You need to know the density of the metal and the edge length of the unit cell.

**EXAMPLE:**

Sodium has a density of 0.968 g/cm³ and a unit cell edge length #a# of 429 pm. How many sodium atoms are there per unit cell?

**Solution:**

**Step 1.** Calculate the volume of the unit cell.

The volume of a cube is given by

#V = a^3#, where #a# is the edge length of the cube.

#V_"cell" = a^3 = (429 cancel("pm"))^3 × ((1 cancel("m"))/(10^12 cancel("pm")))^3 × ("100 cm"/(1 cancel("m")))
^3 =#

#7.895 × 10^-23 "cm"^3#

**Step 2.** Determine the volume of an #"Na"# atom

#V_"Na" = 1 cancel("Na atom") × (1 cancel("mol Na atoms"))/(6.022 × 10^23 cancel("Na atoms")) × (22.99 cancel("g Na"))/(1 cancel("mol Na atoms")) × ("1 cm"^3)/(0.968 cancel("g Na")) = 3.944 × 10^-23"cm"^3#

**Step 3.** Calculate the number of atoms per unit cell

#"No. of atoms" = (7.895 × 10^-23 cancel("cm³"))/"1 unit cell" × "1 Na atom"/(3.944 × 10^-23 cancel("cm³")) ="2.00 Na atoms/unit cell" ≈"2 Na atoms/unit cell"#