h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

My friends' chem prof didn't adequately explain how to find the density of a metal based on its atomic radius and packing structure. So we're given the atomic radius and the type of structure, like face-centered cubic, and they have to find the density. I can't remember how to do that, so does anyone know what's going on?
Thanks

Never mind, we figured it out.

Come back Gnomon,

I don't know and I'm nosy. Please tell me.

Well, let's see. The metal atoms can be considered to be hard spheres of the given radius. Let's call it r.

The face-centred cubic lattice means that the crystal consist of cubes with an atom at each corner, with the centre of the atom at the corner of the cube. In the centre of each face of the cube, there is another atom.

If we cut one cube out of the lattice, we'll get one eight of an atom inside the cube at each corner, making up one full atom, and one half of an atom in the centre of each of the six faces, making up another three atoms. So each cube has four atoms inside it: an average density of four atoms per cube. All we need now is to work out the size of the cube.

The size of the cube is as small as possible, so that the atom in the centre of each face touches each of the four atoms at the corners of the face. They are arranged like the five dots representing the five on a dice, but with the atoms closer together so that the central one touches the other four. If we call the side of the cube d, it is easy to work out that d = r time square root of 8. So given r, we can calculate d, then calculate d cubed. This is the volume of one cube and we know that there are 4 atoms per cube. So we have the density in atoms per cubic metre. If we want the density in kilograms per cubic metre, we just look up the weight of one atom of the metal and multiply it by this.

For body centred cubic, I reckon there are two atoms per cube, and the side of the cube is given by d = 4r / sqrt(3).

hcp metals require a bit of trig.

B

What's hcp?

hexagonal close packing

along with face centred cubic it's the densest possible form of packing for spheres.

I'd make a wild guess from your comments that it is the same density as the face centred packing, so the formula given above will work for it too, and no further trig is needed.

yeah, I don't think there's a difference between hexagonal close pack and face-centered cubic. The inorganic, solid state chemists sometimes denote them "ABA" and "ABC" for whether the relative stacking of the layers, but that doesn't change the density...

Indeed - if you were to try the above using hcp as your basis though you would as it doesn't have a cubic unit cell.

Upshot being, use fcc - the maths is easier

Ah, that ABA and ABC brings it all back to me. Strangely enough, one of the two hexagonal close packings does have a cubic structure as well, and it is face-centred. I might write an entry about that.

Not so strange - hold a cube so one diagonal is vertical and slice it across the middle horizonatally. Voila - the cross section is a regular hexagon.