Problem: The density of solid Fe is 7.87 g/cm3. How many atoms are present per cubic centimeter of Fe? As a solid, Fe adopts a body-centered cubic unit cell. How many unit cells are present per cubic centimeter of Fe? What is the volume of a unit cell of this metal? What is the edge length of a unit cell of Fe?

molar mass Fe = 55.845 g/mol

1 mole = 6.022x10²³ entities
entities = atoms, ions, molecules, formula units

density of solid Fe = 7.87 g/cm³

Calculate the atoms present per cubic centimeter of Fe

$\frac{\mathbf{atoms}}{{\mathbf{cm}}^{\mathbf{3}}}\mathbf{=}\frac{\mathbf{7}\mathbf{.}\mathbf{87}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{Fe}}}{{\mathbf{cm}}^{\mathbf{3}}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{Fe}}}{\mathbf{55}\mathbf{.}\mathbf{845}\mathbf{ }\cancel{\mathbf{g}\mathbf{ }\mathbf{Fe}}}\mathbf{\times }\frac{\mathbf{6}\mathbf{.}\mathbf{022}\mathbf{\times }{\mathbf{10}}^{\mathbf{23}}\mathbf{ }\mathbf{Fe}\mathbf{ }\mathbf{atoms}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }\mathbf{Fe}}}$

atoms/cm³ = 8.486x10²² Fe atoms/cm³

Calculate the number of unit cells are present per cubic centimeter of Fe

recall the # of atoms present per 1 BCC unit cell: corner atoms contribute 1/8 and the center atom contribute 1

$\mathbf{\#}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{atoms}\mathbf{=}\left(\mathbf{8}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{8}}\right)\mathbf{+}\mathbf{1}$

# of atoms = 2 per 1 unit cell