Mole Concept Explained with Examples
Introduction
The mole concept is a cornerstone of modern quantitative chemistry, enabling precise measurement of substances at the atomic and molecular scale. Whether calculating reactant quantities for a chemical reaction or interpreting analytical data, a clear understanding of the mole ensures accuracy in both academic and competitive exam environments like JEE, NEET, and IIT JAM. In this post, we will dissect the mole concept through definitions, detailed explanations, practical examples, and revision tables, equipping students with the tools to master this fundamental topic.
Definition and Core Concepts
– What Is a Mole?
A mole (mol) is defined as the amount of substance containing exactly 6.022×10236.022 \times 10^{23} elementary entities (Avogadro’s number). This standard provides a bridge between the microscopic world of atoms and the macroscopic world of grams.
| Quantity | Symbol | Value |
|---|---|---|
| Avogadro’s number | Nₐ | 6.022 × 10²³ entities/mol |
| One mole of carbon-12 | — | Exactly 12 grams |
| Unit of amount of substance | mol | Base SI unit |
– Molar Mass and Molar Volume
- Molar mass (MM) is the mass of one mole of a substance, expressed in g/mol.
- Molar volume is the volume occupied by one mole of a gas at standard temperature and pressure (STP: 273.15 K, 1 atm), equal to 22.414 L.
| Substance | Molar Mass (g/mol) | Molar Volume at STP (L/mol) |
|---|---|---|
| H₂ | 2.016 | 22.414 |
| O₂ | 32.00 | 22.414 |
| CO₂ | 44.01 | 22.414 |
Key Points and Formulas
- Number of moles:
n = m / M
where m = mass (g), M = molar mass (g/mol) - Ideal gas law:
n = PV / RT - Particles to moles:
n = N / Nₐ
where N = number of particles - Concentration and moles:
n = C × V
where C = concentration (mol/L), V = volume (L)
Important Notes
- Molar mass = sum of atomic masses.
- STP = 273.15 K and 1 atm.
- Always check for unit consistency using dimensional analysis.
Classification of Calculations
| Calculation Type | Formula | Typical Use Case |
|---|---|---|
| Mass-to-moles | n = m / M | Gravimetric analysis |
| Volume-to-moles (gas) | n = V / 22.414 | Gas stoichiometry |
| Particles-to-moles | n = N / Nₐ | Nanomaterial calculations |
| Solutions (conc. based) | n = C × V | Titration or dilution setups |
Some Examples
Example 1: Water Formation
Q: Calculate the mass of water produced when 5.00 g of hydrogen reacts completely with oxygen:
Reaction: 2H₂ + O₂ → 2H₂O
Solution:
- Moles of H₂:
n = 5.00 / 2.016 = 2.479 mol - From reaction:
2 mol H₂ → 2 mol H₂O →
Moles of H₂O = 2.479 mol - Mass of H₂O = 2.479 × 18.016 = 44.65 g
Example 2: Volume of Nitrogen at STP
Q: What is the volume of 0.500 mol of N₂ gas at STP?
Solution:
V = 0.500 × 22.414 = 11.207 LQuick Revision Table
| Concept | Formula / Note |
|---|---|
| Avogadro’s number | 6.022 × 10²³ entities/mol |
| Moles (mass-based) | n = m / M |
| Gas at STP | 1 mol = 22.414 L |
| Solution concentration | n = C × V |
| Number of particles | n = N / Nₐ |
Important Formulas
Molar mass = Σ (atomic masses of all atoms in the molecule)
- n = m / M
- n = PV / RT
- n = N / Nₐ
- n = C × V
Mole Concept: Solved Questions with Detailed Solutions
Class 11/12 Level Questions
Q1. How many atoms are present in 2 moles of magnesium?
Solution:
We know that 1 mole of any substance contains Avogadro’s number of particles =6.022 × 10²³.
Number of atoms in 2 moles of Mg
= 2 × 6.022 × 10²³
= 1.2044 × 10²⁴ atoms.
Q2. Calculate the number of moles in 88 g of CO₂.
