Name: ________________________
Hour: ____ Date: _______________
Chemistry: Percentage Composition and Empirical & Molecular Formula
Solve the following problems. Show your work, and always include units where needed.
1. A compound is found to contain 36.5% Na, 25.4% S, and 38.1% O. Find its empirical formula.
2. Find the empirical formula of a compound that is 53.7% iron and 46.3% sulfur.
3. Analysis of a sample of a compound indicates that is has 1.04 g K, 0.70 g Cr, and 0.86 g O. What is its
empirical formula?
4. If 4.04 g of nitrogen combine with 11.46 g of oxygen to produce a compound with a molar mass of 108.0g,
what is the molecular formula of this compound?
5. The molar mass of a compound is 92 g. Analysis of the sample indicates that it contains 0.606 g N and
1.390 g O. Find the compound’s molecular formula.
6. An acid commonly used in the automotive industry is shown to be 31.6% phosphorous, 3.1% hydrogen,
and 63.5% oxygen. Determine the empirical formula of this acid.
7. A solvent is found to be 50.0% oxygen, 37.5% carbon, and 12.5% hydrogen. What is the empirical
formula of this solvent.
8. A particular sugar is determined to have the following composition: 40.0% carbon, 6.7% hydrogen, and
53.5% oxygen. Determine the empirical formula of this sugar molecule.
9. If the molar mass of the sugar in question #8 is 180.0 g, find the molecular formula of the sugar.
10. Ethene, a gas used extensively in preparing plastics and other polymers, has a composition of 85.7%
carbon and 14.3% hydrogen. Its molar mass is 28 g. Find the molecular formula for ethane.
Answers:
1.
2.
3.
4.
5.
Na2SO3
Fe2S3
K2CrO4
N2O5
N2O4
6. H3PO4
7. CH4O (actually, CH3OH, which is methanol)
8. CH2O
9. C6H12O6
10. C2H4
KEY
Chemistry: Percentage Composition and Empirical & Molecular Formula
Solve the following problems. Show your work, and always include units where needed.
1. A compound is found to contain 36.5% Na, 25.4% S, and 38.1% O. Find its empirical formula.
 1mol Na 
  1.59 mol Na  0.79 mol  2 Na
36.5 g Na
 23 g Na 
 1mol S 
  0.79 mol S  0.79 mol  1 S
25.4 g S 
 32 g S 
Na2SO3
 1mol O 
  2.38 mol O  0.79 mol  3 O
38.1 g O 
 16 g O 
2. Find the empirical formula of a compound that is 53.7% iron and 46.3% sulfur.
 1 mol Fe 
  0.959 mol Fe  0.959 mol  1Fe  2  2 Fe
53.7 g Fe
 56 g Fe 
 1mol S 
  1.45 mol S  0.959 mol  1.51 S  2  3 S
46.3 g S 
 32 g S 
Fe2S3
3. Analysis of a sample of a compound indicates that is has 1.04 g K, 0.70 g Cr, and 0.86 g O. What is its
empirical formula?
 1mol K 
  0.027 mol K  0.0135 mol  2 K
1.04 g K 
 39 g K 
 1mol Cr 
  0.0135 mol Cr  0.0135 mol  1 Cr
0.70 g Cr 
 52 g S 
K2CrO4
 1mol O 
  0.054 mol O  0.0135 mol  4 O
0.86 g O 
 16 g O 
4. If 4.04 g of nitrogen combine with 11.46 g of oxygen to produce a compound with a molar mass of 108.0g,
what is the molecular formula of this compound?
 1 mol N 
  0.289 mol N  0.289 mol  1N  2  2 N
4.04 g N
 14 g N 
 1 mol O 
  0.716 mol O  0.289 mol  2.48 O  2  5 O
11.46 g O 
 16 g O 
2 N  14 g/mol
5 O  16 g/mol
 28 g
 80 g
Molecular Mass  108 g
N2O5
KEY – Page 2
Chemistry: Percentage Composition and Empirical & Molecular Formula
5. The molar mass of a compound is 92 g. Analysis of the sample indicates that it contains 0.606 g N and
1.390 g O. Find the compound’s molecular formula.
 1 mol N 
  0.433 mol N  0.433 mol  1N
0.606 g N
 14 g N 
 1mol O 
  0.0869 mol O  0.0433 mol  2 O
1.390 g O 
 16 g O 
1N  14 g/mol  14 g
2 O  16 g/mol  32 g
NO2
2
46 g 92 g  2NO 2   N2 O 4 (molecular formula)
Molecular Mass  46 g
6. An acid commonly used in the automotive industry is shown to be 31.6% phosphorous, 3.1% hydrogen,
and 63.5% oxygen. Determine the empirical formula of this acid.
 1mol P 
  1.02 mol P  1.02 mol  1P
31.6 g P
 31 g P 
 1mol H 
  3.1mol H  1.02 mol  3 H
3.1 g H 
 1g H 
H3PO4
(phosphoric acid)
 1mol O 
  4.08 mol O  3.125 mol  4 O
65.3 g O 
 16 g O 
7. A solvent is found to be 50.0% oxygen, 37.5% carbon, and 12.5% hydrogen. What is the empirical
formula of this solvent.
 1mol O 
  3.33 mol O  3.125 mol  1 O
50 g O
 16 g O 
 1mol C 
  3.125 mol C  3.125 mol  1 C
37.5 g C 
 12 g C 
CH3OH or CH4O (methanol)
 1mol H 
  12 .5 mol H  3.125 mol  4 H
12.5 g H 
 1g H 
8. A particular sugar is determined to have the following composition: 40.0% carbon, 6.7% hydrogen, and
53.5% oxygen. Determine the empirical formula of this sugar molecule.
