MHT-CET Chemistry · Solid State

Density and Crystal Structure Calculations

One relation, ρ = nM/(a³N_A), links density, molar mass, particles per cell and the cell volume; MHT-CET hands you a lumped product such as a³N_A or ρ·a³ so that the answer is one multiplication or division.

All Solid State notes MHT-CET Chemistry strategy

Why this matters

43 PYQs, 2 HARD — the largest subtopic in the chapter. Every row is the same formula solved for a different unknown: density, molar mass, cell volume, the number of particles per cell (which then names the structure), or the number of cells or atoms in a given mass or volume. Read what is given as a LUMP (a³N_A, ρN_A, ρa³) and the question collapses to arithmetic on two numbers.

Concept 1 of 5

The Density Formula and the Lumped Constants

Intuition

Mass of the cell is n atoms × (M/N_A) grams; volume is a³. Their ratio is the density. The paper rarely gives a and M separately — it gives a³·N_A (cm³ mol⁻¹), or ρ·N_A, or ρ·a³ (the mass of one cell), so the arithmetic is two numbers and n.

Definition

$\rho = \dfrac{n\,M}{a^3 N_A}$ with $n = 1, 2, 4$ for sc, bcc, fcc and $a$ in cm.
Given $a^3 N_A$ : $\rho = \dfrac{nM}{a^3 N_A}$ directly. E.g. fcc, $M = 197$ , $a^3 N_A = 40$ : $\rho = \dfrac{4 \times 197}{40} = 19.7\ \text{g cm}^{-3}$ .
Given $a$ in Å or pm: convert to cm, cube, multiply by $6.022 \times 10^{23}$ . $a = 4\ \text{Å}$ : $a^3 N_A = 64 \times 10^{-24} \times 6.022 \times 10^{23} = 38.5$ .
$\rho \cdot a^3$ is the MASS of one unit cell; $\rho \cdot a^3 / n$ is the mass of one atom.

Density of a cubic crystal

\rho = \frac{n\,M}{a^3\,N_A}

Worked example

Sodium (

M = 23

) is bcc with

a = 4.29

Å. Find its density.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Solid StateEASY

Calculate the density of a metal having molar mass

197\,\text{g}\,\text{mol}^{-1}

if it forms fcc structure.

a^3 \times N_A = 40\,\text{cm}^3\,\text{mol}^{-1}

[Q84 · 10th May Shift 2 · 2023]

Using n = 4 for bcc

The wrong n doubles or halves the answer, and that wrong answer is always among the options. bcc is 2; fcc is 4; simple cubic is 1.

Drill 7 more on the density formula and the lumped constants

Concept 2 of 5

Molar Mass From Density

Intuition

Rearrange:

M = \rho\,a^3 N_A / n

. With the lump

a^3 N_A

given, it is density × lump ÷ n. If the mass of the cell is given instead, divide by n for the mass of one atom and multiply by

N_A

Definition

$M = \dfrac{\rho \cdot a^3 N_A}{n}$ . bcc, $\rho = 5.6$ , $a^3 N_A = 75$ : $M = \dfrac{5.6 \times 75}{2} = 210\ \text{g mol}^{-1}$ .
From the cell mass: $M = \dfrac{\rho a^3}{n} \times N_A$ . fcc cell of mass $1.8 \times 10^{-22}$ g: one atom $= 4.5 \times 10^{-23}$ g, $M = 27$ .
From $a$ and $\rho$ : bcc, $\rho = 10$ , $a = 4 \times 10^{-8}$ cm: $M = \dfrac{10 \times 6.4 \times 10^{-23} \times 6.022 \times 10^{23}}{2} = 193$ .

Molar mass

M = \frac{\rho\,a^3 N_A}{n}

Worked example

An fcc metal has density

8.9\ \text{g cm}^{-3}

and

a^3 N_A = 28.4\ \text{cm}^3\ \text{mol}^{-1}

. Find its molar mass.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Solid StateMODERATE

A metal crystallises in bcc structure with edge length

4\times10^{-8}

cm. If density of unit cell is

10\ \text{g cm}^{-3}

. What is its molar mass?

[Q97 · 11th May Shift 2 · 2023]

Dividing by N_A when the lump already contains it

a^3 N_A

in cm³ mol⁻¹ has Avogadro's number built in. Multiply

\rho

by it and divide by n — nothing else. A second

N_A

sends the exponent off by 23.

Drill 8 more on molar mass from density

Concept 3 of 5

Unit Cell Volume From Density and Molar Mass

Intuition

a^3 = nM/(\rho N_A)

. The paper gives

\rho N_A

as one lump (of order

10^{24}

), so the cell volume is n × M ÷ lump — a number of order

10^{-23}

cm³.

Definition

$a^3 = \dfrac{n\,M}{\rho\,N_A}$ . fcc, $M = 27$ , $\rho N_A = 16.0 \times 10^{23}$ : $a^3 = \dfrac{4 \times 27}{1.6 \times 10^{24}} = 6.75 \times 10^{-23}\ \text{cm}^3$ .
Without the lump: bcc Na, $M = 23$ , $\rho = 1$ : $a^3 = \dfrac{2 \times 23}{6.022 \times 10^{23}} = 7.6 \times 10^{-23}\ \text{cm}^3$ .
Sanity check: a cubic cell volume is $10^{-23}$ to $10^{-22}$ cm³ (edges 200–500 pm). An answer of $10^{-21}$ means a lost exponent.

Cell volume

a^3 = \frac{n\,M}{\rho\,N_A}

Worked example

An element of molar mass 52 forms bcc cells with

\rho N_A = 4.3 \times 10^{24}\ \text{g cm}^{-3}\ \text{mol}^{-1}

. Find the cell volume.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Solid StateEASY

Calculate the volume of unit cell when metal having density

1\,\text{g\,cm}^{-3}

and molar mass

23\,\text{g\,mol}^{-1}

crystallises to form bcc structure.

