(Skip to the bottom for my questions). I'm just sounding off my thoughts above. 1. The problem statement, all variables and given/known data Find the density of aluminum, which crystallizes in a face-centered cubic unit cell. The atomic radius is 143 pm. 2. Relevant equations edge length of FCC is (4/sqrt2)r. 3. The attempt at a solution Okay. I need to express density in grams per centimeter cubed so here goes. 1) There are 4 atoms per face-centered cubic unit cell. Therefore there are 4 aluminum atoms. What is the mass of these atoms? The mass of one mole of Al is 26.98 grams. 26.98 grams divided by the number of atoms in a mole gives me the mass of each individual atom. 26.98 / (6.022 * 10^23) = 4.480 * 10^-23 grams per atom of Al. Multiply by 4 and I have the mass of one unit cell of Al (1.79 * 10^-22 grams per unit cell). 2) I have mass. Now I need to find the volume. 10^12 pm = 1 m. Therefore 10^10 pm = 1 cm. I can cube both sides and now I have 10^30 pm^3 = 1 cm^3. There we go. A measure of volume. 3) Volume of the unit cell. Using the formula given above, the edge length of a unit cell is 404.46 pm, or 4.04 * 10^-8 cm. Therefore the volume of a unit cell is simply the edge length (given above) cubed. Or 6.61 * 10^-23 cm^3. 4) We have grams and we have cm^3. We can divide them and get the (simplified) final result. (1.79 * 10^-22 grams per unit cell) / (6.61 * 10^-23 cm^3) = 2.705 g/cm^3. Questions: 1) I know my answer is correct. The second step was an intermediary step; I wasn't familiar with working with picometers. Is there, however, a simplified formula for the densities of the various unit cells? 2) How come dividing molar mass / moles = mass of one atom of the element? g / moles = molar mass right? (g/moles) / moles = g/moles^2 right? Intuitively, however, dividing molar mass by moles seems to work. The mass of Avagadro's number of atoms divided by Avagadro's number is ... the mass of one atom.