We’re being asked to calculate the density of Fe that crystallizes in a body-centered cubic unit cell.

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 **density** has a formula of:

$\overline{){\mathbf{density}}{\mathbf{=}}\frac{\mathbf{mass}}{\mathbf{volume}}}$

**mass = gvolume = cm ^{3}**

We’re going to calculate for the density of Fe using the following steps:

**Step 1**:Calculate the edge length from the atomic radius**Step 2**:** **Calculate the volume of the 1 unit cell**Step 3**: Calculate the density of the Fe

**Step 1:**** **Calculate the edge length from the atomic radius

$\overline{){\mathbf{a}}{\mathbf{=}}\frac{\mathbf{4}\mathbf{r}}{\sqrt{\mathbf{3}}}}\phantom{\rule{0ex}{0ex}}\mathbf{a}\mathbf{=}\frac{\mathbf{4}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{124}\mathbf{}\overline{)\mathbf{nm}}\mathbf{)}}{\sqrt{\mathbf{3}}}\mathbf{\times}\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{}\mathbf{cm}}{\mathbf{1}\mathbf{}\overline{)\mathbf{nm}}}$

**a= 2.86x10 ^{-8} cm**

**Calculate the volume of the unit cell.**

**In a BCC unit cell:**

**Given:** **edge length (a) = 5.025 ****Å**

$\mathbf{volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{unit}\mathbf{}\mathbf{cell}\mathbf{=}\mathbf{volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{cube}\phantom{\rule{0ex}{0ex}}\overline{){\mathbf{volume}}{\mathbf{}}{\mathbf{of}}{\mathbf{}}{\mathbf{unit}}{\mathbf{}}{\mathbf{cell}}{\mathbf{=}}{{\mathbf{a}}}^{{\mathbf{3}}}}\phantom{\rule{0ex}{0ex}}\mathbf{volume}\mathbf{}\mathbf{of}\mathbf{}\mathbf{unit}\mathbf{}\mathbf{cell}\mathbf{}\mathbf{=}{(2.86\times {10}^{-8}\mathrm{cm})}^{\mathbf{3}}$

**volume of unit cell = 2.35 x 10 ^{-23} cm^{3}**

**Step 3: Calculate the density of the Fe**

Iron (Fe) has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. What is its theoretical density in g/cm^{3}?

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