Question

Determine the empirical formula of an oxide of iron, which has 69.9% iron and
30.1%
dioxygen by mass​

Answer 1

Explanation:

Empirical formula can be calculated with help of following table:

Element % mass Atomic mass

Relative no. of moles

(% mass/ atomic mass)

Simple ratio

(Relative no. of moles/ smallest relative no. of moles)

Simplest whole no. ratio

Fe 69.9 56 69.9/ 56 = 1.25 1.25/ 1.25= 1 1 × 2 = 2

O 30.1 16 30.1/ 16 = 1.89 1.89/ 1.25= 1.5 1.5 × 2= 3

Therefore, empirical formula is Fe2O3.

Answer 2

The iron oxide has 69.9% iron and 30.1% dioxygen by mass.

Thus, 100 g of iron oxide contains 69.9 g iron and 30.1 g dioxygen.

The number of moles of iron present in 100 g of iron oxide are

55.8

69.9

=1.25.

The number of moles of dioxygen present in 100 g of iron oxide are

32

30.1

=0.94.

The ratio of the number of oxygen atoms to the number of carbon atoms present in one formula unit of iron oxide is

1.25

2×0.94

=1.5:1=3:2.

Hence, the formula of the iron oxide is Fe

2

O

3