[Q54 · 9th May Shift 1 · 2023]

Trusting a printed exponent over the order of magnitude

One 2024 paper printed

d N_A = 120 \times 10^{21}

; the working gives

6 \times 10^{-21}

, the key says

6.00 \times 10^{-23}

. Match the mantissa to the options and let the sanity range (

10^{-23}

) settle the exponent.

Drill 5 more on unit cell volume from density and molar mass

Concept 4 of 5

Which Structure? Solve for n

Intuition

Solve the density relation for the particles per cell:

n = \rho a^3 N_A / M

, or, when the cell mass and atom mass are given,

n = \rho a^3 / m_{\text{atom}}

. The answer rounds to 1, 2 or 4 and names the cell: simple, body-centred, face-centred.

Definition

$n = \dfrac{\rho \cdot a^3 N_A}{M}$ . $\rho = 8.6$ , $a^3 N_A = 21.5$ , $M = 92$ : $n = 2$ → bcc.
$n = \dfrac{\text{mass of cell}}{\text{mass of one atom}} = \dfrac{\rho a^3}{m}$ . $1.792 \times 10^{-22} / 4.4 \times 10^{-23} = 4.07$ → fcc.
$n = 1$ : simple cubic. $n = 2$ : bcc. $n = 4$ : fcc/ccp. Round — the data are rarely exact.

Particles per cell

n = \frac{\rho\,a^3 N_A}{M} = \frac{\rho\,a^3}{m_{\text{atom}}}

Worked example

An element (

M = 60

) has density

6.23\ \text{g cm}^{-3}

and

a = 400

pm. Identify the cell.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Solid StateMODERATE

Unit cell of an element has edge length of

5

Å with density

4\text{ g cm}^{-3}

, if its atomic mass is 149, identify the crystal structure.

[Q77 · 15th May Shift 2 · 2023]

Stopping at n and not naming the cell

Half these questions ask for the STRUCTURE, not the number. n = 2 is 'body-centred cubic'; n = 4 is 'face-centred cubic'. hcp is never the answer of a cubic-cell calculation.

Drill 5 more on which structure? solve for n

Concept 5 of 5

Counting Unit Cells and Atoms in a Mass or a Volume

Intuition

Divide the sample by one cell. In a mass: cells = mass ÷ (ρ·a³), the mass of one cell; atoms = n × cells. In a volume: cells = V ÷ a³. One mole of a simple-cubic metal is

N_A

cells because each cell holds one atom.

Definition

Cells in mass $w$ : $\dfrac{w}{\rho a^3}$ . $0.9$ g, $\rho a^3 = 3 \times 10^{-22}$ : $3 \times 10^{21}$ cells.
Atoms in mass $w$ : $n \times \dfrac{w}{\rho a^3}$ . bcc, $0.3$ g, $\rho a^3 = 3 \times 10^{-22}$ : $2 \times 10^{21}$ atoms.
Cells from moles: $\dfrac{w}{M} \times \dfrac{N_A}{n}$ . $0.60$ g, $M = 60$ , fcc: $0.01 \times 6.022 \times 10^{23} / 4 = 1.5 \times 10^{21}$ .
Cells in volume $V$ : $\dfrac{V}{a^3}$ . $1\ \text{cm}^3$ , $a = 2 \times 10^{-8}$ cm: $1.25 \times 10^{23}$ .
Simple cubic, 1 mol: $6.022 \times 10^{23}$ cells (one atom per cell).

Counting cells

N_{\text{cells}} = \frac{w}{\rho\,a^3} = \frac{V}{a^3} = \frac{w}{M}\cdot\frac{N_A}{n},\qquad N_{\text{atoms}} = n\,N_{\text{cells}}

Worked example

How many atoms are in 2.0 g of an fcc metal whose unit cell weighs

2.5 \times 10^{-22}

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Solid StateHARD

Calculate the number of atoms in 5-gram metal that crystallises to form simple cubic unit cell structure having edge length 336 pm. (Density of metal = 9.4 g cm

^{-3}

)

[Q92 · Shift 1 · 2022]

Counting cells when atoms are asked

'Number of atoms in 0.3 g of a bcc metal' is cells × 2. The cell count

10^{21}

is offered as option (A) for the student who stops one step early.

Drill 13 more on counting unit cells and atoms in a mass or a volume

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

The Density Formula and the Lumped Constants
Density of a cubic crystal
$\rho = \frac{n\,M}{a^3\,N_A}$
Molar Mass From Density
Molar mass
$M = \frac{\rho\,a^3 N_A}{n}$
Unit Cell Volume From Density and Molar Mass
Cell volume
$a^3 = \frac{n\,M}{\rho\,N_A}$
Which Structure? Solve for n
Particles per cell
$n = \frac{\rho\,a^3 N_A}{M} = \frac{\rho\,a^3}{m_{\text{atom}}}$
Counting Unit Cells and Atoms in a Mass or a Volume
Counting cells
$N_{\text{cells}} = \frac{w}{\rho\,a^3} = \frac{V}{a^3} = \frac{w}{M}\cdot\frac{N_A}{n},\qquad N_{\text{atoms}} = n\,N_{\text{cells}}$

Watch out for (5)

Using n = 4 for bcc
→ The Density Formula and the Lumped Constants
Dividing by N_A when the lump already contains it
→ Molar Mass From Density
Trusting a printed exponent over the order of magnitude
→ Unit Cell Volume From Density and Molar Mass
Stopping at n and not naming the cell
→ Which Structure? Solve for n
Counting cells when atoms are asked
→ Counting Unit Cells and Atoms in a Mass or a Volume

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