⚛️ Atoms and Molecules – Full Breakdown

AMU Class 11 Science Chemistry June 2026
This chapter is the FOUNDATION of all chemistry. AMU asks questions from Dalton's theory, laws of chemical combination, atomic/molecular mass calculations, and mole concept. The numericals are straightforward if you know the three mole formulas. I've broken down everything step-by-step with the exact patterns that appear in AMU papers. Trust me — master this chapter and half your chemistry is done.

1) Laws of Chemical Combination — AMU Direct Questions

Law of Conservation of Mass — Lavoisier (1789)

Statement: Mass can neither be created nor destroyed in a chemical reaction. Total mass of reactants = Total mass of products.

Example: 5 g Na₂CO₃ + 7.2 g CH₃COOH → 4.2 g CO₂ + 3 g H₂O + 5 g CH₃COONa. Check: 5 + 7.2 = 12.2 g reactants. Products = 4.2 + 3 + 5 = 12.2 g. Proved.

⚠️ AMU Direct: "Which postulate of Dalton's atomic theory explains the law of conservation of mass?" Answer: "Atoms can neither be created nor destroyed."

Law of Constant / Definite Proportions — Proust (1799)

Statement: A chemical compound always contains the same elements in the same proportion by mass, regardless of its source.

Example: Water (H₂O) always has H:O = 1:8 by mass. Whether it's from rain, river, or tap — ratio is always 1:8.

⚠️ AMU Gold: "3 g of carbon burns in 8 g oxygen to give 11 g CO₂. What mass of CO₂ is formed when 3 g carbon is burnt in 40 g oxygen?" Answer: 11 g CO₂ (only 8 g O₂ reacts, 32 g O₂ remains unreacted). Law of constant proportion applies — always C:O = 3:8 in CO₂.

Law of Multiple Proportions — Dalton (1803)

Statement: When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in a simple whole number ratio.

Example: CO (C:O = 12:16 = 3:4) and CO₂ (C:O = 12:32 = 3:8). Mass of O per fixed mass of C: 4:8 = 1:2 (simple whole number).

Law of Reciprocal Proportions — Richter

If two elements A and B combine separately with a fixed mass of C, then the ratio of their masses when they combine directly is a simple multiple of the ratio.

2) Dalton's Atomic Theory (1808)

Five Postulates:

  1. All matter is made of tiny indivisible particles called atoms.
  2. Atoms of the same element are identical in mass and chemical properties.
  3. Atoms of different elements have different masses and properties.
  4. Atoms combine in simple whole number ratios to form compounds.
  5. Atoms can neither be created nor destroyed in a chemical reaction.
🧠 Trick to remember — "AIDSC" mnemonic:
A → Atoms are indivisible
I → Identical atoms for same element
D → Different atoms for different elements
S → Simple whole number ratio
C → Cannot be created/destroyed
⚠️ AMU Pattern: "Which postulate was later proved wrong?" Answer: Postulate 1 — "Atoms are indivisible." We now know atoms are divisible into protons, neutrons, electrons (subatomic particles). Also postulate 2 was modified — atoms of same element can have different masses (isotopes).

3) Atomic Mass, Molecular Mass & Formula Mass

Atomic Mass Unit (amu / u)

1 amu = 1/12 × mass of one atom of Carbon-12
1 amu = 1.66 × 10⁻²⁴ g

Atomic mass of H = 1 u, C = 12 u, O = 16 u, N = 14 u, Na = 23 u, Cl = 35.5 u

Average Atomic Mass

Most elements exist as isotopes. The atomic mass on the periodic table is a weighted average.

Example — Chlorine: Cl-35 (75%) + Cl-37 (25%) → Average = (35 × 75 + 37 × 25)/100 = (2625 + 925)/100 = 35.5 u

⚠️ AMU Favourite: "Why is atomic mass of chlorine 35.5 u?" Answer: It has two isotopes: Cl-35 and Cl-37 in 3:1 ratio. The average gives 35.5 u.

Molecular Mass

Sum of atomic masses of all atoms in a molecule.

H₂O = 2×1 + 16 = 18 u. CO₂ = 12 + 2×16 = 44 u. CaCO₃ = 40 + 12 + 3×16 = 100 u.

Formula Mass (for ionic compounds)

Used instead of molecular mass for ionic compounds (they don't have molecules).

NaCl = 23 + 35.5 = 58.5 u. CaCl₂ = 40 + 2×35.5 = 111 u.

4) Mole Concept — The Three Formulas

1. n = m / M (mass to moles)
2. n = N / Nₐ (particles to moles)
3. n = V / 22.4 (gas volume at STP to moles)

Where:
n = number of moles
m = given mass (in g)
M = molar mass (g/mol)
N = number of particles
Nₐ = 6.022 × 10²³ (Avogadro's constant)
V = volume of gas at STP (L)
🧠 TOPPER TRICK — The Mole Triangle:
Draw a triangle: Top = MOLE. Bottom left = MASS/M. Bottom right = PARTICLES/Nₐ. Bottom centre = VOLUME/22.4.
Cover the one you want to find — that's the formula. This is literally ALL you need for mole numericals.

