Question d4817

Apr 10, 2015

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 $\text{Na}$ atom

${V}_{\text{Na" = 1 cancel("Na atom") × (1 cancel("mol Na atoms"))/(6.022 × 10^23 cancel("Na atoms")) × (22.99 cancel("g Na"))/(1 cancel("mol Na atoms")) × ("1 cm"^3)/(0.968 cancel("g Na")) = 3.944 × 10^-23"cm}}^{3}$

Step 3. Calculate the number of atoms per unit cell

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