A 10 g mixture of Cu2S and CuS was treated with 400 ml of 0.4 M - MnOą- in acid solution producing SO2, Cu2+ and Mn2+. The SO2 was boiled off and the excess of MnO4 was titrated with 200 ml of 1 M - Fe2+ solution. The percentage of CuS in original mixture is (Cu = 64)

Key Concepts to Master

1. Permanganate ($MnO_4^-$) in Acidic Medium

   • It is a strong oxidising agent:
     $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$
   • Accepts 5 electrons per mole when it gets reduced to $Mn^{2+}$.

2. Oxidation of Sulphide‐containing Compounds

   • Sulphur in $S^{2-}$ goes to $+4$ in $SO_2$.
     Change = $+6$ oxidation numbers ⇒ 6 electrons given away per S atom.
   • Extra twist for $Cu_2S$: Each $Cu^+$ becomes $Cu^{2+}$ → 1 electron lost per Cu (2 electrons per formula unit).

3. Back-Titration Logic

   • We add excess $MnO_4^-$. Whatever doesn’t get used is titrated by $Fe^{2+}$:
     $MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$
   • That gives a handle on how many moles of $MnO_4^-$ truly oxidised the sulphide mixture.

4. Stoichiometric Equations in Electron Terms

   • Let $x$ = moles of $CuS$
   • Let $y$ = moles of $Cu_2S$
   • Electrons released: $6x + 8y$
     (6 from each $CuS$, 6 + 2 from each $Cu_2S$)
   • $MnO_4^-$ moles used: $\frac{6x+8y}{5}$ (because 5 e⁻ per $MnO_4^-$)

5. Mass Balance

   • $96x + 160y = 10 \text{ g}$
     (because $M_{CuS}=64+32=96$ and $M_{Cu_2S}=2\times64+32=160$)

With these two simultaneous equations (electron balance & mass balance) you can solve for $x$ and $y$ and hence for the required percentage.

What’s Happening (Kid-Style!)

Imagine you have two kinds of copper-sulphur rocks in a jar. One rock is CuS (call it the single-copper rock) and the other is Cu₂S (the double-copper rock). You smash these rocks in a purple cleaning potion (permanganate, $MnO_4^-$) that loves to clean away electrons.

1. The potion eats up the sulphur, turning it into a smelly gas ($SO_2$).
2. Whatever potion is left over is measured by feeding it with iron juice ($Fe^{2+}$). Iron gives electrons to the left-over potion and tells us how much potion was unused.
3. By seeing how much potion was really used to clean the rocks, we can figure out how many of each rock were in the jar.

In the end we discover that 8 g out of the 10 g mixture was the single-copper rock CuS, so the mixture is 80 % CuS.

Step-by-Step Calculation

1. Moles of Permanganate Added
$V = 400\,\text{mL} = 0.400\,\text{L},\; M = 0.4\,\text{M}$
$n_{\text{added}}(MnO_4^-) = 0.400 \times 0.4 = 0.16\,\text{mol}$

2. Moles of Permanganate Left Unreacted (Back-Titrated)
Fe²⁺ titration data: $V = 200\,\text{mL}=0.200\,\text{L},\; M = 1.0\,\text{M}$
$n(Fe^{2+}) = 0.200 \times 1.0 = 0.20\,\text{mol}$ Reaction: $MnO_4^- + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+}$
Therefore $n_{\text{left}}(MnO_4^-) = \frac{0.20}{5} = 0.04\,\text{mol}$

3. Moles of Permanganate that Reacted with the Sulphide Mixture
$n_{\text{consumed}} = 0.16 - 0.04 = 0.12\,\text{mol}$

4. Electron Balance for the Mixture Let $x$ = moles of $CuS$
$y$ = moles of $Cu_2S$

Electrons produced: $6x\; (\text{from each }CuS) \quad \text{and} \quad 8y\; (6 \text{ from S plus }2 \text{ from two }Cu^+)$ Total electrons released: $6x + 8y$ Each $MnO_4^-$ accepts 5 e⁻, so \frac{6x + 8y}{5} = 0.12 \tag{1}

5. Mass Balance of the 10 g Sample 96x + 160y = 10 \tag{2}

6. Solve the Simultaneous Equations From (1): $6x + 8y = 0.60$ Divide by 2: 3x + 4y = 0.30 \tag{1'}

Express $x$ from (2): $x = \frac{10 - 160y}{96}$

Plug into (1'): $3\left(\frac{10 - 160y}{96}\right) + 4y = 0.30$ $\frac{30 - 480y}{96} + 4y = 0.30$ $0.3125 - 5y + 4y = 0.30$ $0.3125 - y = 0.30$ $y = 0.0125\,\text{mol}$

Now $x$: $x = \frac{10 - 160(0.0125)}{96} = \frac{10 - 2}{96} = \frac{8}{96} = 0.08333\,\text{mol}$

7. Mass of $CuS$ $m(CuS) = x \times 96 = 0.08333 \times 96 = 8.0\,\text{g}$

8. Percentage of $CuS$ in the Mixture $\%CuS = \frac{8.0}{10.0} \times 100 = 80\%$

[ \boxed{80,%} ]