$$ \begin{array}{lcl} & \mathrm{Fe} & \mathrm{O} \\ \text { Mass\% } & \frac{69.9}{} & 30.1 \\ \text { Relative No.of Moles : } & \frac{69.9}{56}=1.25 & \frac{30.1}{16}=1.88 \\ \text { Simplest atomic ratio }=1.25 / 1.25 & 1.88 / 1.25 \\ \text { Molar Ratio : } 1: 1.5 & \\ \text { Molar Ratio : } 2: 3 & \\ \text { Answer : } \mathbf{F e}_2 \mathbf{O}_3 & \end{array} $$