mg2+
2.103
(a)
(d)
Ba2+
Sr2+
(b)
(e)
O2–
Rb+
(c)
F–
2.104
(a)
(d)
KNO3
Fe2(CO3)3
(b)
(e)
Ca(C2H3O2)2
Mg3(PO4)2
(c)
NH4Cl
2.105
(a)
(d)
Zn(OH)2
Rb2SO4
(b)
(e)
Ag2CrO4
LiHCO3
(c)
BaSO3
2.106
(a)
(d)
PbO and PbO2
FeO and Fe2O3
(b)
(e)
SnO and SnO2
Cu2O and CuO
(c)
MnO and Mn2O3
2.107
(a)
(d)
CdCl2
NiCl2
(b)
AgCl
(c)
ZnCl2
2.108
(a)
(d)
NaBr
MgBr2
(b)
(e)
KI
BaF2
(c)
BaO
2.109
(a)
(b)
(c)
(d)
(e)
CrCl2 and CrCl3
FeCl2 and FeCl3
MnCl2 and MnCl3
CuCl and CuCl2
ZnCl2
2.110
(a)
(c)
(e)
calcium sulfide
sodium phosphide
rubidium sulfide
(b)
(d)
aluminum bromide
barium arsenide
2.111
(a)
(c)
(e)
sodium fluoride
lithium nitride
potassium selenide
(b)
(d)
magnesium carbide
aluminum oxide
2.112
(a)
(c)
silicon dioxide
tetraphosphorus decaoxide
(b)
(d)
xenon tetrafluoride
dichlorine heptaoxide
2.113
(a)
(c)
chlorine trifluoride
dinitrogen pentaoxide
(b)
(d)
disulfur dichloride
arsenic pentachloride
2.114
(a)
(c)
sodium nitrite
magnesium sulfate heptahydrate
(b)
(d)
potassium permanganate
potassium thiocyanate
2.115
(a)
(c)
potassium phosphate
iron(III) carbonate
(b)
(d)
ammonium acetate
sodium thiosulfate pentahydrate
2.116
(a)
(c)
iron(II) sulfide
tin(IV) oxide
(b)
(d)
copper(II) oxide
cobalt(II) chloride hexahydrate
2.117
(a)
(c)
manganese(III) oxide
lead(II) sulfide
(b)
(d)
mercury(I) chloride
chromium(III) chloride tetrahydrate
26
Chapter 2
2.118
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
ionic
molecular
ionic
molecular
ionic
molecular
ionic
ionic
ionic
molecular
chromium(II) chloride
disulfur dichloride
ammonium acetate
sulfur trioxide
potassium iodate
tetraphosphorous hexaoxide
calcium sulfite
silver cyanide
zinc(II) bromide
hydrogen selenide
2.119
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
ionic
ionic
ionic
ionic
molecular
ionic
ionic
molecular
molecular
molecular
vanadium(III) nitrate
cobalt(II) acetate
gold(III) sulfide
gold(I) sulfide
germanium tetrabromide
potassium chromate
iron(II) hydroxide
diiodine tetraoxide
tetraiodine nonaoxide
tetraphosphorus triselenide
2.120
(a)
(d)
(g)
(NH4)2S
MoS2
P4S6
(b)
(e)
Cr2(SO4)3·6H2O
SnCl4
(c)
(f)
SiF4
H2Se
2.121
(a)
(d)
(g)
Hg(C2H3O2)2
Ca3P2
XeF4
(b)
(e)
Ba(HSO3)2
Mg(H2PO4)2
(c)
(f)
BCl3
CaC2O4
2.122
(a)
(d)
(g)
Na2HPO4
S2F10
SbF5
(b)
(e)
Li2Se
Ni(CN)2
(c)
(f)
Cr(C2H3O2)3
Fe2O3
2.123
(a)
(d)
(g)
Al2Cl6
Cu(HSO4)2
I2O5
(b)
(e)
As4O10
NH4SCN
(c)
(f)
Mg(OH)2
K2S2O3
2.124
diselenium hexasulfide and diselenium tetrasulfide
2.125
diphosphorous pentasulfide
Additional Exercises
2.126
(24.31)1.6605389 × 10–24 g = 4.037 × 10–23 g
(55.85)1.6605389 × 10–24 g = 9.274 × 10–23 g
Comparing answers, we see that both numbers are on the order of 1 × 10–23. We would expect 6 × 1023
atoms of Ca in 40.078 g of Ca.
