Iron (II) oxide has a cubic structure and each unit cell has side 5A. If the density of the oxide is 4 g cm^–3, calculate the number of Fe²⁺ and O^2– ions present in each unit cell. (Molar mass of FeO = 72 g mol^–1, N_A = 6.02 x 10²³ mol^–1).

Solution:
We  have given
volume of the unit cell = (5 x 10^-8cm)³ = 1.25 x 10^-22cm³

Density of FeO = 4g cm^-3

Density,

$$\begin{gathered} \mathrm{\rho } = \frac{\mathrm{z}\times \mathrm{M}}{{\mathrm{a}}^3\times {\mathrm{N}}_{\mathrm{A}}} \\ 4= \frac{\mathrm{z}\times 72}{(5\times {10}^{-8}{)}^3\times 6.02\times {10}^{23}} \\ \mathrm{z}= \frac{4\times 1.25\times {10}^{-22} \times 6.02\times {10}^{23}}{72} = 4.18\cong 4 \end{gathered}$$

Each unit cell has four units of FeO. So it has four Fe²⁺ and four O^2– ions.