Atoms and Molecules Class 9 Free Notes and Mind Map (Free PDF Download)

This chapter explains laws of chemical combination, Dalton’s atomic theory, the ideas of atoms, molecules, ions, and how to write chemical formulae and calculate atomic, molecular and formula unit mass. These ideas form the basic language of chemistry for higher classes.

3.1 Laws of Chemical Combination

By the end of the 18th century, scientists clearly identified elements and compounds and wanted to know how and why elements combine. Lavoisier and Proust established two important laws to describe how substances react by mass.

3.1.1 Law of Conservation of Mass

When two solutions react in a closed container and are weighed before and after the reaction:

• The total mass of the flask and its contents remains the same.
• This shows that there is no loss or gain of mass in a chemical reaction.

Statement:

• Law of conservation of mass: Mass can neither be created nor destroyed in a chemical reaction.

3.1.2 Law of Constant Proportions (Definite Proportions)

Lavoisier and others observed that compounds always have fixed composition by mass:

• In water, the mass ratio of hydrogen : oxygen is always 1 : 8, no matter the source of water.
• If 9 g of water is decomposed, 1 g hydrogen and 8 g oxygen are always obtained.iesc103.pdf​
• In ammonia, nitrogen and hydrogen are always present in the ratio 14 : 3 by mass, regardless of how it is prepared.

Statement (Proust):

• In a chemical substance, the elements are always present in definite proportions by mass. This is called the law of constant proportions or law of definite proportions.

These laws needed explanation, which came from Dalton’s atomic theory.

Dalton’s Atomic Theory

John Dalton used the idea of indivisible particles (atoms) to explain the laws of conservation of mass and constant proportions. He proposed that all matter is made of tiny, indivisible atoms and explained how they combine in simple ratios.

Postulates of Dalton’s Atomic Theory

1. All matter is made of very tiny particles called atoms, which participate in chemical reactions.
2. Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
3. Atoms of a given element are identical in mass and chemical properties.
4. Atoms of different elements have different masses and chemical properties.
5. Atoms combine in the ratio of small whole numbers to form compounds.
6. The relative number and kinds of atoms are constant in a given compound.

Later, it was discovered that atoms themselves are made of smaller particles (protons, electrons, neutrons), but Dalton’s theory still gives a useful basic picture.

Questions (3.1) and Solutions

Q1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium acetate. Show that these observations agree with the law of conservation of mass.

• Given:
  • Reactants:
    • Sodium carbonate = 5.3 g
    • Acetic acid = 6.0 g
    • Total mass of reactants = 5.3 + 6.0 = 11.3 g
  • Products:
    • Carbon dioxide = 2.2 g
    • Water = 0.9 g
    • Sodium acetate = 8.2 g
    • Total mass of products = 2.2 + 0.9 + 8.2 = 11.3 giesc103.pdf​
• Conclusion:
  • Total mass of reactants = total mass of products = 11.3 g.
  • So, mass is conserved, which verifies the law of conservation of mass.iesc103.pdf​

Q2. Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas is required to react completely with 3 g of hydrogen gas?

• Given ratio by mass (H : O) = 1 : 8.
• For 1 g hydrogen → 8 g oxygen.
• For 3 g hydrogen → 3×8=243 \times 8 = 243×8=24 g oxygen.
• Answer: 24 g of oxygen is required.

Q3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?iesc103.pdf​

• The postulate: Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.
• Because atoms are not created or destroyed, total mass remains constant.

Q4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

• The postulate: Atoms combine in the ratio of small whole numbers and the relative number and kinds of atoms are constant in a compound.
• This fixed atomic ratio gives definite mass ratio in a compound (law of constant proportions).

3.2 What is an Atom?

An atom is the basic building block of matter, just like a brick for a wall or a grain of sand for an ant-hill. All substances are made up of atoms.

Size of Atoms

• Atoms are extremely small, much smaller than we can see or imagine.
• Millions of atoms stacked would form a layer only as thick as a sheet of paper.
• Atomic radius is measured in nanometres (nm):
  • 1 nm=10−9 m1 \text{ nm} = 10^{-9} \text{ m}1 nm=10−9 m
  • 1 m=109 nm1 \text{ m} = 10^{9} \text{ nm}1 m=109 nm.iesc103.pdf​

Modern instruments (like electron microscopes) can produce magnified images of surfaces showing atoms, for example on silicon.

3.2.1 Modern Day Symbols of Atoms of Different Elements

Dalton first used symbols for elements and meant “one atom” by each symbol. Later, Berzelius suggested using one or two letters of the element’s name.iesc103.pdf​

Rules for Symbols (IUPAC System)

• Use one or two letters.
• The first letter is capital (uppercase).
• The second letter is small (lowercase).

