How do you find atomic radius given the density?

1 Answer
Jun 1, 2016

Answer:

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 = #"9.32 g/cm"^3#, and exists in a simple cubic unit cell.

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We see that there is 1 atom per unit cell (#1/8 "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_"eff" = a^3 = (2r)^3 = 8r^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_"atom"/V_"eff" = 0.5236#

#V_"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.