A sample of an oxide of iron was analyzed and found to contain 70\%...

Let the total mass of the oxide sample be $100\text{ g}$.

Mass of iron ($\text{Fe}$) $= 70\text{ g}$
Mass of oxygen ($\text{O}$) $= 100\text{ g} - 70\text{ g} = 30\text{ g}$

Number of moles of $\text{Fe} = \dfrac{70}{56} = 1.25\text{ mol}$
Number of moles of $\text{O} = \dfrac{30}{16} = 1.875\text{ mol}$

Molar ratio of $\text{Fe} : \text{O} = 1.25 : 1.875$

Dividing by the smaller value ($1.25$):
Ratio $= 1 : 1.5$

Multiplying by $2$ to obtain a whole number ratio:
Ratio $= 2 : 3$

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

Answer: $\text{Fe}_2\text{O}_3$