Question

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

Answer 1

Answer: Given

Mass % of iron = 69.9 % 

Mass % of oxygen = 30.1 % 

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

Converting in simple ratio we get

Fe : O = 2 : 3

Hence, the empirical formula of the iron oxide is Fe₂O_3.