Q.Silver forms ccp lattice and X-ray studies of its crystals show that the edge length of its unit cell is $408.6$ pm. Calculate the density of silver. (Atomic mass $= 107.9$ u)

What is the Density of a Unit Cell?

Imagine you have a brick wall. The wall is made of thousands of identical bricks, each with a certain mass and volume. If you know the mass of one brick and its exact dimensions, you can find the density of the brick itself. But the density of the wall is the same as the density of one brick — because the wall is just bricks repeating.

A crystal is exactly like that wall. It is built from tiny, repeating boxes called unit cells. The entire crystal's density is the same as the density of a single unit cell. So if you can find the mass and volume of one unit cell, you have the density of the whole crystal.

The Intuition First

Density is always:

For a unit cell:

• Volume is easy: if the edge length is $a$, then volume $V = a^3$.
• Mass is trickier. A unit cell is not just one atom — it contains a specific number of atoms (or ions) depending on its type (simple cubic, BCC, FCC). That number is called $z$, the number of atoms per unit cell.

So the mass inside one unit cell is:

The mass of one atom is its molar mass $M$ divided by Avogadro's number $N_A$:

Putting it together:

And therefore:

What Each Symbol Means

Why This Formula Works — The Logic Chain

1. The crystal is perfectly periodic — every unit cell is identical.
2. The density of the bulk crystal equals the density of one unit cell.
3. The unit cell's volume is $a^3$.
4. The unit cell's mass is the sum of the masses of all atoms inside it.
5. The mass of one atom is $M / N_A$.
6. Multiply by $z$ to get the total mass inside the cell.

That's it. No extra assumptions.

A Quick Example

Problem: Copper crystallizes in an FCC lattice. Its edge length $a = 361 \, \text{pm}$ and molar mass $M = 63.5 \, \text{g/mol}$. Find its density.

Step 1: For FCC, $z = 4$.

Step 2: Convert $a$ to cm:

Step 3: Volume:

Step 4: Apply formula: …