Question

A metal crystallizes in the HCP structure. It has a density of 12.87 g/cm3 and a lattice constant of
a = 0.564 nm. The height of the unit cell can be found from the following relationship, c = 1.633a.
What is the atomic mass of this element?

Answer 1

The formula we are using here is given as:

d = (n * M) / (V * NA)

a is the lattice parameter = 0.564 nm = 0.564 * 10-7 cm

Given, c = 1.633 * a

Where, d is the density = 12.87 g/cm3

n is number of atoms in HCP = 6

M is atomic mass of the material = unknown

V is the volume = (3 * (3)1/2 * a2 * c) / 2  

V = (3 * (3)1/2 * a2 * 1.633 * a) / 2  

V = (3 * (3)1/2 * a3 * 1.633) / 2  

V = 4.18 * (0.564 * 10-7 cm)3

V = 7.5 * 10-22 cm3

NA is the Avogadro number = 6.02 * 1023 atom/mol

Putting all values in the formula

d = (n * M) / (V * NA)

12.87 g/cm3 = (6 * M) / (7.5 * 10-22 cm3 * 6.02 * 1023 atom/mol)

12.87 g/cm3 = (6 * M) / (7.5 * 10-22 cm3 * 6.02 * 1023 atom/mol)

12.87 g/cm3 = (6 * M) / (451.5 cm3 .atom/mol)

(6 * M) = 12.87 g/cm3 * 451.5 cm3 .atom/mol

M = 968.4 g/mol