KEY
Chemistry
Percent Composition & Empirical Formula
Identify the following as molecular formulas, empirical formulas or both.
a. Ribose, C5H10O5, a sugar molecule in RNA.
b. Ethyl butanoate, C6H12O2, a cmpd w/ the odor of pineapple.
both
molecular (emp: C3H6O)
c. Chlorophyll, C55H72MgN4O5, part of photosynthesis.
both
d. DEET, C12H17ON, an insect repellent.
both
e. Oxalic acid H2C2O4, found in spinach and tea.
molecular (empirical: HCO2)
Name or provide the formula for each of the following. Calculate the percent composition
for the following compounds.
1. Cr2O3
Cr2O3:
Cr:
2 * 52.00 = 104.00 /
O:
3 * 16.00 = 48.00 /
152.00
68.42%
31.58%
check: 100.00%
2. Ca3(PO4)2
Ca3(PO4)2
Ca:
3 * 40.08 = 120.24 /
P:
2
30.97 = 61.94 /
O:
8 * 16.00 = 128.00 /
310.18
38.76%
19.97%
41.27%
check: 100.00%
3. Iron (III) oxide
Fe2O3
Fe:
O:
2 * 55.85 = 111.70 /
3 * 16.00 = 48.00 /
159.70
69.94%
30.06%
check: 100.00%
F:\2014-2015\330_ModChem\330_sections\330_07_Chemical_Formulas_and_Compounds\330.07.03.Percent Composition\330.07.03.a1.Empirical Formulas_KEY.docx
Chemistry
Empirical Formulas
p. 2
Calculate the empirical formula for each compound, and provide the name of each
compound.
4. A compound contains 0.0130 mol carbon, 0.0390 mol hydrogen and 0.0065 mol oxygen.
C: 0.0130
mol / 0.0065
=
2.0
C2H60
H: 0.0390
mol / 0.0065
=
6.0
O: 0.0065
mol / 0.0065
=
1.0
5. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass.
Mg:72.2
g
1
mol
=
3.0
Mg3N2
24.31 g
N: 27.8
g
1
mol
14.01 g
=
2.0
6. A glucose contains 40% carbon, 6.7% hydrogen, and 53.3% oxygen by mass.
C: 40
mol = 3.3 /
g 1
1.0
CH20
12.01 g
mol = 6.6 /
g
2.0
O: 53.5 g 1
mol = 3.3 /
16.00 g
1.0
H: 6.7
g 1
1.01
7. Phosphoric acid is found in some soft drinks. A sample of phosphoric acid contains 0.3086
g hydrogen, 3.161 g of phosphorus, and 6.531 g of oxygen.
H: 0.3086 g
1 mol = 0.3
1.01 g
P:
3.161 g
1 mol = 0.1
30.974 g
O:
6.531 g
1 mol = 0.4
16.00 g
H3 PO4
correct formula for
phosphoric acid
Chemistry
Empirical Formulas
p. 3
8. A compound containing 94.1% O, 5.9% H.
O: 94.1 g 1
mol = 5.9
HO
16.00 g
H: 5.9
g 1
1.01
mol = 5.8
g
(e.g., H2O2)
9. A compound containing 79.9% C, 20.1% H.
C: 79.9 g 1
mol =
6.7 / 1.0
12.01 g
CH3
H: 20.1 g 1
1.01
mol = 19.9 / 3.0
g
(e.g., C2H6)
10. The compound methyl butanoate smells like apples. Its percent composition is 58.8% C,
9.8% H, and 31.4% O. If its gram molecular mass is 102 g/mole, what is its molecular
formula?
C: 58.8 g 1
mol = 4.9
C5 H100 2
12.01 g
H: 9.8
g 1
1.01
mol = 9.7
g
O: 31.4 g 1
mol = 2.0
16.00 g
11. A compound of carbon and hydrogen has the composition of 92.25% carbon and 7.75%
hydrogen by mass. What is the empirical formula of this composition?
C: 92.25 g 1
mol =
12.01 g
7.7
CH
H: 7.75
g 1
1.01
mol =
g
7.7
(e.g., C2H2)
Chemistry
Empirical Formulas
p. 4
Determine the empirical formula from the percent composition for each of the following:
12. 36.48 % Na; 25.44 % S; 38.08 % O
Na: 36.48 g 1
mol
= 1.6 /
2.0
Na2S03
22.99 g
S: 25.44 g 1
mol
32.06 g
= 0.8 /
1.0
O: 38.08 g 1
mol
16.00 g
= 2.4 /
3.0
sodium sulfite
13. 49.99 % C; 5.61 % H; 44.40 % O
C: 49.99 g 1
mol
= 4.2 /
2.8 *
1.5 =
3
C3H402
12.01 g
mol
g
= 5.6 /
2.8 *
2.0 =
4
O: 44.40 g 1
mol
16.00 g
= 2.8 /
2.8 *
1.0 =
2
H: 5.61
g 1
1.01