Class 9 Science Notes Chapter 3 (Chapter 3) – Examplar Problem (English) Book

Alright class, let's dive deep into Chapter 3, 'Atoms and Molecules'. This is a fundamental chapter, not just for your Class 9 understanding, but also forms the bedrock for chemistry concepts you'll encounter later, especially if you're preparing for competitive government exams. We'll focus on the core ideas and practice applying them, much like the problems you see in your Exemplar book.

Chapter 3: Atoms and Molecules - Detailed Notes for Exam Preparation

1. Laws of Chemical Combination

These laws were established through careful experimentation and form the basis of stoichiometry (calculation of reactants and products in chemical reactions).

• Law of Conservation of Mass:

  • Stated by: Antoine Lavoisier (often called the 'Father of Modern Chemistry').
  • Statement: Mass can neither be created nor destroyed in a chemical reaction.
  • Meaning: The total mass of the reactants (substances that react) is always equal to the total mass of the products (substances formed) in a chemical reaction.
  • Exam Relevance: Questions might involve simple calculations verifying this law or identifying it from a description.
  • Example: If 5.3 g of sodium carbonate reacts with 6 g of ethanoic acid, the products are 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate.
    Reactants mass = 5.3 g + 6 g = 11.3 g
    Products mass = 2.2 g + 0.9 g + 8.2 g = 11.3 g. Mass is conserved.

• Law of Constant Proportions (or Law of Definite Proportions):

  • Stated by: Joseph Proust.
  • Statement: In a chemical substance, the elements are always present in definite proportions by mass.
  • Meaning: A pure chemical compound, regardless of its source or method of preparation, always contains the same elements combined together in the same fixed ratio by mass.
  • Exam Relevance: Calculating the mass ratio of elements in a compound, or identifying this law from experimental data showing constant composition.
  • Example: Water (H₂O). Atomic mass H=1u, O=16u. In one molecule of H₂O, there are 2 H atoms and 1 O atom.
    Mass ratio H : O = (2 × 1u) : (1 × 16u) = 2 : 16 = 1:8.
    This means 9 g of water always contains 1 g of Hydrogen and 8 g of Oxygen.

2. Dalton's Atomic Theory

John Dalton provided the first scientific theory about the nature of matter, explaining the laws of chemical combination.

• Key Postulates:
  1. All matter is composed of extremely small particles called atoms.
  2. Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. (Explains Law of Conservation of Mass).
  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. (Explains Law of Constant Proportions).
  6. The relative number and kinds of atoms are constant in a given compound.
• Limitations (Modern View):
  • Atoms are divisible (into protons, neutrons, electrons).
  • Atoms of the same element can have slightly different masses (isotopes, e.g., ¹²C, ¹³C, ¹⁴C).
  • Atoms of different elements can sometimes have the same mass (isobars).
• Exam Relevance: Know the postulates and how they explain the laws. Be aware of the limitations.

3. Atoms

• Definition: The smallest particle of an element that retains the chemical properties of that element and can take part in a chemical reaction.
• Size: Atomic radius is measured in nanometers (nm). 1 nm = 10⁻⁹ m. Atoms are incredibly small.
• Symbols: Symbols of elements are abbreviations (often the first one or two letters of the English or Latin name), e.g., H (Hydrogen), O (Oxygen), Na (Sodium - from Natrium), Fe (Iron - from Ferrum). The first letter is always uppercase, the second (if present) is lowercase.
• Atomic Mass:
  • Since atoms are very light, their mass is expressed relative to a standard.
  • Standard: One atom of the Carbon-12 isotope (¹²C).
  • Atomic Mass Unit (u): Defined as exactly 1/12th the mass of one atom of Carbon-12. (1 u ≈ 1.66 × 10⁻²⁴ g).
  • Definition: The relative atomic mass of an element is the average mass of its atoms compared to 1/12th the mass of one Carbon-12 atom. (Average is used due to isotopes).
  • Examples: H = 1u, C = 12u, N = 14u, O = 16u, Na = 23u, Mg = 24u, S = 32u, Cl = 35.5u, Ca = 40u. (Memorize common ones).
• Existence: Atoms of most elements are highly reactive and do not exist independently. They exist as molecules or ions. Noble gases (He, Ne, Ar etc.) are exceptions and exist as single atoms.

