Chapter 23, Problem 62AP

A 0.450-g sample of steel contains manganese as an impurity. The sample is dissolved in acidic solution and the manganese is oxidized to the permanganate ion ${\text{MnO}}_{\text{4}}$. The ${\text{MnO}}_{\text{4}}$ ion is reduced to ${\text{Mn}}^{\text{2}}$ ^- by reacting with 50.0 mL of 0.0800 M ${\text{ FeSO}}_{\text{4}}$ solution. The excess ${\text{Fe}}^{\text{2-}}$ ions are then oxidized to ${\text{Fe}}^{\text{3-}}$ by 22.4 mL of 0.0100 M ${\text{K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_7$. Calculate the percent by mass of manganese in the sample.

Interpretation Introduction

Interpretation:

The mass by percent of manganese in the sample is to be calculated with given mass, volume, and concentration.

Concept introduction:

The reactions in which one reactant gets oxidized while the other gets reduced are called oxidation-reduction reactions or redox reactions.

In an acidic solution, permanganate (VII) is reduced to the colorless $+2$ oxidation state of the manganese (II) ion.

The original amount of iron (II) is calculated as:

$\text{n=}\text{V×}\frac{\text{mol}\text{of}{\text{Fe}}^{2+}}{\text{1000}\text{mL}\text{soln}}$

The excess iron (II) is calculated as:

$\text{m}\text{=}\text{V}\text{×}\frac{\text{moles}\text{of}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}}{\text{1000}\text{ml}\text{solution}}\text{×}\frac{\text{mol}\text{of}{\text{Fe}}^{\text{2+}}}{\text{1}\text{mol}\text{of}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}}$

The amount of iron(II) consumed is calculated as:

${\text{m}}_{F{e}^{2+}}\text{=}\text{original}\text{amount-}\text{excess}\text{amount}$

The mass of manganese is calculated as:

${\text{m}}_{\text{Mn}}\text{=}{\text{m}}_{{\text{Fe}}^{\text{2+}}}\text{×}\frac{\text{mol}{\text{MnO}}_{\text{4}}{}^{\text{-}}}{\text{mol}{\text{Fe}}^{\text{2+}}}\text{×}\frac{\text{1mol}\text{Mn}}{\text{1}\text{mol}{\text{MnO}}_{\text{4}}{}^{\text{-}}}\text{×}\frac{\text{mass}\text{of}\text{}\text{Mn}}{\text{1}\text{mol}\text{Mn}}$

The mass percent of Mn is calculated as:

$\text{mass}\text{percent}\text{=}\left(\frac{\text{mass}\text{of}\text{Mn}}{\text{given}\text{mass}}\right)\text{×100}$

Answer to Problem 62AP

Solution: $6.49\%$

Explanation of Solution

Given information: A $\text{0}\text{.450 g}$

sample of steel contains manganese as an impurity. The permanganate ion is reduced to ${\text{Mn}}^{\text{2+}}$ by reacting with $\text{50 ml of 0}\text{.0800 M}$. The excess ${\text{Fe}}^{\text{2+}}$ ions are then oxidized to ${\text{Fe}}^{\text{3+}}$

by $\text{22}\text{.4 mL}$

of $\text{0}{\text{.0800 M K}}_2{\text{Cr}}_2{\text{O}}_7$.

The balanced equation for the permanganate iron (II) reaction is represented below.

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

Thus, one mole of permanganate is stoichiometrically equivalent to five moles of iron(II).

The original amount of iron(II) is calculated as follows.

$\text{n=}\text{V×}\frac{\text{mol}\text{of}{\text{Fe}}^{2+}}{\text{1000}\text{mL}\text{soln}}$

$\begin{array}{c}{n}_{F{e}^{2+}}=\left(\text{50}\text{.0}\text{mL}\text{×}\frac{\text{0}\text{.0800}\text{mol}{\text{Fe}}^{\text{2+}}}{\text{1000}\text{mL}\text{soln}}\right)\\ \text{=}\text{4}\text{.00}\text{×}{\text{10}}^{-\text{3}}\text{mol}{\text{Fe}}^{\text{2+}}\end{array}$

The excess amount of iron (II) is determined by using the balanced equation represented below.

