# Density of a Cubic Crystal

We have already studied the structure of a cubic crystal. This structure will define the properties of the solid. It will also help us determine the density of a cubic crystal. Let us see how this calculation is done.

## Density of a Cubic Crystal

The first important thing to understand is that a cubic crystal is made up a crystal lattice or a space lattice structure. And as we have studied before, a space lattice is made up of a repeating arrangement of unit cells. A unit cell is the most basic structure of a crystalline solid.

Hence if we are able to find the density of a unit cell, it can be said that we have found the density of the cubic crystal itself. With this in mind, let us determine how to arrive at the density of a cubic crystal cell.

## Calculations of a Unit Cell

Now as we know a unit cell also has a cubic structure. It has one, two or four atoms located at various lattice points. Now with the help of geometry, some basic calculations and certain attributes of this cubic structure we can find the density of a unit cell. Let us start with the basic formula for the density of any solid. This formula is

Density = $$\frac{Mass}{Volume}$$

The density of a Unit Cell will be

D = $$\frac{Mass of Unit Cell}{Volume of Unit Cell}$$

### Mass of a Unit Cell

Now to calculate the mass of one unit cell we add up the mass of all the atoms contained in that particular cell. The number of atoms will depend on the kind of cell it is. So to obtain the mass of a unit cell we multiply the number of atoms “n” into the mass of each atom”m”.

Mass of Unit Cell = m × n

But now the question remains, what is the mass of an atom? Well, this can be represented in terms of its Avogadro Number (NA)., i.e. number of units in one mole of any substance and the molar mass of an atom. So the mass of an atom can be calculated as

Mass of an Atom = $$\frac{Molar Mass}{Avogadro Number}$$

Mass of an Atom = $$\frac{M}{N_A}$$

So now the formula for Mass of Unit Cell is as follows

Mass of Unit Cell = n × $$\frac{M}{N_A}$$

### Volume of Unit Cell

As we already know, the unit cell is a cubic structure. So the Volume of a cube is the cube of the length of the side, Assume that the length of the side of the cube is “A”.

Volume of Unit Cell = a3

### Density of Unit Cell

And now we finally arrive at the actual formula for the density of a unit cell

Density of a Unit Cell = $$\frac{Mass of Unit Cell}{Volume of Unit Cell}$$

Density of a Unit Cell = $$\frac{n × M}{a^3 × N_A}$$

## Solved Example for You

Q: An atomic solid crystallizes in a body center cubic lattice and the inner surface of the atoms at the adjacent corner are separated by 60.3 pm. If the atomic weight of A is48, then the density of the solid is nearly:

1. 2.7 g/cc
2. 5.07 g/cc
3. 3.5 g/cc
4. 1.75 g/cc

Sol: The correct answer is “D”. According to given condition,

0.13 a=60.3 ⟹ a=463.8

So Density = $$\frac{n × M}{a^3 × N_A}$$ = $$\frac{2 × 48 × 10^3}{4(63.8)^3 × 6.023 × (10)^{23}}$$ =  1.75 g/cc

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