Download CHAPTER TWO ATOMS, MOLECULES, AND IONS

CHAPTER TWO
ATOMS, MOLECULES, AND IONS
Questions
15.
a. Atoms have mass and are neither destroyed nor created by chemical reactions. Therefore, mass
is neither created nor destroyed by chemical reactions. Mass is conserved.
b. The composition of a substance depends on the number and kinds of atoms that form it.
c. Compounds of the same elements differ only in the numbers of atoms of the elements forming
them, i.e., NO, N2 O, NO2 .
16.
Some elements exist as molecular substances. That is, hydrogen normally exists as H2 molecules, not
single hydrogen atoms. The same is true for N2 , O2 , F2 , Cl2 , etc.
17.
Deflection of cathode rays by magnetic and electric fields led to the conclusion that they were
negatively charged. The cathode ray was produced at the negative electrode and repelled by the
negative pole of the applied electric field.
18.
J. J. Thomson discovered electrons. Henri Becquerel discovered radioactivity. Lord Rutherford
proposed the nuclear model of the atom. Dalton's original model proposed that atoms were indivisible
particles (that is, atoms had no internal structure). Thomson and Becquerel discovered subatomic
particles, and Rutherford's model attempted to describe the internal structure of the atom composed
of these subatomic particles. In addition, the existence of isotopes, atoms of the same element but with
different mass, had to be included in the model.
19.
The proton and neutron have similar mass with the mass of the neutron slightly larger than that of the
proton. Each of these particles has a mass approximately 1800 times greater than that of an electron.
The combination of the protons and the neutrons in the nucleus makes up the bulk of the mass of an
atom, but the electrons make the greatest contribution to the chemical properties of the atom.
20.
If the plum pudding model were correct (a diffuse positive charge with electrons scattered throughout),
then alpha particles should have traveled through the thin foil with very minor deflections in their path.
This was not the case as a few of the alpha particles were deflected at very large angles. Rutherford
reasoned that the large deflections of these alpha particles could be caused only by a center of
concentrated positive charge that contains most of the atom’s mass (the nuclear model of the atom).
21.
The atomic number of an element is equal to the number of protons in the nucleus of an atom of that
element. The mass number is the sum of the number of protons plus neutrons in the nucleus. The
18
CHAPTER 2
ATOMS, MOLECULES, AND IONS
19
atomic mass is the actual mass of a particular isotope (including electrons). As we will see in Chapter
Three, the average mass of an atom is taken from a measurement made on a large number of atoms.
The average atomic mass value is listed in the periodic table.
22.
A family is a set of elements in the same vertical column. A family is also called a group. A period
is a set of elements in the same horizontal row.
23.
A compound will always contain the same numbers (and types) of atoms. A given amount of hydrogen
will react only with a specific amount of oxygen. Any excess oxygen will remain unreacted.
24.
The halogens have a high affinity for electrons, and one important way they react is to form anions of
the type X-. The alkali metals tend to give up electrons easily and in most of their compounds exist
as M+ cations. Note: These two very reactive groups are only one electron away (in the periodic table)
from the least reactive family of elements, the noble gases.
Exercises
Development of the Atomic Theory
25.
a. The composition of a substance depends on the numbers of atoms of each element making up the
compound (i.e., on the formula of the compound) and not on the composition of the mixture from
which it was formed.
b. Avogadro’s hypothesis implies that volume ratios are equal to molecule ratios at constant
temperature and pressure. H2 (g) + Cl2 (g) ÷ 2 HCl(g). From the balanced equation (2 molecules
of HCl are produced per molecule of H2 or Cl2 reacted), the volume of HCl produced will be twice
the volume of H2 (or Cl2 ) reacted.
26.
27.
From Avogadro’s hypothesis, volume ratios are equal to molecule ratios at constant temperature and
pressure. Therefore, we can write a balanced equation using the volume data, Cl2 + 3 F2 ÷ 2 X. Two
molecules of X contain 6 atoms of F and two atoms of Cl. The formula of X is ClF3 for a balanced
equation.
=1.000;
= 1.999;
= 2.999
The masses of fluorine are simple ratios of whole numbers to each other, 1:2:3.
28.
