Unit Cell Calculations

A crystal is not just a random pile of atoms — it is a precisely repeating three-dimensional arrangement. The smallest repeating unit of this arrangement is called the unit cell. By analysing what fraction of each atom actually belongs inside one unit cell, chemists can calculate the effective number of atoms per unit cell (denoted Z) and use that to connect measurable macroscopic quantities — like density and lattice parameter — to atomic-scale properties. The three most important cubic unit cells encountered in high school and introductory university chemistry are the Simple Cubic (SC), Body-Centred Cubic (BCC), and Face-Centred Cubic (FCC) cells. Each differs in how many atoms are present and where they sit, leading to different values of Z and different geometric relationships between the edge length a of the cube and the atomic radius r. Unit cell calculations are powerful because they link X-ray crystallography data (which gives a precisely) to atomic radii and to the density of a solid. Once you master the geometry of sharing atoms between adjacent cells, you can solve problems that relate crystal structure, molar mass, Avogadro's number, and macroscopic density — a beautiful bridge between the atomic and bulk worlds.

Key takeaways

Atom-Sharing Fractions
Simple Cubic (SC): Z = 1
Body-Centred Cubic (BCC): Z = 2
Face-Centred Cubic (FCC): Z = 4
Density from Unit Cell Parameters

An atom sitting at a corner of a cube is shared equally among 8 adjacent unit cells, so it contributes 1/8 to each. An atom on a face is shared between 2 cells, contributing 1/2. An atom on an edge is shared among 4 cells, contributing 1/4. An atom entirely inside the body of the cell is not shared at all, contributing 1 full atom.

Simple Cubic (SC): Z = 1

The SC cell has one atom at each of its 8 corners. Each corner atom contributes 1/8, so Z = 8 × (1/8) = 1. The atoms along each edge just touch, so the edge length a equals two radii: a = 2r. Packing efficiency is only 52.4%, making SC rare in nature (polonium is the classic example).

Body-Centred Cubic (BCC): Z = 2

The BCC cell has 8 corner atoms plus 1 atom entirely at the body centre. Z = 8 × (1/8) + 1 × 1 = 2. The atoms touch along the body diagonal of the cube. The body diagonal has length a√3. Since two radii from the corner atom plus two radii from the body-centre atom span the full diagonal: 4r = a√3, giving a = 4r/√3. Packing efficiency ≈ 68%. Many metals (Fe, Cr, W, Na) adopt BCC.

Face-Centred Cubic (FCC): Z = 4

The FCC cell has 8 corner atoms and 6 face-centre atoms. Z = 8 × (1/8) + 6 × (1/2) = 1 + 3 = 4. Atoms touch along the face diagonal, which has length a√2. Four radii span the face diagonal: 4r = a√2, giving a = 2√2 · r. FCC is also called cubic close-packed (CCP) and has the highest packing efficiency of the three: 74.05%. Most metallic elements (Cu, Al, Au, Ag, Ni) and NaCl-type ionic compounds adopt FCC-based structures.

Density from Unit Cell Parameters

Once Z is known, the theoretical density ρ of a crystalline solid can be calculated. The mass of one unit cell equals Z atoms multiplied by the mass of one atom (M/Nₐ, where M is molar mass and Nₐ is Avogadro's number). The volume of the cubic unit cell is a³. Therefore ρ = ZM / (Nₐ · a³). This equation is central to X-ray crystallography: measuring a with X-rays and knowing M and Z lets chemists determine ρ, or conversely, measuring ρ and knowing M and Z lets them determine a and hence r.

The Key Equation

\rho = \frac{Z \cdot M}{N_A \cdot a^3}

\rho — Density (g cm⁻³): Mass per unit volume of the crystal
Z — Number of atoms (formula units) per unit cell (dimensionless): 1 for SC, 2 for BCC, 4 for FCC
M — Molar mass (g mol⁻¹): Mass of one mole of the substance
N_A — Avogadro's number (mol⁻¹): 6.022 × 10²³ entities per mole
a — Edge length of unit cell (cm or pm): Length of one side of the cubic unit cell

Worked Example

Problem. Copper crystallises in an FCC lattice. The density of copper is 8.96 g cm⁻³ and its molar mass is 63.55 g mol⁻¹. (a) Calculate the edge length a of the unit cell. (b) Calculate the atomic radius of copper.

a = 361.4 \text{ pm},\quad r = \frac{a\sqrt{2}}{4} = 127.8 \text{ pm}

Answer. Edge length a = 361.4 pm; Atomic radius r = 127.8 pm

Figure

Diagram

Origins · 1 minWilliam Lawrence Bragg and how this was discovered

William Lawrence Bragg

1890–1971 · Australian-British

Bragg, along with his father William Henry Bragg, developed Bragg's Law (nλ = 2d sinθ) in 1913, which describes how X-rays diffract off crystal planes. This law made it possible to measure the lattice parameter a of any crystal with extraordinary precision, directly enabling the unit cell calculations that link atomic radii to crystal density. W. L. Bragg remains the youngest Nobel Prize laureate in Physics (1915), receiving the honour at age 25. Bragg used X-ray crystallography to determine the crystal structure of table salt (NaCl) — one of the first ionic structures ever solved — proving that NaCl consists of alternating Na⁺ and Cl⁻ ions in an FCC-based arrangement, not discrete NaCl molecules.

Crystal Lattice Types and Bravais Lattices — Unit cell calculations are grounded in understanding the 14 Bravais lattices; SC, BCC, and FCC are three of the seven cubic variants.
Packing Efficiency and Void Spaces — Packing efficiency is calculated directly from Z and the a–r relationship derived in unit cell calculations.
Bragg's Law and X-Ray Diffraction — X-ray diffraction provides the experimental value of a, which is the key input to all unit cell density and radius calculations.
Ionic Crystal Structures (NaCl, ZnS, CsCl) — Ionic compounds adopt FCC-based or BCC-based unit cells; extending Z calculations to two-ion systems requires the same corner/face/body counting method.