4. Molecules

• Definition: The smallest particle of an element or a compound that can exist independently under ordinary conditions and shows all the properties of that substance. It is a group of two or more atoms held together by chemical bonds.
• Molecules of Elements: Formed when atoms of the same element combine.
  • Atomicity: The number of atoms present in one molecule of an element.
    • Monoatomic: Atomicity 1 (e.g., He, Ne, Ar).
    • Diatomic: Atomicity 2 (e.g., H₂, N₂, O₂, F₂, Cl₂, Br₂, I₂).
    • Triatomic: Atomicity 3 (e.g., Ozone O₃).
    • Polyatomic: Atomicity > 2 (e.g., Phosphorus P₄, Sulphur S₈).
• Molecules of Compounds: Formed when atoms of different elements combine in a fixed, whole-number ratio.
  • Examples: H₂O (Water), CO₂ (Carbon Dioxide), NH₃ (Ammonia), CH₄ (Methane), C₆H₁₂O₆ (Glucose). The formula shows the type and number of atoms of each element in one molecule.

5. Ions

• Definition: An atom or a group of atoms that has acquired a net electrical charge by losing or gaining electrons.
• Formation:
  • Cations: Positively charged ions formed by the loss of one or more electrons (e.g., Na → Na⁺ + e⁻; Ca → Ca²⁺ + 2e⁻). Metals tend to form cations.
  • Anions: Negatively charged ions formed by the gain of one or more electrons (e.g., Cl + e⁻ → Cl⁻; O + 2e⁻ → O²⁻). Non-metals tend to form anions.
• Types:
  • Simple Ions: Formed from single atoms (e.g., Na⁺, Mg²⁺, Al³⁺, Cl⁻, O²⁻, N³⁻).
  • Polyatomic Ions (Compound Ions): A group of covalently bonded atoms that acts as a single unit with an overall charge (e.g., NH₄⁺ (Ammonium), OH⁻ (Hydroxide), NO₃⁻ (Nitrate), CO₃²⁻ (Carbonate), SO₄²⁻ (Sulphate), PO₄³⁻ (Phosphate)). Memorize the formulae and charges of common polyatomic ions.
• Valency:
  • The combining capacity of an element or ion.
  • For simple ions, the valency is equal to the magnitude (numerical value) of the charge on the ion. (Valency of Mg²⁺ is 2; Valency of Cl⁻ is 1).
  • Crucial for writing chemical formulae.

6. Writing Chemical Formulae

A chemical formula is a symbolic representation of the composition of a compound.

• Rules (Criss-Cross Method):
  1. Write the symbols/formulae of the cation (positive ion) first, followed by the anion (negative ion).
  2. Write the valency (charge magnitude) of each ion below it.
  3. Criss-cross the valencies. The valency of the first ion becomes the subscript of the second, and vice versa.
  4. Write the subscripts. Ignore the charges. If a subscript is 1, it is not written.
  5. If the subscripts have a common factor, divide by it to get the simplest ratio.
  6. If a polyatomic ion takes a subscript greater than 1, enclose the ion in parentheses before writing the subscript.
• Examples:
  • Magnesium chloride: Mg²⁺ Cl¹⁻ → Mg₁Cl₂ → MgCl₂
  • Aluminium oxide: Al³⁺ O²⁻ → Al₂O₃
  • Calcium nitrate: Ca²⁺ (NO₃)¹⁻ → Ca₁(NO₃)₂ → Ca(NO₃)₂
  • Ammonium sulphate: (NH₄)¹⁺ (SO₄)²⁻ → (NH₄)₂(SO₄)₁ → (NH₄)₂SO₄
  • Sodium carbonate: Na¹⁺ (CO₃)²⁻ → Na₂(CO₃)₁ → Na₂CO₃

7. Molecular Mass and Formula Unit Mass

• Molecular Mass:
  • The sum of the atomic masses of all the atoms present in one molecule of a substance (used for covalent compounds).
  • Unit: atomic mass units (u).
  • Calculation: Multiply the atomic mass of each element by the number of its atoms in the molecule and add them up.
  • Example (H₂SO₄): (2 × Atomic mass of H) + (1 × Atomic mass of S) + (4 × Atomic mass of O) = (2 × 1u) + (1 × 32u) + (4 × 16u) = 2u + 32u + 64u = 98u.
• Formula Unit Mass:
  • The sum of the atomic masses of all atoms in a formula unit of an ionic compound. (Ionic compounds exist as crystal lattices, not discrete molecules, so 'formula unit' is the simplest ratio of ions).
  • Calculation method is identical to molecular mass.
  • Unit: atomic mass units (u).
  • Example (CaCl₂): (1 × Atomic mass of Ca) + (2 × Atomic mass of Cl) = (1 × 40u) + (2 × 35.5u) = 40u + 71u = 111u.

