Page 60 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 3] ATOMS AND ATOMIC MASSES 49

3.32. The ratio of atoms of Y to atoms of X in a certain compound is 1:3. The mass of X in a certain sample is 3.00 g, and
the mass of Y in the sample is 4.00 g. What is the ratio of their atomic masses?
Ans. Let N = number of atoms of Y; then 3N = number of atoms of X. A Y and A X represent their atomic masses.
So
N(A Y ) 4.00 g
=
3N(A X ) 3.00 g
The Ns cancel, and the ratio of atomic masses is A Y /A X = 4/1.

3.33. Two oxides of nitrogen have compositions as follows:
Compound 1 46.68% N 53.32% O
Compound 2 36.86% N 63.14% O
Show that these compounds obey the law of multiple proportions.
Ans. The ratio of the mass of one element in one compound to the mass of that element in the other compound—for
a fixed mass of the other element—must be in a ratio of small whole numbers. Let us calculate the mass of
oxygen in the two compounds per gram of nitrogen. That is, the fixed mass will be 1.000 g N.
53.32gO 63.14gO

1.000gN = 1.142gO 1.000gN = 1.713gO
46.68gN 36.86gN
Is the ratio of the masses of oxygen a ratio of small integers?
1.713 1.50 3
= =
1.142 1 2
The ratio of 1.50:1 is equal to the ratio 3:2, and the law of multiple proportions is satisfied. Note that the
ratio of masses of nitrogen to oxygen is not necessarily a ratio of small integers; the ratio of mass of oxygen
in one compound to mass of oxygen in the other compound is what must be in the small integer ratio.

3.34. (a) Compare the number of seats in a baseball stadium during a midweek afternoon game to the number during a
weekend doubleheader (when two games are played successively). (b) Compare the number of locations available
for electrons in the second shell of a hydrogen atom and the second shell of a uranium atom.
Ans. (a) The number of seats is the same at both times. (The number occupied probably is different.)
(b) The number of locations is the same. In uranium, all 8 locations are filled, whereas in hydrogen they
are empty.

3.35. Show that the following compounds obey the law of multiple proportions.
% Cobalt % Oxygen
Compound 1 71.06 28.94
Compound 2 78.65 21.35
Does the law require that the ratio (mass of cobalt)/(mass of oxygen) be integral for each of these compounds?
Ans. The ratio of mass of cobalt to mass of oxygen is not necessarily integral. Take a constant mass of cobalt, for
example, 1.000 g Co. From the ratios given, there are

28.94gO
Compound 1 1.000gCo = 0.4073 g O
71.06gCo
21.35gO

Compound 2 1.000gCo = 0.2715 g O
78.65gCo
The ratio of masses of O (per 1.000 g Co) in the two compounds is
0.4073 1.500 3
= =
0.2715 1.000 2
The ratio of masses of oxygen in the two compounds (for a given mass of Co) is the ratio of small integers,
as required by the law of multiple proportions. The ratio of mass of cobalt to mass of oxygen is not
integral.