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Q.

If an iron oxide has 69.9 mass% of Fe $(molar mass = 56.0 g {mol}^{- 1})$ and 30.1 mass% of oxygen, its molecular formula will be

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a

${Fe}_{2} O_{6}$

b

${Fe}_{2} O_{3}$

c

${Fe}_{3} O_{4}$

d

$FeO$

answer is B.

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Detailed Solution

$Fe : O :: \frac{69.9}{56.0} : \frac{30.1}{16.0} :: 1.248 : 1.881 :: 1 : \frac{1.881}{1.248} :: 1 : 1.5 :: 2 : 3$ Molecular formula ${Fe}_{2} O_{3}$

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