By knowing the density of an unknown metal and the dimension of its unit cell, the atomic mass of the metal can be determined.

Let ‘a’ be the edge length of a unit cell of a crystal, ‘d’ be the density of the metal, ‘m’ be the atomic mass of the metal and ‘z’ be the number of atoms in the unit cell.

$\mathrm{Now}, \mathrm{density} \mathrm{of} \mathrm{the} \mathrm{unit} \mathrm{cell}=\frac{\mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{unit} \mathrm{cell}}{\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{unit} \mathrm{cell}}$
$\Rightarrow \mathrm{d}=\frac{\mathrm{zm}}{{\mathrm{a}}^3} \left(\mathrm{i}\right)$

[Since mass of the unit cell = Number of atoms in the unit cell × Atomic mass]

[Volume of the unit cell = (Edge length of the cubic unit cell)³]

From equation (i), we have:

$\mathrm{m}=\frac{{\mathrm{da}}^3}{\mathrm{z}} \left(\mathrm{ii}\right)$

$\mathrm{Now}, \mathrm{mass} \mathrm{of} \mathrm{the} \mathrm{metal} \left(\mathrm{m}\right)=\frac{\mathrm{Atomic} \mathrm{mass} \left(\mathrm{M}\right)}{\mathrm{Avogadro}\text{'}\mathrm{s} \mathrm{number} \left({\mathrm{N}}_{\mathrm{A}}\right)}$
$\mathrm{M}=\frac{{\mathrm{da}}^3{\mathrm{N}}_{\mathrm{A}}}{\mathrm{Z}} \left(\mathrm{iii}\right)$

Therefore,

If the edge lengths are different (say a, b and c), then equation (ii) becomes:

$\mathrm{m}=\frac{\mathrm{d} \left(\mathrm{abc}\right) {\mathrm{N}}_{\mathrm{A}}}{\mathrm{z}} \left(\mathrm{iv}\right)$

From equations (iii) and (iv), we can determine the atomic mass of the unknown metal.