 1 mol C 
  3.33 mol C  3.33 mol  1 O
40 g C
 1c g O 
 1 mol H 
  6.7 mol H  3.33 mol  2 H
6.7 g H 
 1 gH 
 1 mol O 
  3.34 mol O  3.33 mol  1 O
53.5 g O 
 16 g O 
CH2O
(methanol)
KEY – Page 3
Chemistry: Percentage Composition and Empirical & Molecular Formula
9. If the molar mass of the sugar in question #8 is 180.0 g, find the molecular formula of the sugar.
CH2O
1 C  12 g/mol
1 O  16 g/mol
2 H  1 g/mol
 12 g
 16 g
 2g
6
30 g 180 g  6CH 2O  C 6H12 O 6 (molecular formula)
Molecular Mass  30 g
C6H12O6
10. Ethene, a gas used extensively in preparing plastics and other polymers, has a composition of 85.7%
carbon and 14.3% hydrogen. Its molar mass is 28 g. Find the molecular formula for ethane.
 1 mol C 
  7.14 mol C  7.14 mol  1 C
85.7 g C
 12 g C 
 1 mol H 
  14 .3 mol H  7.14 mol  2 H
14.3 g H 
 1g H 
1 C  12 g/mol
2 H  1 g/mol
 12 g
 2g
CH2
2
14 g 28 g  2CH 2   C 2H4 (molecular formula)
Molecular Mass  14 g
Answers:
1.
2.
3.
4.
5.
Na2SO3
Fe2S3
K2CrO4
N2O5
N2O4
6. H3PO4
7. CH4O (actually, CH3OH, which is methanol)
8. CH2O
9. C6H12O6
10. C2H4
KEY
Percentage Composition and Empirical & Molecular Formula
1.
 1mol Na 
  1.59 mol Na  0.79 mol  2 Na
36.5 g Na
 23 g Na 
 1mol S 
  0.79 mol S  0.79 mol  1 S
25.4 g S 
 32 g S 
Na2SO3
 1mol O 
  2.38 mol O  0.79 mol  3 O
38.1 g O 
 16 g O 
2.
 1 mol Fe 
  0.959 mol Fe  0.959 mol  1Fe  2  2 Fe
53.7 g Fe
 56 g Fe 
 1mol S 
  1.45 mol S  0.959 mol  1.51 S  2  3 S
46.3 g S 
 32 g S 
Fe2S3
3.
 1mol K 
  0.027 mol K  0.0135 mol  2 K
1.04 g K 
 39 g K 
 1mol Cr 
  0.0135 mol Cr  0.0135 mol  1 Cr
0.70 g Cr 
 52 g S 
K2CrO4
 1mol O 
  0.054 mol O  0.0135 mol  4 O
0.86 g O 
 16 g O 
4.
 1 mol N 
  0.289 mol N  0.289 mol  1N  2  2 N
4.04 g N
 14 g N 
 1 mol O 
  0.716 mol O  0.289 mol  2.48 O  2  5 O
11.46 g O 
 16 g O 
2 N  14 g/mol
5 O  16 g/mol
N2O5
 28 g
 80 g
Molecular Mass  108 g
5.
 1 mol N 
  0.433 mol N  0.433 mol  1N
0.606 g N
 14 g N 
 1mol O 
  0.0869 mol O  0.0433 mol  2 O
1.390 g O 
 16 g O 
1N  14 g/mol  14 g
2 O  16 g/mol  32 g
Molecular Mass  46 g
NO2
2
46 g 92 g  2NO 2   N2 O 4 (molecular formula)
KEY – Page 2
Percentage Composition and Empirical & Molecular Formula
6.
 1mol P 
  1.02 mol P  1.02 mol  1P
31.6 g P
 31 g P 
 1mol H 
  3.1mol H  1.02 mol  3 H
3.1 g H 
 1g H 
H3PO4
(phosphoric acid)
 1mol O 
  4.08 mol O  3.125 mol  4 O
65.3 g O 
 16 g O 
7.
 1mol O 
  3.33 mol O  3.125 mol  1 O
50 g O
 16 g O 
 1mol C 
  3.125 mol C  3.125 mol  1 C
37.5 g C 
 12 g C 
CH3OH or CH4O (methanol)
 1mol H 
  12 .5 mol H  3.125 mol  4 H
12.5 g H 
 1g H 
8.
 1 mol C 
  3.33 mol C  3.33 mol  1 O
40 g C
 1c g O 
 1 mol H 
  6.7 mol H  3.33 mol  2 H
6.7 g H 
 1 gH 
CH2O
(methanol)
 1 mol O 
  3.34 mol O  3.33 mol  1 O
53.5 g O 
 16 g O 
9.
CH2O
1 C  12 g/mol
1 O  16 g/mol
2 H  1 g/mol
 12 g
 16 g
 2g
6
30 g 180 g  6CH 2O  C 6H12 O 6 (molecular formula)
Molecular Mass  30 g
C6H12O6
10.
 1 mol C 
  7.14 mol C  7.14 mol  1 C
85.7 g C
 12 g C 
 1 mol H 
  14 .3 mol H  7.14 mol  2 H
14.3 g H 
 1g H 
1 C  12 g/mol
2 H  1 g/mol
 12 g
 2g
Molecular Mass  14 g
CH2
2
14 g 28 g  2CH 2   C 2H4 (molecular formula)