# Cubic Unit Cell Calculator (SC, BCC, FCC)

URL: https://speedycalculator.com/calculators/chemistry/cubic-cell/
Last updated: 2026-10-30

> The cubic unit cell calculator relates the edge length (lattice constant) and atomic radius of simple cubic, body-centred cubic, face-centred cubic and diamond lattices, and gives atoms per cell, coordination number, packing efficiency, cell volume and the theoretical density from the molar mass.

## Inputs

- Lattice type
- I know the
- Atomic radius (r) (Å, nm, pm, m)
- Edge length (a) (Å, nm, pm, m)
- Density (ρ) (g/mL, g/cm³, kg/m³, kg/L, lb/ft³)
- Molar mass of the element (M) (g/mol, kg/mol, Da (u), kDa)

## Results

- Atoms per unit cell (Z)
- Edge length (lattice constant) a
- Atomic radius r
- Atomic radius r
- Edge–radius relation
- Coordination number
- Packing efficiency (atomic packing factor)
- Empty space in the cell
- Nearest-neighbour distance (2r)
- Unit cell volume (a³)
- Theoretical density
- Atoms per cm³ (× 10²²)

## Formula

```
SC: a = 2r, Z = 1, APF = π/6 · BCC: a = 4r/√3, Z = 2, APF = π√3/8 · FCC: a = 2√2·r, Z = 4, APF = π/(3√2) · diamond: a = 8r/√3, Z = 8 · ρ = Z·M ÷ (N_A·a³)
```

## Explanation

Metals and many simple solids crystallise in cubic lattices, where the repeating unit – the unit cell – is a cube of edge length a. In a simple cubic (SC) cell the atoms sit only at the eight corners and touch along the edges. A body-centred cubic (BCC) cell adds one atom in the middle, and the atoms touch along the body diagonal. A face-centred cubic (FCC) cell puts an atom in the centre of each face, and they touch along the face diagonals; FCC is the same arrangement as cubic close packing.

Because corner atoms are shared by eight cells and face atoms by two, an SC cell contains 1 atom, BCC 2 and FCC 4. The diamond structure (carbon, silicon, germanium) is an FCC lattice with a second, shifted set of atoms, giving 8 per cell. The geometry of touching spheres fixes the relation between a and the atomic radius r, and from it the atomic packing factor – the fraction of space filled by atoms: 52.4% (SC), 68.0% (BCC), 74.0% (FCC) and only 34.0% for diamond.

Once the cell size is known, the density follows directly: the mass of Z atoms divided by the volume a³. Turned around, a measured density and molar mass give the lattice constant and radius – a classic way to check X-ray data or estimate Avogadro's number.

### How to use the Cubic Unit Cell Calculator

- **Lattice type**
- **I know the**
- **Atomic radius (r)**
- **Edge length (a)**
- **Density (ρ)**
- **Molar mass of the element (M)**

Results update as you type, every field has its own unit menu, and everything is calculated in your browser.

### How it works

Choose the lattice and what you know: the atomic radius, the edge length, or the density. The calculator applies the touching-sphere relation for that lattice, then computes the cell volume, packing efficiency and density with ρ = Z·M/(N_A·a³), with a converted to centimetres.

### Worked example

Aluminium is FCC with a metallic radius of 1.43 Å. The edge length is a = 2√2 × 1.43 = **4.0447 Å** (the measured value is 4.0495 Å), and the cell volume is 66.17 ų = 6.617 × 10⁻²³ cm³. With 4 atoms per cell and M = 26.982 g/mol, ρ = 4 × 26.982 ÷ (6.02214 × 10²³ × 6.617 × 10⁻²³) = 2.709 g/cm³, close to the handbook density of 2.70 g/cm³. For BCC iron with r = 1.241 Å, a = 4 × 1.241 ÷ √3 = 2.866 Å and ρ = 2 × 55.845 ÷ (6.02214 × 10²³ × (2.866 × 10⁻⁸)³) ≈ 7.88 g/cm³.

