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SOLUTIONS - CHAPTER 2 Problems
1) Protons are far more massive than electrons. Using the information in the textbook,
find mp/me, the ratio of the mass of a proton to that of an electron.
mp/me = 1.672623 x 10-24 g = 1836.153
9.109387 x 10-28 g
So protons are about 2000 times more massive than electrons.
2) (Burdge, 2.13) Explain the meaning of each term in the symbol AZX.
A = Mass number = number of protons + number of neutrons
Z = Atomic number = number of protons (usually omitted)
X = Symbol for element
3) (Burdge, 2.18) Write the appropriate symbol for each of the following isotopes:
23
23
a) Z = 11; A = 23
11Na = Na
64
64
b) Z = 28, A = 64
28Ni = Ni
115
115
c) Z = 50, A = 115
Sn
50Sn =
42
42
d) Z = 20; A = 42
20Ca = Ca
4) There are two common isotopes of bromine, 79Br and 81Br. Give the number of
protons, the number of neutrons, and the number of electrons for one atom of each of
these two isotopes.
79
Br
#protons = atomic number = 35
#electrons = #protons = 35
#neutrons = mass number - atomic number = 79 - 35 = 44
81
Br
#protons = atomic number = 35
#electrons = #protons = 35
#neutrons = mass number - atomic number = 81 - 35 = 46
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5) (Burdge, 2.22) The following radioactive isotopes are used in medicine for imaging
organs, studying blood circulation, treating cancer, and so on. Give the number of
protons, neutron, and electrons present in each isotope:
The number of protons, Z, is found in the periodic table with the symbol for the
element. The atoms are neutral, and so the number of electrons is equal to the number of
protons. The mass number, A, is given as a superscript to the left of the symbol for the
element. The number of neutrons is A - Z.
a) 198Au
b) 47Ca
c) 60Co
d) 18F
e) 125I
f) 131I
g) 42K
h) 43K
i) 24Na
j) 32P
k) 85Sr
l) 99Tc
79 protons, 119 neutrons, 79 electrons
20 protons, 27 neutrons, 20 electrons
27 protons, 33 neutrons, 27 electrons
9 protons, 9 neutrons, 9 electrons
53 protons, 72 neutrons, 53 electrons
53 protons, 78 neutrons, 53 electrons
19 protons, 23 neutrons, 19 electrons
19 protons, 24 neutrons, 19 electrons
11 protons, 13 neutrons, 11 electrons
15 protons, 17 neutrons, 15 electrons
38 protons, 47 neutrons, 38 electrons
43 protons, 56 neutrons, 43 electrons
6) What is meant by the term "radioactive"?
The nucleus of an atom is radioactive if it spontaneously emits particles or
radiation.
7) (Burdge, 2.31) What is the mass (in amu) of a carbon-12 atom? Why is the atomic
mass of carbon listed as 12.01 amu in the table on the inside front cover of the book?
The mass of one 12C atom, by definition, is exactly 12. amu. The mass for carbon
in the periodic table is the average mass of naturally occurring carbon. It is slightly
larger than 12.00 because of the small amount of 13C and 14C that occurs in nature.
8) (Burdge, 2.34) The atomic masses of 35Cl (75.78 percent) and 37Cl (24.22 percent) are
34.969 amu and 36.956 amu, respectively. Calculate the average atomic mass of
chlorine. The percentages in parentheses denote the relative abundances.
Average mass = (0.7578)(34.969 amu) + (0.2422)(36.956 amu) = 35.45 amu
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9) (Burdge, 2.36) There are only two isotopes of the element thallium (Tl) found in
nature, 203Tl and 205Tl with masses 202.972320 amu and 204.974401 amu, respectively.
Calculate the natural abundances (percentages) of these two isotopes. The average
atomic mass of thallium is 204.3833 amu.
Let x = fraction of 203Tl. Then (1 - x) = fraction of 205Tl.
So
204.3833 = x (202.972320) + (1 - x) (204.974401)
204.3833 = 202.972320 x + 204.974401 - 204.974401 x
204.3833 = - 2.002081 x + 204.974401
So
204.3833 - 204.974401 = - 2.002081 x
- 0.591101 = - 2.002081 x
x = ( - 0.591101) = 0.2952
( - 2.002081)
So natural abundance is 29.52% 203Tl and (100.00% - 29.52%) = 70.48% 205Tl.
