Question

Determine the empirical formula of an oxide of iron , which has 69.9% of Fe and 30.1% Oxygen ?​

Answer 1

Explanation:

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.

Answer 2

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.

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