Finding composition of mixture of Fe3O4 and Fe2O3

[Titan97]

Homework Statement

An $\mathrm{8~g}$ sample of ${Fe_3O_4}$ and ${Fe_2O_3}$ containing some inert impurity was treated with excess of $\mathrm{aq}$ ${KI}$ in acidic medium. which converted all iron to $Fe^{2+}$. The resulting solution was then diluted to $\mathrm{50~ml}$.

$\mathrm{10~ml}$ of it was taken and the liberated iodine required $\mathrm{7.2~ml}$ of $\mathrm{1M}$ sodium thiosulphate to reduce all iodine.

Another $\mathrm{25~ml}$ was taken and the iodine was removed. The remaining solution required $\mathrm{4.2~mL}$ of $\mathrm{1M}$ ${KMnO_4}$ to oxidize all $Fe^{2+}$.

Calculate the percentage of composition of the mixture.
The initial solution already contains $Fe^{2+}$ from $FeO$ whic does not react until the second titration.

Homework Equations

equivalence concept

The Attempt at a Solution

[/B]
Working backwards, milli equivalence of ${Fe^{2+}}$ is $\mathrm{5\times 4.2\times 1=21}$ in $\mathrm{25~mL}$ solution. So in $\mathrm{50~mL}$, the amount of ${Fe^2+}$ is $\mathrm{42}$ m.eq.

Similarly, the m.eq of iodine from titration with thiosulphate is $\mathrm{36}$

From the first reaction (with iodide), $$\mathrm{m.eq~of~iodine~liberated=m.eq~of~Fe^{2+}~formed}$$
The excess ${Fe^{2+}}$ is from ${FeO}$ which is present in ${Fe_3O_4}$

amount of ${FeO}$ is $\mathrm{6~m.eq}$.
Hence its weight is $\frac{6}{1000}\times\frac{72}{2}=\mathrm{0.216~g}$ since n-factor is $\mathrm{2}$.

Since $\mathrm{1~mol}$ of ${Fe_3O_4}$ conatains $\mathrm{72~g}$ of ${FeO}$, the amount of ${Fe_3O_4}$ is $\mathrm{0.6945~g}$. But this answer is wrong. Is there anything wrong in my procedure?

 

[Borek]

I haven't checked other things, as this caught my attention first:

From the first reaction (with iodide), $$\mathrm{m.eq~of~iodine~liberated=m.eq~of~Fe^{2+}~formed}$$

Are you sure you are not tricking yourself with equivalents?

 

[Titan97]

During the calculation, I got different values for iodine and fe2+. This is because a part of the fe2+ is from FeO which was already present. When KI is added, the meq of iodine formed will be equal to the meq of Fe3+ formed=meq of Fe2+ reacted.

 

[Borek]

I strongly suggest you redo the calculations using moles.

 

[Titan97]

OK. But before that, can you help with the reaction of Fe3Or? Should I take fe3o4 as one mole of FeO and one mole of Fe2O3.

 

[Borek]

Should I take fe3o4 as one mole of FeO and one mole of Fe2O3.

Yes, it is a mixed oxide.