QCC Notes
CLASS 11 · CHEMISTRY JEE MAIN × NEET हिंदी
§ 1.8NCERT Class 11 · Chemistry · Chapter 1

Mole Concept: Formulas, Numericals & the Complete Mole Map

From Avogadro’s number to gas volumes — NCERT-aligned theory, the traps NTA actually sets, an interactive mole-map converter, and JEE/NEET numericals solved the way toppers write them.

01

What is Mole Concept? — Complete Theory

Chemistry happens at a scale where counting is impossible and weighing particles individually is absurd. The mole (symbol: mol) is the SI’s answer — the base unit for amount of substance. One mole of any substance contains exactly 6.02214076 × 10²³ elementary entities — atoms, molecules, ions, electrons, whatever you are counting. Think of it as a chemist’s dozen: the way “a dozen” always means 12, a mole always means 6.022 × 10²³ of anything.

Where does that strange number come from? It is chosen so that the invisible atomic scale and the laboratory scale agree perfectly: the molar mass of a substance in g/mol is numerically equal to its atomic or molecular mass in u. One ¹²C atom has a mass of exactly 12 u; one mole of ¹²C atoms has a mass of exactly 12 g. That single bridge sentence is the heart of the entire mole concept — learn it as an idea, not as a formula.

The molar mass (M) is the mass of one mole of a substance, in g mol⁻¹, built straight from the formula: M(H2O) = 2(1.008) + 16.00 ≈ 18 g mol⁻¹, M(CO2) = 44 g mol⁻¹. For ionic compounds like NaCl there are no molecules — we count formula units and use formula mass (58.5 u for NaCl, hence 58.5 g mol⁻¹).

Every mole problem is one of three conversions — and all of them pass through n, the number of moles:

From massn = m / M From particlesn = N / NA From gas volume, STPn = V / 22.4
Table 1 — One mole of different substances
SubstanceMolar mass1 mol containsNatureVol. of 1 mol at STP
Carbon-12 (¹²C)12 g6.022 × 10²³ atomsSolid—
Water (H2O)18 g6.022 × 10²³ moleculesLiquid—
Oxygen gas (O2)32 g6.022 × 10²³ moleculesGas22.4 L
Carbon dioxide (CO2)44 g6.022 × 10²³ moleculesGas22.4 L
Sodium chloride (NaCl)58.5 g6.022 × 10²³ formula unitsSolid—
Table 2 — Terms students confuse (and papers exploit)
TermMeaningUnit
Atomic massMass of one atom relative to ¹²Cu
Molecular massSum of atomic masses in one moleculeu
Formula massSum for ionic compounds (NaCl)u
Molar massMass of one mole of entitiesg mol⁻¹
Gram atomic massAtomic mass expressed in grams = mass of 1 mol atomsg

The percentage composition of a compound is each element’s share of the molar mass: % element = (number of atoms × atomic mass ÷ molar mass) × 100. Water: %H = (2 × 1)/18 × 100 = 11.11%, %O = 88.89%. Run the logic backwards and percentage data gives you the empirical formula (simplest whole-number ratio), which you scale to the molecular formula using n = molar mass ÷ empirical-formula mass.

Why do chemists obsess over moles? Because a balanced equation is a mole recipe: C + O2 → CO2 reads “1 mol C reacts with 1 mol O2 to give 1 mol CO2” — never grams directly. Convert what’s given into moles, use the coefficient ratio, convert back: that is the entire game of stoichiometry (Section 1.10).

02

Visualising the Mole Map

One hub, three roads. Whatever you’re given and whatever is asked, the journey always changes trains at n.

The mole map: n in the centre; mass, particles and gas volume around it, each linked by its conversion factor FIG. 1 — THE MOLE MAP × NA ÷ NA × M ÷ M × 22.4 ÷ 22.4 MOLE n amount · mol PARTICLES N entities MASS m grams (g) VOLUME V gas · STP (L) click any node → try it live below
FIG. 1 — n is the hub of the map: × M carries you between mass and moles, × NA between moles and particles, × 22.4 between moles and gas volume at STP (22.7 at 1 bar). The road between any two outer points is just the two roads through n.
Try it live

Mole Map Converter

g
mol
count
L

M values follow NCERT exam convention (H₂O = 18, CO₂ = 44 …). Particle input accepts e-notation (6.022e23). Molar volume: 22.4 L/mol at 1 atm · 22.7 L/mol at 1 bar.

03

Solved Examples (Step-by-Step)

Given → formula → substitute → verify. Write these same steps on your rough sheet — that is what earns full method marks.

