By the end of this chapter you'll be able to…

  • 1State Dalton's atomic theory and identify which postulates have been refined by modern science
  • 2Apply the laws of conservation of mass and constant proportions to numerical problems
  • 3Calculate atomic mass, molecular mass and formula unit mass of any compound
  • 4Write the correct chemical formula of any binary compound using valencies (cross-multiplication method)
  • 5Define mole, Avogadro's number and molar mass; interconvert mass, moles, and particle count using n = m/M = N/Nₐ
  • 6Identify cations, anions and polyatomic ions from a given list
  • 7Compute atomicity of common molecules and the percentage composition of elements in a compound
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Why this chapter matters
This is where chemistry stops being qualitative ('this dissolves, that fizzes') and becomes quantitative. The mole concept and the formula n = m/M = N/Nₐ is the most-used idea in all of school + college chemistry — every later reaction calculation rests on it.

Atoms and Molecules — Class 9 (CBSE)

Hand a child two cups, one of salt water and one of sugar water. They look identical. Smell them — identical. Yet one is salty, one sweet. The difference is invisible: different atoms arranged into different molecules. This chapter is about understanding stuff at the level where stuff becomes itself.


1. The story — from philosophers to Dalton

In §1 of the previous chapter we met Maharishi Kanada (~600 BCE) and Democritus (~400 BCE), both of whom proposed that matter is made of tiny indivisible particles ("anu" and "atomos" — meaning uncuttable). For 2200 years that idea remained philosophy, not science. Then in 1808, English chemist John Dalton put together quantitative experimental results from Lavoisier, Proust and others, and formalised the first scientific atomic theory.

Dalton's 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 small whole-number ratios to form compounds.
  5. In chemical reactions, atoms are neither created nor destroyed — only rearranged.

Three of these turned out to be slightly wrong:

  • Atoms ARE divisible (electrons, protons, neutrons discovered in next chapter).
  • Atoms of the same element CAN have different masses (isotopes).
  • Nuclear reactions DO change atoms into other atoms (fusion, fission).

But for ordinary chemistry — making and breaking bonds, mixing reagents in beakers — Dalton's theory is essentially right. It's the foundation on which all of high-school chemistry is built.


2. Laws of chemical combination

Before Dalton, two laws — established by careful weighing experiments — already pointed at the existence of atoms.

Law of conservation of mass (Lavoisier, 1789)

Mass is neither created nor destroyed in a chemical reaction.

If you start with 10 g of reactants in a sealed container, you end with exactly 10 g of products. The atoms are just rearranged.

Example:

Law of constant proportions / definite proportions (Proust, 1799)

A pure chemical compound always contains the same elements in the same proportion by mass, regardless of where it came from.

Water from a Himalayan glacier, water from your kitchen tap, water synthesised in a lab — all have hydrogen : oxygen = 1 : 8 by mass (or H : O = 2 : 1 by atoms).

Both laws were neatly explained by Dalton's "atoms in fixed whole-number ratios" postulate — and that's why his theory caught on.


3. Atomic mass — and why it's relative

We can't put a single atom on a weighing scale; it's far too small. So instead, chemists use relative atomic masses: how heavy an atom of one element is compared with an atom of a chosen standard.

Three standards in history

  1. Hydrogen (atomic mass = 1) — used initially because H is the lightest.
  2. Oxygen (atomic mass = 16) — used in the late 19th and early 20th century.
  3. Carbon-12 (atomic mass = 12) — current standard since 1961.

The modern definition: one atomic mass unit (u, also called dalton) is 1/12 the mass of one atom of carbon-12.

Atomic masses you must memorise (Class 9 essentials)

ElementSymbolAtomic mass (u)
HydrogenH1
CarbonC12
NitrogenN14
OxygenO16
SodiumNa23
MagnesiumMg24
AluminiumAl27
SulphurS32
ChlorineCl35.5
PotassiumK39
CalciumCa40
IronFe56
CopperCu63.5
ZincZn65
SilverAg108

Why is chlorine 35.5? Because natural chlorine is a 75:25 mix of Cl-35 and Cl-37 — its "atomic mass" is the weighted average.


