Question

Aluminum has a face-centered cubic lattice with all atoms at lattice points. Its atomic mass is 26.982 amu. (4 r = 1(2)^1/2 ; Avogadro's # is 6.022 x 10^23)

a) If each aluminum atom has a radius of 143 pm, what is the edge length of the unit cell?

b) Calculate the density of aluminum metal.

Answer 1

(a) The edge length of the unit cell is 286 pm.

(b) The  the density of aluminum metal is 1.27 x 10⁻²³ g/cm³.

What is edge length of the aluminum atom?

The edge length of the aluminum atom is calculated as follows;

a = 2r

where;

• r is the radius of the atom
• a is edge length

a = 2 x 143 pm = 286 pm

Volume of the aluminum atom

V = a³

V = (286 x 10⁻¹²)³

V = 2.34 x 10⁻²⁹ m³

Density of the aluminum metal

ρ = ZM/VN

where;

• Z is 4 for face-centered cubic
• M is mass of aluminum atom (g/mol), 26.982 amu = (1.66 x 10⁻²⁴ x 26.982) = 4.479 x 10⁻²³ g/mol
• V is volume
• N is Avogadro's number

ρ = (4 x 4.479 x 10⁻²³) / ( 2.34 x 10⁻²⁹ x 6.023 x 10²³)

ρ = 1.27 x 10⁻¹⁷ g/m³

ρ = 1.27 x 10⁻¹⁷ g/m³  x (1 m³ / 10⁶ cm³)

ρ = 1.27 x 10⁻²³ g/cm³

Thus, the edge length of the unit cell is 286 pm and the  the density of aluminum metal is 1.27 x 10⁻²³ g/cm³.

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