Question

A metal having a cubic structure has a density of 2.6g/cm³, an atomic weight of 87.62 g/mol, and a lattice parameter of 6.0849 A. One atom is associated with each lattice point.
Determine the crystal structure of the metal.

Answer 1

Answer:

Face Centered Cubic

Explanation:

In this question our strategy should be to determine the number of atoms in the unit cell since we know that the cubic structure can be:

primitive            = 1  atom/ unit cell  )

body centered  = 2 atoms/ unit cell )

face centered   = 4 atoms/ unit cell )

We know the density ( d= m/V ) ,  the volume can be computed and the molecular weight, so we have all the information required.

m unit cell  = # atoms in the unit cell x mass of 1 atom

Need to convert the lattice parameter, a,  in Angstroms  to cm for unit consitency with the density.

1 A = 10⁻⁸  cm

a = 6.0849  A x 1x 10⁻⁸ cm/A

a = 6.0849 A x  1.x 10⁻⁸ cm

⇒ V = a³ = ( 6.0849 x 10⁻⁸ cm)³ = 2.3 x 10⁻²² cm³

m  of the unit cell =  d x V  = 2.6 g/cm³ x  2.3 x 10⁻²² cm³ =  5.98 x 10⁻²² g

We need to convert the atomic weight per mol to atomic weight per atom:

87.62 g/ mol x (1 mol / 6.022 x ²³ atom ) = 1.45 x 10⁻²²  g/atom

Then this mass of 5.98 x 10⁻²² g must be equal to

5.98 x 10⁻²² g = # atoms/ unit cell x  1.45 x 10⁻²² g/atom

4.11  = # atoms/ unit cell

Rounding off to integer  # atoms/unit cell  = 4

Therefore the crystal structure of the cell is face centered cubic.