## Chemistry: Molecular Approach (4th Edition)

Since iron (III) oxide is 69.94% iron by mass, each 100% of this compound has that value of iron. So, we can use this as a conversion factor to calculate the mass of iron (III) oxide: $58.7g (Iron)\times \frac{100\% (Iron(III)-Oxide)}{69.94\% (Iron)} = 83.9$ $g (Iron(III)-Oxide)$