8. Mole Concept

The mole is the SI unit for the amount of substance. It provides a way to connect mass with the number of particles.

• Avogadro's Number (N<0xE2><0x82><0x90>): The fixed number of particles (atoms, molecules, ions, electrons, etc.) present in one mole of any substance.

  • N<0xE2><0x82><0x90> = 6.022 × 10²³ mol⁻¹ (particles per mole).

• Definition of Mole:

  • One mole is the amount of substance that contains exactly 6.022 × 10²³ elementary entities (particles).
  • Alternatively, one mole of any species (atoms, molecules, ions) is that quantity whose mass in grams is numerically equal to its atomic/molecular/formula unit mass.

• Molar Mass (M):

  • The mass of one mole of a substance.
  • Unit: grams per mole (g/mol).
  • Relationship: Molar mass (in g/mol) is numerically equal to the atomic/molecular/formula unit mass (in u).
  • Examples:
    • Atomic mass of Na = 23u → Molar mass of Na = 23 g/mol (This mass contains 6.022 × 10²³ Na atoms).
    • Molecular mass of H₂O = 18u → Molar mass of H₂O = 18 g/mol (This mass contains 6.022 × 10²³ H₂O molecules).
    • Formula unit mass of NaCl = 58.5u → Molar mass of NaCl = 58.5 g/mol (This mass contains 6.022 × 10²³ NaCl formula units).

• Key Relationships & Formulae:
  Let:

  • n = Number of moles
  • m = Given mass (in grams)
  • M = Molar mass (in g/mol)
  • N = Given number of particles
  • N<0xE2><0x82><0x90> = Avogadro's number (6.022 × 10²³)
  1. n = m / M (Moles = Given mass / Molar mass)
  2. n = N / N<0xE2><0x82><0x90> (Moles = Given number of particles / Avogadro's number)
  3. m = n × M (Mass = Moles × Molar mass)
  4. N = n × N<0xE2><0x82><0x90> (Number of particles = Moles × Avogadro's number)
  5. m / M = N / N<0xE2><0x82><0x90> (Relates mass directly to number of particles)

• Exam Relevance: Mole concept calculations are very common. Be comfortable converting between mass, moles, and number of particles (atoms, molecules, ions). Pay attention to whether the question asks for atoms or molecules.

  • Example: Calculate the number of oxygen atoms in 90 g of water (H₂O).
    1. Molar mass of H₂O = 18 g/mol.
    2. Moles of H₂O = n = m / M = 90 g / 18 g/mol = 5 mol.
    3. Number of H₂O molecules = N = n × N<0xE2><0x82><0x90> = 5 × 6.022 × 10²³.
    4. Each H₂O molecule contains 1 oxygen atom.
    5. Number of oxygen atoms = 1 × (Number of H₂O molecules) = 5 × 6.022 × 10²³ = 3.011 × 10²⁴ atoms.

Now, let's test your understanding with some Multiple Choice Questions, similar to what you might encounter.

Multiple Choice Questions (MCQs)

1. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
   (a) Atoms of a given element are identical in mass and chemical properties.
   (b) Atoms combine in the ratio of small whole numbers to form compounds.
   (c) The relative number and kinds of atoms are constant in a given compound.
   (d) Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.

