Some Basic Concepts of Chemistry Class 11 Notes | CBSE Chemistry Chapter 1

Table of Contents

• Key Concepts - Matter, laws of combination, mole concept, stoichiometry, concentration terms
• Weightage in Board & Entrance Exams
• Important Definitions
• Solved Examples
• Important Questions for Board Exams
• Quick Revision Points

Key Concepts

1. Matter and Its Classification

Matter is anything that has mass and occupies space. At the macroscopic level it exists in three physical states - solid, liquid and gas - that can be interconverted by changing temperature and pressure.

Chemically, matter is classified into mixtures and pure substances. A mixture (e.g., air, sugar solution) contains two or more components in any ratio, while a pure substance (e.g., water, gold) has a fixed composition.

• Homogeneous mixture: uniform composition throughout (e.g., salt dissolved in water).
• Heterogeneous mixture: non-uniform composition (e.g., a mixture of sand and iron filings).
• Elements: contain only one kind of atom (e.g., Na, O₂).
• Compounds: two or more elements combined in a fixed ratio by mass (e.g., H₂O, CO₂).

2. Properties and Their Measurement (SI Units)

Physical properties (e.g., density, melting point) can be measured without changing the identity of the substance, while chemical properties (e.g., combustibility) involve a chemical change.

Chemistry uses the SI system of seven base units. The two you use most often are the kilogram (kg) for mass and the mole (mol) for amount of substance.

Precision, Accuracy and Significant Figures

• Accuracy: how close a measured value is to the true value.
• Precision: how close repeated measurements are to one another.
• Significant figures are the meaningful digits known with certainty plus one uncertain digit. All non-zero digits are significant; leading zeros are not; trailing zeros after a decimal are significant.

Rules for calculation: In multiplication/division, the answer has as many significant figures as the term with the fewest. In addition/subtraction, the answer has as many decimal places as the term with the fewest.

3. Laws of Chemical Combination

These five laws govern how elements combine to form compounds and laid the experimental basis for Dalton’s atomic theory.

The Five Laws

• Law of Conservation of Mass (Lavoisier): mass can neither be created nor destroyed in a chemical reaction - total mass of reactants equals total mass of products.
• Law of Definite (Constant) Proportions (Proust): a given compound always contains the same elements in the same fixed proportion by mass (e.g., water is always 1:8 H:O by mass).
• Law of Multiple Proportions (Dalton): when two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole-number ratio (e.g., in CO and CO₂ the oxygen ratio is 1:2).
• Gay Lussac’s Law of Gaseous Volumes: when gases react, they do so in volumes that bear a simple whole-number ratio to one another and to the products, at constant T and P.
• Avogadro’s Law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

4. Dalton’s Atomic Theory

In 1808 John Dalton proposed the first modern atomic theory, explaining the laws of chemical combination.

• Matter consists of indivisible particles called atoms.
• Atoms of the same element are identical in mass and properties; atoms of different elements differ.
• Atoms combine in simple whole-number ratios to form compounds.
• Atoms are neither created nor destroyed in a chemical reaction.

Key idea: The first two postulates were later modified - atoms are divisible (electrons, protons, neutrons) and isotopes of an element differ in mass.

5. Atomic and Molecular Mass

Atomic mass is the mass of one atom expressed in atomic mass units (u), where 1 u = 1/12 the mass of one carbon-12 atom (1 u = 1.66 × 10⁻²⁴ g).

Average atomic mass accounts for the natural abundance of isotopes. For example, chlorine (75% ³⁵Cl, 25% ³⁷Cl) has average atomic mass = (35 × 0.75) + (37 × 0.25) = 35.5 u.

• Molecular mass: sum of the atomic masses of all atoms in a molecule (e.g., H₂O = 2 × 1 + 16 = 18 u).
• Formula mass: used for ionic compounds that exist as lattices, not molecules (e.g., NaCl = 23 + 35.5 = 58.5 u).

6. The Mole Concept and Molar Mass

One mole is the amount of substance that contains exactly 6.022 × 10²³ elementary entities (atoms, molecules, ions). This number is Avogadro’s number (Nₐ).

The molar mass is the mass of one mole of a substance in grams. Numerically it equals the atomic/molecular mass in u (e.g., molar mass of H₂O = 18 g/mol).

The Master Conversion Formulae

• Number of moles, n = given mass (m) / molar mass (M)
• Number of particles = n × Nₐ = (m/M) × 6.022 × 10²³
• For a gas at STP (273.15 K, 1 bar): volume = n × 22.7 L (22.4 L at 1 atm).

Key idea: The mole is the bridge that connects the mass you weigh in the lab to the number of atoms reacting at the particle level.

7. Percentage Composition

The percentage composition tells you the mass percent of each element in a compound, useful for checking purity and finding formulae.

Mass % of an element = (mass of element in 1 mole of compound / molar mass of compound) × 100

For example, in water (M = 18): mass % of H = (2/18) × 100 = 11.1%, and mass % of O = (16/18) × 100 = 88.9%.

8. Empirical and Molecular Formula

The empirical formula gives the simplest whole-number ratio of atoms in a compound. The molecular formula gives the actual number of atoms in one molecule.

Molecular formula = n × empirical formula, where n = molecular mass / empirical formula mass.

Steps to Find the Empirical Formula

• Convert the mass % of each element to moles (divide by atomic mass).
• Divide all mole values by the smallest one to get the simplest ratio.
• If needed, multiply to obtain whole numbers - these become the subscripts.

For example, a compound that is 40% C, 6.7% H and 53.3% O gives the empirical formula CH₂O; if its molecular mass is 180, then n = 180/30 = 6, so the molecular formula is C₆H₁₂O₆ (glucose).

9. Stoichiometry and Stoichiometric Calculations

Stoichiometry is the quantitative relationship between reactants and products in a balanced chemical equation. The coefficients give the mole ratio in which species react.

Consider: CH₄ + 2O₂ → CO₂ + 2H₂O. This reads: 1 mole of CH₄ reacts with 2 moles of O₂ to give 1 mole of CO₂ and 2 moles of H₂O.

Steps for a Stoichiometry Problem

• Write and balance the chemical equation.
• Convert the given quantity to moles.
• Use the mole ratio from the equation to find moles of the required species.
• Convert moles back to mass, volume, or number of particles as asked.

10. Limiting Reagent

The limiting reagent is the reactant that is completely consumed first and therefore decides the maximum amount of product formed. The other reactant(s) are left over in excess.

Method: Find the moles of each reactant, divide each by its stoichiometric coefficient, and the smallest value identifies the limiting reagent.

Key idea: Product yield is always calculated from the limiting reagent, never from the excess reactant - a common board-exam trap.

11. Concentration Terms (Methods of Expressing Concentration)

The concentration of a solution tells you how much solute is dissolved in a given amount of solvent or solution. CBSE focuses on four temperature-independent or temperature-dependent terms.

Note: Molality and mole fraction are independent of temperature because they use mass, not volume; molarity changes with temperature because volume expands on heating. The sum of mole fractions of all components equals 1.