Class 11 Chemistry · Academic Guide

Mole Concept & Stoichiometry (Class 11 Chemistry): Formulas + Solved Numericals

The three mole formulas every Class 11 student must know, with worked examples on mass, particles and limiting reagents.

Class 11 · Chemistry · Physical Chemistry · Updated 22 August 2026

Quick summary: A mole is simply a counting unit — 6.022 × 10²³ particles (Avogadro’s number). Almost every Class 11 numerical uses one of three links: moles from mass, moles from number of particles, or moles from gas volume at STP.

The three formulas

n = mass / molar mass | n = N / N_A | n = V / 22.4 L (gas at STP)

Where n = moles, N = number of particles, N_A = 6.022 × 10²³, and 22.4 L is the molar volume of an ideal gas at STP.

You are given	Use	To find
Mass in grams	n = m/M	Moles
Number of atoms/molecules	n = N/N_A	Moles
Gas volume at STP	n = V/22.4	Moles

Worked example 1 — mass to molecules

Q. How many molecules are in 36 g of water?

Molar mass of H₂O = 18 g/mol, so n = 36/18 = 2 mol. Molecules = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules.

Worked example 2 — limiting reagent

Q. 4 g H₂ reacts with 32 g O₂ to form water (2H₂ + O₂ → 2H₂O). Which is limiting?

n(H₂) = 4/2 = 2 mol; n(O₂) = 32/32 = 1 mol. The ratio needed is 2:1, and here it is exactly 2:1 — so both are fully consumed with none in excess, producing 2 mol (36 g) of water.

Common mistakes

Using 22.4 L for non-STP conditions.
Confusing atoms with molecules (1 mol O₂ = 2 mol O atoms).
Forgetting to balance the equation before comparing mole ratios.

FAQs

Is the mole concept important for NEET and JEE too?

Yes — it is foundational for physical chemistry across boards and entrance exams.

What is molar mass in one line?

The mass in grams of one mole of a substance, numerically equal to its molecular/atomic mass in u.

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