4. Molecules and ions

A molecule is the smallest particle of an element or compound that can exist on its own and still retain that substance's chemical properties.

Molecules of elements

Most non-metals exist as molecules of fixed size:

  • , , , , , , diatomic (2 atoms).
  • (ozone) — triatomic.
  • (white phosphorus) — tetra-atomic.
  • — octa-atomic.
  • Inert/noble gases (He, Ne, Ar) — monatomic.

This count is the atomicity of the element.

Molecules of compounds

Atoms of different elements bonded together: H₂O, CO₂, NH₃, CH₄, H₂SO₄, NaCl (debated — see ions below).

Ions — charged particles

When atoms gain or lose electrons, they become charged → ions.

  • Cation (positive): lost electrons. Na → Na⁺ + e⁻.
  • Anion (negative): gained electrons. Cl + e⁻ → Cl⁻.

Compounds like NaCl, CaCO₃, KNO₃ are made of ions held together by electrostatic forces — they don't form discrete molecules, but you'll still write "formulas" for them (these are called formula units).

Common ions to memorise

CationSymbolValencyAnionSymbolValency
SodiumNa⁺1ChlorideCl⁻1
PotassiumK⁺1BromideBr⁻1
HydrogenH⁺1IodideI⁻1
SilverAg⁺1HydroxideOH⁻1
AmmoniumNH₄⁺1NitrateNO₃⁻1
Copper(I)Cu⁺1Hydrogen carbonateHCO₃⁻1
MagnesiumMg²⁺2OxideO²⁻2
CalciumCa²⁺2SulphideS²⁻2
ZincZn²⁺2SulphateSO₄²⁻2
Iron(II)Fe²⁺2CarbonateCO₃²⁻2
AluminiumAl³⁺3PhosphatePO₄³⁻3
Iron(III)Fe³⁺3NitrideN³⁻3

Valency rules for writing formulas

Cross-multiply the valencies and simplify:

Example 1 — sodium chloride:

  • Na is +1, Cl is −1. Cross-multiply: Na₁Cl₁ → NaCl.

Example 2 — aluminium oxide:

  • Al is +3, O is −2. Cross-multiply: Al₂O₃ → Al₂O₃.

Example 3 — calcium phosphate:

  • Ca is +2, PO₄ is −3. Cross-multiply: Ca₃(PO₄)₂ → Ca₃(PO₄)₂.
  • Use brackets when the polyatomic ion is taken more than once.

5. Molecular mass and formula unit mass

Molecular mass = sum of atomic masses of all atoms in one molecule.

Example — water, H₂O:

Example — sulphuric acid, H₂SO₄:

Formula unit mass is the same calculation but used for ionic compounds (which don't form discrete molecules):

Example — sodium chloride:

Example — calcium carbonate, CaCO₃:

The numerical work is identical to molecular mass; the terminology distinguishes ionic from molecular compounds.


6. The mole concept — the heart of chemistry

Why we need a "mole"

In a chemistry lab you weigh out grams. But chemical reactions happen at the level of individual atoms and molecules. We need a unit that bridges "grams I can weigh" and "atoms I can count."

That bridge is the mole.

Definition

One mole of any substance contains particles of that substance. This is Avogadro's number ().

Why this specific number? It's defined so that 1 mole of carbon-12 atoms weighs exactly 12 grams. Same number trick:

  • 1 mole of H atoms weighs 1 g.
  • 1 mole of O atoms weighs 16 g.
  • 1 mole of H₂O molecules weighs 18 g.

The mass in grams of 1 mole = atomic/molecular mass expressed in u. This is the molar mass.

The three relationships you'll use everywhere

Where:

  • = number of moles
  • = mass in grams
  • = molar mass (g/mol)
  • = number of particles
  • = Avogadro's number ()

Three quantities, three formulas, easy to remember:

  • moles from mass: .
  • moles from particle count: .
  • combine to compute one from the other: .

