Problem: Chromium crystallizes with a body-centered cubic unit cell. The radius of a chromium atom is 125 pm. Calculate the density of solid crystalline chromium in grams per cubic centimeter.

A body-centered cubic (BCC) unit cell is composed of a cube with one atom at each of its corners and one atom at the center of the cube. 

Recall that the edge length (a) of a BCC unit cell can be calculated using the equation:

$\boxed{\mathbf{a}\mathbf{=}\frac{\mathbf{4}\mathbf{r}}{\sqrt{\mathbf{3}}}}$

Step 1: Calculate the volume of 1 unit cell using density

• molar mass Cr = 51.99 g/mol
• 1 mole = 6.022x10²³ entities (Avogadro' number)
  entities = atoms, ions, molecules, formula units

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

• Solving volume of 1 unit cell:

$$\begin{gathered} \mathbf{volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}\mathbf{=}\mathbf{volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{cube} \\ \boxed{\mathbf{volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}\mathbf{=}{\mathbf{a}}^{\mathbf{3}}} \\ \mathbf{v}\mathbf{olume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{unit}\mathbf{ }\mathbf{cell}\mathbf{ }\mathbf{=}{\mathbf{(}\frac{\mathbf{4}}{\sqrt{\mathbf{3}}}\mathbf{\times }\mathbf{125}\mathbf{ }\cancel{\mathbf{pm}}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{ }\cancel{\mathbf{m}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{pm}}}\mathbf{)}}^{\mathbf{3}}\mathbf{\times }{\mathbf{\left(}\frac{\mathbf{1}\mathbf{ }\mathbf{cm}}{{\mathbf{10}}^{\mathbf{-}\mathbf{2}}\mathbf{ }\cancel{\mathbf{m}}}\mathbf{\right)}}^{\mathbf{3}} \end{gathered}$$

volume of unit cell = 2.406 x 10^-23 cm³