### Cubic lattices compared

LatticeAtoms per cellCoordinationa in terms of rPacking efficiencyExamplesSimple cubic162r52.36%α-PoBody-centred cubic284r/√3 ≈ 2.309r68.02%Fe, Cr, W, Mo, NaFace-centred cubic4122√2·r ≈ 2.828r74.05%Al, Cu, Ni, Ag, Au, PbDiamond cubic848r/√3 ≈ 4.619r34.01%C, Si, Ge

### Inputs explained

FieldWhat to enterLattice typeChoose one: Simple cubic (SC), Body-centred cubic (BCC), Face-centred cubic (FCC / ccp), Diamond cubicI know theChoose one: Atomic radius r, Edge length (lattice constant) a, Density ρAtomic radius (r)Units: Å, nm, pm, mEdge length (a)Units: Å, nm, pm, mDensity (ρ)Units: g/mL, g/cm³, kg/m³, kg/L, lb/ft³Molar mass of the element (M)Units: g/mol, kg/mol, Da (u), kDa

### What each result means

- **Edge length (lattice constant) a** = [edge]
- **Atomic radius r** = [radius]
- **Atomic radius r** = [radius]
- **Edge–radius relation** = [Lattice type] - 1
- **Packing efficiency (atomic packing factor)** = ([Atoms per unit cell (Z)]×4÷3×pi×[radius]^3÷[edge]^3)×100
- **Empty space in the cell** = (1 - [Atoms per unit cell (Z)]×4÷3×pi×[radius]^3÷[edge]^3)×100
- **Nearest-neighbour distance (2r)** = 2×[radius]
- **Unit cell volume (a³)** = [edge]^3
- **Theoretical density** = [Atoms per unit cell (Z)]×[Molar mass of the element (M)]÷(6.02214076e23×([edge]×100)^3)
- **Atoms per cm³ (× 10²²)** = [Atoms per unit cell (Z)]÷([edge]×100)^3÷1e22

## Worked example

Inputs — Lattice type: Face-centred cubic (FCC / ccp); I know the: Atomic radius r; Atomic radius (r): 1.43 Å; Molar mass of the element (M): 26.982 g/mol.

Results — Atoms per unit cell (Z) = 4; **Edge length (lattice constant) a = 4.04465 Å**; Edge–radius relation = a = 2√2·r – atoms touch along the face diagonal; Coordination number = 12; Packing efficiency (atomic packing factor) = 74.05 %; Empty space in the cell = 25.95 %; Nearest-neighbour distance (2r) = 2.86 Å; Unit cell volume (a³) = 66.1673 ų; Theoretical density = 2.7086 g/mL; Atoms per cm³ (× 10²²) = 6.0453.

## Key terms

- **Unit cell**: The smallest repeating block whose translations rebuild the whole crystal.
- **Lattice constant (a)**: The edge length of the cubic unit cell.
- **Atomic packing factor**: The fraction of the cell volume occupied by atoms modelled as hard spheres.
- **Coordination number**: The number of nearest neighbours of each atom.
- **Close packing**: Arrangement of equal spheres that fills 74% of space (FCC and HCP).

## FAQ

**How many atoms are in SC, BCC and FCC unit cells?**
1, 2 and 4. Corner atoms count ⅛ (8 × ⅛ = 1), body-centre atoms count 1, and face atoms count ½ (6 × ½ = 3).

**What is the packing efficiency of FCC?**
π/(3√2) = 74.05%, the densest possible packing of equal spheres – the same as hexagonal close packing. BCC fills 68.02% and simple cubic 52.36%.

**How do I get the edge length from the atomic radius?**
a = 2r for SC, a = 4r/√3 ≈ 2.309r for BCC and a = 2√2·r ≈ 2.828r for FCC.

**Which metals are FCC and which BCC?**
FCC: aluminium, copper, nickel, silver, gold, platinum, lead. BCC: α-iron, chromium, tungsten, molybdenum, sodium, potassium. Only α-polonium is simple cubic.

**How do I calculate density from the unit cell?**
ρ = Z·M/(N_A·a³). For copper (FCC, a = 3.615 Å, M = 63.546 g/mol): 4 × 63.546 ÷ (6.022 × 10²³ × 4.724 × 10⁻²³ cm³) ≈ 8.93 g/cm³.

**Why is my calculated density slightly off?**
Metallic radii are rounded and depend on coordination number; real crystals also contain vacancies and impurities, and lattice constants change with temperature.

**What is the coordination number?**
The number of nearest neighbours touching each atom: 6 in SC, 8 in BCC, 12 in FCC and 4 in diamond.


## Sources

- [OpenStax – Chemistry 2e](https://openstax.org/details/books/chemistry-2e)
- [International Union of Crystallography – Online Dictionary of Crystallography](https://dictionary.iucr.org/)
- [LibreTexts Chemistry](https://chem.libretexts.org/)