Copper crystallizes in an FCC unit cell with cell edge of  $3.608\times {10}^{-8}\mathrm{cm}.$ The density of copper is  $\mathrm{8}.92 \mathrm{g}/{\mathrm{cm}}^{\mathrm{3}}$, Calculate the atomic mass of copper.

The number of effective atoms in a FCC unit cell, $\mathrm{Z}=4\hspace{0.17em}\mathrm{atoms}$

$\mathrm{M}=\mathrm{ }\frac{\mathrm{d}\times \mathrm{ }{\mathrm{N}}_0\times \mathrm{ }{\mathrm{a}}^3 }{\mathrm{Z}}$

Where,

$\mathrm{M}$ is atomic mass of copper

$\mathrm{d}$ is density of copper= $8\mathrm{ }.92 \mathrm{g}/\mathrm{c}{\mathrm{m}}^3,$

${\mathrm{N}}_0$ is the Avogadro number $=6.022\times \mathrm{ }{10}^{23} {\mathrm{mol}}^{-1}$
$\mathrm{a}$ is the edge length $=3.608\times {10}^{-8}\text{cm}$  

Substituting the values in the formula,

$\Rightarrow \mathrm{M}=\frac{8.92\mathrm{g}/{\mathrm{cm}}^{\mathrm{3}}\times 6.022\times {10}^{23}\mathrm{atom}/\mathrm{mol}\times {(3.608\times {10}^{-8}\mathrm{cm})}^3}{4\mathrm{\hspace{0.17em}}\mathrm{atoms}}$

 $\Rightarrow \mathrm{M}=63.1\mathrm{g}/\mathrm{mol}$

Atomic mass of copper $=63.1\hspace{0.17em}\mathrm{u}$