Mole concept for JEE Main

The mole concept rarely gets a long question of its own in JEE Main, but it sits inside almost every physical chemistry calculation. Equilibrium, electrochemistry, solutions and thermodynamics all start by converting grams or litres into moles. A candidate who is slow here is slow everywhere, and a slip in the first line carries into the final answer.

The official 2026 syllabus lists the basic concepts in its first unit: Dalton's atomic theory, the laws of chemical combination, atomic and molecular masses, the mole concept and molar mass, percentage composition, empirical and molecular formulas, and chemical equations and stoichiometry. The concentration terms (molality, molarity, mole fraction and percentage by mass and by volume) are listed in the solutions unit, and they belong here in practice.

The mole and its conversions

One mole contains Nₐ = 6.022 × 10²³ particles. Every conversion passes through moles:

• moles = mass/molar mass = particles/Nₐ
• for a gas at STP, moles = volume/22.4 L
• molar mass in g/mol equals the molecular mass in u

Be clear about which particle you are counting. 1 mol of O₂ contains Nₐ molecules but 2Nₐ atoms, and 1 mol of H₂SO₄ contains 4 mol of oxygen atoms.

Laws of chemical combination

Empirical and molecular formula

1. Divide each percentage by the atomic mass to get relative moles.
2. Divide by the smallest to get a ratio; multiply through if the ratio is not whole (1.5 means double, 1.33 means triple).
3. The molecular formula is n times the empirical formula, where n = molar mass/empirical formula mass.

Stoichiometry, limiting reagent, purity and yield

• Limiting reagent: divide the moles of each reactant by its coefficient. The smallest value identifies the limiting reagent, and it decides the amount of product.
• Percentage purity = (mass of pure substance/mass of sample) × 100.
• Percentage yield = (actual yield/theoretical yield) × 100.

Concentration terms

Useful relations:

• Dilution: M₁V₁ = M₂V₂
• Mixing solutions of the same solute: M = (M₁V₁ + M₂V₂)/(V₁ + V₂)
• Molarity to molality: m = 1000M/(1000d − M × M₂), with d the solution density in g/mL and M₂ the solute's molar mass
• the mole fractions of all components add up to 1

Worked problems

Problem 1: limiting reagent. 10 g of H₂ reacts with 64 g of O₂ to form water. Find the mass of water formed and the reactant left over.

2H₂ + O₂ → 2H₂O. Moles: H₂ = 10/2 = 5; O₂ = 64/32 = 2.
Moles ÷ coefficient: H₂ 5/2 = 2.5; O₂ 2/1 = 2. O₂ is limiting.
Water = 2 × 2 = 4 mol = 72 g. H₂ used = 4 mol, so 1 mol = 2 g of H₂ is left.
Check: 10 + 64 = 74 g = 72 + 2 g.

Problem 2: molecular formula. A compound contains 40.0% C, 6.67% H and 53.33% O. Its molar mass is 180 g/mol. Find its molecular formula.

Relative moles: C 40/12 = 3.33; H 6.67/1 = 6.67; O 53.33/16 = 3.33.
Ratio 1 : 2 : 1, so the empirical formula is CH₂O, of mass 30.
n = 180/30 = 6, so the molecular formula is C₆H₁₂O₆.

Problem 3: molarity to molality and mole fraction. An aqueous NaOH solution is 2 M, with density 1.08 g/mL. Find its molality and the mole fraction of NaOH.

Take 1 L of solution: mass = 1,080 g. Solute = 2 × 40 = 80 g, so solvent = 1,000 g.
Molality = 2 mol/1 kg = 2 mol/kg.
Moles of water = 1,000/18 = 55.6. Mole fraction of NaOH = 2/(2 + 55.6) = 0.035.

Problem 4 (numerical answer): percentage purity. 20 g of an impure sample of CaCO₃ gives 3.36 L of CO₂ at STP on heating. Impurities do not decompose. Find the percentage purity.

CaCO₃ → CaO + CO₂. Moles of CO₂ = 3.36/22.4 = 0.15.
Moles of CaCO₃ = 0.15; mass = 0.15 × 100 = 15 g.
Purity = 15/20 × 100 = 75.

Traps to watch

• Molarity uses the volume of solution, not the volume of solvent.
• Molality uses the mass of solvent in kg, not the mass of solution.
• Diatomic gases. 16 g of oxygen is 1 mol of O atoms but 0.5 mol of O₂ molecules.
• Mixing changes volume. When two solutions are mixed, divide by the total volume.
• Molar volume. Use the value the question gives; do not mix 22.4 and 22.7.

Practice set

1. How many atoms are present in 4.4 g of CO₂?
2. Find the mass of 3.011 × 10²² molecules of NH₃.
3. 4 g of NaOH is dissolved to make 250 mL of solution. Find the molarity.
4. 200 mL of 0.5 M HCl is mixed with 300 mL of 0.2 M HCl. Find the molarity of the mixture.
5. 18 g of glucose (M = 180) is dissolved in 500 g of water. Find the molality.
6. Find the mole fraction of ethanol in a solution of 46 g of ethanol and 36 g of water.
7. A hydrocarbon has 85.7% C and 14.3% H, and molar mass 56 g/mol. Find its molecular formula.
8. What mass of CaCO₃ must be heated to produce 5.6 L of CO₂ at STP?

Answers

1. 4.4/44 = 0.1 mol of molecules, each with 3 atoms: 0.3 × 6.022 × 10²³ = 1.81 × 10²³ atoms.
2. 0.05 mol × 17 = 0.85 g.
3. 0.1 mol/0.25 L = 0.4 M.
4. (100 + 60) mmol/500 mL = 0.32 M.
5. 0.1 mol/0.5 kg = 0.2 mol/kg.
6. Ethanol 1 mol, water 2 mol: x = 1/3 ≈ 0.33.
7. C 85.7/12 = 7.14; H 14.3/1 = 14.3; ratio 1 : 2, so CH₂ (mass 14); n = 56/14 = 4: C₄H₈.
8. 5.6/22.4 = 0.25 mol of CO₂, so 0.25 mol of CaCO₃ = 25 g.

What to do next

• Solve 20 limiting-reagent and purity problems, writing moles ÷ coefficient each time.
• Practise converting between molarity, molality and mole fraction until each takes under a minute.
• Go on to concentration terms and stoichiometry and limiting reagent for more practice.
• Then start atomic structure.