Mole Concept Full Notes | Class 11 Chemistry

Introduction

The Mole Concept is a fundamental topic in Class 11 Chemistry. It helps in converting mass to moles, determining formulas, and solving chemical calculations.

In this post, you will find step-by-step explanations, formulas in fraction/code style, solved examples, and a PDF download link for full notes.

1. Laws of Chemical Composition

1.1 Law of Conservation of Mass

Mass can neither be created nor destroyed in a chemical reaction.

Example:          2 H2 + O2 → 2 H2O

Total mass of reactants = Total mass of products

1.2 Law of Definite Proportion (Proust’s Law)

A chemical compound always contains the same proportion of elements by mass.

Example: Water (H2O) always has 2 g H per 16 g O → 1:8 mass ratio

1.3 Law of Multiple Proportions (Dalton)

When 2 elements form more than one compound, the masses of one element combined with a fixed mass of the other are in small whole number ratios.

Example:     CO: C=12, O=16 → C:O = 12:16

CO2: C=12, O=32 → C:O = 12:32 → ratio 16:32=1:2

1.4 Law of Reciprocal Proportions

If element A combines with B and C separately, the ratio in which B and C combine with each other is related to the ratio of B and C with A.

1.5 Gay-Lussac’s Law of Gaseous Volumes

Volumes of gases reacting together or produced in a chemical reaction are in small whole number ratios at the same temperature and pressure.

Example:    2 volumes H2 + 1 volume O2 → 2 volumes H2O

1.6 Avogadro’s Hypothesis

Equal volumes of gases at the same temperature and pressure contain equal number of molecules.

Implication: Volume of gas ∝ Number of moles

• Example:

1 L of H₂ gas and 1 L of O₂ gas at the same T & P contain equal number of molecules.

2. Mole Concept

2.1 Mole Definition

• 1 Mole = 6.022 × 10²³ particles (Avogadro Number)

• Used to count atoms, molecules, or ions in chemistry.

• Example:    1 mole of H₂O contains 6.022 × 10²³ water molecules.

       

2.2 Mass → Moles Conversion

Formula :      n = mass / molar mass

This formula is used to calculate number of moles (n) when the mass of a substance and its molar mass are known.

Example: 12 g of Carbon →    n = 12 / 12 = 1 mole

This means 12 g of carbon contains 1 mole of carbon atoms.

2.3 Empirical Formula

•The simplest whole number ratio of atoms in a compound.

Steps to Calculate:

* Convert % composition to grams.

* Convert grams to moles.

* Divide each mole value by the smallest number of moles.

* Write the formula using whole numbers.

• Example:

 

Given: 40% C, 6.7% H, 53.3% O

Convert % → grams: C=40 g, H=6.7 g, O=53.3 g

Convert grams → moles:

C: 40/12 ≈ 3.33

H: 6.7/1 ≈ 6.7

O: 53.3/16 ≈ 3.33

Divide by smallest:

C: 3.33/3.33 = 1

H: 6.7/3.33 ≈ 2

O: 3.33/3.33 = 1

Empirical formula: CH₂O

2.4 Molecular Formula

• Gives the actual number of atoms in a molecule.

• Molecular formula = (Empirical formula)×n

• n = Molar mass / Empirical formula mass

Example:

Empirical formula: CH2O

Molar mass = 180 →        Molecular formula = C6H12O6

3. Concentration Terms & Calculations

3.1 Weight by Weight (w/w %)

w/w % = (mass of solute / mass of solution) * 100

•Example:

10 g NaCl in 100 g solution →     w/w % = (10/100)*100 = 10%

3.2 Weight by Volume (w/v %)

w/v % = (mass of solute / volume of solution) * 100

•Example:

5 g NaOH in 100 mL solution →   w/v % = (5 / 100) * 100 = 5%

3.3 Volume by Volume (v/v %)

v/v % = (volume of solute / volume of solution) * 100

•Example:

50 mL ethanol in 200 mL solution →    v/v % = (50 / 200) * 100 = 25%

3.4 Molarity (M)

M = moles of solute / volume of solution in L

•Example:

3 moles of NaOH in 1.5 L solution →    M = 3 / 1.5 = 2 M

3.5 Molality (m)

m = moles of solute / mass of solvent in kg

•Example:

2 moles of solute in 0.5 kg solvent →   m = 2 / 0.5 = 4 m

3.6 Normality (N)

N = equivalents of solute / volume of solution in L

Relation: N = M * n_factor

•Example:

1 mole H₂SO₄ (n-factor = 2) in 1 L solution →    N = 1 * 2 / 1 = 2 N

3.7 Mole Fraction (X)

X_A = n_A / (n_A + n_B + …)

•Example:

2 moles solute A, 3 moles solute B →   X_A = 2 / (2+3) = 0.4

3.8 ppm & ppb

ppm = (mass of solute / mass of solution) * 10^6

ppb = (mass of solute / mass of solution) * 10^9

•Example:

1 mg solute in 1 kg solution →   ppm = (0.001 / 1) * 10^6 = 1 ppm

3.9 Equivalent Mass & n-factor

Equivalent mass = Molar mass / n factor

3.10 Mixtures & Relations

M1×V1 + M2×V2 = Mf×Vf

Relation between Molarity & Normality: N = M × n factor

•Example:

1 L 1 M NaOH + 2 L 2 M NaOH → Mf = ?

Mf = (1*1 + 2*2) / (1+2) = 5/3 ≈ 1.67 M

For full solved examples, practice problems, and detailed explanations,

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