2.127
(0.0580 × 53.9396 u) + (0.9172 × 55.9349 u) + (0.0220 × 56.9354 u) +
(0.0028 × 57.9333 u) = 55.85 u
27
Chapter 2
2.128
(a)
(b)
(c)
(d)
(e)
Metal
55
25 Mn
30 neutrons
25 electrons
54.9380
= 4.578 times heavier than a C-12 atom
12.000
2.129
Using the mass ratio of the compound containing O2X, find the atomic mass of X:
1.00 g X ⎛ u X ⎞⎛ 1 atom X ⎞
=
⎟
1.14 g O ⎜⎝ 16.00 u O ⎟⎜
⎠⎝ 2 atom O ⎠
u X = 28.0 u
Now, using the ratio of the masses of X to Y and the ratio of the atoms, X and Y, in the compound,
calculate the atomic mass of Y:
1.00 g X ⎛ 28.0 u X ⎞ ⎛ 1 atom X ⎞
=
5.07 g Y ⎜⎝ u Y ⎟⎠ ⎜⎝ 4 atom Y ⎟⎠
u Y = 35.49 u
From the periodic table, Cl has an atomic mass of 35.45 u. The element Y is chlorine.
2.130
(a)
Ca(s) + Br2(l) J CaBr2(s)
(b)
C(s) + 2Cl2(g) J CCl4(l)
(c)
2Al(s) + 3S(g) J Al2S3(s)
2.131
(a)
(c)
(e)
(g)
(i)
auric sulfate
plumbic oxide
cupric sulfate
cobaltous hydroxide
stannic sulfide
2.132
Let x equal the abundance of 79Br and y equal the abundance of 81Br.
We know that x + y = 1 and x(78.9183) + y(80.9163) = 79.904.
Substituting y = 1 – x we get x(78.9183) + (1 – x)80.9163 = 79.904.
Solving for x we get x = 0.5067 and y = 0.4933.
2.133
ferric nitrate nonahydrate
2.134
N2O5(g) + 3SO2(g) J 3SO3(g) + 2NO(g)
The small whole number ratio for oxygen in the nitrogen oxides is 5 to 1.
The small whole number ratio for oxygen in the sulfur oxides is 3 to 2.
2.135
The mass ratio of Fe to O in the first compound is 2.325 g Fe to 1.000 g O and the atom ratio is 2 Fe atoms
to 3 O atoms. Dividing the masses by the number of atoms gives the mass ratios of Fe to O
Mass of Br 79.904 × 2
=
= 3.99
Mass of Ca
40.078
Mass of Cl 35.453 × 4
=
= 11.8
Mass of C
12.0107
Mass of S
32.065 × 2
=
= 0.792
Mass of Al
26.9815 × 3
(b)
(d)
(f)
(h)
auric nitrate
mercurous chloride
mercuric chloride
stannous chloride
2.325 g Fe
= 1.162 g Fe/atom Fe
2 atoms Fe
1.000 g O
= 0.3333 g O/ atom O
3 atoms O
28
Chapter 2
The second compound has a mass ratio of 2.616 g Fe to 1.000 g O. Dividing the mass of Fe by the
mass/atom ratio determined for Fe and dividing the mass of O by the mass/atom ratio determined for O
gives the ratio of number of atoms in the compound.
2.616 g Fe
= 2.251 atom Fe
1.162 g Fe/atom Fe
1.000 g O
= 3.000 atom O
0.3333 g O/atom O
Multiplying the atomic ratios by 4 gives whole numbers to the subscripts in the formula: Fe9O12. Both 9
and 12 are divisible by 3 to give: Fe3O4.
2.136
(a)
(b)
(c)
(d)
(e)
(f)
(g)
CaO
CuCl2
Cl2
Cl2
CaO
Cl2
CaO
HBr
NO2
HBr
CaO
CuCl2
HBr
CuCl2
2.137
(a)
(b)
(c)
(d)
(e)
(f)
(g)
copper(II) bromide
copper(I) iodide
iron(II) sulfate
chromium(II) chloride
mercury(I) nitrate
manganese(II) sulfate
lead(II) acetate
2.138
Hg2(NO3)2·2H2O
CuCl2
AsH3
AsH3
HBr
NO2
AsH3
NaNO3
NO2
CuBr2
CuI
FeSO4
CrCl2
Hg2(NO3)2
MnSO4
Pb(C2H3O2)2
Hg(NO3)2·H2O
29
NO2