Ferrous oxide (FeO) crystal has a cubic structure and each edge of the unit cell is 5.0 $\stackrel{\mathrm{o}}{\mathrm{A}}$. Taking density of the oxide as 4.0 g cm^-3. The number of FeO formula units present in each unit cell is: 

$\text{ Volume of the unit cell, }{a}^3={\left(5\times {10}^{-8}\right)}^3$

$=1.25\times {10}^{-22}{\mathrm{cm}}^3(\because \mathrm{a}=5\text{Å ) }$

$\text{ Density }\left(\mathrm{\rho }\right)=4.0\mathrm{g}{\mathrm{cm}}^{-3}$

$\text{ Mass of unit cell }=\mathrm{V}\times \mathrm{\rho }=1.25\times {10}^{-22}\times 4.0$

$=5.0\times {10}^{-22}\mathrm{g}$

$\text{ Mass of one molecule of }\mathrm{FeO}=\frac{\text{ Mol. mass (in grams) }}{\text{ Avogadro's number }}$

$=\frac{72}{6.022\times {10}^{23}}=1.195\times {10}^{-22}\mathrm{g}$

$\text{ Number of FeO molecules per unit cell }=\frac{5.0\times {10}^{-22}\mathrm{g}}{1.195\times {10}^{-22}\mathrm{g}}$

$=4.19\approx 4$

Thus, there are four FeO formula units in each unit cell.