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(B) Mass percent of iron $(Fe) = 69.9 \%$.Mass percent of oxygen $(O) = 30.1 \%$.Number of moles of iron $= \frac{69.9}{55.85} = 1.25 \, mol$.Number o

Vedclass · Accepted Answer

$Fe_2O_3$

Vedclass · Answer

(B) Mass percent of iron $(Fe) = 69.9 \%$.Mass percent of oxygen $(O) = 30.1 \%$.Number of moles of iron $= \frac{69.9}{55.85} = 1.25 \, mol$.Number of moles of oxygen $= \frac{30.1}{16.0} = 1.88 \, mol$.Ratio of iron to oxygen $= 1.25 : 1.88 = 1 : 1.5 = 2 : 3$.Empirical formula $= Fe_2O_3$.Empirical formula mass $= 2(55.85) + 3(16.00) = 159.7 \, g \, mol^{-1}$.$n = \frac{	ext{Molar mass}}{	ext{Empirical formula mass}} = \frac{159.69}{159.7} \approx 1$.Molecular formula $= n 	imes 	ext{Empirical formula} = 1 	imes Fe_2O_3 = Fe_2O_3$.

Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are $69.9$ and $30.1$ respectively. Given that the molar mass of the oxide is $159.69 \, g \, mol^{-1}$.

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