# Density of Crystal

Theoretical density of the crystal

If the length of edge of the unit cell is known we can calculate the density of the crystal as follows:

Let length of edge of the unit cell be l.

Volume of the unit cell = $l^3 = V cm^3$

Density of the unit cell = $\dfrac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$

Mass number of the unit cell = number of atoms in a unit cell × Mass of one atom

= n × mg

But mass of one atom(m) = $\dfrac{\text{atomic masses}}{\text{avogadro numbers}}$

$= \dfrac{M}{N_0}$

$\therefore \text{Density of the unit cell} = \dfrac{n \times M}{V \times N_0} g cm^{-3}$ $\rho$

Mass of the unit cell Number of atoms in a unit cell Mass of one atom

Example 1. An element (atomic mass = 60) having face-centred cubic crystal has a density of $6.23 g cm^{-3}$ . What is the edge length of the unit cell. (Avogadro constant,$N_0 = 6.02 \times 10^{23} mol^{-1}$).

Solution: since element has fcc structure hence there are 4 atoms in a unit cell (or n = 4). Atomic mass is 60 (or M = 60) $N_0 = 6.02 \times 10^{23} \text{and} \rho = 6.23 g cm^{-3}$

$\therefore \rho = \dfrac{n \times M}{V \times N_0}$

$or V = \dfrac{n \times M}{\rho N_0}$

$= \dfrac{(4)(60 g mol^{-1})}{6.23 g cm^{-3})(6.02 \times 10^{23} mol^{-1})}$

$= \dfrac{240 \times 10^{-23} cm^{3}}{37.5046}$

$6.4 \times 10^{-23} cm^3 = 6.4 \times 10^{-24} cm^3$

Let ‘l’ be the length of the edge of the unit cell.

$\therefore l^3 = V = 64 \times 10^{-24} cm^3 \\ \text{or} l = 4.0 \times 10^{-8} cm$

Example 2. The density of KBr is $2.75 g / cm^3$ . The length of the edge of the unit cell is 654 pm. Show that KBr has a face centred cubic structure.

Solution,

Length of the edge of the unit cell = 654 pm = $6.54 \times 10^{-8}cm$

$\therefore$ Volume (V) of the unit cell = $(6.54 \times 10^{-8}cm)^3$

Molecular mass of KBr = 39 + 80 = $119 g mol^{-1}$

Density of KBr = $2.75 g cm^{-3}$

$\because \rho = \dfrac{n \times M}{V \times N_0}$

$\text{or} n = \dfrac{\rho \times V \times N_0}{M}$ [/latex]

$= \dfrac{(2.75 g cm^{-3})(6.54 \times 10^{-8} cm)^3 (6.02 \times 10^{23} mol^{-1})}{119 g mol^{-1}}$

$= \dfrac{2.75 \times (654)^3 (6.023)(10^{-1})}{119} \\[3mm] = 3.09 = 4$

Since number of atoms (or ions) in a unit cell is four hence the crystal must be face centred cubic.

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