Some Basic Concepts of Chemistry – ICSE Class 10 Chemistry Notes & Guide

In chemistry, everything around us is made of matter. Understanding its basic properties and how we measure it forms the foundation for all later topics. This guide covers the essential ideas you need to solve numerical problems and answer theory questions in the ICSE Class 10 exam.

Key concepts

Matter and its classification

Matter is anything that has mass and occupies space. It is classified into:

• Pure substances – elements (made of one type of atom) and compounds (two or more elements chemically combined).
• Mixtures – a physical combination of two or more substances. Mixtures may be:
• Homogeneous (uniform throughout, e.g., salt solution) or
• Heterogeneous (visible different parts, e.g., sand‑water mixture).

Understanding the difference helps in separating components and in writing correct chemical equations.

Atomic and molecular masses

• Relative atomic mass (Ar) of an element is the weighted average mass of its naturally occurring isotopes, expressed relative to ¹²C = 12.
• Relative molecular mass (Mr) of a molecule is the sum of Ar values of all atoms present.

Both are dimensionless numbers; they are used to convert between mass and number of atoms/molecules.

Mole concept and stoichiometry

• Mole (mol) is the amount of substance containing as many entities as there are atoms in 12 g of carbon‑12 (Avogadro’s number, 6.022 × 10²³).
• Molar mass (M) is the mass of one mole of a substance (in g mol⁻¹) and numerically equals its relative atomic or molecular mass.
• Stoichiometry uses balanced chemical equations to relate moles of reactants and products. The mole ratio from the equation tells how many moles of each substance participate.

Empirical and molecular formulas

• Empirical formula shows the simplest whole‑number ratio of atoms in a compound.
• Molecular formula gives the actual number of atoms of each element in a molecule. It is a multiple of the empirical formula.

To find them, use percent composition data, convert to masses, then to moles, and simplify the ratio.

Percent composition

Percent composition of an element in a compound = (mass of element in 1 mol of compound ÷ molar mass of compound) × 100 %. It is useful for determining empirical formulas and for checking the correctness of a given formula.

Important definitions and formulas

• Mole: 1 mol = 6.022 × 10²³ entities.
• Molar mass (M) = relative atomic/molecular mass (Ar or Mr) in g mol⁻¹.
• Number of moles (n) = mass (m) ÷ molar mass (M).
• Mass‑mole relationship: m = n × M.
• Stoichiometric conversion: Use mole ratios from balanced equation.
• Empirical formula: Convert % composition → mass → moles → divide by smallest mole value → round to nearest whole number.
• Molecular formula: (Molar mass of compound ÷ molar mass of empirical formula) = integer n; multiply empirical formula by n.
• Percent composition: % = (mass of element in 1 mol ÷ molar mass of compound) × 100.

Solved examples

Example 1 – Finding moles from given mass

Problem: 18 g of water (H₂O) is taken. How many moles of water are present? Solution:

1. Calculate molar mass of H₂O: H = 1 × 2 = 2, O = 16 → Mr = 18 g mol⁻¹.
2. Use n = m ÷ M = 18 g ÷ 18 g mol⁻¹ = 1 mol.

Thus 18 g of water corresponds to 1 mol.

Example 2 – Empirical formula from percent composition

Problem: A compound contains 40 % C, 6.7 % H and 53.3 % O by mass. Find its empirical formula. Solution:

1. Assume 100 g of compound → 40 g C, 6.7 g H, 53.3 g O.
2. Convert to moles:

• C: 40 g ÷ 12 g mol⁻¹ = 3.33 mol
• H: 6.7 g ÷ 1 g mol⁻¹ = 6.70 mol
• O: 53.3 g ÷ 16 g mol⁻¹ = 3.33 mol

1. Divide by the smallest mole value (3.33):

• C: 3.33 ÷ 3.33 = 1
• H: 6.70 ÷ 3.33 ≈ 2
• O: 3.33 ÷ 3.33 = 1

1. Empirical formula = CH₂O.

Example 3 – Percent composition and molecular formula

Problem: A compound has empirical formula CH₂O and a molar mass of 180 g mol⁻¹. Find its molecular formula and percent composition of carbon. Solution:

1. Molar mass of empirical formula CH₂O = 12 + (2×1) + 16 = 30 g mol⁻¹.
2. n = (Molar mass of compound) ÷ (Molar mass of empirical) = 180 ÷ 30 = 6.
3. Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.
4. Percent C = (mass of C in one mole) ÷ (molar mass) × 100 = (6×12) ÷ 180 × 100 = 72 ÷ 180 × 100 = 40 %.

Common mistakes to avoid

• Forgetting to use the smallest whole‑number ratio when deriving empirical formulas; always check if rounding is needed.
• Mixing up relative atomic mass (Ar) with molar mass (M) – the former is dimensionless, the latter carries g mol⁻¹.
• Ignoring the need for balanced equations before applying stoichiometric ratios.
• Using incorrect atomic masses from outdated tables; stick to the values given in the ICSE syllabus.
• Treating percentages as absolute masses without first assuming a 100 g sample.

Exam tips

• Memorise the atomic masses of C, H, N, O, S, Cl – they appear frequently.
• Always write down the balanced chemical equation before starting stoichiometry.
• When a question gives percent composition, assume a 100 g sample to simplify calculations.
• After finding the empirical formula, always verify the molecular formula by comparing the calculated molar mass with the given one.
• Keep a small table of common conversions (mass ↔ moles ↔ molecules) at the back of your notebook for quick reference.

Frequently asked questions

What is the difference between empirical and molecular formula?

Empirical formula shows the simplest whole‑number ratio of atoms; molecular formula shows the actual number of atoms in a molecule and is a whole‑number multiple of the empirical formula.

How do I convert percent composition to empirical formula?

Assume 100 g of the compound, convert each percentage to grams, then to moles using atomic masses, divide all mole values by the smallest one, and round to the nearest whole number.

Why is the mole concept important in chemistry?

The mole links the macroscopic world (grams) with the microscopic world (atoms, molecules). It allows us to use balanced equations to predict quantities of reactants and products accurately.