🔢 Mole Concept & Stoichiometry – Full Breakdown

ICSE Class 10 Chemistry June 2026

        The mole concept is where most students start sweating because numbers get big and formulas get confusing. But here's the truth — there are only THREE formulas you actually need. Once you understand how to connect them, every numerical becomes the same problem with different numbers. I'll walk through each concept slowly, with reasons, and then give you a step-by-step method to solve any numerical. Written by someone who scored 93.4% — and trust me, if I could crack this chapter, anyone can.
    

1) Gay Lussac's Law of Combining Volumes (1808)

Statement: When gases react, they do so in volumes which bear a simple whole number ratio to each other and to the volume of the product(s), all measured at the same temperature and pressure.

Gay Lussac's Law — Volume Ratios of Gaseous Reactions
H₂(g) + Cl₂(g) → 2HCl(g) (Ratio = 1:1:2)
N₂(g) + 3H₂(g) → 2NH₃(g) (Ratio = 1:3:2)
2H₂(g) + O₂(g) → 2H₂O(g) (Ratio = 2:1:2)
NOTES: Only for gases at same T & P. Volumes are in simple whole number ratios.

Limitation: Only works for gaseous reactants and products at same T & P. Can't apply to solids or liquids. Also, the volumes must be measured at same conditions.

2) Avogadro's Law (1811)

Statement: Equal volumes of all gases under the same conditions of temperature and pressure contain the equal number of molecules.

This was the missing piece that explained Gay Lussac's law. If 1 vol H₂ + 1 vol Cl₂ → 2 vol HCl, then according to Avogadro: same volume = same number of molecules. So 1 molecule H₂ + 1 molecule Cl₂ → 2 molecules HCl. That means H₂ and Cl₂ must be diatomic (H₂ + Cl₂ → 2HCl).

Molar Volume (Important for Numericals)

At STP (Standard Temperature & Pressure: 0°C, 1 atm): 1 mole of any gas = 22.4 litres
At RTP (Room Temperature & Pressure: 20°C, 1 atm): 1 mole of any gas = 24 litres (some books use 22.7 L at 25°C — check which one your board follows)

⚠️ ICSE usually uses STP (22.4 L) unless specified otherwise. Always check the question for "at STP" or "at RTP" or "at room conditions". If nothing is mentioned and the gas is involved, assume STP.

3) Vapour Density & Its Relation to Molecular Mass

Definition: Vapour density (VD) of a gas is the ratio of the mass of a certain volume of the gas to the mass of the same volume of hydrogen under the same conditions of T & P.

Vapour Density = Mass of V volume of gas / Mass of V volume of H₂ OR VD = (Density of gas) / (Density of H₂) [at same T & P]

The golden relation:

Molecular Mass = 2 × Vapour Density

Why? Because of Avogadro's Law — equal volumes have equal molecules. If V volume of gas has mass M_gas and same volume of H₂ has mass M_H₂, then M_gas / M_H₂ = mass of 1 molecule of gas / mass of 1 molecule of H₂ = molecular mass of gas / 2. So VD = MM / 2.

Examples: If VD of a gas = 16, then MM = 32. If MM = 44 (CO₂), VD = 22.

4) Atomic Mass, Molecular Mass & Gram Formula Mass

Relative Atomic Mass (RAM)

Mass of one atom of the element compared to 1/12th mass of one atom of C-12 (which is taken as 12 atomic mass units).

Examples: H = 1, C = 12, O = 16, Na = 23, Cl = 35.5, Ca = 40, Fe = 56

📖 The C-12 scale: 1 atomic mass unit (amu) = 1/12 × mass of one C-12 atom ≈ 1.66 × 10⁻²⁴ g. This is how we can say H has RAM 1 even though it doesn't weigh 1g. RAM is a ratio, not an actual weight in grams.

Relative Molecular Mass (RMM)

Sum of the atomic masses of all atoms in a molecule.

Examples: H₂O = 2×1 + 16 = 18. CO₂ = 12 + 2×16 = 44. CaCO₃ = 40 + 12 + 3×16 = 100.

Gram Atomic Mass & Gram Molecular Mass

Just RAM/RMM expressed in grams.

Gram atomic mass of O = 16 g
Gram molecular mass of H₂O = 18 g

Formula Mass (for ionic compounds)

Ionic compounds like NaCl don't have molecules — they have formula units. So we use "formula mass" instead of molecular mass. NaCl formula mass = 23 + 35.5 = 58.5.

5) The Mole — The Three Formulas That Solve Everything

A mole is simply a counting unit, like a dozen (12) or a gross (144). One mole = 6.022 × 10²³ particles (Avogadro's constant, Nₐ).

