Question

A fcc element (atomic mass = 60) has a cell edge of 400pm. What is the density?

Answer 1

Hint: Calculate the number of atoms in face-centred cubic structure and use of the formula for calculating density in terms of molecular mass and Avogadro’s number

Complete answer:
The number of atoms in a face-centred cubic structure is calculated as:
Number of corners in each cube = g
But each corner atom is shared with 8 other cubes, so its contribution to one cube is \[{\left( {\dfrac{1}{8}} \right)^{th}}\].
Number of faces in a cube = 6
Each atom on a face-centre is shared by two cubes, therefore, the contribution of each face centred atom is \[\dfrac{1}{2}\].
Hence, total number of atoms in a cube
= Contribution of corner atoms + Contribution of face - centred atoms
\[= 8 \times \dfrac{1}{8} + 6 \times \dfrac{1}{2} = \;4\;atoms\]
Use the formula, \[ = \dfrac{{z.M}}{{{N_A}.\;{a^3}}}\]; where d is the density, z is the number of atoms per unit cell, M is the molecular or atomic mass in g/mol, \[{N_A}\] is the Avogadro’s number and ‘a’ is the edge length.
\[d = \dfrac{{4 \times 60 \times \dfrac{g}{{mol}}}}{{6.022 \times {{10}^{23}} \times \dfrac{1}{{mol}} \times {{\left( {400\;pm} \right)}^3} \times \dfrac{{1c{m^3}}}{{{{\left( {{{10}^{10}}pm} \right)}^3}}}}}\]
\[ = > d = \dfrac{{4 \times 60}}{{6.022 \times {{10}^{23}} \times {{400}^3} \times {{10}^{ - 30}}}}.\dfrac{g}{{c{m^3}}}\]
\[ = > \;d = \;6.22\;g\;c{m^{ - 3}}\; \sim 6gc{m^{ - 3}}\]

Therefore the density of the unit cell is 6 g$cm^{-3}$.

Note: The number of atoms per unit cell, \[z\], must be taken into account to calculate the total mass of atoms per unit cell. The mass of each atom is Molar mass divided by Avogadro’s number. The units must be converted to a single system (for example C.G.S) using appropriate conversion factors as shown above. The answer must be reported in significant digits equal to the quantity containing the least amount of significant digits in the question.
\[Density = \dfrac{Z}{{{N_0}}} \times \dfrac{M}{V}\]
Z = 4, V = \[{(4 \times {10^{ - 8}})^3}c{m^3}\]
\[(1pm = {10^{ - 12}}m = {10^{ - 10}}cm)\]
Substituting in the formula we get,
\[Density = \dfrac{{4 \times 60}}{{6.023 \times {{10}^{23}} \times 64 \times {{10}^{ - 24}}}} = 6.23gc{m^{ - 3}}\]