2. The formula for Aluminium Sulphate is Al₂(SO₄)₃. What is the valency of Aluminium (Al) and the Sulphate ion (SO₄) respectively?
   (a) 2 and 3
   (b) 3 and 2
   (c) 3 and 6
   (d) 2 and 2

3. Calculate the number of moles in 11 g of CO₂. (Atomic mass: C=12u, O=16u)
   (a) 0.25 mol
   (b) 0.5 mol
   (c) 1 mol
   (d) 4 mol

4. Which of the following has the largest number of atoms?
   (a) 18 g of H₂O (Molar mass = 18 g/mol)
   (b) 16 g of O₂ (Molar mass = 32 g/mol)
   (c) 4.4 g of CO₂ (Molar mass = 44 g/mol)
   (d) 16 g of CH₄ (Molar mass = 16 g/mol)

5. The chemical symbol 'Fe' stands for:
   (a) Fluorine
   (b) Fermium
   (c) Iron
   (d) Francium

6. In a reaction, 5 g of calcium carbonate (CaCO₃) on heating gave 2.8 g of calcium oxide (CaO) and 2.2 g of carbon dioxide (CO₂). This observation is in agreement with the:
   (a) Law of Constant Proportions
   (b) Law of Conservation of Mass
   (c) Law of Multiple Proportions
   (d) Avogadro's Law

7. What is the mass of 0.5 moles of Ammonia (NH₃)? (Atomic mass: N=14u, H=1u)
   (a) 17 g
   (b) 8.5 g
   (c) 34 g
   (d) 1.7 g

8. The atomicity of Ozone (O₃), Phosphorus (P₄), and Sulphur (S₈) are respectively:
   (a) 1, 4, 8
   (b) 3, 4, 8
   (c) 3, 8, 4
   (d) 8, 4, 3

9. Which pair of elements will form an ionic compound with the formula XY₂? (X=Cation, Y=Anion)
   (a) X with valency 1, Y with valency 2
   (b) X with valency 2, Y with valency 1
   (c) X with valency 2, Y with valency 2
   (d) X with valency 1, Y with valency 1

10. How many molecules are present in 34 g of NH₃? (N<0xE2><0x82><0x90> = 6.022 × 10²³)
    (a) 6.022 × 10²³
    (b) 3.011 × 10²³
    (c) 1.2044 × 10²⁴
    (d) 2 × 6.022 × 10²³

Answer Key for MCQs:

1. (d) - Dalton's postulate about atoms being neither created nor destroyed directly explains mass conservation.
2. (b) - From the criss-cross method applied in reverse: Al₂(SO₄)₃ implies Al had valency 3 and SO₄ had valency 2.
3. (a) - Molar mass of CO₂ = 12 + (2 × 16) = 44 g/mol. Moles = m/M = 11 g / 44 g/mol = 0.25 mol.
4. (d) - Calculate total atoms in each:
   (a) 18g H₂O = 1 mol H₂O = 1 mol × 3 atoms/molecule × N<0xE2><0x82><0x90> = 3 N<0xE2><0x82><0x90> atoms.
   (b) 16g O₂ = 0.5 mol O₂ = 0.5 mol × 2 atoms/molecule × N<0xE2><0x82><0x90> = 1 N<0xE2><0x82><0x90> atoms.
   (c) 4.4g CO₂ = 0.1 mol CO₂ = 0.1 mol × 3 atoms/molecule × N<0xE2><0x82><0x90> = 0.3 N<0xE2><0x82><0x90> atoms.
   (d) 16g CH₄ = 1 mol CH₄ = 1 mol × 5 atoms/molecule × N<0xE2><0x82><0x90> = 5 N<0xE2><0x82><0x90> atoms. (Largest)
5. (c) - Fe comes from the Latin name 'Ferrum'.
6. (b) - Mass of reactant (CaCO₃) = 5 g. Mass of products (CaO + CO₂) = 2.8 g + 2.2 g = 5 g. Mass is conserved.
7. (b) - Molar mass of NH₃ = 14 + (3 × 1) = 17 g/mol. Mass = n × M = 0.5 mol × 17 g/mol = 8.5 g.
8. (b) - Atomicity is the number of atoms in the molecule: O₃ (3), P₄ (4), S₈ (8).
9. (b) - For XY₂, X must have valency 2 and Y must have valency 1 (e.g., Mg²⁺ and Cl¹⁻ form MgCl₂).
10. (c) - Molar mass of NH₃ = 17 g/mol. Moles = m/M = 34 g / 17 g/mol = 2 mol. Number of molecules = n × N<0xE2><0x82><0x90> = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴.

Make sure you understand the reasoning behind each answer, especially the calculation-based ones. Go through these notes thoroughly, practice writing chemical formulae, and work on mole concept problems from your textbook and Exemplar. Let me know if any part needs further clarification!