Percent Composition and Empirical Formula Review
Solutions
1. Calculate the percent by mass of Cu2O
127.092
15.999
= 88.819% Cu
= 11.181% O
143.091
143.091
2. Which of the following pairs of compounds have the same empirical formula?
a. C2H2 and C6H6
Both CH
b. C2H6 and C4H10
CH3 and C2H5
3. A compound was analyzed and was found to contain the following
percentages of the elements by mass: Barium, 98.55%; Hydrogen 1.447%.
Determine the empirical formula of the compound.
98.55
1.447
1.436

= 0.7176
= 1.436
=2
137.33
1.0079
0.7176
EF = BaH2
4. A compound with empirical formula CH was found by experiment to have a
molar mass of approximately 78g. What is the molecular formula of the
compound?
MM
78
MF = C6H6
=
=6
EM 13.0189
5. A component of protein called serine has an approximate molar mass of 105
g/mol. If the percent composition is as follows, what is the empirical and
molecular formula of serine?
C = 34.95 %
H = 6.884 %
N = 13.59 %
O = 46.56 %
34.95
6.884
13.59
46.56
= 2.909833
= 6.830043
= 0.970229
= 2.910182
12.011
1.0079
14.007
15.999
2.909833
6.830043
2.910182
1
=3
=7
=3
0.970229
0.970229
0.970229
EF = C3H7NO3
MM / EM = 105 / 105.09 = 1
MF = C3H7NO3
6. What is the empirical mass of C4H10?
EF = C2H5 so EM = 29.062 g/mol
7. Calculate the percent composition of Fe2O3
111.69
47.997
= 69.943% Fe
= 30.057% O
159.687
159.687
8. A compound with empirical formula C2H5O was found in a separate
experiment to have a molar mass of approximately 90 g. What is the
molecular formula of the compound?
90
MF = C4H10O2
=2
45.0605
9. A 50.51 g sample of a compound made from phosphorus and chlorine is
decomposed. Analysis of the products showed that 11.39 g of phosphorus
atoms were produced. What is the empirical formula of the compound?
If 11.39 g of a 50.51 g sample is P then 50.51 – 11.39 = 39.12 g is Cl
P
Cl
11.39
39.12
= 0.36773
= 1.10343
30.974
35.453
0.36773
1.10343
=1
=3
0.36773
0.36773
EF = PCl3
10. Which of the following pairs of compounds have the same empirical formula?
a. C12H10O and C6H5OH
b. NO2 and N2O4
11. The percent composition of an unknown acid is found to be 39.9% C, 6.7% H,
and 53.4% O and the molar mass of the compound is determined to be 60.0
g/mol. What is the molecular formula for the compound?
C
H
O
39.9
6.7
53.4
= 3.32195
= 6.64748
= 3.3377
12.011
1.0079
15.999
EF = CH2O
60.0
=2
30.0258
MF = C2H4O2
12. Analysis of a certain compound yielded the following percentages by mass:
nitrogen 29.16%, hydrogen, 8.392%; carbon 12.50%; oxygen, 49.95%.
Determine the empirical formula of the compound.
29.16
8.392
12.50
49.995
= 2.0818
= 8.326
= 1.1407
= 4.4196
14.007
1.0079
12.011
15.999
2.0818
=2
1.0407
8.326
=8
1.0407
1.0407
=1
1.0407
EF = N2H8CO3
4.4196
=3
1.0407