An oxide of iron contains 69.9% iron. Its emp | vedclass

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(B) $1$. Percentage of $Fe = 69.9\%$. Therefore, percentage of $O = 100\% - 69.9\% = 30.1\%$.$2$. Calculate moles of each element:Moles of $Fe = \frac

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$Fe_2O_3$

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(B) $1$. Percentage of $Fe = 69.9\%$. Therefore, percentage of $O = 100\% - 69.9\% = 30.1\%$.$2$. Calculate moles of each element:Moles of $Fe = \frac{69.9}{56} = 1.248 	ext{ mol}$.Moles of $O = \frac{30.1}{16} = 1.881 	ext{ mol}$.$3$. Determine the simplest molar ratio:Divide by the smallest value $(1.248)$:$Fe = \frac{1.248}{1.248} = 1$.$O = \frac{1.881}{1.248} \approx 1.5$.$4$. Convert to whole numbers:Multiply by $2$ to get $Fe : O = 2 : 3$.$5$. Thus, the empirical formula is $Fe_2O_3$.

An oxide of iron contains $69.9\%$ iron. Its empirical formula is: (Given: Molar mass of $Fe$ and $O$ are $56$ and $16 \text{ g mol}^{-1}$ respectively.)

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