Chemistry Physical How to Solve: Mole Concept & Stoichiometry (Limiting Reagent, Yield, Purity) – IIT JEE Guide

How to Solve: Mole Concept & Stoichiometry (Limiting Reagent, Yield, Purity) – IIT JEE Guide

Introduction

Mastering mole concept and stoichiometry unlocks 30-40% of IIT JEE Chemistry marks—from balancing equations to calculating yields in real-world reactions like drug synthesis or fuel production. If you can’t solve limiting reagent problems, you’ll lose easy 5-6 marks in JEE Main and 10+ marks in JEE Advanced.

WHAT YOU NEED TO KNOW FIRST

1. Balanced chemical equations – Coefficients = mole ratios.
2. Mole calculations – Mass → Moles (using molar mass), Moles → Particles (using Avogadro’s number).
3. Percentage composition – How to find % of an element in a compound.

KEY TERMS & FORMULAS

1. Mole Concept

• Formula: n = m / M
• n = number of moles (MEMORISE THIS)
• m = mass (g)

• M = molar mass (g/mol)

• Avogadro’s Number: 6.022 × 10²³ particles = 1 mole (MEMORISE THIS)

2. Stoichiometry (Mole Ratios)

• Balanced equation: aA + bB → cC + dD
• Mole ratio = a : b : c : d (MEMORISE THIS)

3. Limiting Reagent

• Definition: The reactant that runs out first, stopping the reaction.
• How to find:
• Convert all reactants to moles.
• Divide moles by their coefficient in the balanced equation.
• The smallest value = limiting reagent.

4. Theoretical Yield

• Formula: Theoretical Yield = (Moles of limiting reagent) × (Mole ratio) × (Molar mass of product)
• MEMORISE THIS – Always calculate from the limiting reagent.

5. Percentage Yield

• Formula: % Yield = (Actual Yield / Theoretical Yield) × 100 (MEMORISE THIS)

6. Purity of Reactants

• Formula: Purity (%) = (Mass of pure substance / Total mass of sample) × 100
• Given on exam sheet, but understand how to apply it.

STEP-BY-STEP METHOD

Step 1: Write the Balanced Equation

• If not given, balance the equation first.
• Example: N₂ + 3H₂ → 2NH₃

Step 2: Convert All Given Quantities to Moles

• Use n = m / M for solids/liquids.
• For gases: Use n = PV / RT (if volume/pressure given).

Step 3: Identify the Limiting Reagent

• Divide moles of each reactant by its coefficient.
• The smallest value = limiting reagent.

Step 4: Calculate Theoretical Yield

• Use the limiting reagent’s moles × mole ratio × molar mass of product.

Step 5: Adjust for Purity (If Given)

• If reactant is impure, multiply mass by purity % before converting to moles.
• Example: 5 g of 80% pure CaCO₃ → 5 × 0.8 = 4 g pure CaCO₃.

Step 6: Calculate Percentage Yield (If Asked)

• % Yield = (Actual Yield / Theoretical Yield) × 100

WORKED EXAMPLES

Example 1 – Basic (Limiting Reagent & Yield)

Question: 2.0 g H₂ reacts with 10.0 g O₂ to form H₂O. What is the theoretical yield of H₂O?

Solution:
1. Balanced equation: 2H₂ + O₂ → 2H₂O
2. Moles of H₂: n = 2.0 / 2 = 1.0 mol
3. Moles of O₂: n = 10.0 / 32 = 0.3125 mol
4. Limiting reagent check: - H₂: 1.0 / 2 = 0.5 - O₂: 0.3125 / 1 = 0.3125 → O₂ is limiting
5. Theoretical yield of H₂O: - Mole ratio (O₂ : H₂O) = 1 : 2 - Moles of H₂O = 0.3125 × 2 = 0.625 mol - Mass of H₂O = 0.625 × 18 = 11.25 g

What we did and why: - Converted masses to moles → compared mole ratios → found limiting reagent → calculated yield from it.

Example 2 – Medium (Purity & Yield)

Question: 5.0 g of 90% pure CaCO₃ is heated to form CaO and CO₂. If 2.0 g CaO is obtained, what is the % yield?

Solution:
1. Balanced equation: CaCO₃ → CaO + CO₂
2. Pure CaCO₃ mass: 5.0 × 0.9 = 4.5 g
3. Moles of CaCO₃: n = 4.5 / 100 = 0.045 mol
4. Theoretical yield of CaO: - Mole ratio (CaCO₃ : CaO) = 1 : 1 - Moles of CaO = 0.045 mol - Mass of CaO = 0.045 × 56 = 2.52 g
5. % Yield: (2.0 / 2.52) × 100 = 79.37%

What we did and why: - Adjusted for purity → calculated theoretical yield → compared with actual yield.

Example 3 – Exam-Style (Disguised Problem)

Question: A mixture of 4.0 g CH₄ and 16.0 g O₂ is ignited. What mass of CO₂ is formed? (Assume complete combustion.)

Solution:
1. Balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O
2. Moles of CH₄: n = 4.0 / 16 = 0.25 mol
3. Moles of O₂: n = 16.0 / 32 = 0.5 mol
4. Limiting reagent check: - CH₄: 0.25 / 1 = 0.25 - O₂: 0.5 / 2 = 0.25 → Both are limiting (stoichiometric ratio)
5. Theoretical yield of CO₂: - Mole ratio (CH₄ : CO₂) = 1 : 1 - Moles of CO₂ = 0.25 mol - Mass of CO₂ = 0.25 × 44 = 11.0 g

What we did and why: - Recognized stoichiometric ratio → both reactants fully consumed → calculated yield from either.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 30-second crash course for mole concept and stoichiometry:
1. Balance the equation first—coefficients = mole ratios.
2. Convert everything to moles—mass → moles (n = m/M), volume → moles (PV/RT).
3. Find the limiting reagent—divide moles by coefficients, smallest wins.
4. Calculate yield from the limiting reagent—mole ratio × molar mass.
5. Adjust for purity—multiply mass by purity % before moles.
6. Percentage yield = (actual / theoretical) × 100. That’s it. No shortcuts—follow the steps. Now go crush that exam!