# How do you find atomic radius given the density?

Jun 1, 2016

Here's one way to do it.

#### Explanation:

For a metal, you need the density, the molar mass, and the crystal structure.

Let's calculate the atomic radius of polonium, which has molar mass = 209 g/mol, density = ${\text{9.32 g/cm}}^{3}$, and exists in a simple cubic unit cell.

We see that there is 1 atom per unit cell ($\frac{1}{8} \text{atom}$ at each corner) and that the edge length of the cell ($a$) is twice the atomic radius ($r$).

The volume of one atom is

V_"atom" = 4/3πr^3.

The atom occupies the whole cube, so its effective volume is

${V}_{\text{eff}} = {a}^{3} = {\left(2 r\right)}^{3} = 8 {r}^{3}$

The packing efficiency is

V_"atom"/V_"eff" = (4/3πcolor(red)(cancel(color(black)(r^3))))/(8color(red)(cancel(color(black)(r^3)))) = (4/3π)/8 = π/6 = 0.5236

That is, the atom fills only 52.36 % of the unit cell.

Experimentally, the effective volume of an atom is

V_"eff" = ("1 cm"^3 color(white)(l)"Po")/(9.32 color(red)(cancel(color(black)("g Po")))) × (209 color(red)(cancel(color(black)("g Po"))))/(1 color(red)(cancel(color(black)("mol Po")))) × (1 color(red)(cancel(color(black)("mol Po"))))/(6.022 × 10^23color(white)(l) "atoms Po")

= 3.723 × 10^"-23"color(white)(l) "cm"^3"/atom Po"

But ${V}_{\text{atom"/V_"eff}} = 0.5236$

${V}_{\text{atom" = 0.5236V_"eff" = 0.5236 × 3.723 × 10^"-23"color(white)(l) "cm"^3 = 1.950 × 10^"-23"color(white)(l) "cm}}^{3}$

4/3πr^3 = 1.950 × 10^"-23"color(white)(l) "cm"^3

r^3 = 3/(4π) ×1.950 × 10^"-23"color(white)(l) "cm"^3 = 4.655 × 10^"-24"color(white)(l) "cm"^3

r = 1.67 × 10^"-8" color(white)(l)"cm" = 1.67 × 10^"-10"color(white)(l) "m" = 167 × 10^"-12"color(white)(l) "m" = "167 pm"

The radius of a polonium atom is 167 pm.