EXAMPLE 01Foundation · NEET

Mass → moles → molecules

How many moles of water and how many molecules of water are present in 36 g of pure water? (M = 18 g mol⁻¹, NA = 6.022 × 10²³ mol⁻¹)

  1. Givenm = 36 g, M(H2O) = 18 g mol⁻¹. Find n and N.
  2. Formulan = m / M
  3. Substituten = 36 ÷ 18 = 2 mol
  4. ConvertN = n × NA = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴

n = 2 mol · N = 1.204 × 10²⁴ molecules

EXAMPLE 02JEE Main · Atomicity twist

Atoms, not molecules — count twice

The total number of atoms present in 4.25 g of NH3 is approximately: (a) 1.5 × 10²³ (b) 2.5 × 10²³ (c) 6.022 × 10²³ (d) 4 × 6.022 × 10²³

  1. Givenm = 4.25 g, M(NH3) = 17 g mol⁻¹ — NH3 has 4 atoms per molecule.
  2. Molesn = 4.25 ÷ 17 = 0.25 mol → molecules = 0.25 × NA
  3. Atomicityatoms = 4 × 0.25 × NA = 1.0 × NA — multiply by 4 — this is the step the marks are on.

Option (c) — 6.022 × 10²³ atoms

EXAMPLE 03JEE Advanced-friendly · % composition

From percentage data to empirical formula

An oxide of iron contains 69.9% Fe and 30.1% oxygen by mass. Determine its empirical formula. (Fe = 56, O = 16)

  1. AssumeTake exactly 100 g of oxide: Fe = 69.9 g, O = 30.1 g.
  2. Molesn(Fe) = 69.9 ÷ 56 = 1.248 · n(O) = 30.1 ÷ 16 = 1.881
  3. RatioDivide by the smaller: Fe = 1.000, O = 1.507 ≈ 1.5
  4. Whole no.× 2 → Fe : O = 2 : 3
  5. VerifyM(Fe2O3) = 160 → %Fe = 112/160 × 100 = 70% ≈ 69.9% ✓

Empirical formula: Fe₂O₃

05

Key Formulas & Takeaways

Formula card

Eight lines that solve this chapter

n = m / MMoles from mass — m in grams, M in g mol⁻¹.
n = N / NAMoles from count of atoms / molecules / ions.
N = n × NAParticles from moles — multiply by 6.022 × 10²³.
V = n × 22.4Gas volume at STP (1 atm); use 22.7 L mol⁻¹ at 1 bar.
mentity = M / NAMass of one molecule or atom, in grams.
% element = (atoms × a × 100) / MPercentage composition; a = atomic mass of that element.
n = M / (empirical mass)Multiplier from empirical to molecular formula.
Natoms = n × NA × atomicityAtoms vs molecules — the step where marks are won.

NA = 6.02214076 × 10²³ mol⁻¹ (exact, SI 2019)  ·  1 u = 1.6605 × 10⁻²⁴ g  ·  Molar volume = 22.4 L mol⁻¹ (1 atm) | 22.7 L mol⁻¹ (1 bar)  ·  STP = 273.15 K

  1. A mole is a count, not a weight — 6.022 × 10²³ of anything. The gram-value arrives only through molar mass.
  2. Every road passes through n — mass ↔ n ↔ particles / gas volume. Two formulas do 90% of the numericals.
  3. Molar volume is a gas-only, STP-only shortcut — 22.4 L at 1 atm, 22.7 L at 1 bar. Liquids and solids: no entry.
  4. Atoms vs molecules is where marks die — multiply by atomicity before you box the answer.
06

FAQs

What is a mole in chemistry?

One mole is the amount of substance that contains exactly 6.02214076 × 10²³ elementary entities — atoms, molecules, ions or electrons. It is the SI base unit of amount of substance (symbol: mol) and works like a chemist’s dozen.

How do you calculate the number of moles from mass?

Divide the given mass by the molar mass: n = m ÷ M. For example, 36 g of water (M = 18 g/mol) contains 36 ÷ 18 = 2 moles of water molecules.

Why is 22.4 L called the molar volume of a gas?

At STP (273.15 K, 1 atm) one mole of any ideal gas occupies 22.4 L. The latest NCERT defines STP at 1 bar, where the molar volume is 22.7 L — so always check which standard pressure a question assumes.

How many particles are there in one mole?

Exactly 6.02214076 × 10²³ particles — Avogadro’s constant. So 0.5 mol of O₂ contains 3.011 × 10²³ O₂ molecules, which is 6.022 × 10²³ oxygen atoms.

04

Practice Questions

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QCC Notes — Class 11 Chemistry

Strictly NCERT-aligned notes for JEE Main & NEET, prepared by QCC Notes (Padho Likho JEE). Content follows the latest NCERT edition and current NTA exam pattern.

Last updated
29 Aug 2026