# What mass of "Fe" is in "316.80 g Fe"_2"O"_3"?

Nov 10, 2015

${\text{316.80 g Fe"_2"O}}_{3}$ contains $\text{221.56 g Fe}$.

#### Explanation:

One mole $\text{Fe"_2"O"_3}$ contains two moles $\text{Fe}$ atoms.

The molar mass of iron is 55.845 g/mol. Two moles of iron atoms equals $2 \times 55.845 \text{g/mol Fe"="111.69 g/mol Fe}$.

Determine the percentage of iron in one mole of iron(III) oxide, which has a molar mass of $\text{159.70 g/mol}$.

(111.69cancel"g/mol Fe")/(159.70cancel"g/mol Fe"_2"O"_3)xx100="69.937% Fe"

In order to determine the grams of iron in 316.80 g of iron(III) oxide, multiply the percentage of $\text{Fe}$ times $316.80 \text{g Fe"_2"O"_3}$.

$0.69937 \text{Fe"xx316.80"g Fe"_2"O"_3="221.56 g Fe}$