What is the Mole Concept?
The mole is the SI unit for amount of substance. One mole of any substance contains exactly 6.022 × 1023 particles (Avogadro's number, NA). This bridges the atomic/molecular world with the measurable macroscopic world. The molar mass (in g/mol) numerically equals the substance's atomic or molecular weight. At Standard Temperature and Pressure (STP), one mole of any ideal gas occupies 22.4 litres.
Core Mole Concept Formulas
| Symbol | Quantity | Unit | Value / Notes |
|---|
| n | Amount of substance (moles) | mol | Fundamental SI unit for amount |
| Nₐ | Avogadro's number | mol−1 | 6.022 × 1023 |
| M | Molar mass | g/mol | Numerically = molecular/atomic weight |
| m | Mass of substance | g | m = n × M |
| N | Number of particles | dimensionless | N = n × Nₐ |
| VSTP | Volume at STP (gas) | L | 22.4 L/mol at old STP; 22.711 at new STP |
| C or M | Molarity | mol/L (M) | Moles of solute per litre of solution |
| N | Normality | eq/L (N) | N = M × n-factor |
Solved Examples (NCERT / JEE Level)
Example 01 — Mole Conversions (Water)
Calculate the number of molecules in 36 g of water (H₂O). Also find the volume it would occupy if converted to steam at STP. Molar mass of H₂O = 18 g/mol.
n = m / M = 36 / 18 = 2 mol
Number of molecules:
N = n × Nₐ = 2 × 6.022 × 1023
N = 1.204 × 1024 molecules
Number of atoms (each H₂O has 3 atoms):
= 3 × 1.204 × 1024 = 3.613 × 1024 atoms
Volume at STP (steam):
V = n × 22.4 = 2 × 22.4 = 44.8 L
n = 2 mol | N = 1.204 × 1024 molecules | V = 44.8 L at STP
Example 02 — Molarity & Normality
4.9 g of H₂SO₄ is dissolved in water to make 500 mL of solution. Calculate the molarity and normality. (Molar mass of H₂SO₄ = 98 g/mol; n-factor = 2)
Moles of H₂SO₄ = m / M = 4.9 / 98 = 0.05 mol
Molarity = n / V(L) = 0.05 / 0.500 = 0.1 M
Normality = Molarity × n-factor
N = 0.1 × 2 = 0.2 N
Dilution check: If diluted to 1000 mL:
M₁V₁ = M₂V₂
0.1 × 500 = M₂ × 1000
M₂ = 0.05 M
Molarity = 0.1 M | Normality = 0.2 N | After dilution = 0.05 M
Example 03 — Limiting Reagent
In the reaction: 2H₂ + O₂ → 2H₂O, 4 g of H₂ reacts with 32 g of O₂. Find the limiting reagent and theoretical yield of water.
Moles of H₂ = 4 / 2 = 2 mol
Moles of O₂ = 32 / 32 = 1 mol
Stoichiometric ratio: H₂ : O₂ = 2 : 1
For 2 mol H₂, we need 1 mol O₂ → exactly available
→ Both are consumed completely (no excess)
Moles of H₂O produced = 2 mol (from 2 mol H₂)
Mass of H₂O = n × M = 2 × 18 = 36 g
Neither is strictly limiting (exact ratio) | Theoretical yield = 36 g H₂O
Frequently Asked Questions (FAQs)
What is Avogadro's number and why is it exactly 6.022 × 1023?
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Avogadro's number (Nₐ = 6.022 × 1023) is the number of entities (atoms, molecules, ions) in one mole of a substance. It is defined such that one mole of carbon-12 atoms has a mass of exactly 12 grams. In 2019, the SI redefined the mole with Nₐ = 6.02214076 × 1023 mol−1 exactly. The number is enormous because atoms are incredibly small — it bridges the sub-atomic scale with the laboratory scale.
What is the difference between molarity and normality?
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Molarity (M) is the number of moles of solute per litre of solution. Normality (N) is the number of equivalents per litre, where equivalents account for the reactive capacity (n-factor) of the solute. N = M × n-factor. For HCl, n-factor = 1 (releases 1 H&sup+;), so N = M. For H₂SO₄, n-factor = 2, so N = 2M. For redox reactions, the n-factor is the change in oxidation state per molecule (e.g., KMnO₄ in acidic medium: n = 5).
How do you identify the limiting reagent?
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The limiting reagent is the reactant that is completely consumed first, limiting the amount of product formed. To find it: (1) Convert all reactant masses to moles. (2) Divide each by its stoichiometric coefficient. (3) The reactant with the smallest quotient is the limiting reagent. The other reactant is in excess. Product yield is calculated based on the moles of the limiting reagent only. The excess reactant's leftover mass can also be calculated.
What is the molar volume of a gas at STP?
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At Old STP (0°C, 1 atm = 101.325 kPa), one mole of ideal gas occupies 22.4 L. This is the value used in most NCERT and JEE problems. At New STP (0°C, 100 kPa, IUPAC 1982 definition), it is 22.711 L. At RTP (Room Temperature and Pressure: 25°C, 1 atm), it is approximately 24.465 L. Always check which standard your problem specifies.
What is percentage yield and why is it rarely 100%?
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Percentage yield = (Actual yield / Theoretical yield) × 100. The theoretical yield assumes complete conversion of the limiting reagent under ideal conditions. In practice, yield is less than 100% because of: (1) Side reactions producing unwanted by-products. (2) Reversible reactions that don't go to completion. (3) Physical losses during transfer, filtration, or purification. (4) Impure reagents. Industrial processes aim to maximise yield while minimising waste.