Question

A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to Fe²⁺ in acid and then titrating the Fe²⁺ with MnO₄^-. A 1.1081-g sample was dissolved in acid and then titrated with 39.32mL of 0.03190M KMnO₄. The balanced equation is

$\mathbf{8}{\mathbf{H}}^{\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{5}{\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{MnO}}_{\mathbf{4}}^{\mathbf{-}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{\to }\mathbf{5}{\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}{\mathbf{Mn}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{aq}\mathbf{\right)}\mathbf{+}\mathbf{4}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$

Calculate the mass percent of iron in the ore.

Short answer

You need to calculate the mass percent of iron.

Step-by-step solution

Balanced chemical equation

Titration of Fe²⁺ with KMnO₄ produces Fe³⁺ and Mn²⁺. The balanced chemical equation is like

$8{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+5{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)\to 5{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Calculation of moles of MnO4-

According to the question,

Volume of MnO₄^- = 39.32mL = 0.03932L

Molarity of MnO₄^- = 0.03190M

Again you know,

$\mathrm{moles}=\mathrm{molarity}\times \mathrm{volume}$

Now, moles of MnO₄^-

$\begin{array}{l}=(0.03190\times 0.03932)\left(\mathrm{mol}\right)\\ =0.001254\left(\mathrm{mol}\right)\end{array}$

Hence, moles of MnO₄^- are 0.001254mol.

Calculation of moles of Fe

According to the balanced equation

$8{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+5{\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+{\mathrm{MnO}}_4^{-}\left(\mathrm{aq}\right)\to 5{\mathrm{Fe}}^{3+}\left(\mathrm{aq}\right)+{\mathrm{Mn}}^{2+}\left(\mathrm{aq}\right)+4{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

1 mole of MnO₄^- reacts with 5 moles Fe²⁺

And 5moles of Fe²⁺ = 5 moles of Fe

Now, moles of Fe

$\begin{array}{l}=0.001254\times \frac{5}{1}\left(\mathrm{mol}\right)\\ =0.00627\left(\mathrm{mol}\right)\end{array}$

Hence, moles of Fe are 0.00627mol.

Calculation of mass of Fe

Moles of Fe = 0.00627moles

Molecular mass of Fe = 55.85g/mol

Again you know,

$\mathrm{mass}=\mathrm{moles}\times \mathrm{mass}\left(\mathrm{molar}\right)$

Mass of Fe

$\begin{array}{l}=0.00627\times 55.85\left(\mathrm{g}\right)\\ =0.35018\left(\mathrm{g}\right)\end{array}$

Hence, mass of Fe is 0.35018g.

Calculation of mass percent of Fe

Again you know,

Mass percent of Fe

$\begin{array}{l}=\frac{\mathrm{mass}\left(\mathrm{Fe}\right)}{\mathrm{mass}\left(\mathrm{sample}\right)}\times 100\mathrm{\%}\\ =\frac{0.35018}{1.1081}\times 100\mathrm{\%}\\ =31.60\mathrm{\%}\end{array}$

Hence, mass percent of Fe is 31.60%.

Conclusion

Hence, mass percent of Fe in the ore is 31.60%.