Do you know that some substances float while others sink in water? If you answered yes, you should be familiar with the idea of density and why it is important in determining the properties of matter, whether solids, liquids, or gases. We’ll look at the Density of Unit Cells and their formula in this article.

The smallest group of atoms with the general symmetry of a crystal is known as a unit cell. It’s a three-dimensional construction that may be utilised to construct a full lattice. It’s important to remember that the constituent particles of crystalline solids have a regular and recurring pattern. You must be able to visualise the entire structure of a unit cell before you can compute its density.

Before studying the concept of density of unit cells, let us understand what a crystal lattice and unit cell are. Continue reading to know more.

Crystal Lattices or Space Lattices and Unit Cell

Space lattice is a regular arrangement of constituent atoms or ions or molecules in three-dimensional space. A small part of the lattice, which produces the complete space lattice when repeated in different directions can be selected to describe the space lattice completely. This small portion of the lattice is called the ‘unit cell.’

Characteristics of a Crystal Lattice

The following are some of the features of a crystal lattice:

i. A lattice point or lattice site is a point in a lattice.

ii. Each crystal lattice point represents a constituent particle, which could be an atom, a molecule, or an ion. Lattice points are joined by lines to represent the geometry of the lattice.

Unit Cell

The smallest three-dimensional portion of a complete space lattice, when repeated in different directions, produces the complete space lattice.

The lattice can be constructed by stacking the unit cells therefore, the unit cell may be considered the fundamental building block of the crystal lattice.

Density

Density is a basic fundamental concept that has a strong relationship with the mass of an object.

Density refers to the measurement of relative compactness that can be expressed as the amount of mass of a substance per unit volume. Densities vary with different materials or substances. It is donated by the Greek letter rho \({\rm{(\rho )}}\) The formula of the density is as follows-

\({\rm{\rho  = }}\frac{{\rm{m}}}{{\rm{V}}}\)

S.I. Unit of Density

When the unit of mass is taken in kilograms and the unit of volume is taken in meters, then the \({\rm{S}}{\rm{.I}}\) unit is- \(\frac{{{\rm{kg}}}}{{{{\rm{m}}^{\rm{2}}}}}\)

When the unit of mass is taken in grams, and the unit of volume is taken in centimeters, then the \({\rm{C}}{\rm{.G}}{\rm{.S}}{\rm{.}}\) unit is- \(\frac{{\rm{g}}}{{{\rm{c}}{{\rm{m}}^{\rm{2}}}}}\)

Density of Unit Cells

The density of the crystal lattice is the same as that of its unit cells. Therefore,

\({\rm{The}}\,{\rm{density}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell}}{\mkern 1mu} \,{\rm{ = }}{\mkern 1mu} \,\rho {\mkern 1mu} \,{\rm{ = }}\,{\mkern 1mu} \frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell}}}}{{{\rm{The}}\,{\rm{volume}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell}}}}\)

Where,

\({\rm{Mass}}\,{\rm{of}}\,{\rm{unit}}\,{\rm{cell}}\,{\rm{ = }}\,{\rm{Rank}}\,{\rm{of}}\,{\rm{unit}}\,{\rm{cell}}\, \times {\rm{Mass}}\,{\rm{of}}\,{\rm{each}}\,{\rm{particle}}\,\,{\rm{z}}\,{\rm{ \times }}\,{\rm{m}}\)

For atoms, the molar mass is equal to the mass of one Avogadro number of atoms, and for molecules, the molar mass is equal to the mass of one Avogadro number of the molecules. Thus,

\({\rm{Mass}}\,{\rm{of}}\,{\rm{each}}\,{\rm{particle}}\,\,{\rm{ = }}\,\frac{{{\rm{The}}\,{\rm{molar}}\,{\rm{mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{particle}}}}{{{\rm{Avogadro’s}}\,{\rm{number}}}}\)

\({\rm{Mass}}{\mkern 1mu} {\rm{of}}{\mkern 1mu} {\rm{a}}{\mkern 1mu} {\rm{each}}{\mkern 1mu} {\rm{particle}}{\mkern 1mu} \,{\rm{ = }}\,{\mkern 1mu} {\rm{m}}\, = \,\frac{{\rm{M}}}{{{{\rm{N}}_{\rm{A}}}}}\)

The volume of the unit cell \({\rm{V}}\,{\rm{ = }}\,{{\rm{a}}^{\rm{3}}}{\rm{,}}\) where a \( = \) edge length of the unit cell.

\(\therefore \,{\rm{The}}\,{\rm{density}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell = }}\,\frac{{{\rm{Rank}}\,{\rm{of}}\,{\rm{unit}}\,{\rm{cell}}\,{\rm{ \times }}\,{\rm{Mass}}\,{\rm{of}}\,{\rm{one}}\,{\rm{particle}}}}{{{\rm{The}}\,{\rm{volume}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell}}}}\)

\({\rm{\rho  = }}\,\frac{{{\rm{z}}\,\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

Applications of the Density \(\left( {\rm{\rho }} \right)\)

The relation between the density and the dimensions of the unit cell contains five terms. Therefore, it can be used to evaluate any one quantity provided the other four are known.

