Q22.40P - Chemistry: Molecular Nature Of Matter And Change

Question

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

Step-by-Step Solution

Answer

The mass $\text{\% }$ are as follows:

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

b) The mass $\text{\% }$ of iron in ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{ is 72.36\% }$ .

c) The mass $\text{\% }$ of iron in ${\text{FeS}}_{\text{2}}\text{ is 46.55\% .}$ 

Step 1: Concept Introduction

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  $\mathbf{55}\mathbf{.}\mathbf{845}\mathbf{ }\mathbf{g}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}$.

Step 2: Calculating the Mass Percentage of Iron in Fe 2 O 3

a) The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{ is 159.687 g/mol}$ 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 ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\text{ is 69.94\% .}$ 

Step 3: Calculating the Mass Percentage of Iron in Fe 3 O 4

b) Similarly, the molar mass of  ${\text{Fe}}_{\text{3}}{\text{O}}_{\text{4}}\text{ is 231.533 g/mol}$ . 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 $\text{72.359\% }$.

Step 4: Calculating the Mass Percentage of Iron in Fe 3 S 2

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

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 $46.545\%$.