The unit cell of copper corresponds to a face centered cube of edge length 3.596 Aᵒ with one copper atom at each lattice point. The calculated density of copper in kg/m³ is ___________. [Molar mass of Cu : 63.54 g; Avogadro Number = 6.022 × 10²³]

Solution

Chemistry ma’am · 4 steps

An fcc cell holds 4 atoms, so the density is the mass of 4 Cu atoms divided by a³, all in SI units.

1. Atoms per cell. In a face-centred cube, 8 corners give 8 × 1/8 = 1 atom and 6 face centres give 6 × 1/2 = 3 atoms. So Z = 4.

   ρ = (Z M)/(N_A a³)

2. Convert to SI. M = 63.54 × 10⁻³ kg mol⁻¹ and a = 3.596 × 10⁻¹⁰ m. Then a³ = 46.50 × 10⁻³⁰ = 4.650 × 10⁻²⁹ m³.
3. Put in numbers. Mass of one cell is (4 × 63.54 × 10⁻³)/(6.022 × 10²³) = 4.2205 × 10⁻²⁵ kg. Divide by the cell volume: ρ = (4.2205 × 10⁻²⁵)/(4.650 × 10⁻²⁹) = 9.076 × 10³ kg m⁻³.
4. So. The density of copper is about 9.08 × 10³ kg m⁻³. That matches the key of 9077 within rounding.

The common slip: 2269. Taking Z = 1 instead of 4 for the fcc cell gives about 2269, one quarter of the right density.

The idea to keep. For any cubic cell, density is ZM/(N_A a³). Get Z right first (sc 1, bcc 2, fcc 4). Then put every quantity in the units the answer asks for.

Answer from the NTA final answer key. Solution by PrepWiser.