Question

An element with density 11.2 gm cm−3 forms fcc lattice with edge length \[4 \times {10^{ - 8}}cm\]. Calculate the atomic mass of the element. (Given \[{N_A} = 6.022 \times {10^{23}}mo{l^{ - 1}}\])

Answer 1

Hint: A unit cell refers to the smallest representation of the whole crystal. The cubic crystal system in crystallography refers to the crystal system in which the unit cell acquires the shape of a cube which is the most common or the simplest shape in case of minerals or crystals. The three main types of crystals include Primitive cubic (i.e. cP), Body-centered cubic (i.e. bcc) and Face-centered cubic (i.e. fcc).

Complete answer
The bcc (i.e. body-centered cubic) lattice possesses a coordination number of 8 and comprises 2 atoms per unit cell. On the other hand, the fcc (i.e. face-centered cubic) lattice possesses a coordination number of 12 and comprises 4 atoms per unit cell. The maximum packing density exists when the atoms possess a radius equal to one quarter of the diagonal of one of the faces of the unit cell. The information provided in the question is stated below:
Density, \[d = 11.2{\text{ }}g{\text{ }}c{m^{ - 3}}\]
Edge length, \[a = 4 \times {10^{ - 8}}cm\]
Avogadro number, \[{N_A} = 6.022 \times {10^{23}}mo{l^{ - 1}}\]
Number of atoms per unit cell, \[n\left( {fcc} \right) = 4\]
We have to find out the atomic mass i.e. M
Now with the aid of geometry, basic calculations and with certain attributes of the cubic structure we can derive the density formula of a unit cell.
$
  Density{\text{ }}of{\text{ }}unit{\text{ }}cell = \dfrac{{mass{\text{ }}of{\text{ }}unit{\text{ }}cell}}{{volume{\text{ }}of{\text{ }}unit{\text{ }}cell}} \\
  Mass{\text{ }}of{\text{ }}unit{\text{ }}cell = m \times n \\
 $
Where, m = mass of each atom, n = number of atoms
$
  m = \dfrac{{Molar{\text{ }}mass}}{{Avogadro{\text{ }}Number}} = \dfrac{M}{{{N_A}}} \\
  \therefore Mass{\text{ }}of{\text{ }}unit{\text{ }}cell = n \times \dfrac{M}{{{N_A}}} \\
 $
\[
  Volume{\text{ }}of{\text{ }}unit{\text{ }}cell = {a^3} \\
  \therefore {\text{Density }}of{\text{ }}unit{\text{ }}cell = \dfrac{{n \times M}}{{{a^3} \times {N_A}}} \\
 \]
From the density formula, we can calculate the value of M as stated below:
$
  M = \dfrac{{d \times {a^3} \times {N_A}}}{Z} \\
 M = \dfrac{{11.2 \times 64 \times {{10}^{ - 24}} \times 6.022 \times {{10}^{23}}}}{4} = 107.91g \\
$

Hence, the atomic mass of the element is 107.91 g.

Note:
FCC crystal structure exhibits more ductility (deforms more readily) in comparison to BCC crystal structure. Actually, bcc lattice, though cubic, is not closely packed and thus, forms strong metals. On the other hand, fcc lattice is both cubic as well as closely packed and thus, forms more ductile materials.