Question

Aluminum crytallizes with a face-centered cubic cell.

a) What is the net number of Al atoms inside the unit cell?

b) If the edge length of the unit cell is 4.045x10^-10 m, what is the density of Al in grams per cm^3?

Answer 1

First of all, you need to know how many atoms are there in that unit cell. It was stated that the unit cell is a face-centered cubic which contains 4 atoms/unit cell.


To start, convert the cell edge from pm to cm. 1 m = 10^-12 m

404 pm x ( 10^-12 m/ 1pm)( 100 cm/1m) = 4.04 x 10 ^ -8 cm


Now, get the volume of the cube. Since it is a cube, the length of all the edges are the same.

V of cube = L x W x H = (4.04 x 10 ^ -8 cm ) ^3
V of cube = 6.59 x 10 ^ -23 cu. cm per unit cell


Take the reciprocal of the answer to get the amount of unit cell per cubic cm.

Unit cells = 1.52 x 10 ^ 22 unit cells per cu. cm.


Calculate now the number of atoms in that unit cells. As stated earlier, 1 unit cell contains 4 atoms of aluminum.

number of atoms per cu. cm = (1.52 x 10 ^ 22)(4) = 6.07 x 10 ^ 22 atoms per cu.cm.


Convert that atoms into moles by dividing the answer with Avogadro's Number.

moles of Al = 6.07 x 10 ^ 22 / 6.022 x 10 ^ 23

moles of Al = 0.1008 moles/cm3


Lastly, multiply the number of moles with atomic mass of Al to get the density.

Density = (0.1008 moles/cm3) x ( 26.98 g/mol)

Density = 2.72 g cm^3