You only need to remember three connections to a mole:

n = m / M n = N / Nₐ n = V / Vₘ Where: n = number of moles m = given mass (in grams) M = molar mass (g/mol) N = number of particles (atoms / molecules / ions) Nₐ = 6.022 × 10²³ V = volume of gas (in litres) Vₘ = molar volume (22.4 L at STP, 24 L at RTP)

🧠 Topper's method — the Mole Triangle: Draw a triangle. At the top write "MOLE". At bottom left write "MASS / MOLAR MASS". At bottom right write "PARTICLES / Nₐ". Below write "VOLUME / 22.4". Cover the one you want to find. This is literally the only thing you need for 90% of numericals.

                MASS (m)
                    ╱│╲
            m/M  ╱  │  ╲  N/Nₐ
                ╱   │   ╲
            MOLE ───┼─── PARTICLES
               ╲    │    ╱
           V/22.4 ╲ │ ╱
                     V
              VOLUME (at STP)
        

6) Percentage Composition

% of element = (Mass of element in 1 molecule / Molecular mass) × 100

Example — Find % of H₂O in CuSO₄·5H₂O:

Molar mass of CuSO₄·5H₂O = 63.5 + 32 + 4×16 + 5×18 = 63.5 + 32 + 64 + 90 = 249.5 g/mol
Mass of H₂O = 5 × 18 = 90 g
% of H₂O = (90 / 249.5) × 100 ≈ 36.07%

⚠️ Board loves this: Calculating water of crystallisation percentage or finding the number of water molecules given the % mass. E.g., "A hydrated salt has molar mass 246 and water of crystallisation contributes 36.6%. Find the number of water molecules." → 36.6% of 246 = 90, 90/18 = 5 → 5H₂O.

7) Empirical & Molecular Formula

Empirical Formula

The simplest whole number ratio of atoms of different elements in a compound.

Steps to Find Empirical Formula

Write the % of each element (if mass given directly, use that)
Divide each by its atomic mass → gives mole ratio
Divide all values by the smallest number
If you don't get whole numbers, multiply by appropriate factor (2, 3, etc.) to get whole numbers
Write the empirical formula

Molecular Formula

Molecular Formula = n × (Empirical Formula) Where n = Molecular Mass / Empirical Formula Mass

Full Example: A compound has C = 40%, H = 6.67%, O = 53.33%. Its molecular mass is 60. Find the molecular formula.

Step 1: C = 40/12 = 3.33, H = 6.67/1 = 6.67, O = 53.33/16 = 3.33
Step 2: Divide by smallest (3.33): C = 1, H = 2, O = 1 → Empirical formula = CH₂O
Step 3: Empirical mass = 12 + 2 + 16 = 30. n = 60/30 = 2
Step 4: Molecular formula = C₂H₄O₂

8) Stoichiometry — Calculations from Chemical Equations

This is the practical application of everything above. A balanced equation tells you the mole ratio between reactants and products. Everything else follows from that.

The 4-Step Method (works for every numerical)

Write and balance the chemical equation
Write the mole ratio from the coefficients
Convert given data to moles (using n = m/M or n = V/22.4)
Use ratio to find required moles → convert back to whatever the question asks

Type 1: Mass-Mass Problems

Q: What mass of CO₂ is produced when 50 g of CaCO₃ is heated?

CaCO₃ → CaO + CO₂
1 mol → 1 mol
Moles of CaCO₃ = 50 / 100 = 0.5 mol
Moles of CO₂ produced = 0.5 mol
Mass of CO₂ = 0.5 × 44 = 22 g

Type 2: Mass-Volume Problems

Q: What volume of CO₂ at STP is produced when 100 g of CaCO₃ reacts with excess HCl?

CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O
1 mol → 1 mol
Moles of CaCO₃ = 100 / 100 = 1 mol
Moles of CO₂ = 1 mol
Volume at STP = 1 × 22.4 = 22.4 L

Type 3: Volume-Volume Problems (Gay Lussac)

Q: What volume of NH₃ at STP is produced when 30 L of N₂ reacts with excess H₂?

N₂ + 3H₂ → 2NH₃
1 vol → 2 vol
Volume of NH₃ = 30 × 2 = 60 L
(No need to convert to moles — Gay Lussac directly applies since all are gases, same T & P)

Type 4: Limiting Reagent

Q: 10 g of H₂ reacts with 80 g of O₂. Which is the limiting reagent? How much H₂O is formed?