Examples (English name based):

• Hydrogen – H
• Aluminium – Al (not AL)
• Cobalt – Co (not CO)

Some symbols are taken from Latin / German / Greek names:

• Iron – Fe (from ferrum)
• Sodium – Na (from natrium)
• Potassium – K (from kalium)

Each element has a name and a unique chemical symbol approved by IUPAC.

Table: Symbols for Some Elements

3.2.2 Atomic Mass

Dalton proposed that each element has a characteristic atomic mass. Because measuring the mass of a single atom is difficult, scientists used relative atomic mass.

Example: In carbon monoxide (CO),

• 3 g carbon combines with 4 g oxygen.
• So, carbon combines with 43\frac{4}{3}34 times its mass of oxygen.

To standardise, scientists chose a reference atom and defined atomic mass unit (u):

• Earlier: 1/16th of the mass of an oxygen atom.
• Now (IUPAC, 1961): 1 atomic mass unit (u) = 1/12th of the mass of one carbon-12 atom.​

Definition:

• Relative atomic mass of an element = average mass of its atoms compared to 1/12th of carbon-12.

Atomic Masses of Some Elements

Questions (3.2) and Solutions

Q1. Define the atomic mass unit.

• Atomic mass unit (u): It is 1/12th of the mass of one atom of carbon-12 isotope.
• It is used as a standard unit for expressing atomic and molecular masses.

Q2. Why is it not possible to see an atom with naked eyes?

• Atoms are extremely small, with radii of the order of 10⁻¹⁰ m, so millions of atoms would form a layer as thin as a paper sheet.
• This size is far below the resolving power of the human eye, so atoms cannot be seen without powerful instruments.iesc103.pdf​

3.3 What is a Molecule?

A molecule is a group of two or more atoms held together by chemical bonds. It is the smallest particle of an element or a compound that:

• Can exist independently.
• Shows all the properties of that substance.

Atoms of the same or different elements can join to form molecules.

3.3.1 Molecules of Elements

• Molecules of an element consist of same type of atoms.
• Some elements exist as single-atom molecules (monoatomic):
  • Helium (He), Argon (Ar), Neon (Ne).
• Many non-metals exist as diatomic molecules (two atoms):
  • Oxygen (O₂), Hydrogen (H₂), Nitrogen (N₂), Chlorine (Cl₂).
• Three oxygen atoms can combine to form ozone (O₃).

Atomicity: Number of atoms present in a molecule.

Atomicity of Some Elements

Metals and some elements like carbon have very large networks of atoms, not simple fixed atomicity.

3.3.2 Molecules of Compounds

Atoms of different elements join in fixed ratios to form molecules of compounds.

Table: Molecules of Some Compounds

From mass ratios and atomic masses, we can calculate number ratios of atoms in molecules (e.g., H:O = 2:1 in water).

3.3.3 What is an Ion?

Ions are charged particles formed when atoms or groups of atoms gain or lose electrons.

• Cation: Positively charged ion (loss of electrons).
• Anion: Negatively charged ion (gain of electrons).

Example: In sodium chloride (NaCl):

• Sodium forms Na⁺ (cation).
• Chlorine forms Cl⁻ (anion).

Polyatomic ion: A group of atoms carrying a net charge, e.g., SO₄²⁻, CO₃²⁻, NH₄⁺.

Table: Some Ionic Compounds and Mass Ratios

Table 3.6 – Names, Symbols and Valencies of Some Ions

Cations (positive):

Anions (negative, non-metallic):

Polyatomic anions:

Some elements show more than one valency (e.g., Fe²⁺, Fe³⁺, Cu⁺, Cu²⁺). Roman numerals denote the valency.

3.4 Writing Chemical Formulae

The chemical formula of a compound symbolically shows:

• Which elements are present.
• The number of atoms of each element (per molecule or formula unit).

To write formulae, we use:

• Symbols of elements/ions.
• Their valencies/charges.

Valency

• Valency is the combining capacity of an element or ion.
• It can be imagined as the number of bonds/arms an atom can use to combine.

Rules for Writing Chemical Formulae

1. The total positive and negative charges must balance (overall charge = 0).
2. In a compound of metal + non-metal, the metal symbol is written first, e.g., NaCl, CaO, FeS.
3. For polyatomic ions, if more than one ion is needed, enclose the ion in brackets and write the number outside, e.g., Ca(OH)₂, (NH₄)₂SO₄.
   • If only one polyatomic ion is present, no bracket is used, e.g., NaNO₃.

