Density of Unit Cell

Packing efficiency tells us what fraction of a unit cell is occupied by atoms, but chemists more often need the

crystal's actual, measurable density — the mass per unit volume, usually in $\text{g cm}^{-3}$. Because a unit

cell is representative of the whole crystal (the same arrangement repeats throughout), the density calculated for

one unit cell equals the density of the bulk crystal.

Density is mass divided by volume. The mass of the contents of one unit cell is the number of atoms per cell,

$Z$, multiplied by the mass of a single atom. The mass of a single atom is the molar mass $M$ divided by Avogadro's

number $N_A$ (since $M$ is the mass of $N_A$ atoms). The volume of a cubic unit cell of edge length $a$ is

simply $a^3$. Putting these together gives the standard density formula:

where $\rho$ is the density, $Z$ is the number of atoms (or formula units) per unit cell, $M$ is the molar mass,

$N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}$ is Avogadro's number, and $a$ is the edge length of the unit cell —

which must be converted to centimetres (usually from picometres, $1\ \text{pm} = 10^{-10}\ \text{cm}$) so that the

density comes out in the standard units of $\text{g cm}^{-3}$.

Worked derivation — copper (fcc). Copper crystallises in a face-centred cubic lattice ($Z = 4$) with edge

length $a = 361\ \text{pm} = 361\times10^{-10}\ \text{cm} = 3.61\times10^{-8}\ \text{cm}$, and molar mass

$M = 63.5\ \text{g mol}^{-1}$. First, $a^3 = (3.61\times10^{-8})^3 = 4.70\times10^{-23}\ \text{cm}^3$. Then,

which matches copper's well-known experimental density of about $8.96\ \text{g cm}^{-3}$, confirming both that

copper's edge length was correctly used and that fcc is indeed its crystal structure.

The same equation, rearranged, is equally useful for finding a unit cell's edge length when the density is known

experimentally: $a^3 = \dfrac{Z \cdot M}{\rho \cdot N_A}$, followed by taking the cube root — this is exactly the …