Determine Empirical Formula of Iron Oxide

To find the empirical formula of the oxide of iron, we first determine the mass percentages of each element present in a $100\text{ g}$ sample of the compound:

• Mass percentage of Iron ($\text{Fe}$) = $69.9\%$
• Mass percentage of Oxygen ($\text{O}$) = $100\% - 69.9\% = 30.1\%$

Next, we calculate the number of moles of each element in a $100\text{ g}$ sample using their given molar masses ($M_{\text{Fe}} = 56\text{ g mol}^{-1}$ and $M_{\text{O}} = 16\text{ g mol}^{-1}$):

$\text{Moles of Fe } (n_{\text{Fe}}) = \frac{69.9\text{ g}}{56\text{ g mol}^{-1}} \approx 1.248\text{ mol}$

$\text{Moles of O } (n_{\text{O}}) = \frac{30.1\text{ g}}{16\text{ g mol}^{-1}} \approx 1.881\text{ mol}$

Now, we determine the simplest molar ratio by dividing the number of moles of each element by the smallest number of moles obtained ($1.248$):

$\text{Ratio of Fe} = \frac{1.248}{1.248} = 1$

$\text{Ratio of O} = \frac{1.881}{1.248} \approx 1.5$

To convert this into the simplest whole number ratio, we multiply both values by $2$:

$\text{Fe} : \text{O} = (1 \times 2) : (1.5 \times 2) = 2 : 3$

Thus, the empirical formula of the oxide of iron is $\text{Fe}_2\text{O}_3$.

Therefore, the correct option is B.