Hydrazine: 1.44 × 10-1 g H/g N; Ammonia: 2.16 × 10-1 g H/g N
Hydrogen azide: 2.40 × 10-2 g H/g N
Let's try all of the ratios:
= 6.00;
= 9.00;
= 1.50 =
20
CHAPTER 2
ATOMS, MOLECULES, AND IONS
All the masses of hydrogen in these three compounds can be expressed as simple whole number ratios.
The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios 6:9:1.
29.
To get the atomic mass of H to be 1.00, we divide the mass of hydrogen that reacts with 1.00 g of
oxygen by 0.126, i.e.,
= 1.00. To get Na, Mg and O on the same scale, we do the same
division.
Na:
= 22.8; Mg:
Relative Value
Accepted Value
= 11.9; O:
= 7.94
H
O
Na
Mg
1.00
7.94
22.8
11.9
1.008
16.00
22.99
24.31
The atomic masses of O and Mg are incorrect; the atomic masses of H and Na are close to the values
in the periodic table. Something must be wrong about the assumed formulas of the
compounds.
It turns out the correct formulas are H2 O, Na2 O, and MgO. The smaller discrepancies result from
the error in the atomic mass of H.
30.
If the formula is InO, then one atomic mass of In would combine with one atomic mass of O, or:
, A = atomic mass of In = 76.54
If the formula is In2 O3 , then two times the atomic mass of In will combine with three times the atomic
mass of O, or:
, A = atomic mass of In = 114.8
The latter number is the atomic mass of In used in the modern periodic table.
The Nature of the Atom
31.
Density of hydrogen nucleus (contains one proton only):
Vnucleus =
d=
(3.14) (5 × 10-14 cm)3 = 5 × 10-40 cm3
= 3 × 1015 g/cm3
Density of H-atom (contains one proton and one electron):
Vatom =
(3.14) (1 × 10-8 cm)3 = 4 × 10-24 cm3
CHAPTER 2
ATOMS, MOLECULES, AND IONS
= 0.4 g/cm3
d=
32.
21
Since electrons move about the nucleus at an average distance of about 1 × 10-8 cm, then the diameter
of an atom is about 2 × 10-8 cm. Let's set up a ratio:
, Solving:
diameter of model = 2 × 105 mm = 200 m
33.
5.93 × 10-18 C ×
34.
First, divide all charges by the smallest quantity, 6.40 × 10-13.
= 37 negative (electron) charges on the oil drop
= 4.00;
= 12.00;
= 6.00
Since all charges are whole number multiples of 6.40 × 10-13 zirkombs, then the charge on one electron
could be 6.40 × 10-13 zirkombs. However, 6.40 × 10-13 zirkombs could be the charge of two electrons
(or three electrons, etc.). All one can conclude is that the charge of an electron is
6.40 × 10-13 zirkombs or an integer fraction of 6.40 × 10-13 zirkombs.
35.
sodium - Na; beryllium - Be; manganese - Mn; chromium - Cr; uranium - U
36.
fluorine - F; chlorine - Cl; bromine - Br; sulfur - S; oxygen - O; phosphorus - P
37.
Sn - tin; Pt - platinum; Co - cobalt; Ni - nickel; Mg - magnesium; Ba - barium; K - potassium
38.
As - arsenic; I - iodine; Xe - xenon; He - helium; C - carbon; Si - silicon
39.
The noble gases are He, Ne, Ar, Kr, Xe, and Rn (helium, neon, argon, krypton, xenon, and radon).
Radon has only radioactive isotopes. In the periodic table, the whole number enclosed in parentheses
is the mass number of the longest-lived isotope of the element.
40.
promethium (Pm) and technetium (Tc)
41.
a. Eight; Li to Ne
b. Eight; Na to Ar
42.
c. Eighteen; K to Kr
a. Six; Be, Mg, Ca, Sr, Ba, Ra
d. Five; N, P, As, Sb, Bi
b. Five; O, S, Se, Te, Po
c. Four; Ni, Pd, Pt, Uun
d. Six; He, Ne, Ar, Kr, Xe, Rn
43.
a.
Pu: 94 protons, 238 - 94 = 144 neutrons
b.
Cu: 29 protons, 65 - 29 = 36 neutrons
22
44.
CHAPTER 2
ATOMS, MOLECULES, AND IONS
c.
Cr: 24 protons, 28 neutrons
d.
He: 2 protons, 2 neutrons
e.
Co: 27 protons, 33 neutrons
f.
a.