3.4.1 Formulae of Simple (Binary) Compounds

Binary compounds consist of two different elements.
Method: Write symbols with valencies, then criss-cross the valencies to get subscripts.

Examples

1. Hydrogen chloride
   • Symbols: H, Cl
   • Valencies: H = 1, Cl = 1
   • Formula: HCl.
2. Hydrogen sulphide
   • Symbols: H, S
   • Valencies: H = 1, S = 2
   • Criss-cross → H₂S.
   • Formula: H₂S.
3. Carbon tetrachloride
   • Symbols: C, Cl
   • Valencies: C = 4, Cl = 1
   • Criss-cross → CCl₄.
   • Formula: CCl₄.
4. Magnesium chloride
   • Mg²⁺, Cl⁻ → MgCl₂ (2 Cl⁻ for each Mg²⁺).
   • Formula: MgCl₂.
5. Aluminium oxide
   • Al³⁺, O²⁻ → Al₂O₃.
   • Formula: Al₂O₃.
6. Calcium oxide
   • Ca²⁺, O²⁻ → CaO (2 and 2 reduced to 1:1).
   • Formula: CaO (not Ca₂O₂).
7. Sodium nitrate
   • Na⁺, NO₃⁻ → NaNO₃.
   • Formula: NaNO₃.
8. Calcium hydroxide
   • Ca²⁺, OH⁻ → Ca(OH)₂.
   • Brackets show two OH groups per Ca²⁺.
   • Formula: Ca(OH)₂.
9. Sodium carbonate
   • Na⁺, CO₃²⁻ → Na₂CO₃.
   • Formula: Na₂CO₃.
10. Ammonium sulphate
    • NH₄⁺, SO₄²⁻ → (NH₄)₂SO₄.
    • Formula: (NH₄)₂SO₄.

Questions (3.4) and Solutions

Q1. Write the formulae of:
(i) Sodium oxide (ii) Aluminium chloride (iii) Sodium sulphide (iv) Magnesium hydroxide.

• (i) Sodium oxide
  • Na⁺, O²⁻ → Na₂O
  • Na₂O
• (ii) Aluminium chloride
  • Al³⁺, Cl⁻ → AlCl₃
  • AlCl₃
• (iii) Sodium sulphide
  • Na⁺, S²⁻ → Na₂S
  • Na₂S
• (iv) Magnesium hydroxide
  • Mg²⁺, OH⁻ → Mg(OH)₂
  • Mg(OH)₂.iesc103.pdf​

Q2. Write names of compounds represented by:
(i) Al₂(SO₄)₃ (ii) CaCl₂ (iii) K₂SO₄ (iv) KNO₃ (v) CaCO₃.

• (i) Aluminium sulphate
• (ii) Calcium chloride
• (iii) Potassium sulphate
• (iv) Potassium nitrate
• (v) Calcium carbonate.

Q3. What is meant by the term chemical formula?

• A chemical formula is a symbolic representation of a compound that shows:
  • The elements present, and
  • The number of atoms of each element in one molecule or formula unit of the compound.

Q4. How many atoms are present in: (i) H₂S molecule (ii) PO₄³⁻ ion?

• (i) H₂S:
  • 2 hydrogen atoms + 1 sulphur atom = 3 atoms.
• (ii) PO₄³⁻:
  • 1 phosphorus atom + 4 oxygen atoms = 5 atoms

3.5 Molecular Mass

Molecular mass is the sum of atomic masses of all atoms present in a molecule of a substance.iesc103.pdf​

• It is expressed in atomic mass units (u).

Example 3.1

(a) Relative molecular mass of water (H₂O):

• Atomic masses: H = 1 u, O = 16 u.
• Water has 2 H and 1 O:
  • Molecular mass = 2×1+1×16=182 × 1 + 1 × 16 = 182×1+1×16=18 u.

(b) Molecular mass of HNO₃:

• H = 1 u, N = 14 u, O = 16 u.
• HNO₃ has 1 H, 1 N, 3 O:
  • Molecular mass = 1+14+3×16=1+14+48=631 + 14 + 3×16 = 1 + 14 + 48 = 631+14+3×16=1+14+48=63 u.

3.5.1 Formula Unit Mass

For ionic compounds, we talk of formula units instead of molecules.iesc103.pdf​

• Formula unit mass = sum of atomic masses of all atoms in a formula unit of an ionic compound.

Example: NaCl

• Na = 23 u, Cl = 35.5 u
• Formula unit mass = 23+35.5=58.523 + 35.5 = 58.523+35.5=58.5 u.iesc103.pdf​

Example 3.2: Formula unit mass of CaCl₂:

• Ca = 40 u, Cl = 35.5 u
• Mass = 40+2×35.5=40+71=11140 + 2×35.5 = 40 + 71 = 11140+2×35.5=40+71=111 u.

