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Chapter 1: Some Basic Concepts of Chemistry

Grade 11 Science | Chapter 1

Some Basic Concepts of Chemistry

Chemistry begins with counting and weighing matter. This chapter builds the laws of chemical combination, the mole concept and the stoichiometry that underpins every calculation in the subject.

Core Concepts

Key Principles

Worked Examples

Practice Sets

Contents

Introduction: Measuring Matter

Laws of Chemical Combination

Atomic and Molecular Mass

The Mole Concept

Percentage Composition and Formulae

Stoichiometry and Concentration

Key Reasoning (Principles)

Worked Examples (10)

Practice Sets A to D

10.

Summary and Exam Quick-Check

1. Introduction: Measuring Matter

Chemistry studies matter and its changes, and almost every change is described by how much of each substance takes part. Because atoms are far too small and too numerous to count, chemists weigh matter and convert mass into number using a counting unit called the mole. This chapter sets up that bridge between the mass we can weigh and the number of particles actually reacting.

We begin with the experimental laws of chemical combination that revealed how elements combine, then build atomic and molecular masses, the mole, and the stoichiometry that lets us predict the amounts in a reaction.

Core idea

Atoms combine in fixed, whole-number ratios. The mole links the mass we weigh to the number of particles, so a balanced equation can be read in moles, masses or particles.

2. Laws of Chemical Combination

Several laws, found by careful weighing, describe how substances combine. The law of conservation of mass states that mass is neither created nor destroyed in a chemical change. The law of definite proportions states that a pure compound always contains the same elements in the same fixed proportion by mass, so water is always one part hydrogen to eight parts oxygen by mass.

Diagram 1 – Conservation of Mass

A balance showing the mass of reactants equal to the mass of products

Fig 1. In a chemical reaction the total mass of the reactants equals the total mass of the products; nothing is gained or lost.

3. Atomic and Molecular Mass

The relative atomic mass compares the mass of an atom with a standard, so hydrogen is about 1, carbon 12 and oxygen 16. The molecular mass is the sum of the atomic masses in a formula, so water (H two O) is 1 + 1 + 16 = 18. For substances that are not simple molecules the same sum is called the formula mass.

Substance

Formula

Molecular or Formula Mass

Water

H₂O

Carbon dioxide

CO₂

Methane

CH₄

Sodium chloride

NaCl

58.5

4. The Mole Concept

One mole of any substance contains Avogadro’s number of particles, about 6.022 × 10²³, and has a mass in grams equal to its molecular or formula mass. The link is simple: number of moles = mass ÷ molar mass. For a gas, one mole occupies about 22.4 litres at standard temperature and pressure.

Diagram 2 – What One Mole Connects

One mole linked to mass in grams, Avogadro's number of particles, and gas volume at STP

Fig 2. One mole links three things: a mass equal to the molar mass in grams, Avogadro’s number of particles, and for a gas a volume of about 22.4 litres at STP.

5. Percentage Composition and Formulae

The percentage composition gives the mass of each element as a percentage of the whole. From it we find the empirical formula, the simplest whole-number ratio of atoms, by dividing each element’s mass percentage by its atomic mass and simplifying. The molecular formula is a whole-number multiple of the empirical formula, found using the molecular mass.

6. Stoichiometry and Concentration

Stoichiometry uses the balanced equation to relate the amounts of reactants and products in moles, and hence in mass. The concentration of a solution is often given as molarity, the number of moles of solute per litre of solution. These tools let a chemist predict exactly how much product a reaction will give.

Watch out

Always balance the equation before doing a stoichiometry calculation. The coefficients give the mole ratio, and the whole calculation rests on that ratio being correct.

7. Key Reasoning (Principles)

Principle 1: Mass is conserved

In any chemical change the total mass stays the same, so a chemical equation must balance, with equal numbers of each kind of atom on both sides.

Principle 2: The mole links mass and number

Number of moles = mass ÷ molar mass, and one mole always holds Avogadro’s number of particles. This is the bridge between the laboratory and the particle world.

Principle 3: Equations are read in moles

The coefficients of a balanced equation give the ratio of moles, which can then be turned into masses or volumes. Stoichiometry always starts from this ratio.

8. Worked Examples

Example 1

Q: Find the molar mass of water, H two O. (H = 1, O = 16)

▶ Show Solution

1 + 1 + 16.

= 18 g/mol.

Answer: 18 g/mol.

Example 2

Q: How many moles are in 36 g of water?

▶ Show Solution

Moles = mass ÷ molar mass = 36 ÷ 18.

= 2 mol.

Answer: 2 mol.

Example 3

Q: How many molecules are there in 2 mol of water?

▶ Show Solution

Molecules = moles × Avogadro’s number = 2 × 6.022 × 10²³.

= 1.204 × 10²⁴ molecules.

Answer: 1.204 × 10²⁴ molecules.

Example 4

Q: Find the percentage by mass of hydrogen in water.

▶ Show Solution

Mass of H = 2; total = 18.

(2 ÷ 18) × 100 = 11.1%.

Answer: 11.1%.

Example 5

Q: Find the percentage by mass of oxygen in water.

▶ Show Solution

(16 ÷ 18) × 100.

= 88.9%.

Answer: 88.9%.

Example 6

Q: Find the molar mass of carbon dioxide, CO two. (C = 12, O = 16)

▶ Show Solution

12 + 16 + 16.

= 44 g/mol.

Answer: 44 g/mol.

Example 7

Q: How many moles are in 88 g of carbon dioxide?

▶ Show Solution

88 ÷ 44.

= 2 mol.

Answer: 2 mol.

Example 8

Q: Find the mass of 0.5 mol of sodium chloride. (NaCl = 58.5)

▶ Show Solution

Mass = moles × molar mass = 0.5 × 58.5.

