Answer to Question #306033 in General Chemistry for kaltook

Question

Rhodium has an FCC crystal structure, an atomic radius of 0.1345 nm and atomic weight of 102.9 g/mol. Calculate the density of rhodium.

Answer 1

$\rho$ = (nA_v) / (V_cN_A)

n - atoms in unit cell,

A_v - atomic weight,

V_c - unit cell volume,

N_A - Avogadro's number

In a FCC lattice there are 8 atoms at eight corners and 6 at face centers. Now each corner contributes to eight cells so per unit cell contribution is 1/8×8=1 atom. Now similarly each face center contributes to two unit cells so contribution per unit cell by six face centers is equal to 1/2​×6=3 atoms. Hence total number of atoms per unit cell of FCC lattice is n = (1+3) = 4 atoms.

A_v = 102.9 g/mol

For FCC unit cell volume V_c = 16R³√2 = (16)(0.1345 × 10^-9 m)(√2) = 5.5 × 10^-29 m³ = 5.5 × 10^-23 cm³

N_A = 6.022 × 10²³

$\rho$ = [(4 atoms/unit cell)(102.9 g/mol)] / [(5.5 × 10^-23 cm³)(6.022 × 10²³ atoms/mol)] = 12.43 g/cm³