A8.8 Density of Pure metal
The density of a pure metal can be determined by using the measurements of the unit cell for that metal.
You need various pieces of information:
1. The mass of the atom (divide the relative atomic mass by the Avogadro constant).
2. The type of unit cell – and from this the number of atoms per unit cell.
3. The cell dimensions (see diagrams below for different options).

Worked example 🖊

Vanadium has a body centred structure. Its unit cell has dimensions of 303pm
$\textsf{(3.3}\times\textsf{10}{^{-10}}\textsf{m)}$
. Calculate the density of vanadium metal.
1. Ar of vanadium = 50.94 So 1 atom has a mass of
$\dfrac{\textsf{50.94}}{\textsf{6.02}\times\textsf{10}{^\textsf{23}}}\textsf{ = 8.46}\times\textsf{10}{^\textsf{-23}} \textsf{g}$
2. BCC structure has 2 atoms per unit cell So total mass of 1 unit cell
$\textsf{8.46}\times\textsf{10}{^\textsf{-23}} \times\textsf{ 2}\textsf{ = 1.692}\times\textsf{10}{^\textsf{-22}} \textsf{g}$
3. Volume of the unit cell =
$\textsf{(3.3}\times\textsf{10}{^\textsf{-10}}\textsf{)}^\textsf{3} \textsf{= 3.594}\times\textsf{10}^\textsf{-29}\textsf m^\textsf{-3}$
Density = mass/volume
$=\dfrac{\textsf{ 1.692}\times \textsf{10} {^\textsf{-22}}} {\textsf{ 3.594}\times \textsf{10}{^\textsf{-29}}} \textsf{ = 4.704}\times \textsf{ 10}{^\textsf{6}} \textsf{ g m} {^{-3}}$
Since there are
$\textsf{10}{^\textsf{6}} \ \textsf{cm}{^\textsf{3}}$
in
$\textsf{1m}{^{3}}$
$\textsf{gcm}{^\textsf{-3}}$
$\textsf {4.6 g cm}{^\textsf{-3}}$