Br: 35 protons, 79 - 35 = 44 neutrons. Since the charge of the atom is neutral, the number
Cr: 24 protons, 30 neutrons
of protons = the number of electrons = 35.
b.
c.
Pu: 94 protons, 145 neutrons, 94 electrons
d.
Cs: 55 protons, 78 neutrons, 55 electrons
e.
f.
45.
Br: 35 protons, 46 neutrons, 35 electrons
H: 1 proton, 2 neutrons, 1 electron
Fe: 26 protons, 30 neutrons, 26 electrons
a. Element #5 is boron.
B
b. Z = 7; A = 7 + 8 = 15;
c. Z = 17; A = 17 + 18 = 35;
e. Z = 6; A = 14;
46.
C
d. A = 92 + 143 = 235;
f.
Z = 15; A = 31;
a. Element 8 is oxygen. A = mass number = 9 + 8 = 17;
b. Chlorine is element 17.
Cl
d. Z = 26; A = 26 + 31 + 57;
f. Lithium is element 3.
47.
Cl
U
P
O
c. Cobalt is element 27.
Fe
N
e. Iodine is element 53.
Co
I
Li
Atomic number = 63 (Eu); Charge = +63 - 60 = +3; Mass number = 63 + 88 = 151;
Symbol:
Eu3+
Atomic number = 50 (Sn); Mass number = 50 + 68 = 118; Net charge = +50 - 48 = +2; The symbol
is
Sn2+ .
48.
49.
Atomic number = 16 (S); Charge = +16 - 18 = -2; Mass number = 16 + 18 = 34; Symbol:
S2-
Atomic number = 16 (S); Charge = +16 - 18 = -2; Mass number = 16 + 16 = 32; Symbol:
S2-
CHAPTER 2
ATOMS, MOLECULES, AND IONS
23
Number of protons in
nucleus
Number of neutrons
in nucleus
Number of
electrons
Net
charge
As3+
33
42
30
3+
Te2-
52
76
54
2-
S
16
16
16
0
Tl+
81
123
80
1+
Pt
78
117
78
0
Symbol
50.
Number of protons in
nucleus
Number of neutrons in
nucleus
Number of
electrons
Net
charge
92
146
92
0
Ca2+
20
20
18
2+
V3+
23
28
20
3+
Y
39
50
39
0
Br-
35
44
36
1-
P3-
15
16
18
3-
Symbol
U
51.
Metals: Mg, Ti, Au, Bi, Ge, Eu, Am. Nonmetals: Si, B, At, Rn, Br.
52.
Si, Ge, B, At. The elements at the boundary between the metals and the nonmetals are: B, Si, Ge, As,
Sb, Te, Po, At. Aluminum has mostly properties of metals.
53.
a and d. A group is a vertical column of elements in the periodic table. Elements in the same family
24
CHAPTER 2
ATOMS, MOLECULES, AND IONS
(group) have similar chemical properties.
54.
a. transition metals
c. alkali metals
e. halogens
55.
Carbon is a nonmetal. Silicon and germanium are metalloids. Tin and lead are metals. Thus, metallic
character increases as one goes down a family in the periodic table.
56.
The metallic character decreases from left to right.
57.
Metals lose electrons to form cations, and nonmetals gain electrons to form anions. Group 1A, 2A and
3A metals form stable +1,+2 and +3 charged cations, respectively. Group 5A, 6A and 7A nonmetals
form -3, -2 and -1 charged anions, respectively.
58.
b. alkaline earth metals
d. noble gases
a. Lose 1 e- to form Na+ .
b. Lose 2 e- to form Sr2+.
c. Lose 2 e- to form Ba2+ .
d. Gain 1 e- to form I-.
e. Lose 3 e- to form Al3+ .
f.
a. Lose 2 e- to form Ra2+.
b. Lose 3 e- to form In3+ .
c. Gain 3 e- to form P3-.
d. Gain 2 e- to form Te2-.
e. Gain 1 e- to form Br-.
f.
Gain 2 e- to form S2-.
Lose 1 e- to form Rb+ .
Nomenclature
59.
60.
61.
a. sodium chloride
b. rubidium oxide
c. calcium sulfide
d. aluminum iodide
a. mercury(I) oxide
b. iron(III) bromide
c. cobalt(II) sulfide
d. titanium(IV) chloride
a. chromium(VI) oxide
b. chromium(III) oxide
d. sodium hydride
e. calcium bromide
f.
