Homework #3-2: Problems pg. 106 #3.36, 3.39, 3.40, 3.48, 3.50, the following hydrate problem:
The general formula of Epson salts can be written as MgSO4 · x H2O. When 5.061 g of this hydrate is heated
to 250ºC, all the water of hydration is lost, leaving 2.472 g of MgSO4. What is the value of x?
36
Describe how the knowledge of the percent composition by mass of an unknown compound
can help us identify the compound.
For a pure material, the percent composition is constant. So, to determine what a sample is, the percent
composition will provide evidence of its formula.
39
Tin (Sn) exists in Earth’s crust as SnO2. Calculate the percent composition by mass of Sn and O
in SnO2.
Molar mass of SnO2 = (118.7 g) + 2(16.00 g) = 150.7 g
%Sn =
40
118.7 g/mol
× 100% = 78.77%
150.7 g/mol
%O =
(2)(16.00 g/mol)
× 100% = 21.23%
150.7 g/mol
Calculate the percent composition of chloroform (CHCl3).
The molar mass of CHCl3 = 12.01 g/mol + 1.008 g/mol + 3(35.45 g/mol) = 119.4 g/mol. The percent
by mass of each of the elements in CHCl3 is calculated as follows:
%C =
%Cl =
12.01 g/mol
× 100% = 10.06%
119.4 g/mol
%H =
1.008 g/mol
× 100% = 0.8442%
119.4 g/mol
3(35.45) g/mol
× 100% = 89.07%
119.4 g/mol
Check: The sum of the percentages is (10.06% + 0.8442% + 89.07%) = 99.97%. The small discrepancy
from 100% is due to the way we rounded off.
48
What is the mass of F, in grams, in 24.6 g of tin(II) fluoride (SnF2)?
First, we must find the mass % of fluorine in SnF2. Then, we multiply by the mass of the compound
(24.6 g), to find the mass of fluorine in 24.6 g of SnF2.
mass % F =
mass of F in 1 mol SnF2
2(19.00 g)
× 100% =
× 100% = 24.25% F
molar mass of SnF2
156.7 g
! 24.25% $
? g F in 24.6 g SnF2 = #
& ( 24.6 g ) = 5.97 g F
" 100% %
Check: As a ball-park estimate, note that the mass percent of F is roughly 25 percent, so that a quarter
of the mass should be F. One quarter of approximately 24 g is 6 g, which is close to the answer.
Note: This problem could have been worked by the following conversions:
g of SnF2 → mol of SnF2 → mol of F → g of F
50
What are the empirical formulas of the compounds with the following compositions?
(a) 40.1 percent C, 6.6 percent H, 53.3 percent O
If we have 100 g of the compound, then each percentage can be converted directly to grams. In this
sample, there will be 40.1 g of C, 6.6 g of H, and 53.3 g of O. Let n represent the number of moles
of each element so that
2
HOMEWORK #3-2 ANSWER KEY
1 mol C
= 3.34 mol C ÷ 3.33 ≈ 1
12.01 g C
1 mol H
n H = 6.6 g H ×
= 6.5 mol H ÷ 3.33 ≈ 2
1.008 g H
n C = 40.1 g C ×
This gives the empirical
formula, CH2O.
1 mol O
= 3.33 mol O ÷ 3.33 = 1
16.00 g O
(b) 18.4 percent C, 21.5 percent N, 60.1 percent K
n O = 53.3 g O ×
n C = 18.4 g C ×
1 mol C
= 1.53 mol C
12.01 g C
n N = 21.5 g N ×
1 mol N
= 1.53 mol N
14.01 g N
n K = 60.1 g K ×
1 mol K
= 1.54 mol K
39.10 g K
Dividing by the smallest
number of moles (1.53
mol) gives the empirical
formula, KCN.
Hydrate Problem:
The key to solving hydrate problems is to realize that when heated, what is left behind is only the
anhydrate, whose molar mass and number of moles can be calculated:
Molar Mass MgSO4 = 24.30 g + 32.06 g + 4(16.00 g ) = 120.36 g/mol
Moles MgSO4 = 2.472 g MgSO4 ×
1 mol MgSO4
= 2.05 ×10−2 mol MgSO4
120.36 g
Next, the number of moles of H2O is determined from the amount of water lost:
Mass H2O lost = 5.061 g – 2.472 g = 2.589 g H2O
Molar Mass H 2O = 2(1.008 g)+16.00 g = 18.02 g/mol
Moles H 2O = 2.589 g H 2O ×
1 mol H 2O
= 1.44 ×10−1 mol H 2O
18.02 g H 2O
The value x represents the ratio of the number of moles of water to the number of moles of the
anhydrate, so
1.44 ×10−1 mol H 2O
x=
= 7.01 ≈ 7 (MgSO4 · 7 H2O, magnesium sulfate heptahydrate)
2.05 ×10−2 mol MgSO4