Solution:
Molar mass of CO₂ = 12 (C) + 16×2 (O) = 44 g/mol
Moles = Given mass / Molar mass
= 88 / 44 = 2 moles
Q3. What is the mass of 0.5 mole of O₂ gas?
Solution:
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass = Moles × Molar mass
= 0.5 × 32 = 16 g
Q4. How many molecules are present in 1.8 g of water (H₂O)?
Solution:
Molar mass of water = 2×1 + 16 = 18 g/mol
Moles of water = 1.8 / 18 = 0.1 mol
Molecules = Moles × Avogadro’s number
= 0.1 × 6.022 × 10²³ = 6.022 × 10²² molecules
Q5. Calculate the number of moles of electrons in 1 C of charge. (Given: 1 e = 1.6 × 10⁻¹⁹ C)
Solution:
Number of electrons = Total charge / Charge of one electron
= 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons
Moles of electrons = Number of electrons / Avogadro’s number
= (6.25 × 10¹⁸) / (6.022 × 10²³)
≈ 1.038 × 10⁻⁵ mol
JEE Level Questions
Q6. (JEE Main PYQ)
0.5 mole of a gas occupies 11.2 L at STP. What is the molar volume of the gas at STP?
Solution:
Volume of 0.5 mol = 11.2 L
Molar volume = Volume / Moles
= 11.2 / 0.5 = 22.4 L/mol
This matches the standard molar volume at STP.
Q7. (JEE Advanced PYQ)
If 2.0 g of a gas occupies 1.12 L at STP, find its molar mass.
Solution:
At STP, 1 mole occupies 22.4 L.
Given Volume = 1.12 L
Moles = Volume / 22.4 = 1.12 / 22.4 = 0.05 mol
Molar mass = Mass / Moles = 2.0 / 0.05 = 40 g/mol
Q8. (JEE Main PYQ)
How many oxygen atoms are present in 0.5 mole of Al₂(SO₄)₃?
Solution:
Each Al₂(SO₄)₃ unit has 3 SO₄²⁻, and each SO₄ has 4 oxygen atoms.
So, oxygen atoms per formula unit = 3 × 4 = 12
Number of oxygen atoms = 0.5 × 6.022 × 10²³ × 12
= 3.613 × 10²⁴ atoms
NEET Level Questions
Q9. (NEET PYQ)
Calculate the number of molecules in 36 g of water.
Solution:
Molar mass of water = 18 g/mol
Moles = 36 / 18 = 2 mol
Molecules = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴
Q10. (NEET PYQ)
How many moles are present in 5.0 × 10²⁴ molecules of NH₃?
Solution:
Moles = Number of molecules / Avogadro’s number
= (5.0 × 10²⁴) / (6.022 × 10²³)
≈ 8.3 mol
Q11. (NEET PYQ)
Calculate the mass of 1.204 × 10²⁴ atoms of iron. (Atomic mass of Fe = 56 g/mol)
Solution:
Number of moles = (1.204 × 10²⁴) / (6.022 × 10²³)
= 2 mol
Mass = Moles × Molar mass = 2 × 56 = 112 g
IIT JAM Level Questions
Q12. (IIT JAM PYQ)
What is the mass of 1.5 moles of calcium nitrate, Ca(NO₃)₂?
Solution:
Molar mass:
Ca = 40, N = 14, O = 16
Ca(NO₃)₂ = 40 + 2 × (14 + 48) = 40 + 2 × 62 = 164 g/mol
Mass = 1.5 × 164 = 246 g
Q13. (IIT JAM PYQ)
Calculate the number of formula units in 4.9 g of H₂SO₄.
Solution:
Molar mass = 2 + 32 + 64 = 98 g/mol
Moles = 4.9 / 98 = 0.05 mol
Number of formula units = 0.05 × 6.022 × 10²³
= 3.011 × 10²² units
Q14. (IIT JAM PYQ)
A compound contains 4.8 g of oxygen and 1.2 g of carbon. Calculate the empirical formula.
Solution:
Moles of C = 1.2 / 12 = 0.1 mol
Moles of O = 4.8 / 16 = 0.3 mol
Mole ratio = C:O = 0.1:0.3 = 1:3
Empirical formula = CO₃
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