Question

Question: Iron(II) oxide is nonstoichiometric. A particular sample was found to contain 76.55% iron and 23.45% oxygen by mass.

(a) Calculate the empirical formula of the compound (four significant figures).

(b) What percentage of the iron in this sample is in the +3 oxidation state?

Short answer

Answer:

1. ${\text{Fe}}_{0.9352}\text{O}$
   
2. 13.9% of the iron

Step-by-step solution

Subpart (a) The empirical formula of the compound Iron(II) oxide.

The sample contains 76.55% iron and 23.45% oxygen by mass which means 100g of iron oxide contains 76.55g of iron and 23.45g of oxygen.

As$\text{moles}=\frac{\text{mass}}{\text{Mol.mass}}$

Therefore, the number of moles of iron present in 100 g of iron oxide is:

$\text{moles}=\frac{\text{76.55}}{\text{55.845}}=1.3707$

And the number of moles of oxygen present in 100 g of iron oxide is:

$\text{moles}=\frac{23.45}{16}=1.4656$

The ratio of the number of iron atoms to the number of oxygen atoms present in one formula unit of iron oxide is

$\frac{1.3707}{1.4656}=0.9352:1$

Hence, the empirical formula of iron oxide is${\text{Fe}}_{0.9352}\text{O}$ .

Subpart (b) The percentage of the iron in this sample in the +3-oxidation state.

From the total ions of iron;${\text{Fe}}_{0.9352}\text{O}$

Let, x be the number of ions of${\text{Fe}}^{3+}$ and so the remaining $0.9352-x$will be the ions of ${\text{Fe}}^{2+}$.

For neutral compounds;

$$\begin{gathered} \text{Total positive charge = Total negative charge} \\ \text{2(0.9352 - x) + 3x = 2} \end{gathered}$$

As the charge on negative oxygen is -2.

$$\begin{gathered} \text{1.87 - 2x + 3x = 2} \\ \text{1.87 + x = 2} \end{gathered}$$

Therefore,

$$\begin{gathered} \text{x = 2 - 1.87} \\ \text{x = 0.13} \end{gathered}$$

The percentage of ${\text{Fe}}^{3+}$is;

${\text{\% of Fe}}^{3+}=\frac{0.13}{0.9352}\times 100=13.9\%$