Question

Iron has a body-centred cubic unit cell with a cell dimension of $286.65\,{\text{pm}}$ . Density of iron is ${\text{7}}{\text{.87}}\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}$ . Use this information to calculate Avogadro’s number? (Atomic mass of iron${\text{56}}{\text{.0u}}$,)

Answer 1

Hint: Density depends upon the number of atoms present in a unit cell, the mass of the compound, and the length of a unit cell.

Formula used: $d\, = \dfrac{{z\,m}}{{{N_a}{a^3}}}$

Complete step by step answer:
The formula to calculate the density of cubic lattice is as follows:
$d\, = \dfrac{{z\,m}}{{{N_a}{a^3}}}$
Where,
d is the density.
z is the number of atoms in a unit cell.
m is the molar mass of the metal.
${N_a}$ is the Avogadro number.
a is the length of the unit cell.
The number of atoms present in a body-centred cubic unit cell is 2.
Convert the unit cell dimension in cm from pm as follows:
$1{\text{pm}}\,{\text{ = }}\,{\text{1}}{{\text{0}}^{ - 10}}\,{\text{cm}}$
$286.65\,{\text{pm}}\,\,{\text{ = }}\,286.65\, \times \,{10^{ - 10}}\,{\text{cm}}$
Substitute 2 for number of atoms, ${\text{7}}{\text{.87}}\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}$ for density of the crystal, $56.0\,{\text{u}}$ for molar mass of the iron metal, and $286.65\, \times \,{10^{ - 10}}\,{\text{cm}}$ for unit cell length.
${\text{7}}{\text{.87}}\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}}\, = \dfrac{{2 \times 56.0\,\,{\text{g/mol}}}}{{{N_a}\, \times {{\left( {286.65 \times {{10}^{ - 10}}\,{\text{cm}}} \right)}^3}}}$
${N_a}\,\, = \dfrac{{2 \times 56.0\,\,{\text{g/mol}}}}{{{\text{7}}{\text{.87}}\,{\text{g}}\,{\text{c}}{{\text{m}}^{ - 3}} \times {{\left( {286.65 \times {{10}^{ - 10}}\,{\text{cm}}} \right)}^3}}}$
${N_a}\,\, = \dfrac{{112\,{\text{mo}}{{\text{l}}^{ - 1}}}}{{1.85 \times {{10}^{ - 22}}}}$
${N_a}\,\, = 6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}$

Therefore, Avogadro's number is $6.02 \times {10^{23}}\,{\text{mo}}{{\text{l}}^{ - 1}}$.

Note: The value of the number of atoms depends upon the type of lattice. For body-centred cubic lattice, the number of atoms is two whereas four for face-centred and one for the simple cubic lattice. In the body-centred cubic lattice, eight atoms are present at the corner and one atom is present at the centre of the unit cell. Each atom of corner contribute $1/8$ to a unit cell and each atom of body centre contribute 1 to a unit cell so, the total number of atoms is,
\[ = \left( {\dfrac{1}{8} \times 8} \right)\, + \left( 1 \right)\] = 2