ICSE Class 10 Chemistry • Chapter 05
Statement: When gases react together, they do so in volumes which bear a simple whole number ratio to one another and to the volumes of products, all volumes measured under the same conditions of temperature and pressure.
Example: H₂ + Cl₂ → 2HCl
1 volume + 1 volume → 2 volumes
Ratio = 1 : 1 : 2
Statement: Equal volumes of all gases, under the same conditions of temperature and pressure, contain equal number of molecules.
Avogadro's Number (Nₐ): $6.022 \times 10^{23}$
This is the number of particles in one mole of any substance.
Mole: The amount of substance that contains as many particles (atoms, molecules, ions) as there are atoms in exactly 12 g of carbon-12.
1 mole = 6.022 × 10²³ particles (Avogadro's number)
Definition: The atomic mass of an element expressed in grams.
1 GAM = 1 mole of atoms = 6.022 × 10²³ atoms
Example: GAM of Oxygen = 16 g (contains 6.022 × 10²³ O atoms)
Definition: The molecular mass of a compound expressed in grams.
1 GMM = 1 mole of molecules = 6.022 × 10²³ molecules
Example: GMM of H₂O = 18 g (contains 6.022 × 10²³ H₂O molecules)
At STP (Standard Temperature & Pressure):
1 mole of any gas occupies 22.4 litres
STP conditions: 0°C (273 K) and 1 atm pressure
| Formula | Description |
|---|---|
| $\text{Moles} = \frac{\text{Given Mass}}{\text{Molar Mass}}$ | Calculate moles from mass |
| $\text{Moles} = \frac{\text{Number of particles}}{6.022 \times 10^{23}}$ | Calculate moles from particles |
| $\text{Moles of gas} = \frac{\text{Volume at STP}}{22.4 \text{ L}}$ | Calculate moles from gas volume |
| $\text{Vapour Density} = \frac{\text{Molecular Mass}}{2}$ | Relation between VD and MM |
Master Formula: Molecular Mass = 2 × Vapour Density
| Element | Formula | Atomicity |
|---|---|---|
| Hydrogen | H₂ | 2 (Diatomic) |
| Oxygen | O₂ | 2 (Diatomic) |
| Nitrogen | N₂ | 2 (Diatomic) |
| Chlorine | Cl₂ | 2 (Diatomic) |
| Noble Gases | He, Ne, Ar | 1 (Monoatomic) |
| Ozone | O₃ | 3 (Triatomic) |
$\text{% of element} = \frac{\text{Mass of element in 1 mole of compound}}{\text{Molar mass of compound}} \times 100$
Example: Find % of oxygen in H₂O.
Molar mass of H₂O = 2(1) + 16 = 18 g
Mass of O = 16 g
% of O = $\frac{16}{18} \times 100 = 88.89\%$
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 = n × Empirical Formula
where $n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}}$
Example: CH₂O (Empirical) → C₆H₁₂O₆ (Molecular for glucose)
Empirical mass = 12 + 2 + 16 = 30
Molecular mass = 180
n = 180/30 = 6
Steps for solving stoichiometry problems:
Example: Calculate the volume of CO₂ at STP when 10 g of CaCO₃ decomposes completely.
CaCO₃ → CaO + CO₂
Molar mass of CaCO₃ = 40 + 12 + 48 = 100 g
Moles of CaCO₃ = 10/100 = 0.1 mol
From equation: 1 mol CaCO₃ → 1 mol CO₂
∴ Moles of CO₂ = 0.1 mol
Volume at STP = 0.1 × 22.4 = 2.24 L
| Quantity | 1 Mole = |
|---|---|
| Number of particles | 6.022 × 10²³ |
| Mass of element | Gram Atomic Mass (g) |
| Mass of compound | Gram Molecular Mass (g) |
| Volume of gas at STP | 22.4 litres |
BOARD Calculate the number of molecules in 5.6 litres of CO₂ gas at STP.
BOARD A compound contains 40% carbon, 6.67% hydrogen, and 53.33% oxygen. Its molecular mass is 180. Find empirical and molecular formula.