How big is Avogadro's number?

is unfathomably large. If you had Avogadro's number of seconds, it would last years — about 1.4 million times longer than the age of the universe. One mole of marbles would cover the entire Earth to a depth of 80 km.

Yet 1 mole of water is just 18 g — about a tablespoonful. Atoms really are that tiny.


7. Worked example walkthrough

Question: How many molecules are there in 36 g of water?

Step-by-step solution:

Step 1 — Find molar mass.

Step 2 — Find moles.

Step 3 — Number of molecules.

Answer: molecules.


8. Closing thought

What this chapter did, beautifully:

  • Took the invisible atom and made it weighable (via relative atomic mass).
  • Took the uncountable atom and made it countable (via the mole).
  • Took gross mass measurements in a lab and translated them into particle counts at atomic scale.

Every later chapter — chemical reactions, stoichiometry, gas laws, thermodynamics, electrochemistry — depends on this translation. Master the formula and you have the working language of chemistry.

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Atomic mass unit (u)
1 u = (1/12) × mass of C-12 atom = 1.66 × 10⁻²⁷ kg
Modern standard since 1961.
Molecular mass
M = Σ (atomic mass × number of atoms)
Add all atomic masses in one molecule.
Formula unit mass
Same as molecular mass, but for ionic compounds
NaCl, CaCO₃, etc.
Avogadro's number
Nₐ = 6.022 × 10²³ particles per mole
The bridge between gram and atom.
Moles from mass
n = m / M
m in grams, M in g/mol, n is unitless.
Moles from particle count
n = N / Nₐ
N = number of atoms/molecules/ions.
Mass from particle count
m = (N / Nₐ) × M
Combination of the above two.
Number of particles from mass
N = (m / M) × Nₐ
Inverse direction.
Percentage by mass of element
(mass of element / mass of compound) × 100
Verify chemical formulas this way.
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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Forgetting to multiply by the subscript when computing molecular mass
For H₂O: 2 × 1 + 1 × 16 = 18. NOT 1 + 16 = 17. The 2 in H₂ multiplies the atomic mass.
WATCH OUT
Confusing atomic mass and mass number
Atomic mass is a measured average (often non-integer). Mass number is a count of protons + neutrons (always an integer). They're close but different concepts.
WATCH OUT
Saying NaCl is a molecule
Ionic compounds don't form discrete molecules. NaCl is an extended ionic lattice. Use the term 'formula unit' for NaCl.
WATCH OUT
Cross-multiplying valencies without simplifying
Ca²⁺ and S²⁻ give Ca₂S₂ — simplify to CaS. Always reduce to lowest whole-number ratio.
WATCH OUT
Not using brackets for polyatomic ions
Ca(NO₃)₂ NOT CaNO₃₂. The bracket scopes the subscript over the whole NO₃ unit.
WATCH OUT
Using 6.022 × 10²² instead of 10²³ for Avogadro's number
Memorise: 6.022 × 10²³. Order-of-magnitude errors here cost full marks on numericals.
WATCH OUT
Confusing molar mass and atomic mass numerically
They are numerically equal! Atomic mass = 16 u means molar mass = 16 g/mol. Same number, different units.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Conservation