${\text{Cr}}_{\text{2}}{\text{O}}^{\text{2-}}{}_{\text{7}}\left(aq\right){\text{ + 14H}}^{\text{+}}\left(aq\right){\text{ + 6Fe}}^{\text{2+}}\left(aq\right)\to {\text{2Cr}}^{\text{3+}}\left(aq\right){\text{ + 7H}}_{\text{2}}\text{O}\left(l\right){\text{ + 6Fe}}^{\text{3+}}\left(aq\right)$

Thus, one mole of dichromate is equivalent to six moles of iron (II).

The excess iron (II) is calculated as follows:

$\text{m}\text{=}\text{V}\text{×}\frac{\text{moles}\text{of}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}}{\text{1000}\text{ml}\text{solution}}\text{×}\frac{\text{mol}\text{of}{\text{Fe}}^{\text{2+}}}{\text{1}\text{mol}\text{of}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}}$

$\begin{array}{c}\text{m}=\text{ 22}\text{.4}\text{mL}\text{×}\frac{\text{0}\text{.0100}\text{mol}{\text{Cr}}_2{\text{O}}_{\text{7}}^{\text{2-}}}{\text{1000}\text{mL}\text{soln}}\text{×}\frac{\text{6}\text{mol}{\text{Fe}}^{\text{2+}}}{\text{1}\text{mol}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}^{\text{2-}}}\\ =\text{1}\text{.34}\text{×}{\text{10}}^{-\text{3}}\text{mol}{\text{Fe}}^{\text{2+}}\end{array}$

The amount of iron(II) consumed is calculated as follows:

${\text{m}}_{F{e}^{2+}}\text{=}\text{original}\text{amount-}\text{excess}\text{amount}$

$\begin{array}{c}{\text{m}}_{F{e}^{2+}}\text{= (4}{\text{.00×10}}^{-}{}^{\text{3}}\text{mol)}-\text{(1}{\text{.34×10}}^{-3}\text{mol) }\\ \text{= 2}{\text{.66×10}}^{-\text{3}}\text{mol}\end{array}$

The mass of manganese is calculated as follows:

${\text{m}}_{\text{Mn}}\text{=}{\text{m}}_{{\text{Fe}}^{\text{2+}}}\text{×}\frac{\text{mol}{\text{MnO}}_{\text{4}}{}^{\text{-}}}{\text{mol}{\text{Fe}}^{\text{2+}}}\text{×}\frac{\text{1mol}\text{Mn}}{\text{1}\text{mol}{\text{MnO}}_{\text{4}}{}^{\text{-}}}\text{×}\frac{\text{mass}\text{of}\text{}\text{Mn}}{\text{1}\text{mol}\text{Mn}}$

$\begin{array}{c}{m}_{Mn}=(2.66\times {10}^{-3}\cancel{mol}\cancel{F{e}^{2+}})\times \left(\frac{1\cancel{mol}\cancel{Mn{O}_4^{-}}}{5\cancel{mol}\cancel{F{e}^{2+}}}\right)\times \left(\frac{1\cancel{mol}Mn}{1\cancel{mol}\cancel{Mn{O}_4^{-}}}\right)\times \left(\frac{54.94g\cancel{Mn}}{1\cancel{mol}\cancel{Mn}}\right)\\ =0.0292gMn\end{array}$

Finally, the mass by mass percent of Mn is calculated as follows:

$\text{mass}\text{percent}\text{=}\left(\frac{\text{mass}\text{of}\text{Mn}}{\text{given}\text{mass}}\right)\text{×100}$

$\begin{array}{c}\text{Mn\%}=\left(\frac{\text{0}\text{.0292}\cancel{\text{g}}}{\text{0}\text{.450}\cancel{\text{g}}}\right)\text{×}\text{100\%}\\ \text{=}6.49\%\end{array}$

Conclusion

The mass by percent of manganese is calculated to be $6.49\%$.