Question

Derive the relationship between molar mass, density of the substance, and unit cell edge length.

Hint:

This formula is particularly useful for deriving the density of crystalline materials from X-ray crystallography data.

Answer 1

Density \( (\rho) \) of a crystalline solid is calculated as the mass per unit volume. For a unit cell with edge length \( a \), the volume is given by \( a^3 \). 
Number of Atoms per Unit Cell: \( Z \) represents the number of formula units per unit cell. For instance, \( Z \) equals 1 in a simple cubic lattice, 2 in a body-centered cubic (BCC) lattice, and 4 in a face-centered cubic (FCC) lattice. 
Mass of the Unit Cell: The mass of one formula unit is the molar mass \( M \) divided by Avogadro's number \( N_A \). The mass of the unit cell then is: \[ \text{Mass of unit cell} = \frac{Z M}{N_A} \] 
Volume of the Unit Cell: \[ V_{\text{unit cell}} = a^3 \] 
Density Calculation: Density is mass divided by volume. For the crystalline solid, it is calculated as: \[ \rho = \frac{\text{Mass of unit cell}}{V_{\text{unit cell}}} = \frac{\frac{Z M}{N_A}}{a^3} \] Simplifying this: \[ \rho = \frac{Z M}{N_A a^3} \] The formula ties together molar mass, density, and unit cell dimensions: \[ \rho = \frac{Z M}{N_A a^3} \] 
Details included: 
- \( Z \) = Number of formula units per unit cell,
- \( M \) = Molar mass, 
- \( N_A \) = Avogadro’s number \( (6.022 \times 10^{23} \, \text{mol}^{-1}) \), 
- \( a \) = Edge length of the unit cell.