Question

A mixture of $\mathrm{FeO}$ and ${\mathrm{Fe}}_3{\mathrm{O}}_4$ when heated in air to a constant weight gains 55 in its mass. Find the composition of the initial mixture.

Answer 1

Step 1: Assuming the amount of mixture:

• Let the percent of $\mathrm{FeO}$ be x.
• Let the percent of ${\mathrm{Fe}}_3{\mathrm{O}}_4$ be (100-x)

Step 2: The conversion of $\mathrm{FeO}$ into ${\mathrm{Fe}}_2{\mathrm{O}}_3$ is as follows:

$4\mathrm{FeO}\left(\mathrm{s}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)\to 2{\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$

From the equation, 288 g of $\mathrm{FeO}$ yields 320 g of ${\mathrm{Fe}}_2{\mathrm{O}}_3$

Thus for x g of $\mathrm{FeO}$, the amount of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ will be as follows:

$\frac{320}{288}\times \mathrm{x}\mathrm{g}\mathrm{of}{\mathrm{Fe}}_2{\mathrm{O}}_3$-----------(i)

Step 3: The conversion of ${\mathrm{Fe}}_3{\mathrm{O}}_4$ into ${\mathrm{Fe}}_2{\mathrm{O}}_3$ is as follows:

$2{\mathrm{Fe}}_3{\mathrm{O}}_4\left(\mathrm{s}\right)+\frac{1}{2}{\mathrm{O}}_2\left(\mathrm{g}\right)\to 3{\mathrm{Fe}}_2{\mathrm{O}}_3\left(\mathrm{s}\right)$

From the equation, 446 g of ${\mathrm{Fe}}_3{\mathrm{O}}_4$ yields 480 g of ${\mathrm{Fe}}_2{\mathrm{O}}_3$

So, (100-x) g of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ will yield, $\frac{480}{464}\times (100-\mathrm{x})\mathrm{of}{\mathrm{Fe}}_2{\mathrm{O}}_3$--------------(ii)

Step 4: Calculation for percentages:

From equation (i) and (ii),

The total amount of ${\mathrm{Fe}}_2{\mathrm{O}}_3$ will be:

$\mathrm{Total}\mathrm{amount}\mathrm{of}{\mathrm{Fe}}_2{\mathrm{O}}_3=\frac{320}{288}\times \mathrm{x}\mathrm{g}+\frac{480}{464}\times (100-\mathrm{x})\mathrm{g}$

According to the question,

$$\begin{gathered} \frac{320}{288}\times \mathrm{x}+\frac{480}{464}\times (100-\mathrm{x})=105 \\ \mathrm{x}=20.2 \end{gathered}$$

Thus, the percent of $\mathrm{FeO}$ be 20.2 and the percent of ${\mathrm{Fe}}_3{\mathrm{O}}_4$ be 79.8.