Question

Molecular formula of oxide of iron in which mass percentage of any 69

Answer 1

Answer:

Mass % of iron = 69.9 % [Given]

Mass % of oxygen = 30.1 % [Given]

Element

Atomic mass

Mass %

Mass % / atomic mass

Fe

55.85

69.9

69.9/55.85 = 1.25

O

16.00

30.1

30.1/16.00 = 1.88

Fe : O = 1.25 : 1.88

Convert in simple ratio we get

Fe : O = 2 : 3

The empirical formula of the iron oxide is Fe2O3.

N = Molar mass / Empirical mass …(1)

Empirical formula mass of Fe2O3 = [2×55.85 + 3×16.00] g = 159.70 g

Given that Molar mass of Fe2O3 = 159.69 g

Plug the values in equation (1), we get

N = 159.69/159.70 = 1

Molecular formula = Empirical formula × n = Fe2O3 × 1 = Fe2O3

Hence molecular formula is also Fe2O3

Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the empirical formula of the given oxide is Fe2O 3 and n is 1.

Hence, the molecular formula of the oxide is Fe2O3.