Solid State
6.0 Calculation of density of a cubic crystal from its edge
6.0 Calculation of density of a cubic crystal from its edge
If we know the type of crystal structure possessed by the cubic crystal, so that the number of particles per unit cell are known and the edge length for it is known by X-Ray studies, the density of the crystal can be determined.
Case $I$: For cubic crystals of elements
Let the edge of the unit cell $= a\ pm$
No. of atoms present per unit cell $= Z$
Atomic mass of the element $= M$
Volume of the unit cell is $$ = {(a\;pm)^3} = {a^3}p{m^3} = {a^3} \times {10^{ - 30}}c{m^3}$$
Density of the unit cell is $$\begin{equation} \begin{aligned} \rho = \frac{{{\text{mass of the unit cell}}}}{{{\text{volume of the unit cell}}}} \\ \rho = \frac{{\left( {{\text{No}}{\text{. of atoms in its unit cell}}} \right) \times \left( {{\text{Mass of each atom}}} \right)}}{{{a^3} \times {{10}^{ - 30}}}} \\ \rho = \frac{{Z \times \frac{M}{{{N_a}}}}}{{{a^3} \times {{10}^{ - 30}}}}g/c{m^3} \\ \rho = \frac{{Z \times M}}{{{N_a} \times {a^3} \times {{10}^{ - 30}}}}g/c{m^3} \\\end{aligned} \end{equation} $$
Case $II$: For cubic crystals of ionic compounds
Here, $Z$ is the number of formula units present in one unit cell and $M$ is the formula mass. The formula will remain the same viz.$$\rho = \frac{{Z \times M}}{{{N_a} \times {{10}^{ - 30}}}}g/c{m^3}$$
