CBSE Class 9 Chemistry Atoms and Molecules Notes Set 03

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Access comprehensive revision notes for Atoms and Molecules using the CBSE Class 9 Chemistry Atoms and Molecules Notes Set 03. Designed to align with the 2026-27 academic syllabus for Class 9 Chemistry, these concept summaries help students streamline their exam preparation and review complex topics efficiently.

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Laws of Chemical Combination

Law of conservation of mass: matter is neither created nor destroyed in a reaction. Law of definite proportions: a compound always contains its elements in a fixed ratio by mass. Law of multiple proportions: when two elements form more than one compound, the masses of one that combine with a fixed mass of the other are in a simple whole-number ratio.

Atoms, Molecules and Formulae

Dalton's theory: elements are made of tiny indivisible atoms, all alike for one element and different from other elements, and atoms combine in simple whole-number ratios. A molecule is the smallest particle of a compound or of an element such as hydrogen that can exist freely. A formula is written by balancing the valences of the positive and negative radicals so that the molecule is neutral, for example Al\( ^{3+} \) and SO\( _4^{2-} \) give Al\( _2 \)(SO\( _4 \))\( _3 \).

Masses and the Mole

Relative atomic mass is the mass of an atom compared with 1/12 of a carbon-12 atom. Molecular mass is the sum of the atomic masses in the formula, for example H\( _2 \)O = 2 + 16 = 18 u.
One mole contains \( 6.022 \times 10^{23} \) particles (Avogadro number) and, for a gas at STP, occupies 22.4 litres. Number of moles = mass / molar mass.
Example: 36 g of water is 36/18 = 2 mol, which is \( 12.044 \times 10^{23} \) molecules.

Percentage Composition and Formulae

Percentage of an element = (mass of the element in one mole / molar mass) \( \times \) 100. Dividing the percentage of each element by its atomic mass gives the ratio of atoms, that is the empirical formula. The molecular formula is a whole-number multiple of it, found from the molecular mass.
Example: a compound with 40% C, 6.7% H and 53.3% O has ratio 3.33 : 6.7 : 3.33 = 1 : 2 : 1, so the empirical formula is CH\( _2 \)O (mass 30). If the molar mass is 180, the molecular formula is C\( _6 \)H\( _{12} \)O\( _6 \).

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