2.4 Density and Unit Cell Dimensions
One can determine dimensions of a unit cell from its density, or vice-versa.
2.4.1 Simple Cubic
Given that the edge length of the unit cell for polonium (provided below) is 336 pm,
- Determine the radius (in pm) of a single polonium
- Determine the density (in g cm–3) of polonium
First, recognize that the unit cell given is a simple cubic. There are two adjacent atoms next to each other so the edge length (l) of this cubic is two times the radius (r) of the atoms.
\[l=2r\]
Therefore the radius of one atom is
\[r = \dfrac{1}{2}l = \dfrac{1}{2}\left ( 336~\mathrm{pm} \right ) = 168~\mathrm{pm}\]
The density of the substance can be found by finding the mass of the unit cell and dividing by its volume. The number of equivalent atoms in the unit cell is 1.
First, find the mass of the unit cell.
\[1~\mathrm{Po~unit~cell} \left ( \dfrac{1~\mathrm{Po~atom}}{1~\mathrm{Po~unit~cell}} \right ) \left ( \dfrac{1~\mathrm{mol~Po}}{6.022\times 10^{23}~\mathrm{Po~atoms}} \right ) \left ( \dfrac{208.998~\mathrm{g}}{\mathrm{mol~Po}} \right ) = 3.47\times 10^{-22}~\mathrm{g} \]
Next, find the volume of the unit cell.
\[V = (2r)^3 = l^3 = \left ( 366\times 10^{-10}~\mathrm{cm} \right )^3 = 3.79 \times 10^{-23}~\mathrm{cm^3}\]
Finally, calculate the density of Po.
\[d\mathrm{(Po)} = \dfrac{3.47\times 10^{-22}~\mathrm{g}}{3.79\times 10^{-23}~\mathrm{cm}^3} = 9.16~\mathrm{g~cm^{-3}}\]
2.4.2 Face Centered Cubic
Calcium adopts a face-centered cubic as a solid. The edge length of the unit cell is 558.8 pm.
- What is the atomic radius of Ca (in pm)?
- What is the density (in g cm–3)?
From the image, there is a right-angled triangle with sides, a, and hypotenuse, 4r, where r is the radius of a calcium atom. Therefore, we can solve for the radius of an atom by solving the right-angled triangle equation.
\[\begin{align*} \mathrm{base^2} + \mathrm{height^2} &= \mathrm{hypotenuse^2} \\[1.5ex] a^2 + a^2 &= (4r)^2 \\[1.5ex] \left (558.8~\mathrm{pm} \right )^2 + \left (558.8~\mathrm{pm} \right )^2 &= (4r)^2 \\[1.5ex] r &= \sqrt{\dfrac{(558.8~\mathrm{pm})^2 + (558.8~\mathrm{pm})^2}{16}} = 197.6~\mathrm{pm} \end{align*}\]
Find the mass of the unit cell. There are 4 equivalent atoms in the unit cell.
\[ \mathrm{1~\mathrm{Ca~unit~cell}} \left ( \dfrac{\mathrm{4~Ca~atoms}}{1~\mathrm{Ca~unit~cell}} \right ) \left ( \dfrac{1~\mathrm{mol~Ca}}{6.022\times 10^{23}~\mathrm{mol~Ca~atoms}} \right ) \left ( \dfrac{40.078~\mathrm{g}}{1~\mathrm{mol~Ca}} \right ) = 2.662\times 10^{-22}~\mathrm{g} \]
Find the volume of the unit cell.
\[ V = a^3 = (558.8\times 10^{-10}~\mathrm{cm})^3 = 1.745\times 10^{-22}~\mathrm{cm^{-3}} \]
Finally, determine the density of calcium.
\[ \mathrm{Ca} = \dfrac{2.662\times 10^{-22}~\mathrm{g}}{1.745\times 10^{-22}~\mathrm{cm}^3} = 1.53~\mathrm{g~cm^{-3}} \]