= 29.25 g.

Answer: 29.25 g.

Example 9

Q: Find the molarity of a solution with 1 mol of solute in 2 litres.

▶ Show Solution

Molarity = moles ÷ litres = 1 ÷ 2.

= 0.5 M.

Answer: 0.5 M.

Example 10

Q: A compound is 40% C, 6.7% H and 53.3% O by mass. Find its empirical formula. (C = 12, H = 1, O = 16)

▶ Show Solution

Divide by atomic masses: C 40 ÷ 12 = 3.33, H 6.7 ÷ 1 = 6.7, O 53.3 ÷ 16 = 3.33.

Divide by the smallest (3.33): C 1, H 2, O 1, giving CH₂O.

Answer: CH₂O.

9. Practice Sets A to D

Set A – Multiple Choice (Basic)

1. Avogadro’s number is about: (a) 6.022 × 10²³ (b) 3 × 10⁸ (c) 9.1 × 10^-31 (d) 1.6 × 10^-19

2. The molar mass of CO₂ is: (a) 28 (b) 32 (c) 44 (d) 56

3. Number of moles equals: (a) mass × molar mass (b) mass ÷ molar mass (c) molar mass ÷ mass (d) mass + molar mass

4. One mole of a gas at STP occupies about: (a) 1 L (b) 22.4 L (c) 100 L (d) 6 L

5. Molarity is moles of solute per: (a) gram (b) mole (c) litre of solution (d) kilogram

▶ Reveal Answers

1. (a) 6.022 × 10²³.

2. (c) 44.

3. (b) mass ÷ molar mass.

4. (b) 22.4 L.

5. (c) litre of solution.

Set B – Short Answer (Understanding)

1. State the law of conservation of mass.

2. Define one mole of a substance.

3. What is the difference between empirical and molecular formula?

4. Write the relationship between moles, mass and molar mass.

5. Why must an equation be balanced before a stoichiometry calculation?

▶ Reveal Answers

1. Mass is neither created nor destroyed in a chemical reaction.

2. The amount of a substance containing Avogadro’s number of particles, about 6.022 × 10²³.

3. The empirical formula is the simplest whole-number ratio of atoms; the molecular formula is a whole-number multiple of it.

4. Number of moles = mass ÷ molar mass.

5. Because the coefficients give the mole ratio on which the whole calculation depends.

Set C – Application and Reasoning

1. Find the molar mass of methane, CH four. (C = 12, H = 1)

2. How many moles are in 90 g of water?

3. How many molecules are in 0.5 mol of any substance?

4. Find the percentage by mass of carbon in CO₂.

5. Find the molarity of 2 mol of solute in 4 litres of solution.

▶ Reveal Answers

1. 12 + 4 = 16 g/mol.

2. 90 ÷ 18 = 5 mol.

3. 0.5 × 6.022 × 10²³ = 3.011 × 10²³ molecules.

4. (12 ÷ 44) × 100 = 27.3%.

5. 2 ÷ 4 = 0.5 M.

Set D – Higher Order (Challenge)

1. A compound has empirical formula CH₂O and molecular mass 180. Find its molecular formula.

2. Find the mass of 3 mol of carbon dioxide.

3. A solution contains 49 g of H₂SO₄ (molar mass 98) in 0.5 L. Find its molarity.

4. Calculate the percentage by mass of nitrogen in ammonia, NH₃. (N = 14, H = 1)

5. Explain why the molecular formula is always a whole-number multiple of the empirical formula.

▶ Reveal Answers

1. Empirical mass = 12 + 2 + 16 = 30; 180 ÷ 30 = 6, so the molecular formula is C₆H₁₂O₆.

2. 3 × 44 = 132 g.

3. Moles = 49 ÷ 98 = 0.5 mol; molarity = 0.5 ÷ 0.5 = 1 M.

4. Mass of N = 14; total = 17; (14 ÷ 17) × 100 = 82.4%.

5. Because a molecule contains whole atoms, the actual atom counts must be a whole-number multiple of the simplest ratio.

Chapter Summary

Measuring Matter

Chemistry counts particles by weighing matter and converting through the mole.

Laws of Combination

Mass is conserved, and a pure compound has a fixed proportion by mass.

Atomic and Molecular Mass

Relative masses (H 1, C 12, O 16) add to give molecular or formula mass.

The Mole

One mole has the molar mass in grams and Avogadro’s number of particles.

Formulae

Empirical formula is the simplest ratio; molecular formula is a multiple of it.

Stoichiometry

A balanced equation gives the mole ratio, read as moles, mass or volume.

Quantity

Unit

Symbol

Avogadro’s number

6.022 x 10^23

per mole

Moles

mass ÷ molar mass

mol

Molarity

moles ÷ litres

8-Point Exam Quick-Check

Atoms combine in fixed, whole-number ratios by mass.

Mass is conserved, so a chemical equation must balance.

Molecular mass is the sum of the atomic masses in the formula.

One mole has Avogadro’s number of particles, about 6.022 times 10^23.

Number of moles = mass divided by molar mass.

One mole of a gas occupies about 22.4 litres at STP.

Empirical formula is the simplest ratio; molecular formula is a multiple of it.

A balanced equation gives the mole ratio used in stoichiometry.

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Class 11 Chemistry Chapter 1: Some Basic Concepts of Chemistry, Complete Notes and Practice

This revision guide follows the current NCERT Class 11 Chemistry syllabus and covers the laws of chemical combination, conservation of mass, atomic and molecular mass, the mole concept and Avogadro’s number, percentage composition, empirical and molecular formulae, and stoichiometry and molarity, with two diagrams, ten worked examples and graded practice. Visit SchoolRevise.com to revise, practise and excel.