Question

A 10g mixture of $C{u_2}S$ and CuS was treated with 200mL of 0.75M $Mn{O_4}^ - $ in acid solution producing $S{O_2},M{n^{2 + }},C{u^{2 + }}$. The $S{O_2}$ was boiled off and the excess of $Mn{O_4}^ - $ was titrated with 175mL of 1M $F{e^{2 + }}$ solution. Calculate the percentage of CuS in the original mixture (To the nearest integer).

Answer 1

Hint: First find the amount of $Mn{O_4}^ - $ solution used against 1M $F{e^{2 + }}$ solution. Then find the number of electrons involved in oxidation and reduction reactions. Use mole equivalents to calculate the composition of the given mixture.

Complete answer:
We will first find the amount of 0.75M $Mn{O_4}^ - $ solution that was titrated with a given amount of 1M$F{e^{2 + }}$ solution. Then, we will find the used mole equivalents of $Mn{O_4}^ - $ for the reaction with the given mixture of compounds. From it, we will find the % of CuS in the mixture.
- We are given that 200mL of 0.75M $Mn{O_4}^ - $ solution was added to the mixture. We know that in this compound, manganese undergo reduction as below.
\[M{n^{7 + }} \to M{n^{2 + }} + 5{e^ - }\]
Thus, we can see that 5 electrons are produced in the reaction.
So, we can say that number of mole equivalents of $Mn{O_4}^ - $ present in the solution = Volume $ \times $ Concentration $ \times $ Number of electrons
Mole equivalents of $Mn{O_4}^ - $ = $200 \times 0.75 \times 5 = 750$
Now, it is given that the remaining $Mn{O_4}^ - $ is reacted with $F{e^{2 + }}$ solution. We know that $F{e^{2 + }}$ undergoes oxidation in the following way.
\[F{e^{2 + }} \to F{e^{3 + }} + {e^ - }\]
We can say that number of mole equivalents of $F{e^{2 + }}$ = Volume $ \times $ Concentration $ \times $ Number of electrons
Thus, mole equivalents of $F{e^{2 + }}$ = $175 \times 1 \times 1 = 175$
So, we can say that number of mole equivalents of $Mn{O_4}^ - $ used in the reaction with mixture = total mole equivalents of $Mn{O_4}^ - $ - mole equivalents of $F{e^{2 + }}$
Thus, mole equivalents of $Mn{O_4}^ - $ used in the reaction mixture = 750 – 175 = 575 mole equivalents
Now, we will find the number of electrons required to oxidize both $C{u_2}S$ and CuS.
- For $C{u_2}S$,
\[2C{u^ + } \to 2C{u^{2 + }} + 2{e^ - }\]
\[{S^{2 - }} \to {S^{4 + }} + 6{e^ - }\]
For CuS,
\[{S^{2 - }} \to {S^{4 + }} + 6{e^ - }\]
Now, assume that the masses of $C{u_2}S$ and CuS is x and y respectively.
So, we can say that
\[x + y = 10gm{\text{ }}....{\text{(1)}}\]
To find the mole equivalent of CuS and $C{u_2}S$, we will use the following formula.
\[{\text{Mole equivalents = }}\dfrac{{{\text{Weight }} \times {\text{1000}} \times {\text{Number of electrons involved}}}}{{{\text{Molecular weight}}}}\]
Now,for $C{u_2}S$, we can write the above formula as
\[{\text{Mole equivalents = }}\dfrac{{x \times 1000 \times 8}}{{159.2}}\]
For CuS,
\[{\text{Mole equivalents = }}\dfrac{{y \times 1000 \times 6}}{{95.6}}\]
Now, we can write that
Mole equivalents of $Mn{O_4}$ used = Mole equivalents of $C{u_2}S$ + Mole equivalents of CuS
575 = $\dfrac{{x \times 1000 \times 8}}{{159.2}}$ +$\dfrac{{y \times 1000 \times 6}}{{95.6}}$ ……(2)
Now, as we solve the equation (1) and (2), we obtain that
x = 4.206gm and y = 5.794gm
Now, we can write that % of CuS in the mixture = $\dfrac{{{\text{Mass of CuS}}}}{{{\text{Mass of mixture}}}} \times 100 = \dfrac{{5.794}}{{10}} \times 100 = 57.94\% $

Thus, we can conclude that % of CuS in the mixture is 58%.

Note:
Do not forget that as the number of electrons involved in particular oxidation and reduction reactions are different, we need to multiply the concentration of the compound by the number of electrons involved in order to find the mole equivalents.