Question

determine the empirical formula of an oxide of iron which has 69.9 % iron and 30. 1 %dioxygen by mass​

Answer 1

Answer:

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 Hence, the empirical formula of the iron oxide is Fe2O3.

Explanation:

This may help you ☺️

Thank you