What is the mass percent of iron in each of the following iron ores: ${\mathbf{\text{Fe}}}_{\mathbf{\text{2}}}{\mathbf{\text{O}}}_{\mathbf{\text{3}}}{\mathbf{\text{, Fe}}}_{\mathbf{\text{3}}}{\mathbf{\text{O}}}_{\mathbf{\text{4}}}{\mathbf{\text{, FeS}}}_{\mathbf{\text{2}}}$?

The concentration of an element in a compound or a component in a mixture can be expressed in mass percentage.

To calculate the mass$\mathbf{\text{\%}}$of iron in each of the ores, the following formula can be used:

$\mathbf{\text{mass}}\mathbf{\%}\mathbf{=}\frac{\mathbf{\text{gram}}\mathbf{ }\mathbf{\text{atomic mass of an element Number of moles of element}}}{\mathbf{\text{molar mass of compound}}}\mathbf{\text{100}}$

Where the molar mass of element - iron, is $\mathbf{55}\mathbf{.}\mathbf{845}\mathbf{}\mathbf{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$.

a) The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{is 159.687g/mol}$There are $\text{2 mol of Fe}$per one mole of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.

Then, the mass $\text{\%}$can be calculated:

$$\begin{gathered} \mathrm{mass}\%=\frac{55.845\mathrm{g}/\mathrm{mol}\times 2\mathrm{mol}}{159.687\mathrm{g}/\mathrm{mol}}\times 100\% \\ \mathrm{mass}\%=69.94\% \end{gathered}$$

The mass $\text{\%}$of iron in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{is 69.94\%.}$

b) Similarly, the molar mass of ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{is 231.533g/mol}$. There are $3\mathrm{mol}\mathrm{of}\mathrm{Fe}$per one mole of ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$.

Then, the mass can be calculated:

$$\begin{gathered} \mathrm{mass}\%=\frac{55.845\mathrm{g}/\mathrm{mol}\times 3\mathrm{mol}}{231.533\mathrm{g}/\mathrm{mol}}\times 100\% \\ \mathrm{mass}\%=72.359\% \end{gathered}$$

The mass $\text{\%}$of iron in ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}$is $\text{72.359\%}$.

c) Similarly, the molar mass of ${\text{FeS}}_{\text{2}}\text{is 119.98g/mol}$

There is $\text{1 mole of Fe}$per one mole of ${\text{FeS}}_{\text{2}}$.

Then, the mass $\text{\%}$can be calculated:

$$\begin{gathered} \mathrm{mass}\%=\frac{55.845\mathrm{g}/\mathrm{mol}\times 1\mathrm{mol}}{119.98\mathrm{g}/\mathrm{mol}}\times 100\% \\ \mathrm{mass}\%=46.545\% \end{gathered}$$

The mass $\text{\%}$of iron in ${\text{FeS}}_{\text{2}}$is $46.545\%$.