62.
c. aluminum oxide
zinc chloride (Zinc only forms +2 ions so no Roman numerals are needed for zinc compounds.)
a. cesium fluoride
b. lithium nitride
c. silver sulfide (Silver only forms +1 ions so no Roman numerals are needed.)
63.
64.
d. manganese(IV) oxide
e. titanium(IV) oxide
a. potassium perchlorate
b. calcium phosphate
c. aluminum sulfate
d. lead(II) nitrate
a. barium sulfite
b. sodium nitrite
c. potassium permanganate
d. potassium dichromate
f.
strontium phosphide
CHAPTER 2
65.
66.
67.
68.
69.
70.
25
a. nitrogen triiodide
b. sulfur difluoride
c. phosphorus trichloride
d. dinitrogen tetrafluoride
a. dinitrogen tetroxide
b. iodine trichloride
c. sulfur dioxide
d. diphosphorus pentasulfide
a. copper(I) iodide
b. copper(II) iodide
d. sodium carbonate
e. sodium hydrogen carbonate or sodium bicarbonate
f.
tetrasulfur tetranitride
g. sulfur hexafluoride
i.
barium chromate
j.
c. cobalt(II) iodide
h. sodium hypochlorite
ammonium nitrate
a. acetic acid
b. ammonium nitrite
c. cobalt(III) sulfide
d. iodine monochloride
e. lead(II) phosphate
f.
potassium iodate
g. sulfuric acid
h. strontium nitride
i.
aluminum sulfite
j.
k. sodium chromate
l.
hypochlorous acid
tin(IV) oxide
a. CsBr
b. BaSO4
c. NH4 Cl
d. ClO
e. SiCl4
f.
g. BeO
h. MgF2
ClF3
a. SF2
b. SF6
c. NaH2 PO4
d. Li3 N
e. Cr2 (CO3 )3
f.
SnF2
g. NH4 C2 H3 O2
h. NH4 HSO4
i.
Co(NO3 )3
k. KClO3
l.
NaH
j.
71.
ATOMS, MOLECULES, AND IONS
Hg2 Cl2 ; Mercury(I) exists as Hg2 2+ ions.
a. Na2 O
b. Na2 O2
c. KCN
d. Cu(NO3 )2
e. SeBr4
f.
PbS
g. PbS2
h. CuCl
i.
GaAs (Predict Ga3+ and As3- ions.)
j.
CdSe (Cadmium only forms +2 charged ions in compounds.)
k. ZnS (Zinc only forms +2 charged ions in compounds.)
l.
72.
HNO2
m. P2 O5
a. (NH4 )2HPO4
b. Hg2S
c. SiO2
d. Na2 SO3
e. Al(HSO4 )3
f.
NCl3
g. HBr
h. HBrO2
i.
HBrO4
j. KHS
k. CaI2
l.
CsClO4
26
CHAPTER 2
ATOMS, MOLECULES, AND IONS
Additional Exercises
73.
Yes, 1.0 g H would react with 37.0 g 37Cl and 1.0 g H would react with 35.0 g 35Cl.
No, the mass ratio of H/Cl would always be 1 g H/37 g Cl for 37 Cl and 1 g H/35 g Cl for 35 Cl.
As long as we had pure 37 Cl or pure 35 Cl, the above ratios will always hold. If we have a mixture
(such as the natural abundance of chlorine), the ratio will also be constant as long as the composition
of the mixture of the two isotopes does not change.
74.
: 43 protons and 55 neutrons;
: 43 protons and 56 neutrons
Tc is in the same family as Mn. We would expect Tc to have properties similar to Mn.
permanganate: MnO4 -; pertechnetate: TcO4 -; ammonium pertechnetate: NH4TcO4
75.
76.
a. nitric acid, HNO3
b. perchloric acid, HClO4
d. sulfuric acid, H2 SO4
e. phosphoric acid, H3PO4
c. acetic acid, HC2 H3 O2
a. Fe2+ : 26 protons (Fe is element 26.); protons - electrons = charge, 26 -2 = 24 electrons; FeO is
the formula since the oxide ion has a -2 charge.
b. Fe3+ : 26 protons; 23 electrons; Fe2O3
c. Ba2+ : 56 protons; 54 electrons; BaO
d. Cs+ : 55 protons; 54 electrons; Cs2O
e. S2-: 16 protons; 18 electrons; Al2S3
f.
