CBSE Class 11 Mole Concept — NCERT-Style Problems Solved
"Some Basic Concepts of Chemistry" is the opening chapter of NCERT Class 11 Chemistry, and the mole concept is its heart. Students rarely fail this chapter because it is hard — they fail it because they try to remember dozens of separate formulas instead of the three that actually matter. This article gives you those three relations, then solves seven graded problems in full, showing every step of the arithmetic.
The only three bridges you need
The mole sits in the middle of everything. Mass, number of particles and gas volume never connect to each other directly — they each connect to the mole.
2. Mole ↔ particles: N = n × NA, where NA = 6.022 × 10²³ mol⁻¹
3. Mole ↔ gas volume: V = n × Vm
Here n is the amount in moles, m the mass in grams, M the molar mass in g/mol, and Vm the molar volume of a gas.
A note on molar volume that confuses many students. The current NCERT defines STP as 273.15 K and 1 bar, where Vm = 22.7 L/mol. Older books define STP as 273.15 K and 1 atm, where Vm = 22.4 L/mol. Both appear in question papers. Use the value your question implies, and state which one you used — that sentence alone protects your marks.
Problem 1 — Moles from mass
Q. Calculate the number of moles in 5.85 g of sodium chloride.
M(NaCl) = 22.99 + 35.45 = 58.44 g/mol
n = m / M = 5.85 ÷ 58.44 = 0.100 mol
Sense check: 5.85 g is roughly one-tenth of 58.44 g, so 0.1 mol is right.
Problem 2 — Number of molecules and atoms
Q. How many molecules are present in 8.8 g of CO₂? How many oxygen atoms?
M(CO₂) = 12.011 + 2 × 15.999 = 12.011 + 31.998 = 44.01 g/mol
n = 8.8 ÷ 44.01 = 0.200 mol
N = 0.200 × 6.022 × 10²³ = 1.204 × 10²³ molecules
Each CO₂ molecule has 2 oxygen atoms, so
oxygen atoms = 2 × 1.204 × 10²³ = 2.409 × 10²³ atoms
Problem 3 — Mass from moles
Q. What is the mass of 0.25 mol of sulphuric acid?
M(H₂SO₄) = 2 × 1.008 + 32.06 + 4 × 15.999
= 2.016 + 32.06 + 63.996 = 98.07 g/mol
m = n × M = 0.25 × 98.07 = 24.52 g
Problem 4 — Gas volume at STP
Q. Find the volume occupied by 3.2 g of oxygen gas at STP.
M(O₂) = 2 × 16.00 = 32.00 g/mol
n = 3.2 ÷ 32.00 = 0.100 mol
Using Vm = 22.4 L/mol (273.15 K, 1 atm):
V = 0.100 × 22.4 = 2.24 L
Using Vm = 22.7 L/mol (273.15 K, 1 bar):
V = 0.100 × 22.7 = 2.27 L
Remember that the molar volume applies to any gas — 0.1 mol of H₂, CO₂ or O₂ all occupy the same volume at the same temperature and pressure.
Problem 5 — Empirical and molecular formula
Q. A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. Its molar mass is 180 g/mol. Find its empirical and molecular formula.
Step 1 — take 100 g of the compound, so the percentages become grams: 40.0 g C, 6.7 g H, 53.3 g O.
Step 2 — convert to moles.
C: 40.0 ÷ 12.011 = 3.330 mol
H: 6.7 ÷ 1.008 = 6.647 mol
O: 53.3 ÷ 15.999 = 3.331 mol
Step 3 — divide by the smallest (3.330).
C: 3.330 ÷ 3.330 = 1.00
H: 6.647 ÷ 3.330 = 2.00
O: 3.331 ÷ 3.330 = 1.00
Empirical formula = CH₂O
Step 4 — scale up to the molecular formula.
Empirical formula mass = 12.011 + 2.016 + 15.999 = 30.03
n = 180 ÷ 30.03 = 5.99 ≈ 6
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆ — glucose.
Problem 6 — Limiting reagent
Q. 28 g of N₂ is reacted with 6 g of H₂ to form ammonia. Which is the limiting reagent, and what mass of NH₃ is produced?
Balanced equation: N₂ + 3H₂ → 2NH₃
Step 1 — moles of each reactant.
M(N₂) = 28.02 g/mol → n(N₂) = 28 ÷ 28.02 = 1.00 mol
M(H₂) = 2.016 g/mol → n(H₂) = 6 ÷ 2.016 = 2.98 mol
Step 2 — compare with the equation. One mole of N₂ needs 3 mol of H₂. We have only 2.98 mol of H₂, so hydrogen is the limiting reagent and some nitrogen will be left over.
Step 3 — work from the limiting reagent only. 3 mol H₂ gives
2 mol NH₃, so
n(NH₃) = 2.98 × (2 ÷ 3) = 1.984 mol
Step 4 — convert to mass. M(NH₃) = 14.007 + 3 × 1.008 = 17.03 g/mol
m(NH₃) = 1.984 × 17.03 = 33.8 g
Problem 7 — Molarity of a solution
Q. 4 g of NaOH is dissolved in water and made up to 250 mL. Find the molarity.
M(NaOH) = 22.99 + 15.999 + 1.008 = 40.00 g/mol
n = 4 ÷ 40.00 = 0.100 mol
Volume = 250 mL = 0.250 L
Molarity = 0.100 ÷ 0.250 = 0.400 mol L⁻¹ (0.4 M)
Common mistakes that cost marks
- Dividing volume in millilitres. Molarity needs litres. 250 mL is 0.250 L, not 250.
- Using atomic mass where molecular mass is needed. Oxygen gas is O₂ with M = 32, not 16. This single slip halves or doubles most gas answers.
- Ignoring the limiting reagent. If the question gives you the amounts of both reactants, it is almost certainly a limiting reagent problem. Never calculate the product from the excess reactant.
- Rounding the mole ratio too aggressively. In empirical formula questions, 1.5 means multiply everything by 2 — it does not mean "about 1" or "about 2".
- Mixing up molarity and molality. Molarity uses litres of solution; molality uses kilograms of solvent. Read the unit carefully: mol L⁻¹ versus mol kg⁻¹.
The conversion map in one table
| You have | You want | Do this |
|---|---|---|
| Mass (g) | Moles | Divide by molar mass |
| Moles | Mass (g) | Multiply by molar mass |
| Moles | Molecules / atoms | Multiply by 6.022 × 10²³ |
| Molecules | Moles | Divide by 6.022 × 10²³ |
| Moles of gas | Volume at STP | Multiply by 22.4 L (1 atm) or 22.7 L (1 bar) |
| Moles + volume (L) | Molarity | Divide moles by litres |
Practise the conversions until they are automatic. The free Mass ↔ Mole calculator converts grams to moles and moles to grams for any formula, and shows the molar mass it used — perfect for checking Problems 1, 3 and 4 above and for building speed before a test.
Open the Mass ↔ Mole Calculator →Want this chapter taught properly rather than memorised? ABC Chemistry runs Class 11–12 chemistry coaching at the Gurugram centre and online classes across India, with home tuition available in the Delhi–Gurugram region — details at abcchemistry.in.