Mole Concept and Stoichiometry — JEE Main & Advanced Notes

Build the calculation language of chemistry: mole, molar mass, limiting reagent, empirical formula, concentration, purity and equivalent concept.

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Learning resources and corrections: check assumptions, work through the examples, and report any unclear step.

1. Mole concept: mass, particles and gas volume

A mole counts specified entities: atoms, molecules, ions or formula units. One mole contains exactly 6.02214076 × 10²³ entities. Identify what you are counting before multiplying by Avogadro's constant.

Amount n = mass / molar mass. Use grams with g mol⁻¹ to obtain moles.
Number of entities = n × Nₐ. One mole of water contains one mole of molecules but three moles of atoms.
For an ideal gas, n = PV/(RT). Use kelvin and a gas constant consistent with the pressure and volume units.

Gas-volume condition: at 273.15 K and 1 atm, the ideal molar volume is approximately 22.4 L mol⁻¹. At 273.15 K and 1 bar it is approximately 22.7 L mol⁻¹. Do not use 22.4 L when the question gives different conditions.

2. Stoichiometry and the limiting reagent

Balance the equation, convert reactants to moles, then divide each amount by its coefficient. The smallest ratio determines the reaction extent and the theoretical product. Equal smallest ratios mean those reactants are present in stoichiometric proportions.

For aA + bB → cC, the maximum amount of C is c × min(n(A)/a, n(B)/b), assuming the stated reaction goes to completion. Subtract the amount consumed from the starting amount to find excess reactant.

Apply sample purity before converting mass to moles. Apply percentage yield to the theoretical product afterward: actual product = theoretical product × yield/100.

3. Concentration and formula checklist

Molarity: moles of solute / litres of solution. It can change with temperature because volume changes.
Molality: moles of solute / kilograms of solvent. Do not use the mass of the solution as the denominator.
Mole fraction: moles of a component / total moles of all components; the fractions sum to one.
Dilution: M₁V₁ = M₂V₂ when the solute amount is conserved. V₂ is the final solution volume, not the volume of water added.
Mixing the same solute: add the solute moles and divide by the final volume. Add the initial volumes only when volume additivity is given or assumed.
Equivalents: equivalent mass = molar mass / reaction-specific n-factor. For acid–base reactions, count the H⁺ or OH⁻ actually exchanged; for redox, count electrons transferred per formula unit. Partial neutralisation can change the n-factor.
Empirical formula: convert elemental masses to moles and simplify their ratio. Molecular formula multiplier = molecular molar mass / empirical formula mass.

4. Six worked examples with answers

1. Count molecules and atoms in water

How many molecules and total atoms are in 4.50 g of water? Use a molar mass of 18.0 g mol⁻¹ and Nₐ ≈ 6.022 × 10²³ mol⁻¹.

n = 4.50/18.0 = 0.250 mol. Molecules = 0.250 × Nₐ = 1.51 × 10²³. Each molecule has three atoms, so total atoms = 4.52 × 10²³. Multiplying by two would count only hydrogen atoms.

2. Find the limiting reagent and excess

For N₂ + 3H₂ → 2NH₃, mix 14 g N₂ with 6 g H₂. Use molar masses 28, 2 and 17 g mol⁻¹ respectively. Find the theoretical ammonia mass and hydrogen remaining.

n(N₂) = 14/28 = 0.5 mol; n(H₂) = 6/2 = 3 mol. The coefficient-adjusted amounts are 0.5/1 and 3/3 = 1, so nitrogen limits the reaction. Ammonia = 2 × 0.5 = 1 mol = 17 g. Hydrogen consumed = 3 × 0.5 = 1.5 mol, leaving 1.5 mol = 3 g. Mass check: 14 + 6 = 17 + 3.

3. Combine purity with percentage yield

A 25.0 g limestone sample contains 80.0% CaCO₃. It decomposes by CaCO₃ → CaO + CO₂ with a CO₂ yield of 75.0%. Use molar masses 100 and 44 g mol⁻¹ for CaCO₃ and CO₂.

Pure CaCO₃ mass = 25.0 × 0.800 = 20.0 g, or 0.200 mol. The 1:1 ratio gives a theoretical CO₂ mass of 0.200 × 44 = 8.80 g. Actual CO₂ = 8.80 × 0.750 = 6.60 g.

4. Calculate a dilution

What final volume is needed to dilute 25.0 mL of 0.400 mol L⁻¹ NaCl to 0.100 mol L⁻¹?

V₂ = M₁V₁/M₂ = (0.400 × 25.0)/0.100 = 100 mL. Transfer the original solution and add water until the total solution volume reaches 100 mL. The answer is the final volume, not 100 mL of added water.

5. Derive empirical and molecular formulas

A compound contains 40.0% carbon, 6.67% hydrogen and 53.33% oxygen by mass. Its molar mass is 180 g mol⁻¹. Use atomic masses C = 12, H = 1, O = 16.

For a 100 g sample, mole amounts are 40/12 ≈ 3.333, 6.67/1 = 6.67 and 53.33/16 ≈ 3.333. Divide by 3.333 to obtain 1:2:1. The empirical formula is CH₂O with mass 30 g mol⁻¹. The multiplier is 180/30 = 6, giving C₆H₁₂O₆.

6. Distinguish molality from molarity

Dissolve 9.0 g glucose (molar mass 180 g mol⁻¹) in 100 g water. Calculate molality. Can you also calculate molarity?

Glucose amount = 9.0/180 = 0.050 mol. Solvent mass = 0.100 kg, so molality = 0.050/0.100 = 0.50 mol kg⁻¹. Molarity cannot be determined without the final solution volume or sufficient density information. The 109 g solution mass is not the molality denominator.

5. Check your understanding

Try these before opening the answers.

What is the total amount of atoms in 0.20 mol CO₂?

Each molecule contains three atoms. Total amount of atoms = 3 × 0.20 = 0.60 mol, consisting of 0.20 mol carbon atoms and 0.40 mol oxygen atoms.

Does the reactant with the lowest mass always limit a reaction?

No. Mass must first be converted to moles and divided by the balanced coefficient. Different molar masses and coefficients can reverse a comparison based only on grams.

When can N₁V₁ = N₂V₂ be used?

At equivalence for the specified reaction, with normalities defined using the correct reaction-specific n-factors and consistent volume units. It is not a general formula for mixing arbitrary solutions.

6. Revision route and next chapters

First practise mass-to-mole conversions. Then solve limiting-reagent problems, followed by purity, yield, concentration and empirical formulas. Write the equation and units before substituting values. Use the timed practice on this page to identify which step needs another attempt.

Continue with JEE Chemistry chapter notes and practice, or review Class 9 Atoms and Molecules for the school-level foundations.

Definition references: BIPM: the mole and IUPAC: standard conditions for gases. Explanations and worked problems on this page are written for LearnAtMyPlace.

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