Thus,

a. \({\rm{\rho }}\,{\rm{ = }}\,\frac{{{\rm{z}}{\mkern 1mu} {\mkern 1mu} {\rm{ \times }}{\mkern 1mu} {\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}{\mkern 1mu} {\rm{ \times }}{\mkern 1mu} {{\rm{N}}_{\rm{A}}}}}\)

b. \({{\rm{a}}^{\rm{3}}}\,{\rm{ = }}\,\frac{{{\rm{z}}{\mkern 1mu} {\mkern 1mu} {\rm{ \times }}{\mkern 1mu} {\rm{M}}}}{{{\rm{r}}{\mkern 1mu} {\rm{ \times }}{\mkern 1mu} {{\rm{N}}_{\rm{A}}}}}{\rm{ = }}\) Volume of the unit cell

c. \({{\rm{N}}_{\rm{A}}}\,{\rm{ = }}\,\frac{{{\rm{z}}{\mkern 1mu} {\mkern 1mu} \times {\mkern 1mu} {\rm{M}}}}{{{{\rm{a}}^3}{\mkern 1mu} \times {\mkern 1mu} \rho }} =\) Avogadro’s number

d. \({\rm{M}}\,{\rm{ = }}\,\frac{{{{\rm{a}}^{\rm{3}}} \times {{\rm{N}}_{\rm{A}}} \times \rho }}{{\rm{z}}} =\) Molar Mass

e. \({\rm{z}}{\mkern 1mu} \,{\rm{ = }}{\mkern 1mu} {\mkern 1mu} \,\frac{{{{\rm{a}}^{\rm{3}}} \times {{\rm{N}}_{\rm{A}}} \times \rho }}{{\rm{M}}} =\) Rank of the unit cell

f. \({\rm{m}}{\mkern 1mu} {\mkern 1mu} \,{\rm{ = }}{\mkern 1mu} \,{\mkern 1mu} \frac{{\rm{M}}}{{{{\rm{N}}_{\rm{A}}}}} = \) Mass of one atom

General Expression of Density of Unit Cell for Various Cases

\(1\). Primitive Unit Cell

The number of atoms in a primitive unit cell is one. Therefore, the density can be given as-

\({\rm{\rho  = }}\frac{{{\rm{z}}\,\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

\(\therefore {\rm{\rho  = }}\frac{{1\,\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

\(2\). Body-centered Cubic Unit Cell

The number of atoms in a body-centered unit cell is two. Therefore, the density can be given as-

\({\rm{\rho }}\,{\rm{ = }}\frac{{{\rm{z}}\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

\(\therefore \,\,{\rm{\rho }}\,{\rm{ = }}\frac{{2\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

\(3\). Face-centered Cubic Unit Cell

The number of atoms in a face-centered cubic unit cell is four. Therefore, the density can be given as-

\({\rm{\rho }}\,{\rm{ = }}\frac{{{\rm{z}}\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

\(\therefore \,\,{\rm{\rho }}\,{\rm{ = }}\frac{{4\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

Density of a Unit Cell – Solved Examples

Example 1. An element \({\rm{X}}\) crystallizes as a face-centered cubic lattice with an edge length of \(460\,{\rm{pm}}\) Compute the density of the element when the molar mass of \({\rm{X}}\) atoms is \(60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}\).
Sol: The density and edge length are related by-
\({\rm{\rho }}\,{\rm{ = }}\frac{{{\rm{z}}\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)
For fcc unit cell \({\rm{z}}\,\,{\rm{ = }}\,\,{\rm{4}}\)
The molar mass of atom \({\rm{M}}\,\,{\rm{ = }}\,\,60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ – 1}} = 0.06\,{\rm{kg}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\)
Edge length \( = {\rm{a}}\, = 460\,{\rm{pm}}\,{\rm{ = }}\,{\rm{460}}\,\, \times \,{10^{ – 12}}\,{\rm{m}}\)
Avogadro’s number \( = {{\rm{N}}_{\rm{A}}}\, = 6.022 \times {10^{23}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\)
\(\therefore \,\,{\rm{\rho }}\, = \,\frac{{4 \times 0.06\,{\rm{kg}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}}}{{{{(460\,\, \times \,\,{{10}^{ – 12}}{\rm{m)}}}^3} \times 6.022\, \times \,{{10}^{23}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}}} = 4096\,\,{\rm{kg/}}{{\rm{m}}^3}\)
Hence, the density of element \({\rm{X}}\) is \(4096\,\,{\rm{kg/}}{{\rm{m}}^{\rm{3}}}\)

Example 2. A face-centered cubic element (atomic mass \( = \,60\)) has a unit cell edge of \(400\,{\rm{pm}}\). What is its density?\(({{\rm{N}}_{\rm{A}}} = \,6.\,022\, \times \,{10^{23}}\,{\rm{mo}}{{\rm{l}}^{ – 1}})\)
Sol: The density and edge length are related by-
\({\rm{\rho }}\,{\rm{ = }}\,\frac{{{\rm{z}}\,\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)
\(\therefore \,\,{\rm{\rho }}\,{\rm{ = }}\,\frac{{4\,\, \times \,60\,{\rm{g}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}}}{{(400 \times {\rm{1}}{{\rm{0}}^{ – 10}}{\rm{c}}{{\rm{m}}^{\rm{3}}})\,{\rm{ \times }}\,6.022\, \times \,10{\,^{23}}{\rm{mo}}{{\rm{l}}^{{\rm{ – 1}}}}}} = 6.22\,{\rm{g/c}}{{\rm{m}}^3}\)
Hence, the density of the element is \(6.22\,{\rm{g/c}}{{\rm{m}}^3}\).

