Mole concept for JEE main and advanced 2026
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| Mole concepts quick revision |
Introduction
Mole Concept is one of the most important chapters in Physical Chemistry. It forms the foundation of many JEE topics such as Stoichiometry, Gaseous State, Solutions, Thermodynamics, and Chemical Equilibrium. Every year, JEE Main and JEE Advanced include questions based on this chapter.
This article provides all the essential formulas, key concepts, and quick revision notes for Mole Concept in one place.
1. Core Formulas
Number of Moles
n = m / M
Where:
n = Number of moles
m = Mass of the substance (g)
M = Molar mass (g/mol)
Number of Moles from Particles
n = N / Nₐ
Where:
N = Number of particles
Nₐ = Avogadro's Number = 6.022 × 10²³ particles/mol
2. Mass Relations
Mass from Moles
Mass = n × M
Molar Mass
Molar Mass = Mass / Number of Moles
3. Particle Relations
Number of Particles
N = n × Nₐ
Where:
N = Number of particles
Nₐ = 6.022 × 10²³ particles/mol
4. Gas Relations (At STP)
At Standard Temperature and Pressure (STP):
Volume = n × 22.4 L
or
n = V / 22.4 L
Note: The molar volume of 22.4 L/mol is valid only at STP.
5. Percentage Composition
Percentage of an element in a compound:
% Element = (Mass of Element / Total Mass of Compound) × 100
6. Empirical Formula
The Empirical Formula represents the simplest whole-number ratio of atoms in a compound.
Steps:
Convert percentage composition into mass.
Convert mass into moles.
Divide all mole values by the smallest mole value.
Obtain the simplest whole-number ratio.
7. Molecular Formula
Molecular Formula = n × Empirical Formula
Where,
n = Molecular Mass / Empirical Formula Mass
8. Limiting Reagent
The Limiting Reagent is the reactant that gets completely consumed first and determines the maximum amount of product formed.
Formula
Limiting Reagent = Minimum Value of (Moles / Stoichiometric Coefficient)
9. Equivalent Concept
Equivalent Weight
Equivalent Weight = Molar Mass / n-factor
Examples:
H₂SO₄ = 98/2 = 49
NaOH = 40/1 = 40
10. Concentration Formulae
Molarity (M)
M = Moles of Solute / Volume of Solution (L)
Molality (m)
m = Moles of Solute / Mass of Solvent (kg)
- 11. Density Relation
Density = Mass / Volume
Units:
g/mL
g/L
Common Mistakes
Confusing Molarity with Molality.
Using 22.4 L/mol under non-STP conditions.
Ignoring stoichiometric coefficients while finding the limiting reagent.
Mixing up Molecular Formula and Empirical Formula.
Quick Revision
n = m/M
n = N/Nₐ
N = nNₐ
Mass = nM
Volume (STP) = n × 22.4 L
Molarity = Moles/Litre
Molality = Moles/kg Solvent
Equivalent Weight = Molar Mass/n-factor
Practice Questions
Calculate the number of moles in 18 g of H₂O.
Find the number of molecules present in 2 moles of CO₂.
Calculate the number of moles in 44.8 L of a gas at STP.
Find the equivalent weight of H₂SO₄.
Calculate the mass of 0.5 mole of NaCl.
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