Can someone help me solve this problem?

Question

Clausthalite is a mineral composed of (PbSe). The mineral adopts a NaCl type structure. The density of PbSe at 25C is 8.27g/cm3. Calculate the length of an edge of the PbSe unit cell.

Answer 1

NaCl (and PbSe) crystallize in a face-centered cubic structure. In this structure, there are 4 atoms of Na (or Pb) and 4 atoms of Cl (or Se) in each unit cell.

The atomic masses of Pb and Se are 207.2 gm/mol and 78.96 gm/mol, respectively. The mass of the atoms in each unit cell is therefore:

m = [(207.2 + 78.96)*4 gm/mol]/[6.022*10^23 mol^-1)
m = 1.90*10^-21 gm

The density is the mass divided by the volume, in this case, the volume of a unit cell:

8.27 gm/cm^3 = 1.90*10^-21 gm/V

V = 2.30*10^-22 cm^3

The NaCl structure has cubic symmetry, which means the unit cell is a cube. The volume of a cube is the length of a side raised to the third power, so in this case, the length of the side of the unit cell of PbSe is:

a = V^(1/3) = (2.30*10^-22 cm^3)^(1/3)
a = 6.12*10^-10 meters = 6.12 Angstroms

See source for more detail.

Answer 2

I'll take a stab at it. NaCl forms a cubic unit cell. Look up the density of NaCl and you can probably find the length of an edge of a NaCl unit cell.

Now compare the densities of NaCl and PbSe. If this makes sense, the comparative length of the edges would be inversely proportional to the cube root of the density ratios.