NDA Chemistry · Mole Concept and Stoichiometry

The Mole, Avogadro's Law and Molar Calculations

A mole is a fixed count of particles (6.022 × 10^23 of them); molar mass, molar volume and Avogadro's number are the three bridges that turn grams, litres and molecule-counts into moles and back.

All Mole Concept and Stoichiometry notes NDA Chemistry strategy

Why this matters

This is the engine room of the chapter. Three of five PYQs here are direct one-step conversions (mass of 0.5 mol N2, who proposed the equal-volume law), and the two harder ones (a NOT-correct statement testing all three conversions, and a mass-percent comparison) still reduce to the same mole bridges. Learn n = m/M, N = n*NA and n = V/22.4 cold and you can attempt every question in this subtopic on sight.

Concept 1 of 5: The mole and Avogadro's number

Chemists count atoms by weighing them, but atoms are far too small to count one by one — so they count in bundles. One mole is just a fixed-size bundle: exactly 6.022 x 10^23 particles, the same way a dozen is always 12. The number 6.022 x 10^23 is Avogadro's number.

Definition

Key definitions:

A mole is the amount of a substance that contains as many elementary particles (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12.
That count is Avogadro's number, $N_A = 6.022 \times 10^{23}$ particles per mole.
'Particle' means whatever the substance is made of: for $\text{H}_2$ one mole = $6.022 \times 10^{23}$ molecules (and $2 \times 6.022 \times 10^{23}$ atoms).
The mole is a counting unit, like 'a dozen' — it says nothing about mass on its own; mass depends on the molar mass.

Avogadro's number

N_A = 6.022 \times 10^{23}\ \text{particles per mole}

Worked example

How many oxygen atoms are present in 2 moles of oxygen gas (O2)?

Practice this concept3 quick reps

Molecules and atoms differ for diatomic gases

One mole of

\text{H}_2

\text{O}_2

\text{N}_2

holds

6.022 \times 10^{23}

molecules but twice that many atoms. Read whether the question asks for molecules or atoms.

Concept 2 of 5: Molar mass and moles from mass

Molar mass is the mass of one mole, in grams, and it equals the atomic or molecular mass read straight off the periodic table. To get moles from a given mass, divide the mass by the molar mass; to get mass from moles, multiply.

Definition

Working rules:

Molar mass (M) = mass of one mole in grams; numerically equal to the atomic mass (element) or molecular mass (compound). Example: $M(\text{N}_2) = 28$ g/mol, $M(\text{CO}_2) = 44$ g/mol.
Number of moles from mass: $n = m / M$ .
Mass from moles: $m = n \times M$ .
Build the molecular mass by adding each element's atomic mass times the number of atoms.

Moles from mass

n = \dfrac{m}{M}

nnumber of moles
mgiven mass (g)
Mmolar mass (g/mol)

Worked example

What is the mass of 0.25 mole of calcium carbonate, CaCO3? (Ca = 40, C = 12, O = 16)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2024 · Q60Easy

Example 2 · Mole Concept and Stoichiometry · Mole Concept, Avogadro's Law and Molar Calculations

The mass of 0.5 mole of

\text{N}_2

gas is

Use the molar mass of the WHOLE molecule

For 0.5 mol of

\text{N}_2

the molar mass is 28 g/mol (a nitrogen molecule), not 14. So mass

= 0.5 \times 28 = 14

g — the 14 comes from the calculation, not from using the atomic mass of one N.

Concept 3 of 5: Avogadro's law and molar volume at STP

Avogadro's law says equal volumes of any gases, at the same temperature and pressure, hold equal numbers of molecules. A direct consequence: one mole of ANY gas occupies the same volume at STP — 22.4 litres. So for gases you can bridge straight from volume to moles.

Definition

The gas-phase bridge:

Avogadro's law — equal volumes of all gases at the same temperature and pressure contain an equal number of molecules.
Molar volume — one mole of any gas occupies 22.4 L (22,400 mL) at STP (0 deg C, 1 atm).
Moles from gas volume at STP: $n = V / 22.4$ (V in litres).
This is why half a mole of $\text{N}_2$ measures 11.2 L, and 22.4 L of $\text{CO}_2$ at STP weighs 44 g (one mole).

Moles from gas volume at STP

n = \dfrac{V}{22.4}

nnumber of moles
Vvolume of gas at STP (litres)
22.4molar volume at STP (L/mol)

Worked example

What volume does 0.5 mole of oxygen gas occupy at STP?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

NDA · Sep 2017 · Q85Easy

Example 3 · Mole Concept and Stoichiometry · Mole Concept, Avogadro's Law and Molar Calculations

The proposition 'equal volumes of different gases contain equal numbers of molecules at the same temperature and pressure' is known as

22.4 L only at STP, and only for gases

The molar volume of 22.4 L per mole applies to gases at STP only. It does not apply to liquids or solids, and it does not apply to a gas at room temperature or other pressures.

