Mole Concept and Molar Mass

The mole is the central counting unit of chemistry and the most examined idea in this chapter — expect multiple numericals in CBSE and NEET. Master the interconversions of mass, moles, particles and gas volume.

Core ideas

Mole: one mole = amount of substance containing exactly 6.022 × 10²³ elementary entities (Avogadro's number, Nₐ). One mole of any substance contains Nₐ particles.

Molar mass (M): mass of one mole in grams; numerically equal to atomic/molecular mass in u. Units g mol⁻¹ (H₂O = 18 g mol⁻¹).

Key interconversions:

• Moles = given mass / molar mass → n = m / M
• Number of particles = n × Nₐ
• Volume of a gas at STP = n × 22.4 L (molar volume; 22.7 L at 1 bar STP)

Molar volume: at STP (273.15 K, 1 atm) one mole of any ideal gas occupies 22.4 litres.

Example: 9 g of water = 9/18 = 0.5 mol = 0.5 × 6.022×10²³ = 3.011 × 10²³ molecules, containing 2 × that many H atoms.

Exam Tricks & Tips

• 🎯 Always route through moles: mass ⇄ moles ⇄ particles/volume. Never jump mass → particles directly.
• 🎯 Molar mass in g mol⁻¹ = molecular mass in u — no extra calculation needed.
• 🎯 At STP use 22.4 L/mol (old NCERT) — check whether the paper defines STP at 1 atm (22.4 L) or 1 bar (22.7 L).
• 🎯 To count atoms in a molecule, multiply molecules by the subscript (1 mol H₂O has 2Nₐ H atoms and Nₐ O atoms).
• 🎯 For an ion, 1 mole still contains Nₐ ions — the mole counts any specified entity.
• ❌ Common mistake: using 22.4 L for a substance that is liquid or solid at STP — molar volume 22.4 L applies only to gases.

Expected exam pattern

Expect 2–3 numericals: mass↔moles, moles→number of atoms/molecules, and gas mass↔volume at STP. Often combined ("number of oxygen atoms in 4.4 g CO₂").

Quick recap

1 mole = 6.022×10²³ entities; molar mass (g/mol) = molecular mass (u); n = m/M; particles = n·Nₐ; gas volume at STP = n × 22.4 L. Always convert via moles.

Cover the answer, recall, then check. 12 cards on the mole concept and molar mass.

Q1. What is a mole?
A1. The amount of substance containing 6.022 × 10²³ elementary entities (Avogadro's number).

Q2. State the value of Avogadro's number.
A2. 6.022 × 10²³ per mole.

Q3. Define molar mass.
A3. The mass of one mole of a substance in grams (g mol⁻¹), numerically equal to its atomic/molecular mass in u.

Q4. Formula linking moles, mass and molar mass?
A4. n = mass / molar mass (n = m/M).

Q5. How many molecules are in 0.5 mol of a substance?
A5. 0.5 × 6.022×10²³ = 3.011 × 10²³ molecules.

Q6. Volume of one mole of an ideal gas at STP?
A6. 22.4 litres (22.7 L at 1 bar STP).

Q7. How many moles are in 9 g of water?
A7. 9/18 = 0.5 mol.

Q8. How many hydrogen atoms in one mole of water?
A8. 2 × 6.022×10²³ = 1.204 × 10²⁴ H atoms.

Q9. Number of particles in n moles?
A9. n × 6.022×10²³.

Q10. Does molar volume 22.4 L apply to solids and liquids?
A10. No — only to gases at STP.

Q11. Molar mass of CO₂?
A11. 44 g mol⁻¹.

Q12. How many moles are in 6.022 × 10²² atoms?
A12. 6.022×10²² / 6.022×10²³ = 0.1 mol.

Atoms are absurdly small — a single drop of water holds more than a billion trillion of them. To bridge the invisible atomic world and the gram-scale lab bench, chemists invented one number: the mole. It is the single most used idea in all of chemistry.

Definition: A mole is the amount of substance containing exactly 6.022 × 10²³ elementary entities (atoms, molecules, ions). This number is Avogadro's constant (Nₐ).

Beginner — what a mole counts

Just as "a dozen" means 12 of anything, "a mole" means 6.022 × 10²³ of anything:

• 1 mole of carbon atoms = 6.022 × 10²³ C atoms
• 1 mole of water = 6.022 × 10²³ H₂O molecules
• 1 mole of Na⁺ = 6.022 × 10²³ sodium ions

Intermediate — molar mass, the master link

Molar mass (M) = mass of one mole of a substance, in grams per mole (g/mol). Numerically it equals the atomic/molecular mass in u:

• C: atomic mass 12 u → molar mass 12 g/mol
• H₂O: molecular mass 18 u → molar mass 18 g/mol

The three core interconversions:

• Moles = given mass ÷ molar mass → n = m / M
• Number of particles = moles × Nₐ
• Volume of gas at STP = moles × 22.7 L (22.4 L at old STP 0°C, 1 atm; 22.7 L at current STP 0°C, 1 bar)

Advanced — mole as a hub

Everything routes through moles: mass → moles → particles → gas volume. Molar mass converts mass↔moles; Avogadro's constant converts moles↔particles; molar volume converts moles↔gas volume. Learn to move in both directions along this hub and no numerical can surprise you.

Worked example

How many oxygen atoms are in 36 g of water?

Moles of H₂O = 36 g ÷ 18 g/mol = 2 mol.
Molecules of H₂O = 2 × 6.022 × 10²³ = 1.204 × 10²⁴.
Each H₂O has 1 O atom, so O atoms = 1.204 × 10²⁴.

(Bonus: H atoms = 2 × that = 2.408 × 10²⁴.)

Real-world / exam application

Dosage in pharmacology, reactant weighing in industry, and every titration depend on moles. In JEE/NEET the mole concept underpins ~15–20% of physical-chemistry questions — mass↔mole↔particle conversions, atoms-in-a-sample, and gas-volume problems recur every year.

Exam tricks & mnemonic

Triangle trick: cover what you want in n = m/M (mass over molar mass). For "number of X atoms," always multiply molecules by the count of X per formula unit. Mnemonic "Mass Makes Moles, Moles Make Molecules": m → n → N.

Key formulas

• Avogadro number: N_A = 6.022 × 10²³ per mole.
• Moles: n = mass/molar mass = m/M.
• Number of particles: N = n × N_A.
• Molar volume of a gas at STP: 22.7 L/mol (at 0 °C, 1 bar) [22.4 L at 1 atm].
• Moles of gas: n = V(at STP)/molar volume.
• Molar mass (g/mol) is numerically equal to atomic/molecular mass in u.

The mole links the mass of a substance to the number of particles it contains.

Worked Example

Problem: Calculate (a) the number of moles and (b) the number of molecules present in 22 g of carbon dioxide (CO₂). (Atomic masses: C = 12, O = 16; Avogadro's number = 6.022 × 10²³.)

Solution:
Step 1 — Find the molar mass of CO₂:
M = 12 + 2(16) = 12 + 32 = 44 g/mol.

Step 2 — Number of moles = given mass ÷ molar mass:
n = 22 / 44 = 0.5 mol.

Step 3 — Number of molecules = moles × Avogadro's number:
N = 0.5 × (6.022 × 10²³).

Step 4 — Evaluate:
N = 3.011 × 10²³ molecules.

Answer: 22 g of CO₂ contains 0.5 mol, i.e. 3.011 × 10²³ molecules.