Chapter 21, Problem 21.62QP

Interpretation Introduction

Interpretation:

The percent by mass of manganese in the given sample has to be calculated.

Concept introduction:

Mass percent: To express the concentration of a component in a mixture we can use mass percent.  Dividing the grams of solute by grams of solution and then multiply with 100 to obtain percentage.

The formula to calculate mass percent is

$\text{Mass}\text{percent=}\frac{\text{grams }\text{of}\text{solute}}{\text{grams }\text{of}\text{solution}}\text{×100}$

Answer to Problem 21.62QP

The percent by mass of manganese in the given sample is $6.49\%$

Explanation of Solution

The reaction between iron ion and permanagate and its balanced equation is represented as follows.

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

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

To determine percent by mass of manganese

From the above equation we can say that 1 mole of permanganate is equalent to 5 moles of iron (II) ion.

From this we can calculate the original amount of ion (II) is

$\text{50}\text{.0}\text{mL×}\frac{\text{0}\text{.0800}\text{mol}{\text{Fe}}^{\text{2+}}}{\text{100}\text{mL}\text{solution}}\text{=4}{\text{.00×10}}^{\text{-3}}\text{mol}{\text{Fe}}^{\text{2+}}$

Now, determine the excess amount of iron (II) with the help of balanced equation.

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

From the above equation we can say that 1 mole of dicromate is equalent to 6 moles of iron (II) ion.  The excess amount of iron (II) is calculated as follows

$\text{22}\text{.4}\text{mL×}\frac{\text{0}\text{.0100}\text{mol}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}}{\text{1000}\text{mL}\text{solution}}\text{×}\frac{\text{6}\text{mol}{\text{Fe}}^{\text{2+}}}{\text{1}\text{mol}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}{}^{\text{2-}}}\text{= 1}{\text{.34×10}}^{\text{-3}}\text{mol}{\text{Fe}}^{\text{2+}}$

The consumed amount of iron is

$\left(\text{4}{\text{.00×10}}^{\text{-3}}\text{mol}{\text{Fe}}^{\text{2+}}\right)-\left(\text{1}{\text{.34×10}}^{\text{-3}}\text{mol}{\text{Fe}}^{\text{2+}}\right)=2.66{\text{×10}}^{\text{-3}}\text{mol}{\text{Fe}}^{\text{2+}}$

From the above values, find the mass of manganese

$\left(\text{2}{\text{.66×10}}^{\text{-3}}\text{mol}{\text{Fe}}^{\text{2+}}\right)\text{×}\frac{\text{1}\text{mol}{\text{MnO}}_{\text{4}}{}^{\text{-}}}{\text{5}\text{mol}{\text{Fe}}^{\text{2+}}}\text{×}\frac{\text{1}\text{mol}\text{Mn}}{\text{1}\text{mol}{\text{MnO}}_{\text{4}}{}^{\text{-}}}\text{×}\frac{\text{54}\text{.94}\text{g}\text{Mn}}{\text{1}\text{mol}\text{Mn}}\text{=0}\text{.0292}\text{g}\text{Mn}$

Therefore, the percent by mass of manganese is

$\frac{\text{0}\text{.0292}\text{g}}{\text{0}\text{.450}\text{g}}\text{×100\% = 6}\text{.49\%}$

Conclusion

The percent by mass of manganese in the given sample was $6.49\%$