Practice Problems 4
Chapter 2
CHE 151
Graham/07
How many protons and neutrons are in the nuclei of 204Tl atoms?
1.)
_81 protons___
_123 neutrons_
2.)
What is the atomic number, mass number and symbol for the element that has
45 protons and 58 neutrons?
3.)
Z = 45 protons
A = 103
(Atomic Number)
(Mass Number)
___Rh_____
(symbol)
Write the complete symbol for an atom with the following characteristics
39
a.)
contains 19 electrons and 20 neutrons
_19K_________
15
b.)
nitrogen atom with 8 neutrons
____7N______
c.)
bromine atom with a mass number of 80
_35Br________
d.)
mercury atom that contains 281 subatomic particles
____80Hg____
e.)
contains 20 protons and has a mass number of 42
20Ca_________
f.)
cobalt atom with 33 neutrons
_____27Co____
g.)
iodine atom with 181 subatomic particles
_53I_________
h.)
sulfur atom with 52 subatomic particles
______16S____
i.)
uranium atom with 146 neutrons
_92U_________
j.)
copper atom with 65 nucleons
_______29Cu__
80
201
42
60
128
36
238
65
Magnesium occurs in nature in three isotopic forms:
relative mass
24
Mg (78.70% abundance)
23.985 amu
4.)
26
25
Mg (11.17% abundance)
25.983 amu
Mg (10.13% abundance)
24.986 amu
Calculate the atomic mass of Magnesium from these data
24
Mg:
Mg:
25
Mg
26
0.7870 x 23.985 = 18.876195 =
0.1117 x 25.983 = 2.9023011 =
0.1013 x 24.986 = 2.5310818 =
18.88 amu
2.902 amu
2.531 amu
24.313  24.31 amu
5.)
Calculate the atomic mass of titanium on the basis of the following percent
composition and isotopic mass data for the naturally occurring isotopes.
titanium-46
titanium-47
titanium-48
titanium-49
titanium-50
6.)
=
=
=
=
=
7.93% (45.95263 amu) x 0.0793
7.28% (46.9518 amu) x 0.0728
73.94% (47.94795 amu) x 0.7394
5.51% (48.94787 amu) x 0.0551
5.34% (49.9448 amu) x 0.0534
Natural samples of copper contain two isotopes. 63Cu has a mass of
62.930 amu and 65Cu has a mass of 64.928 amu. The percent abundance of
63
Cu is 69.09%. Calculate the atomic mass of copper.
% abundance 65Cu:
63
65
100.00% - 69.09% = 30.91%
Cu: 62.930 x 0.6909 =
Cu: 64.928 x 0.3091 =
7.)
= 3.64
= 3.42
= 35.45
= 2.70
= 2.67
47.88 amu
43.48
20.05.6%7
63.55 amu
The two naturally occurring isotopes of chlorine are 35Cl with a mass
of 34.9689 amu and 37Cl with a mass of 36.9659 amu. The atomic mass of
elemental chlorine on earth is found to be 35.46 amu. Calculate the percent
abundance of each of the two chlorine isotopes.
34.9689X + 36.9659Y =
35.46 amu
X + Y = 1
34.9689X + 36.9659(1-X) = 35.46 amu
34.9689X + 36.9659 - 36.9659X
34.9689X - 36.9659X
=
Y = (1 - X)
35.46 amu
= 35.46 - 36.9659
 (hundredth place)

-1.9970X = -1.5059 
round
-1.51
X = -1.51 = 0.756 x 100 = 75.6%
-1.997
1 - 0.756 = Y = 0.244 x 100 = 24.4%
37
Cl
35
Cl
8.)
Calculate the atomic mass of an element if 60.4% of the atoms have a mass of
68.9257 amu and the rest have a mass of 70.9249 amu. Identify the element
in the periodic table.
100.0% - 60 4% = 39.6%
0.604 x 68.9257
0.396 x 70.9249
9.)
=
=
41.6
28.1
69.7 amu
= atomic mass of Gallium (Ga)
Europium has two stable isotopes: 151Eu with a mass of 150.9196 amu and
153
Eu with a mass of 152.9209. If elemental Europium is found to have a
mass of 151.96 amu on earth, calculate the percent of each of the two isotopes
(with the correct number of significant figures).
Mass of Eu = (mass 151Eu)(% abundance) + (mass 153Eu)(% abundance)
X
Y
X + Y = 1
Y = 1–X
151.96 amu = (150.9196)X + 152.9209(1 – X)
151.96 amu = 150.9196X + 152.9209 - 152.9209X
151.96 - 152.9209 = 150.9196X - 152.9209X
(2 decimal places) 
-0.9609 = -2.0013X
X = -0.96 = 0.48 x 100 = 48%
-2.0013
151
Y = 1 – X = 1 – 0.48 = 0.52 x 100 = 52%
153
Eu
Eu
10.)
How many electrons are in each of the following ions?
a.) Cu2+ ___27_____
b.) P3- __18______
11.)
What elements readily form cations?
___metals____________
12.)
What elements readily form anions?
__non-metals_________
13.)
Which members of group 14 (IVA) are classified as nonmetals?
________Carbon (C)________________________
14.)
How much oxygen is required when 11.24 g of cadmium is converted to
12.84 g of cadmium oxide?
___1.60 g oxygen_________
Law of Conservation
of mass
15.)
12.84 g CdO
-11.24 g Cd
1.60 g O
The density of gold is 19.3 g/cm3, and the mass of a single gold atom is
3.27 x 10-22 g. How many gold atoms are present in a piece of gold whose
volume is 3.22 cm3 ?
3.22 cm3 x 19.3 g x 1 atom(Au) = 1.90 x 1023 atoms (Au)
1 cm3
3.27 x 10-22g