Mole Concept & Stoichiometry (Class 11 Chemistry): Formulas + Solved Numericals
The three mole formulas every Class 11 student must know, with worked examples on mass, particles and limiting reagents.
The three formulas
Where n = moles, N = number of particles, NA = 6.022 × 10²³, and 22.4 L is the molar volume of an ideal gas at STP.
| You are given | Use | To find |
|---|---|---|
| Mass in grams | n = m/M | Moles |
| Number of atoms/molecules | n = N/NA | Moles |
| Gas volume at STP | n = V/22.4 | Moles |
Worked example 1 — mass to molecules
Q. How many molecules are in 36 g of water?
Molar mass of H₂O = 18 g/mol, so n = 36/18 = 2 mol. Molecules = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules.
Worked example 2 — limiting reagent
Q. 4 g H₂ reacts with 32 g O₂ to form water (2H₂ + O₂ → 2H₂O). Which is limiting?
n(H₂) = 4/2 = 2 mol; n(O₂) = 32/32 = 1 mol. The ratio needed is 2:1, and here it is exactly 2:1 — so both are fully consumed with none in excess, producing 2 mol (36 g) of water.
Common mistakes
- Using 22.4 L for non-STP conditions.
- Confusing atoms with molecules (1 mol O₂ = 2 mol O atoms).
- Forgetting to balance the equation before comparing mole ratios.
FAQs
Is the mole concept important for NEET and JEE too?
Yes — it is foundational for physical chemistry across boards and entrance exams.
What is molar mass in one line?
The mass in grams of one mole of a substance, numerically equal to its molecular/atomic mass in u.
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