Basic Concepts in Chemistry

Master the mole concept, atomic and molecular masses, stoichiometry, and chemical equations for JEE Chemistry.

This chapter covers the fundamental concepts that form the foundation of chemistry - from atomic theory to stoichiometric calculations.

Overview

graph TD
    A[Basic Concepts] --> B[Atomic Theory]
    A --> C[Mole Concept]
    A --> D[Stoichiometry]
    B --> B1[Atoms & Molecules]
    B --> B2[Laws of Combination]
    C --> C1[Avogadro Number]
    C --> C2[Molar Mass]
    D --> D1[Chemical Equations]
    D --> D2[Limiting Reagent]

Atomic and Molecular Masses

Atomic Mass Unit (amu)

$$1 \text{ amu} = \frac{1}{12} \times \text{mass of } ^{12}\text{C atom} = 1.66 \times 10^{-24} \text{ g}$$

Molecular Mass

Sum of atomic masses of all atoms in a molecule.

Example: H₂O = 2(1) + 16 = 18 amu

Mole Concept

Avogadro’s Number

$$N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$$

One mole of any substance contains $N_A$ particles.

Molar Mass

Mass of 1 mole of a substance in grams.

$$\text{Molar mass (g/mol)} = \text{Molecular mass (amu)}$$

Important Relationships

$$\boxed{n = \frac{m}{M} = \frac{N}{N_A} = \frac{V}{22.4} \text{ (at STP for gases)}}$$

where:

  • n = number of moles
  • m = mass (g)
  • M = molar mass (g/mol)
  • N = number of particles
  • V = volume of gas at STP (L)
JEE Tip
At STP: 1 mole of any ideal gas occupies 22.4 L At NTP: 1 mole of any ideal gas occupies 22.7 L

Laws of Chemical Combination

1. Law of Conservation of Mass (Lavoisier)

Mass is neither created nor destroyed in a chemical reaction.

2. Law of Definite Proportions (Proust)

A compound always contains the same elements in the same proportion by mass.

3. Law of Multiple Proportions (Dalton)

When two elements form more than one compound, the ratios of masses are simple whole numbers.

4. Gay-Lussac’s Law

Gases react in simple volume ratios at constant T and P.

5. Avogadro’s Law

Equal volumes of gases at same T and P contain equal number of molecules.

Percentage Composition

$$\text{% of element} = \frac{\text{Mass of element in 1 mol}}{\text{Molar mass}} \times 100$$

Empirical and Molecular Formula

Empirical Formula: Simplest whole number ratio of atoms

Molecular Formula: Actual number of atoms

$$\text{Molecular Formula} = n \times \text{Empirical Formula}$$

where $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}}$

Stoichiometry

Balanced Chemical Equation

Represents a reaction with correct atom balance.

Stoichiometric Calculations

  1. Write balanced equation
  2. Find mole ratio from coefficients
  3. Use given information to find moles
  4. Calculate required quantity

Limiting Reagent

The reactant that is completely consumed first.

Method:

  1. Calculate moles of each reactant
  2. Divide by respective coefficient
  3. Smallest value identifies limiting reagent

Percentage Yield

$$\% \text{ Yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$

Concentration Terms

TermFormulaUnits
Molarity (M)$\frac{\text{moles}}{\text{L solution}}$mol/L
Molality (m)$\frac{\text{moles}}{\text{kg solvent}}$mol/kg
Mole fraction$\frac{n_i}{\sum n}$dimensionless
Mass %$\frac{\text{mass solute}}{\text{mass solution}} \times 100$%

Practice Problems

  1. Calculate the mass of 0.5 mol of glucose (C₆H₁₂O₆).

  2. How many molecules are present in 18 g of water?

  3. 10 g of CaCO₃ is heated. Find the volume of CO₂ produced at STP.

  4. Find the empirical formula of a compound containing 40% C, 6.7% H, and 53.3% O.

Quick Check
Why is the mole concept fundamental to quantitative chemistry?

Further Reading