Aluminum metal crystallizes in a face centered cubic lattice. If each Al atom has a radius of 1.43 A what is the density of aluminum metal?
I'm not sure how to go about this. I know that eventually I will use V=l^3, but I'm not sure how to get there. I also know that there are 4 atoms in a face-centered unit cell, but that's it.
2 answers
I think that the crystal lattice has little to do with it at this point. By saying that the Al atom has a volume, they just want you to divide the mass of an atom by its volume. Without more information on the lattice spacings in 3D, it's hard to say how much space an atom truly occupies.
Yes, the crystal lattice can be used to calculate the density. The theory is that the density of a unit cell is the same as the density of the macroscopic unit.
So the fcc has 4 atoms/unit cell. That's right. The mass of a unit cell then is
4 * 26.98/6.02E23 = mass unit cell.
For a fcc, a(2)^1/2 = 4r.
You know the radius is 1.43 A, I would convert that to cm (1.43E-8 cm), substitute and solve for a which is the length of the unit cell in cm. Cube that to find the volume of the unit cell in cc.
Then density = mass unit cell/volume unit cell. I get about 2.70 g/mL.
So the fcc has 4 atoms/unit cell. That's right. The mass of a unit cell then is
4 * 26.98/6.02E23 = mass unit cell.
For a fcc, a(2)^1/2 = 4r.
You know the radius is 1.43 A, I would convert that to cm (1.43E-8 cm), substitute and solve for a which is the length of the unit cell in cm. Cube that to find the volume of the unit cell in cc.
Then density = mass unit cell/volume unit cell. I get about 2.70 g/mL.
3 answers