Question

The density of aluminum metal is 2.70g/mL, the atomic mass 26.98g/mol, the radius of an aluminum atom is 143 picometer (1 picometer = 1x10^-12m) and the packing density is 74% of theory. a) How many layers of aluminum atoms (not unit cells!) made up the sample of aluminum foil for which we provided the data. b) Briefly outline your strategy for solving this problem (what you computed and why/how, how you used this result to comput what parameter next).

Answer 1

Given that packing density = 74%

It refers FCC lattice.

Given that atomic radius r = 143 pm = 143 x 10^-10 cm

For FCC, edge length a = (8)^1/2 r =  (8)^1/2 x 143 x 10^-10 cm

atomic mass M = 26.98 g /mol

  no of atoms in FCC unit cell = ?

avagadro number = 6.023 x 10²³ mol^-1

density = [M x no of atoms in FCC unit cell] /[ a³ x avogadro number]

2.70g/cm³ = [26.98 g mol^-1 x no of atoms in FCC unit cell ] / [ ( (8)^1/2 x 143 x 10^-10 cm )³ x 6.023 x 10²³ mol^-1]

Then,

   no of atoms in FCC unit cell = 4