Mole Concept and Stoichiometry
Introduction
The mole concept links the microscopic world of atoms and molecules to the macroscopic world of grams and litres. In ICSE Class 10 Chemistry, you learn to perform calculations involving moles, masses, volumes, and the composition of compounds.
Key Definitions
Mole
1 mole = 6.022 × 10²³ particles (Avogadro's number)
The mass of 1 mole of a substance is equal to its gram molecular mass.
- 1 mole of carbon atoms = 12 g = 6.022 × 10²³ atoms
- 1 mole of water molecules = 18 g = 6.022 × 10²³ molecules
Important Formulas
Number of moles (n) = Given mass (m) / Molar mass (M) Number of moles (n) = Volume at STP (L) / 22.4 L Number of particles = n × 6.022 × 10²³
Gay Lussac's Law of Gaseous Volumes
When gases react, they do so in volumes that bear a simple ratio to one another and to the volumes of the products (if gaseous), all volumes measured at the same temperature and pressure.
Example: H₂(g) + Cl₂(g) → 2HCl(g) Volume ratio: 1 : 1 : 2
Avogadro's Law
Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules.
- 1 mole of any gas at STP occupies 22.4 L (molar volume).
- STP: 0°C (273 K) and 1 atm (760 mm Hg) pressure.
Vapour Density (VD)
Vapour density = Mass of a given volume of gas / Mass of same volume of hydrogen (at same T and P)
VD = Molar mass of gas / 2
Molar mass = 2 × Vapour density
Empirical and Molecular Formula
| Term | Definition |
|---|---|
| Empirical formula | Simplest whole-number ratio of atoms in a compound |
| Molecular formula | Actual number of atoms of each element in a molecule |
Molecular formula = (Empirical formula)ₙ n = Molecular mass / Empirical formula mass
Worked Numericals
Example 1: Mole Calculations
Find the number of moles in 22 g of CO₂.
Solution: Molar mass of CO₂ = 12 + 2 × 16 = 44 g/mol Number of moles = 22 / 44 = 0.5 moles
Example 2: Volume at STP
What volume does 2 moles of O₂ occupy at STP?
Solution: Volume = n × 22.4 = 2 × 22.4 = 44.8 L
Example 3: Empirical Formula
A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen. Find its empirical formula. (Given atomic masses: C = 12, H = 1, O = 16)
Solution:
| Element | % | Atomic mass | Moles | Ratio |
|---|---|---|---|---|
| C | 40 | 12 | 40/12 = 3.33 | 1 |
| H | 6.67 | 1 | 6.67/1 = 6.67 | 2 |
| O | 53.33 | 16 | 53.33/16 = 3.33 | 1 |
Empirical formula = CH₂O
If the molecular mass is 60, find the molecular formula. Empirical formula mass = 12 + 2 + 16 = 30 n = 60 / 30 = 2 Molecular formula = C₂H₄O₂
Example 4: Vapour Density
The vapour density of a gas is 32. Find its molar mass.
Solution: Molar mass = 2 × VD = 2 × 32 = 64 g/mol
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| Using 22.4 L for non-STP conditions | 22.4 L is ONLY at STP (0°C, 1 atm) |
| Confusing vapour density with density | VD is relative to H₂; no units |
| Forgetting to simplify the mole ratio to integers | Divide all by the smallest number of moles |
| Wrong molar mass calculation | Sum atomic masses of ALL atoms in the formula |
ICSE Exam Focus
This chapter carries 6–8 marks. Key topics: mole calculations, mass-volume conversions, empirical formula determination, Gay Lussac's law, vapour density.
Marks Blueprint: Mole calculations — 3 marks, Gas volumes (STP) — 2 marks, Empirical/molecular formula — 3 marks.
Self-Test Questions
-
How many moles are present in 40 g of NaOH? (Na = 23, O = 16, H = 1)
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Calculate the volume occupied by 3.2 g of O₂ at STP.
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A compound has C = 54.55%, H = 9.09%, O = 36.36%. Find its empirical formula. If the molecular mass is 88, find the molecular formula.
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The vapour density of a compound is 30. Find its molar mass.
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State Gay Lussac's law and Avogadro's law with examples.
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How many molecules are present in 9 g of water?