2H₂ + O₂ → 2H₂O
Moles of H₂ = 10/2 = 5 mol. Moles of O₂ = 80/32 = 2.5 mol.
From equation: 2 mol H₂ needs 1 mol O₂. So 5 mol H₂ needs 2.5 mol O₂.
Both exactly match — neither is limiting (but if O₂ were less, it would be limiting).
Moles of H₂O = 5 mol. Mass = 5 × 18 = 90 g.

📖 Limiting reagent is not officially in ICSE syllabus, but it appears in some numericals implicitly. The strategy: for each reactant, calculate how much product it would produce. The reactant that gives the least product is the limiting reagent. The amount of product is determined by this reactant only.

9) Common Board Numerical Patterns

⚠️ After solving 10+ years of ICSE papers, these patterns repeat:
1. Find molecular mass from VD (MM = 2 × VD) + then empirical formula
2. Hydrated salt — find x in CuSO₄·xH₂O given % water
3. Mass of product from given mass of reactant (stoichiometry)
4. Volume of gas at STP from given mass of reactant
5. % composition of elements in a compound
6. Balancing equation + mole ratio → find mass/volume of a reactant needed

10) Quick Reference — All Formulas in One Place

Quantity	Formula
Number of moles (n)	n = m / M
Number of moles (from particles)	n = N / Nₐ
Number of moles (from gas volume at STP)	n = V (L) / 22.4
Number of moles (from gas volume at RTP)	n = V (L) / 24
Number of particles	N = n × Nₐ
Mass of substance	m = n × M
Volume of gas at STP	V = n × 22.4
Vapour density & molecular mass	MM = 2 × VD
Empirical formula mass → molecular	n = MM / empirical mass
% composition	% = (mass in formula / total mass) × 100

11) Practice Numericals (Try These)

Q1:

Find the number of moles in 11 g of CO₂. (CO₂ = 44 g/mol)

n = 11/44 = 0.25 mol

Q2:

How many molecules are there in 0.5 mol of H₂O?

N = n × Nₐ = 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules

Q3:

What volume does 2 mol of O₂ occupy at STP?

V = n × 22.4 = 2 × 22.4 = 44.8 L

Q4:

Calculate the percentage of oxygen in KMnO₄. (K=39, Mn=55, O=16)

Molar mass = 39 + 55 + 4×16 = 158. %O = (64/158) × 100 = 40.5%

Q5:

A compound has empirical formula CH₂ and molecular mass 56. Find the molecular formula.

Empirical mass = 14. n = 56/14 = 4. MF = 4 × CH₂ = C₄H₈

Q6:

20 g of CaCO₃ is heated. Calculate the mass of CaO produced. (Ca=40, C=12, O=16)

CaCO₃ → CaO + CO₂. Moles CaCO₃ = 20/100 = 0.2. Moles CaO = 0.2. Mass CaO = 0.2 × 56 = 11.2 g

Q7:

Calculate the volume of NH₃ produced at STP when 42.5 g of NH₄Cl reacts with excess Ca(OH)₂. Ca(OH)₂ + 2NH₄Cl → CaCl₂ + 2NH₃ + 2H₂O

Moles NH₄Cl = 42.5/53.5 = 0.794 mol. Ratio: 2 NH₄Cl → 2 NH₃, so moles NH₃ = 0.794. Volume = 0.794 × 22.4 = 17.8 L

Q8 (Limiting Reagent — Challenge):

6 g of H₂ is mixed with 48 g of O₂ and ignited to form H₂O. Which is the limiting reagent? How much H₂O is produced?

2H₂ + O₂ → 2H₂O. Moles H₂ = 6/2 = 3. Moles O₂ = 48/32 = 1.5.
From equation: 2 mol H₂ needs 1 mol O₂ → 3 mol H₂ needs 1.5 mol O₂. Matches exactly.
Moles H₂O = 3. Mass = 3 × 18 = 54 g.

🧠 From a 93.4% scorer — my final tips:
1. Always start with a balanced equation. Doesn't matter how simple the problem seems — write the equation first. 50% of mistakes happen because people skip this.
2. Track your units. If mass is in grams and molar mass is in g/mol, you're good. If volume is in mL, convert to L first.
3. Use the mole triangle. I literally drew it on my rough sheet at the start of every chemistry exam. Saved me multiple times.
4. For empirical formula questions, always divide by atomic mass first, then by the smallest number. If you get decimals like 1.5, multiply everything by 2. If 1.33, multiply by 3.
5. Don't panic if the numbers seem big. 6.022 × 10²³ just means "a very big number". Treat it like any other number in division.