Question

(4) strontium (sr) has an fcc crystal structure, an atomic radius of 0.215 nm and an atomic weight of 87.62 g/mol. calculate the theoretical density for sr.

Answer 1

2.59 g/cm^3 For a face centered cubic crystal, there is 1 atom at each corner that's shared between 8 unit cells. And since there's 8 corners, that gives 8*1/8 = 1 atom per unit cell. Additionally, there are 6 faces, each with 1 atom, that's shared between 2 cells. So 6*1/2 = 3. So each unit cell has a mass of 1 + 3 = 4 atoms. The size of the unit cell will be equal to either the diameter of one atom along the edge, or the diameter of 2 atoms as the diagonal across one face of the cube, whichever results in the larger unit cell. Taking that into consideration, I will use the value of 2 for the diagonal of a face of the unit cell, resulting in the length of an edge of the unit cell being sqrt(2^2/2) = sqrt(2) = 1.414213562 times the atomic diameter. So 1.414213562 * 2 * 0.215 nm = 0.608 nm So the volume of a single unit cell is (0.608 nm)^3 Avogadro's number of atoms will require 6.0221409x10^23 / 4 = 1.50554x10^23 unit cells and will have a mass of 87.62 grams. The volume will be 1.50554x10^23 * (0.608x10^-7 cm)^3 = 1.50554x10^23 * 0.224755712x10^-22 cm^3 = 33.83776414 cm^3 So the density is approximately 87.62 g/33.83776414 cm^3 = 2.589414585 g/cm^3, when rounded to 3 significant figures is 2.59 g/cm^3.