# Iron crystallizes in a face-centered cubic system. If the radius of an iron atom is 1.26 A (angstroms), what is the edgelength of the unit cell? What is the density of iron if its atomic weight is 55.847 g/mole?

Dec 19, 2015

Here's what I got.

#### Explanation:

In order to be able to calculate the edge length of the unit cell, you need to start from the characteristics of a face-centered cubic system.

As you know, a face-centered cubic system is characterized by a unit cell that has a total of $14$ lattice points

• one lattice point for every one of the eight corners of the unit cell
• one lattice point for every one of the six faces of the unit cell

Now, these lattice points will contain

• $\frac{1}{8} \text{th}$ of an atom in every corner lattice point
• $\frac{1}{2}$ of an atom in every face lattice point

To calculate the edge length of the unit cell, you need to focus on one side of the cube. This face will look like this

Let's say that $r$ represents the radius of an iron atom and $x$ represents the edge length of the unit cell. Notice that you can use Pythagoras' Theorem to express $x$ in terms of the diagonal of the cell, $d$

${d}^{2} = {x}^{2} + {x}^{2}$

${d}^{2} = 2 {x}^{2}$

The diagonal of the cell will be equal to one radius from the top-corner atom, two radii from the face atom, and one radius from the lower-corner atom

$d = r + 2 r + r = 4 r$

This means that you have

${\left(4 r\right)}^{2} = 2 {x}^{2}$

$16 \cdot {r}^{2} = 2 {x}^{2} \implies x = \sqrt{8 {r}^{2}}$

This is equivalent to

x = 2sqrt(2) * r = 2sqrt(2) * "1.26 A" = color(green)("3.56 A")

So, the edge length of the unit cell is equal to $\text{3.56 A}$.

As you know, the density of a substance is defined as mass per unit of volume. This means that your goal now will be to find

• the mass of a unit cell
• the volume of a unit cell

Start by using Avogadro's number to determine the mass of a single atom of iron

$55.847 \text{g"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/(6.022 * 10^(23)"atoms") = 9.274 * 10^(-23)"g/atom}$

Next, use the aforementioned lattice points to determine how many atoms of iron you get per unit cell

$\text{no. of atoms" = overbrace(1/8 xx 8)^(color(blue)("corner lattices")) + overbrace(1/2 xx 6)^(color(red)("face lattices")) = "4 atoms}$

This means that the mass of a unit cell will be

$9.274 \cdot {10}^{- 23} \text{g"/color(red)(cancel(color(black)("atom"))) * (4color(red)(cancel(color(black)("atoms"))))/"unit cell" = 3.710 * 10^(-22)"g/unit cell}$

Next, focus on finding the volume of a unit cell. As you know, the volume of a cube is given by the formula

$\textcolor{b l u e}{V = l \times l \times l = {l}^{3}} \text{ }$, where

$l$ - the edge length of the cube

Now, convert the edge length of the unit cell from Angstroms to meters first, then from meters to centimeters by using the conversion factor

$\text{1 A" = 10^(-1)"m}$

3.56 color(red)(cancel(color(black)("A"))) * (10^(-10)color(red)(cancel(color(black)("m"))))/(1color(red)(cancel(color(black)("A")))) * "1 cm"/(10^(-2)color(red)(cancel(color(black)("m")))) = 3.56 * 10^(-8)"cm"

The volume of the unit cell will thus be

$V = {\left(3.56 \cdot {10}^{- 8}\right)}^{3} {\text{cm}}^{3}$

$V = 4.512 \cdot {10}^{- 23} {\text{cm}}^{3}$

Finally, the density of iron will be equal to

$\textcolor{b l u e}{\rho = \frac{m}{V}}$

$\rho = \left(3.710 \cdot {10}^{- 22} {\text{g")/(4.512 * 10^(-23)"cm"^3) = color(green)("8.22 g/cm}}^{3}\right)$

The answers are rounded to three sig figs, the number of sig figs you have for the radius of an iron atom.