10) Give the symbol for each of the following elements
a) boron
B
b) zinc
Zn
c) sodium
Na
d) titanium
Ti
e) argon
Ar
f) oxygen
O
g) phosphorus P
h) potassium K
11) For each of the following elements give the name of the element, and classify the
elements as either a metal, a nonmetal, or a metalloid.
a) As arsenic
metalloid
b) Fe iron
metal
c) Pb lead
metal
d) F fluorine
nonmetal
e) H hydrogen
nonmetal
f) Ni nickel
metal
g) S sulfur
nonmetal
h) Si silicon
metalloid
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12) For each of the following elements give the name of the element, and classify the
elements as an alkali metal, an alkaline earth metal, a halogen, a noble gas, or a transition
metal.
a) V vanadium
transition metal
b) He helium
noble gas
c) Br bromine
halogen
d) Ca calcium
alkaline earth metal
e) Mg magnesium alkaline earth metal
f) Ag silver
transition metal
g) K potassium
alkali metal
h) Ne neon
noble gas
13) How do metals and nonmetals differ in their conduction of heat and electricity?
Metals are good conductors of heat and electricity; nonmetals are usually poor
conductors of heat and electricity
14) (Burdge, 2.48) Group the following elements in pairs that you would expect to show
similar chemical properties: I, Ba, O, Br, S, Ca.
Pairing is based on the elements being in the same group (column) of the periodic
table
I and Br
Ba and Ca
O and S
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15) (Burdge, 2.78) Show the location of the a) alkali metals; b) alkaline earth metals; c)
halogens; d) noble gases in the given outline of the periodic table. Also draw dividing
lines between metals and metalloids and between metalloids and nonmetals.
a) alkali metals are all the elements in the 1A column except the first element
(which is hydrogen)
b) alkaline earth metals are all the elements in the 2A column
c) halogens are all the elements in the 7A column
d) noble gases are all the elements in the 8A column
The red diagonal represents the metalloids. The elements to the right of the red diagonal
are nonmetals (along with hydrogen). The elements to the left of the red diagonal are
metals (except for hydrogen).
16) How many atoms are there in 5.10 moles of sulfur (S)?
# atoms = 5.10 mol S 6.022 x 1023 atoms = 3.07 x 1024 atoms of S
mol
17) What is the mass in grams of 1.00 x 1012 lead (Pb) atoms?
Mass = 1.00 x 1012 atoms
1 mol
207.2 g = 3.44 x 10-10 g
23
6.022 x 10 atoms mol
18) A chemist prepares a 4.328 g pure sample of aluminum metal. How many atoms of
aluminum are in the sample of metal?
# Al atoms = 4.328 g Al
1 mol Al
6.022 x 1023 atoms Al
26.982 g Al
1 mol Al
= 9.659 x 1022 atoms Al
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19) (Burdge, 2.83) A cube made of platinum (Pt) has an edge length of 1.000 cm. The
density of Pt is 21.45 g/cm3, and the average mass of a single Pt atom is 3.240 x 10-22 g.
a) Calculate the number of Pt atoms in the cube
# atoms = 1.000 cm3 21.45 g
1 atom
1 cm3 3.240 x 10-22 g
= 6.620 x 1022 atoms
b) Atoms are spherical in shape. Therefore, the Pt atoms in the cube cannot fill all
of the available space. If only 74 percent of the space inside of the cube is taken up by Pt
atoms, calculate the radius (in pm = picometers) of a Pt atom. Recall that for a sphere
with radius r is V = 4/3 r3.
Since only 74% of the space inside the cube is occupied the atoms, then the
volume per atom is
V (per atom) = (0.74) (1.000 cm3) = 1.118 x 10-23 cm3
6.620 x 1022 atoms
Since V = 4/3 r3
r = [ 3V/4 ]1/3 = [ 3 (1.118 x 10-23 cm3) / 4 ]1/3 = (2.668 x 10-24 cm3)1/3
= 1.39 x 10-8 cm
In terms of picometers
r = 1.39 x 10-8 cm
1m
1012 pm = 139. pm
100 cm 1 m
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