Examples:

Q: Find moles in 11 g CO₂. → n = 11/44 = 0.25 mol

Q: How many molecules in 0.5 mol H₂O? → N = 0.5 × 6.022×10²³ = 3.011×10²³ molecules

Q: Volume of 2 mol O₂ at STP? → V = 2 × 22.4 = 44.8 L

5) Atomicity & Molecularity

Atomicity: Number of atoms in a molecule.

AtomicityExample
Monoatomic (1 atom)He, Ne, Ar (noble gases)
Diatomic (2 atoms)H₂, O₂, N₂, Cl₂, F₂, Br₂, I₂
Triatomic (3 atoms)O₃ (ozone), H₂O, CO₂
Tetra-atomic (4 atoms)P₄ (phosphorus), NH₃
Polyatomic (>4 atoms)S₈ (sulphur), P₄O₁₀, C₆H₁₂O₆
🧠 Trick — Diatomic elements (7): "Have No Fear Of Ice Cold Beer" → H₂, N₂, F₂, O₂, I₂, Cl₂, Br₂. These 7 elements exist as diatomic molecules naturally.

6) Ions & Ionic Compounds

Ion: A charged particle formed when an atom gains or loses electrons.

Writing Chemical Formulas — Criss-Cross Method

Step 1: Write symbols side by side (cation first, anion second).
Step 2: Write their charges on top.
Step 3: Criss-cross the charges (ignore signs).
Step 4: Simplify if needed.

Example 1: Calcium + Chlorine
Ca²⁺ Cl⁻ → CaCl₂
↙ ↘
(2 goes to Cl, 1 goes to Ca)

Example 2: Aluminium + Sulphate
Al³⁺ SO₄²⁻ → Al₂(SO₄)₃

Example 3: Sodium + Carbonate
Na⁺ CO₃²⁻ → Na₂CO₃

7) Common AMU-Style MCQs with Answers

Q1:

Which of the following is the smallest unit of matter?

(a) Molecule (b) Atom (c) Electron (d) Proton

Answer: (b) Atom (in context of chemical reactions — atom is smallest unit of element)

Q2:

The law of conservation of mass was proposed by:

(a) Dalton (b) Proust (c) Lavoisier (d) Richter

Answer: (c) Lavoisier

Q3:

Atomic mass of chlorine is 35.5 because:

(a) It has isotopes Cl-35 and Cl-37 (b) It has fractional atomic number (c) It is a diatomic molecule (d) None

Answer: (a) It has isotopes Cl-35 and Cl-37

Q4:

Number of moles in 36 g of H₂O is:

(a) 1 (b) 2 (c) 3 (d) 0.5

Answer: (b) 2 (n = 36/18 = 2)

Q5:

Avogadro's constant has a value of:

(a) 6.022 × 10²² (b) 6.022 × 10²³ (c) 6.022 × 10²⁴ (d) 6.022 × 10²¹

Answer: (b) 6.022 × 10²³

Q6:

Which of the following is diatomic?

(a) He (b) Ne (c) O₂ (d) P₄

Answer: (c) O₂

Q7:

What is the mass of 1 mole of CO₂?

(a) 44 g (b) 44 u (c) 22 g (d) 12 g

Answer: (a) 44 g (molar mass = 44 g/mol)

Q8:

Postulate of Dalton's theory that explains law of constant proportion:

(a) Atoms are indivisible (b) Atoms combine in simple whole number ratios (c) Atoms can't be created/destroyed (d) All atoms of same element are identical

Answer: (b) Atoms combine in simple whole number ratios

Q9:

Molecular mass of CaCO₃ is:

(a) 100 u (b) 50 u (c) 74 u (d) 84 u

Answer: (a) 100 u (40 + 12 + 48 = 100)

Q10:

Number of atoms in 1 mole of oxygen gas (O₂) is:

(a) 6.022 × 10²³ (b) 12.044 × 10²³ (c) 3.011 × 10²³ (d) 1

Answer: (b) 12.044 × 10²³ (1 mole O₂ = 6.022×10²³ molecules = 2 × 6.022×10²³ atoms)

8) Quick One-Liner Revision

🧠 From a 2026 AMU topper — my final tips:
1. Mole concept is the MOST important — if you master n = m/M, you can solve 90% of numericals in this chapter.
2. Dalton's postulates are direct MCQ material. Know which postulate explains which law.
3. Atomic mass of Cl = 35.5 is the most common atomic mass trick question in AMU. Remember why (isotopes).
4. Diatomic elements — the "Have No Fear Of Ice Cold Beer" mnemonic saved me in the exam.
5. Practice at least 5-10 mole numericals before the exam. They're free marks if you get the formula right.
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