P3-: 15 protons; 18 electrons; AlP
g. Br-: 35 protons; 36 electrons; AlBr3
h. N3-: 7 protons; 10 electrons; AlN
77.
a. Pb(C2 H3 O2 )2 : lead(II) acetate
b. CuSO4 : copper(II) sulfate
c. CaO: calcium oxide
d. MgSO4 : magnesium sulfate
e. Mg(OH)2 : magnesium hydroxide
f.
CaSO4 : calcium sulfate
g. N2 O: dinitrogen monoxide or nitrous oxide
78.
a. This is element 52, tellurium. Te forms stable -2 charged ions (like other oxygen family
members).
b.
Rubidium. Rb, element 37, forms stable +1 charged ions.
c.
Argon. Ar is element 18.
d.
Astatine. At is element 85.
CHAPTER 2
79.
ATOMS, MOLECULES, AND IONS
27
A chemical formula gives the actual number and kind of atoms in a compound. In all cases, 12
hydrogen atoms are present. For example:
4 molecules H3 PO4 ×
= 12 atoms H
80.
From the XBr2 formula, the charge on element X is +2. Therefore, the element has 88 protons,
which identifies it as radium, Ra. 230 - 88 = 142 neutrons
81.
a. Ca2+ and N3-: Ca3 N2 , calcium nitride
b. K+ and O2-: K2 O, potassium oxide
c. Rb+ and F-: RbF, rubidium fluoride
d. Mg2+ and S2-: MgS, magnesium sulfide
e. Ba2+ and I-: BaI2, barium iodide
f.
Al3+ and Se2-: Al2Se3, aluminum selenide
g. Cs+ and P3-: Cs3 P, cesium phosphide
h. In3+ and Br-: InBr3 , indium(III) bromide. In also forms In+ ions, but one would predict In3+
ions from its position in the periodic table.
82.
These compounds are similar to phosphate (PO4 3-) compounds. Na3 AsO4 contains Na+ ions and AsO4 3ions. The name would be sodium arsenate. H3 AsO4 is analogous to phosphoric acid, H 3 P O 4 .
H3 AsO4 would be arsenic acid. Mg3 (SbO4 )2 contains Mg2+ ions and SbO4 3- ions, and t h e n a me
would be magnesium antimonate.
Challenge Problems
83.
Copper(Cu), silver (Ag) and gold(Au) make up the coinage metals.
84.
Avogadro proposed that equal volumes of gases (at constant temperature and pressure) contain equal
numbers of molecules. In terms of balanced equations, Avogadro’s hypothesis implies that volume
ratios will be identical to molecule ratios. Assuming one molecule of octane reacting, then 1 molecule
of Cx Hy produces 8 molecules of CO2 and 9 molecules of H2 O. Cx Hy + O2 ÷ 8 CO2 + 9 H2 O. Since
all the carbon in octane ends up as carbon in CO2, then octane contains 8 atoms of C. Similarly, all
hydrogen in octane ends up as hydrogen in H2 O, so one molecule of octane contains
9 × 2 = 18 atoms of H. Octane formula = C8 H18 and the ratio of C:H = 8:18 or 4:9.
85.
Compound I:
; Compound II:
The ratio of the masses of R that combine with 1.00 g Q is:
= 2.99 . 3
As expected from the law of multiple proportions, this ratio is a small whole number.
Since Compound I contains three times the mass of R per gram of Q as compared to Compound II
(RQ), then the formula of Compound I should be R3 Q.
28
CHAPTER 2
ATOMS, MOLECULES, AND IONS
86.
The alchemists were incorrect. The solid residue must have come from the flask.
87.
a. Both compounds have C2 H6 O as the formula. Because they have the same formula, their mass
percent composition will be identical. However, these are different compounds with different
properties since the atoms are bonded together differently. These compounds are called isomers
of each other.
b. When wood burns, most of the solid material in wood is converted to gases, which escape. The
gases produced are most likely CO2 and H2 O.
c. The atom is not an indivisible particle, but is instead composed of other smaller particles, e.g.,
electrons, neutrons, protons.
d. The two hydride samples contain different isotopes of either hydrogen and/or lithium. Although
the compounds are composed of different isotopes, their properties are similar because different
isotopes of the same element have similar properties (except, of course, their mass).