In an oxide, $O X^{2 -}$ ions form a ccp lattice of cell edge $408 pm$ . Atoms A (molar mass 24) occupy one-eighth of the tetrahedral voids and atoms B (molar mass 27) occupy half of the octahedral voids. With $O = 16$ and $N_{A} = 6.022 \times 1 0^{23} mo l^{- 1}$ , find the density of the oxide in $g c m^{- 3}$ (two decimals), treating the ccp cell as the unit cell.

Numerical answer type, answer in g/cm^3, tolerance 0.02.

Answer: 3.47 g/cm^3

Per cell: O

= 4

, tetrahedral voids

= 8

, octahedral voids

= 4

. A

= 8/8 = 1

and B

= 4/2 = 2

, so the cell contains

AB X_{2} O X_{4}

with mass

24 + 54 + 64 = 142 g mo l^{- 1}

ρ = \frac{142}{6.022 \times 1 0 ^{23} \times ( 4.08 \times 1 0 ^{- 8} ) ^{3}} = 3.47 g c m^{- 3}