Question

Aluminum metal crystallizes in a face-centered cubic unit cell. If the atomic radius of an Al is 1.43Angstroms, what is the density of aluminum (in g/cm^3)

Answer 1

Convert Angstroms to cm:

1.43 Angstroms x (1 cm/8 pm) = 1.43 x 10¯⁸ cm

Determine the edge length of the unit cell:

Use the Pythagorean Theorem

r = d ÷ 2(√2)

1.43 x 10¯⁸ cm = d ÷ 2(√2)

d = 4.044 x 10¯⁸ cm

Determine the volume of the unit cell:

(4.044 x 10¯⁸ cm)³ = 6.616 x 10¯²³ cm³

Determine mass of 4 atoms of Al in a unit cell

26.98 g/mol divided by 6.022 x 10²³ atoms/mol = 4.479 x 10¯²³ g/atom

4.479 x 10¯²³ g/atom times 4 atoms = 1.791 x 10¯²² g

Determine density:

1.791 x 10¯²² g divided by 6.616 x 10¯²³ cm³ = 2.708 g/cm³