Half a mole of a gas = 11.2 L (this is correct)

In a NOT-correct statement question, 'half mole of nitrogen measures 11.2 L at STP' is a true statement (

0.5 \times 22.4 = 11.2

) — so it is not the wrong one. Check each option's arithmetic separately.

Drill 1 more on avogadro's law and molar volume at stp

Concept 4 of 5: Counting particles from moles

Once you know the moles, the number of actual molecules is just moles times Avogadro's number. Combined with n = m/M, this lets you go all the way from a mass in grams to a raw molecule count.

Definition

The particle bridge:

Number of particles: $N = n \times N_A$ , with $N_A = 6.022 \times 10^{23}$ .
Chained from mass: $N = (m/M) \times N_A$ .
So one mole of a substance always contains $6.022 \times 10^{23}$ molecules — for example, 17 g of $\text{NH}_3$ (M = 17) is exactly one mole and holds $6.022 \times 10^{23}$ molecules.

Particles from moles

N = n \, N_A = \dfrac{m}{M}\, N_A

Nnumber of particles (molecules/atoms)
nnumber of moles
N_AAvogadro's number, $6.022 \times 10^{23}$

Worked example

How many molecules are present in 9 g of water, H2O? (M = 18)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Apr 2025 · Q79Moderate

Example 4 · Mole Concept and Stoichiometry · Mole Concept, Avogadro's Law and Molar Calculations

Which one of the following statements is NOT correct?

4 g of H2 is TWO Avogadro numbers, not one

\text{H}_2

has molar mass 2, so 4 g

= 2

mol

= 2 \times 6.022 \times 10^{23}

molecules. The classic NOT-correct option claims '4 g of hydrogen contains

6.022 \times 10^{23}

molecules' — that is the false statement.

Concept 5 of 5: Mass-percent composition

To find how much of a compound's mass comes from one element, take that element's total mass in the formula and divide by the whole molar mass. The ratio of two elements' mass-percents is fixed by the formula alone — handy for the comparison questions the bank likes.

Definition

Composition by mass:

Mass percent of an element $= \dfrac{(\text{atoms of element}) \times (\text{atomic mass})}{\text{molar mass of compound}} \times 100$ .
The ratio of two elements' mass-percents does not depend on the rest of the formula. For any $\text{C}_6\text{H}_{12}\text{O}_n$ : $\dfrac{\%\text{C}}{\%\text{H}} = \dfrac{6 \times 12}{12 \times 1} = 6$ , so %C is always six times %H.

Mass percent of an element

\%\,\text{element} = \dfrac{a \times A}{M} \times 100

anumber of atoms of the element in the formula
Aatomic mass of the element
Mmolar mass of the whole compound

Worked example

Find the mass percent of carbon in methane, CH4. (C = 12, H = 1)

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

NDA · Sep 2017 · Q84Hard

Example 5 · Mole Concept and Stoichiometry · Mole Concept, Avogadro's Law and Molar Calculations

The compound

\text{C}_6\text{H}_{12}\text{O}_4

contains

Mass-percent ratio ignores the oxygen count

For

\text{C}_6\text{H}_{12}\text{O}_4

versus

\text{C}_6\text{H}_{12}\text{O}_6

, the %C : %H ratio is the same (= 6) because the oxygen mass cancels from the ratio. Changing n only changes the individual percents, not their ratio.

Drill 1 more on mass-percent composition

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 mole and Avogadro's number
Avogadro's number
$N_A = 6.022 \times 10^{23}\ \text{particles per mole}$
Molar mass and moles from mass
Moles from mass
$n = \dfrac{m}{M}$
Avogadro's law and molar volume at STP
Moles from gas volume at STP
$n = \dfrac{V}{22.4}$
Counting particles from moles
Particles from moles
$N = n \, N_A = \dfrac{m}{M}\, N_A$
Mass-percent composition
Mass percent of an element
$\%\,\text{element} = \dfrac{a \times A}{M} \times 100$

Watch out for (6)

Molecules and atoms differ for diatomic gases
→ The mole and Avogadro's number
Use the molar mass of the WHOLE molecule
→ Molar mass and moles from mass
22.4 L only at STP, and only for gases
→ Avogadro's law and molar volume at STP
Half a mole of a gas = 11.2 L (this is correct)
→ Avogadro's law and molar volume at STP
4 g of H2 is TWO Avogadro numbers, not one
→ Counting particles from moles
Mass-percent ratio ignores the oxygen count
→ Mass-percent composition

Test yourself on a real paper

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