Question

A given sample of wüstite has the formula Fe_0.950O_1.00. Calculate the fraction of iron ions present as Fe³⁺. What fraction of the sites normally occupied by Fe²⁺must be vacant in this solid?

Short answer

The given sample of wüstite has the formula Fe_0.950O_1.00.

Step-by-step solution

Given Information

Sample of wüstite has the formula Fe_0.950O_1.00

Fraction of iron ion present

In pure iron oxide (FeO), iron and oxygen are present in a ratio of 1:1.

The number of${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$

Or

Number of Fe²⁺ ions missing =0.07

Each Fe²⁺ ion has 2 positive charges and the total number of charges due to missing (0.07).

$$\begin{gathered} {\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{=}\mathbf{2}\mathbf{\times }\mathbf{0}\mathbf{.}\mathbf{07} \\ {\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{14} \end{gathered}$$

In order to maintain electrical neutrality, a 0.14 positive charge is compensated by the presence of Fe³⁺ions.

Therefore, the replacement of one ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$ ion by one Fe³⁺ increases one positive charge, 0.14 positive charge must be compensated by the presence of 0.14 ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$ions.

In short, 0.93 Fe ions have 0.14 ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$ions

So, in 100 Fe ions have

Fraction of ${\mathbf{Fe}}^{\mathbf{2}\mathbf{+}}$= 0.14/0.93

Fraction of ${\mathbf{Fe}}^{\mathbf{3}\mathbf{+}}$ = 0.79/0.93

Step 3: The given solid

If all ${\mathrm{Fe}}^{2+}$, then 1.000${\mathrm{Fe}}^{2+}$ion/ ${\mathrm{O}}^{2-}$ions,

1.000-9.050=0.050

Vacant sites.

5% of the ${\mathrm{Fe}}^{2+}$ion sites are vacant.