An element of molar mass 50g mol⁻¹ crystallises in a simple cubic lattice with a measured density of 5.0g cm⁻³

Exercise · Q12

50\ \text{g mol}^{-1}

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Concept understanding — Density of Unit Cell

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:

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

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:

\text{Mass of unit cell} = (\text{number of atoms in it}) \times (\text{mass of one atom})

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

\text{Mass of one atom} = \frac{M}{N_A}

Putting it together:

\text{Mass of unit cell} = z \times \frac{M}{N_A}

And therefore:

\text{Density of unit cell} = \frac{z \times M}{a^3 \times N_A}

d = \frac{z \cdot M}{a^3 \cdot N_A}

What Each Symbol Means

Symbol	Meaning	Typical Units
$d$	Density of the crystal	g/cm³ or kg/m³
$z$	Number of atoms per unit cell	dimensionless (1, 2, 4, etc.)
$M$	Molar mass of the element/compound	g/mol
$a$	Edge length of the unit cell	cm (or m)
$N_A$	Avogadro's number	$6.022 \times 10^{23}$ mol⁻¹

Watch out

A very common mistake: using $a$ in picometers or angstroms but forgetting to convert to cm. Since density is usually in g/cm³, convert $a$ to cm first. $1 \, \text{pm} = 10^{-10} \, \text{cm}$ ; $1 \, \text{Å} = 10^{-8} \, \text{cm}$ .

Why This Formula Works — The Logic Chain

The crystal is perfectly periodic — every unit cell is identical.
The density of the bulk crystal equals the density of one unit cell.
The unit cell's volume is $a^3$ .
The unit cell's mass is the sum of the masses of all atoms inside it.
The mass of one atom is $M / N_A$ .
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:

a = 361 \, \text{pm} = 361 \times 10^{-10} \, \text{cm} = 3.61 \times 10^{-8} \, \text{cm}

Step 3: Volume:

a^3 = (3.61 \times 10^{-8})^3 = 4.70 \times 10^{-23} \, \text{cm}^3

Step 4: Apply formula: …

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Starting from $\rho = \dfrac{ZM}{N_A a^3}$ , rearrange for $a^3$ : $a^3 = \dfrac{ZM}{\rho N_A}$ . With $Z=1$ (simple cubic), $M = 50\ \text{g mol}^{-1}$ , $\rho = 5.0\ \text{g cm}^{-3}$ , and $N_A = 6.022\times10^{23}\ \text{mol}^{-1}$ : $a^3 = \frac{1\times50}{5.0\times6.022\times10^{23}} = \frac{50}{3.011\times10^{24}} \approx 1.660\times10^{-23}\ \text{cm}^3$ Taking the cube root: $a^3 = 16.60\times10^{-24}\ \text{cm}^3$ , and $\sqrt[3]{16.60 …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set BZ2 marks
$Q. An element forms ccp lattice. If the length of its edge of unit cell is 408.6 pm, then calculate the density of the element. (Atomic weight = 107.9 u)$
›Reveal solutionSolution

Using $\rho=\dfrac{Z\,M}{a^{3}N_A}$ with $Z=4$ for ccp, the density is about $10.5\ g\ cm^{-3}$ .

Concept: For a cubic crystal, density $\rho=\dfrac{Z\cdot M}{a^{3}\cdot N_A}$ , where $Z$ is the number of atoms per unit cell. A ccp (fcc) lattice has $Z=4$ .

Data: $a = 408.6\ pm = 4.086\times10^{-8}\ cm$ , $M = 107.9\ g\ mol^{-1}$ , $N_A = 6.022\times10^{23}\ mol^{-1}$ .

Volume of cell:

$a^{3} = (4.086\times10^{-8})^{3} = 6.82\times10^{-23}\ cm^{3}$

…

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CBSE 2022Set M2 marks
$^{-3}$
›Reveal solutionSolution

Using $d = \dfrac{ZM}{N_A a^3}$ with $Z=4$ (FCC), $a=556\ \text{pm}$ and $d=1.55\ \text{g cm}^{-3}$ gives $M \approx 40.1\ \text{g mol}^{-1}$ .

Given: FCC lattice $\Rightarrow Z = 4$ atoms per unit cell; edge $a = 556\ \text{pm} = 556\times10^{-10}\ \text{cm} = 5.56\times10^{-8}\ \text{cm}$ ; density $d = 1.55\ \text{g cm}^{-3}$ ; $N_A = 6.022\times10^{23}$ .

Formula:
$d = \frac{Z\,M}{N_A\,a^3} \quad\Rightarrow\quad M = \frac{d\,N_A\,a^3}{Z}$
Volume of the unit cell:
$a^3 = (5.56\times10^{-8})^3 = 1.719\times10^{-22}\ \text{cm}^3$
Substitute: …
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CBSE 2022Set GH2 marks
$408.6$
›Reveal solutionSolution

For ccp (fcc), $Z=4$ ; substituting $M=107.9$ , $a=408.6$ pm into $d=\dfrac{ZM}{a^{3}N_A}$ gives $d\approx10.5\ \text{g cm}^{-3}$ .

Concept — density of a cubic crystal. For a unit cell of edge $a$ containing $Z$ atoms of molar mass $M$ ,

$d=\frac{Z\,M}{a^{3}\,N_A}.$

A cubic close-packed (ccp = fcc) lattice has $Z=4$ atoms per unit cell.

Step 1 — edge length in cm:

$a=408.6\ \text{pm}=408.6\times10^{-10}\ \text{cm}=4.086\times10^{-8}\ \text{cm}$

$a^{3}=(4.086\times10^{-8})^{3}=6.82\times10^{-23}\ \text{cm}^{3}$

Step 2 — substitute ( $N_A=6.022\times10^{23}\ \text{mol}^{-1}$ ): …

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Q.An element of molar mass 50 g mol−150\ \text{g mol}^{-1}50 g mol−1 crystallises in a simple cubic lattice with a measured density of 5.0 g cm−35.0\ \text{g cm}^{-3}5.0 g cm−3. Calculate the edge length of its unit cell.

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Q.An element of molar mass $50\ \text{g mol}^{-1}$ crystallises in a simple cubic lattice with a measured density of $5.0\ \text{g cm}^{-3}$ . Calculate the edge length of its unit cell.

Unlock everything free for 14 days