Irvington High School  AP Chemistry
Mr. Markic
Name______________________________
Number __ Date __/__/__
3  Mass Relationships in Chemical Reactions
P e r c e n t C o m p o s i t i o n &
C h e m i c a l F o r m u l a s
1. 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 
118.7 g/mol
 100%  78.77%
150.7 g/mol
%O 
(2)(16.00 g/mol)
 100%  21.23%
150.7 g/mol
2. Cinnamic alcohol is used mainly in perfumery, particularly in soaps and cosmetics. Its molecular
formula is C9H10O.
a. Calculate the percent composition by mass of C, H, and O in cinnamic alcohol
The molar mass of cinnamic alcohol is 134.17 g/mol.
(a)
%C 
(9)(12.01 g/mol)
 100%  80.56%
134.17 g/mol
%H 
(10)(1.008 g/mol)
 100%  7.51%
134.17 g/mol
%O 
16.00 g/mol
 100%  11.93%
134.17 g/mol
b. How many molecules of cinnamic alcohol are contained in a sample of mass 0.469 g?
(b)
0.469 g C9 H10 O 
1 mol C9 H10 O
6.022  1023 molecules C9 H10O

134.17 g C9 H10O
1 mol C9 H10O
 2.11  1021 molecules C9H10O
3. Allicin is the compound responsible for the characteristic smell of garlic. An analysis of the compound
given the following percent composition by mass: C: 44.4 percent; H: 6.21 percent; S: 39.5 percent; O:
9.86 percent. Calculate its empirical formula. What is its molecular formula given that its molar mass is
about 162 g?
Assume you have exactly 100 g of substance.
nC  44.4 g C 
1 mol C
 3.70 mol C
12.01 g C
nH  6.21 g H 
1 mol H
 6.16 mol H
1.008 g H
nS  39.5 g S 
1 mol S
 1.23 mol S
32.07 g S
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nO  9.86 g O 
1 mol O
 0.616 mol O
16.00 g O
Thus, we arrive at the formula C3.70H6.16S1.23O0.616. Dividing by the smallest number of moles (0.616 mole) gives the
empirical formula, C6H10S2O.
To determine the molecular formula, divide the molar mass by the empirical mass.
molar mass
162 g

1
empirical molar mass
162.3 g
Hence, the molecular formula and the empirical formula are the same, C6H10S2O.
4. The formula for rust can be represented by Fe2O3. How many moles of Fe are present in 24.6 g of the
compound?
1 mol Fe2 O3
2 mol Fe

 0.308 mol Fe
159.7 g Fe2 O3 1 mol Fe2O3
24.6 g Fe2 O3 
5. Calculate the mass in grams of iodine (I2) that will react completely with 20.4 g of aluminum (Al) to
form aluminum oxide (AlI3).
The balanced equation is: 2Al(s)  3I2(s) 
 2AlI3(s)
Using unit factors, we convert: g of Al  mol of Al  mol of I2  g of I2
20.4 g Al 
3 mol I2 253.8 g I 2
1 mol Al


 288 g I 2
26.98 g Al 2 mol Al
1 mol I 2
6. What are the empirical formulas of the compounds with the following compositions?
a. 2.1 percent H, 65.3 percent O, 32.6 percent S
In each case, assume 100 g of compound.
(a)
2.1 g H 
1 mol H
 2.1 mol H
1.008 g H
65.3 g O 
1 mol O
 4.08 mol O
16.00 g O
32.6 g S 
1 mol S
 1.02 mol S
32.07 g S
This gives the formula H2.1S1.02O4.08. Dividing by 1.02 gives the empirical formula, H2SO4.
b. 20.2 percent Al, 79.8 percent Cl
(b)
20.2 g Al 
1 mol Al
 0.749 mol Al
26.98 g Al
79.8 g Cl 
1 mol Cl
 2.25 mol Cl
35.45 g Cl
This gives the formula, Al0.749Cl2.25. Dividing by 0.749 gives the empirical formula, AlCl3.
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7. The anticaking agent added to Morton salt is calcium silicate, CaSiO3. This compound can absorb up to
2.5 times its mass of water and still remains a free-flowing powder. Calculate the percent composition
of CaSiO3.
The molar mass of CaSiO3 is 116.17 g/mol.
%Ca 
40.08 g
 34.50%
116.17 g
%Si 
28.09 g
 24.18%
116.17 g
%O 
(3)(16.00 g)
 41.32%
116.17 g
8. The molar mass of caffeine is 194.19 g. Is the molecular formula of caffeine C4H5N2O or C8H10N4O2?
Find the molar mass corresponding to each formula.
For C4H5N2O:
4(12.01 g)  5(1.008 g)  2(14.01 g)  (16.00 g)  97.10 g
For C8H10N4O2:
8(12.01 g)  10(1.008 g)  4(14.01 g)  2(16.00 g)  194.20 g
The molecular formula is C8H10N4O2.
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