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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