Summary

In this article, we studied in detail the density of a unit cell, its \({\rm{S}}{\rm{.I}}\). unit, and how it is calculated. We also studied the general expression of density of unit cells for Various Cases- Primitive unit cells, body-centered unit cells, and face-centered unit cells. Now we know that the formula of density for different unit cells depends on the number of atoms in the unit cell.

FAQs

We have provided some frequently asked questions here:

Q.1. What is the density of FCC?
Ans: The density of the unit cell is-
\({\rm{\rho }}\,{\rm{ = }}\,\frac{{{\rm{z}}\,\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)
The number of atoms in a face-centered cubic unit cell is four. Therefore, the density can be given as-
\({\rm{\rho }}\,{\rm{ = }}\,\frac{{{\rm{z}}\,\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)
\(\therefore \,\,{\rm{\rho }}\,{\rm{ = }}\,\frac{{4\,\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

Q.2. What is the unit for density?
Ans: The density of the unit cell is-
\(\,{\rm{\rho }}\,{\rm{ = }}\,\frac{{{\rm{z}}\,\,{\rm{ \times }}\,{\rm{M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)
When the unit of mass is taken in kilograms, and the unit of volume is taken in meters, then the \({\rm{S}}{\rm{.I}}\) unit is-
\(\frac{{{\rm{kg}}}}{{{{\rm{m}}^{\rm{3}}}}}\)
When the unit of mass is taken in grams, and the unit of volume is taken in centimeters, then the \({\rm{C}}{\rm{.G}}{\rm{.S}}\) unit is-
\(\frac{{\rm{g}}}{{{\rm{c}}{{\rm{m}}^{\rm{3}}}}}\)

Q.3. What are the different types of crystal lattice?
Ans: There are seven different types of crystal lattice which are as follows-
(i) cubic
(ii) tetragonal
(iii) orthorhombic
(iv) hexagonal
(v) monoclinic
(vi) rhombohedral
(vii) triclinic.

Q.4. What is the unit cell and its type?
Ans: The unit cell is the smallest three-dimensional portion of a complete space lattice that produces the complete space lattice when repeated in different directions.
There are different types of unit cells which are as follows-
(i) Simple / primitive unit cell
(ii) Body centered unit cell
(iii) Face centered unit cell
(iv) End centered unit cell

Q.5. What is the formula for the density of a unit cell ?
Ans: \({\rm{The}}\,{\rm{density}}\,{\rm{of}}\,{\rm{the}}\,\,{\rm{unit}}\,{\rm{cell}}\,{\rm{ = \rho  = }}\frac{{{\rm{Mass}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell}}}}{{{\rm{The}}\,{\rm{volume}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell}}}}\)
Where,
\({\rm{Mass}}\,{\rm{of}}\,{\rm{unit}}\,{\rm{cell}}\,{\rm{ = }}\,{\rm{Rank}}\,{\rm{of}}\,{\rm{unit}}\,{\rm{cell}}\, \times \,{\rm{Mass}}\,{\rm{of}}\,{\rm{each}}\,{\rm{particle}}\,{\rm{ = }}\,{\rm{z}}\,\,{\rm{ \times }}\,{\rm{m}}\)
\(\therefore \,\,{\rm{The}}\,{\rm{density}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell}}\,{\rm{ = }}\frac{{{\rm{Rank}}\,{\rm{of}}\,{\rm{unit}}\,{\rm{cell}}\,{\rm{ \times }}\,{\rm{Mass}}\,{\rm{of}}\,{\rm{one}}\,{\rm{particle}}}}{{{\rm{The}}\,{\rm{volume}}\,{\rm{of}}\,{\rm{the}}\,{\rm{unit}}\,{\rm{cell}}}}\)
\({\rm{\rho }}\,\,{\rm{ = }}\,\,\frac{{{\rm{z}}\,{\rm{ \times M}}}}{{{{\rm{a}}^{\rm{3}}}\,{\rm{ \times }}\,{{\rm{N}}_{\rm{A}}}}}\)

Q.6. What is the volume of a unit cell?
Ans: The volume of a unit cell is equal to the cube of edge length.
\(\therefore \,\,{\rm{V}}\,{\rm{ = }}\,{{\rm{a}}^{\rm{3}}}\)

We hope this detailed article on the Density of Unit Cells will be helpful to you. If you have any doubts about this article, please ping us through the comments section, and we will get back to you as soon as possible.