An oxide of iron,which contains 69.9 % iron a | vedclass

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(A) To determine the empirical formula,we calculate the mole ratio of the elements:$1$. Moles of $Fe = \frac{69.9}{55.85} \approx 1.25$$2$. Moles of $

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$Fe_2O_3$

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(A) To determine the empirical formula,we calculate the mole ratio of the elements:$1$. Moles of $Fe = \frac{69.9}{55.85} \approx 1.25$$2$. Moles of $O = \frac{30.1}{16.00} \approx 1.88$$3$. Divide by the smallest value $(1.25)$:$Fe: \frac{1.25}{1.25} = 1$$O: \frac{1.88}{1.25} = 1.5$$4$. Convert to the simplest whole number ratio by multiplying by $2$:$Fe: 1 	imes 2 = 2$$O: 1.5 	imes 2 = 3$Thus,the empirical formula is $Fe_2O_3$.

An oxide of iron,which contains $69.9 \%$ iron and $30.1 \%$ dioxygen by mass,has the empirical formula:

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