Questions (3.5) and Solutions

Q1. Calculate the molecular masses of: H₂, O₂, Cl₂, CO₂, CH₄, C₂H₂, C₂H₄, NH₃, CH₃OH.

Use atomic masses: H = 1, C = 12, O = 16, N = 14, Cl = 35.5 u.

• H₂: 2 × 1 = 2 u
• O₂: 2 × 16 = 32 u
• Cl₂: 2 × 35.5 = 71 u
• CO₂: 12 + 2 × 16 = 44 u
• CH₄: 12 + 4 × 1 = 16 u
• C₂H₂: 2×12 + 2×1 = 24 + 2 = 26 u
• C₂H₄: 2×12 + 4×1 = 24 + 4 = 28 u
• NH₃: 14 + 3×1 = 17 u
• CH₃OH: (C)12 + (H₄)4 + (O)16 = 12 + 4 + 16 = 32 u.

Q2. Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given: Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, O = 16 u

• ZnO: 65 + 16 = 81 u
• Na₂O: 2×23 + 16 = 46 + 16 = 62 u
• K₂CO₃: 2×39 + 12 + 3×16 = 78 + 12 + 48 = 138 u.

End-of-Chapter Exercises – Questions and Solutions

Q1. A 0.24 g sample of a compound of oxygen and boron was found to contain 0.096 g of boron and 0.144 g of oxygen. Calculate percentage composition by mass.

• Total mass = 0.24 g.
• % of B = (0.096 / 0.24) × 100 = 40%
• % of O = (0.144 / 0.24) × 100 = 60%
• Answer: B = 40%, O = 60%.

Q2. When 3.0 g of carbon is burnt in 8.0 g oxygen, 11.0 g CO₂ is produced. What mass of CO₂ will be formed when 3.0 g carbon is burnt in 50.0 g oxygen? Which law applies?

• In first case: 3 g C + 8 g O → 11 g CO₂.
• C is limiting, O is in excess in second case (50 g O).
• Carbon (3 g) will still combine with only 8 g O to form 11 g CO₂; extra oxygen remains unreacted.
• So, 11 g of CO₂ will form.
• Law: Law of constant proportions (fixed ratio of C:O in CO₂).

Q3. What are polyatomic ions? Give examples.

• Polyatomic ions are groups of two or more atoms that carry an overall positive or negative charge and behave as a single charged unit.
• Examples:
  • Ammonium, NH₄⁺
  • Hydroxide, OH⁻
  • Sulphate, SO₄²⁻
  • Carbonate, CO₃²⁻
  • Nitrate, NO₃⁻.

Q4. Write the chemical formulae of:
(a) Magnesium chloride (b) Calcium oxide (c) Copper nitrate (d) Aluminium chloride (e) Calcium carbonate.

• (a) Mg²⁺, Cl⁻ → MgCl₂
• (b) Ca²⁺, O²⁻ → CaO
• (c) Cu²⁺, NO₃⁻ → Cu(NO₃)₂
• (d) Al³⁺, Cl⁻ → AlCl₃
• (e) Ca²⁺, CO₃²⁻ → CaCO₃.

Q5. Give names of elements present in:
(a) Quick lime (b) Hydrogen bromide (c) Baking powder (d) Potassium sulphate.

• (a) Quick lime – CaO → Calcium, Oxygen
• (b) Hydrogen bromide – HBr → Hydrogen, Bromine
• (c) Baking powder → commonly sodium hydrogen carbonate, NaHCO₃ → Sodium, Hydrogen, Carbon, Oxygen
• (d) Potassium sulphate – K₂SO₄ → Potassium, Sulphur, Oxygen.

Q6. Calculate the molar mass of:
(a) Ethyne, C₂H₂ (b) Sulphur molecule, S₈ (c) Phosphorus molecule, P₄ (d) HCl (e) HNO₃.

Use: C = 12, H = 1, S = 32, P = 31, H = 1, Cl = 35.5, N = 14, O = 16.

• (a) C₂H₂: 2×12 + 2×1 = 24 + 2 = 26 g/mol
• (b) S₈: 8×32 = 256 g/mol
• (c) P₄: 4×31 = 124 g/mol
• (d) HCl: 1 + 35.5 = 36.5 g/mol
• (e) HNO₃: 1 + 14 + 3×16 = 1 + 14 + 48 = 63 g/mol

Download Free Mind Map from the link below

This mind map contains all important topics of this chapter

[Download PDF Here]

Visit our Class 9 Science page for free mind maps of all Chapters