1. 1. What is the empirical formula of a compound
made of g of sodium with 7.74 g of oxygen?
22.26 g Na x 1 mole = moles Na
23.0 g
7.74 g O x 1 mole = moles O
16.0 g
moles Na = = 2
moles O Na2O

2. A compound has an empirical formula of
CH2 and a molecular mass of 84. What is
the molecular formula?
Empirical mass of CH2 = 14
84 = 6 14
molecular formula = C6H12

3. 70.0 g x 1 mole = 1.25 mole Fe 55.8 g 30.0 g x 1 mole = 1.88 mole O
3. What is the empirical formula of a compound
which is 70.0 g iron and 30.0 g oxygen?
70.0 g x 1 mole = 1.25 mole Fe
55.8 g
30.0 g x 1 mole = 1.88 mole O
16.0 g
1.88 mole O = = 3
1.25 mole Fe Fe2O3

4. 21.6% oxygen. What is the empirical formula?
A compound is 67.6% Hg, 10.8% S, and
21.6% oxygen. What is the empirical
formula?
67.6% → 67.6 g
67.6 g x 1 mole = mole Hg (1)
200.6 g
10.8% → 10.8 g
10.8 g x 1 mole = mole S (1)
32.1g
21.6% → 21.6 g
21.6 g x 1 mole = 1.35 mole O (4)
16.0 g HgSO4

5. A compound is 30.4% nitrogen and 69.6%
oxygen with a molecular mass of 92. What
is the molecular formula?
30.4% → g x 1 mol = 2.17 mol N
14.0 g
69.6% → g x 1 mol = 4.35 mol O
16.0 g
= 2 = NO2
Empirical mass NO2 = 46
92 = 2
46 Molecular formula = N2O4

6. A student used electrolysis to break down a
compound to find its formula.
The following data was collected:
Mass container = g
Mass container + compound = g
After electrolysis:
Mass container + iron = g
The other substance produced was iodine.

7. Use the data to determine the following:
1. The mass of iron and the mass of iodine
in the compound.
2. The percentage (by mass) of iron and iodine.
3. The moles of iron and moles of iodine
4. The mole ratio of iodine to iron and the
formula of the compound.
5. Name the resulting compound.

8. Mass container + iron - Mass container = Fe
63.20 g g = g Fe
Mass container + compound – Mass container
66.25 g g = g compound
3.50 g g Fe = g I or
Mass container + compound – Mass container + iron = I
66.25 g g = 3.05 g I

9. % Fe = g x 100 = 13%
3.50 g
 % I = g x 100 = 87.1 = 87%
3.50 g

10. 0.45 g Fe x 1 mole = mole Fe
55.9 g
3.05 g I x 1 mole = = mole I
126.9 g
I mole = FeI3
Fe mole iron (III) iodide