for fcc Z = 4.

d = $${{ZM} \over {{N_A}{a^3}}}$$

$$ \Rightarrow $$ M = $${{d{N_A}{a^3}} \over Z}$$

= $${{2.72 \times 6.023 \times {{10}^{23}} \times {{\left( {404 \times {{10}^{ - 10}}} \right)}^3}} \over 4}$$

= 27 g mol^–1

Total number of carbon atoms per unit cell

= $$8 \times {1 \over 8} + 6 \times {1 \over 2} + 4$$ = 8

Number of atoms in cubic close packing = 4 = O^2-

Number of tetrahedral voids = 2 × N = 2 × 4

Number of A²⁺ ions = 8 $$ \times $$ $${1 \over 4}$$ = 2

Number of octahedral voids = Number of B⁺ ions = N = 4

Ratio of ions will be,

O^2– : A²⁺ : B⁺ = 4 : 2 : 4 = 2 : 1 : 2

Formula of oxide = AB₂O₂

Number of octahedral voids is same as number of atoms.