5.3 g of sodium carbonate reacts with 6 g of acetic acid. The products are 8.2 g of sodium acetate, 2.2 g of CO₂ and 0.9 g of water. Verify the law of conservation of mass.
Show solution
Step 1 — Sum the reactant masses. m(reactants) = 5.3 + 6 = 11.3 g. Step 2 — Sum the product masses. m(products) = 8.2 + 2.2 + 0.9 = 11.3 g. Step 3 — Compare. Mass of reactants = mass of products = 11.3 g. ✓ ✦ Answer: The law of conservation of mass is verified — total mass is unchanged at 11.3 g.
Q2EASY· Atomic mass
Compute the molecular mass of CO₂.
Show solution
Step 1 — List atomic masses. C = 12 u, O = 16 u. Step 2 — Add per atom count. M(CO₂) = 1 × 12 + 2 × 16 = 12 + 32 = 44 u. ✦ Answer: 44 u.
Q3EASY· Formula
Write the chemical formula of calcium chloride.
Show solution
Step 1 — Identify valencies. Ca: +2. Cl: −1. Step 2 — Cross-multiply. Ca₁Cl₂. Simplify (already minimal) → CaCl₂. ✦ Answer: CaCl₂.
Q4EASY· Atomicity
Give the atomicity of: (a) O₃, (b) P₄, (c) H₂SO₄, (d) Argon.
Show solution
Step 1 — Atomicity = total atoms in one molecule. Step 2 — Count for each. (a) O₃ = 3 atoms → triatomic. (b) P₄ = 4 → tetra-atomic. (c) H₂SO₄ = 2 + 1 + 4 = 7 atoms → 7-atomic. (d) Ar = 1 atom → monatomic (noble gas). ✦ Answer: (a) 3, (b) 4, (c) 7, (d) 1.
Q5EASY· Mole
How many moles are in 36 g of water? (M(H₂O) = 18 g/mol)
Show solution
Step 1 — Use n = m/M. Step 2 — Substitute. n = 36 / 18 = 2 mol. ✦ Answer: 2 moles.
Q6MEDIUM· Mole
How many oxygen atoms are present in 9 g of water? (M(H₂O) = 18 g/mol, Nₐ = 6.022 × 10²³)
Show solution
Step 1 — Find moles of water. n = m/M = 9/18 = 0.5 mol. Step 2 — Each H₂O molecule has 1 oxygen atom. Moles of O atoms = 0.5 mol. Step 3 — Number of O atoms. N = 0.5 × 6.022 × 10²³ = 3.011 × 10²³. ✦ Answer: 3.011 × 10²³ oxygen atoms. Watch out: hydrogen atoms would be 6.022 × 10²³ (double the O atoms, since each H₂O has 2 H atoms).
Q7MEDIUM· Mass
Calculate the mass of 0.2 mole of glucose (C₆H₁₂O₆). (Atomic masses: C = 12, H = 1, O = 16)
Show solution
Step 1 — Compute molar mass. M = 6 × 12 + 12 × 1 + 6 × 16 = 72 + 12 + 96 = 180 g/mol. Step 2 — Use m = n × M. m = 0.2 × 180 = 36 g. ✦ Answer: 36 g.
Q8MEDIUM· Composition
Calculate the percentage by mass of nitrogen in urea, CO(NH₂)₂. (Atomic masses: C = 12, O = 16, N = 14, H = 1)
Show solution
Step 1 — Compute molar mass of urea. CO(NH₂)₂: 1 C, 1 O, 2 N, 4 H. M = 12 + 16 + 2 × 14 + 4 × 1 = 12 + 16 + 28 + 4 = 60 g/mol. Step 2 — Mass of nitrogen in 1 mole. Mass(N) = 2 × 14 = 28 g. Step 3 — Percentage. %N = (28 / 60) × 100 = 46.67 %. ✦ Answer: ≈ 46.67 %. Fertiliser context: urea's high N% (~47%) makes it the most popular nitrogenous fertiliser in Indian agriculture.
Q9MEDIUM· Formula writing
Write the chemical formulas for: (a) aluminium nitrate, (b) ammonium sulphate, (c) magnesium hydroxide.
Show solution
Step 1 — Identify ions and valencies. (a) Al³⁺ + NO₃⁻ → cross-multiply: Al(NO₃)₃ (brackets needed; subscript 3 outside scopes the NO₃ group). (b) NH₄⁺ + SO₄²⁻ → cross-multiply: (NH₄)₂SO₄ (brackets around NH₄ because subscript 2 affects all 4 H and the N). (c) Mg²⁺ + OH⁻ → cross-multiply: Mg(OH)₂. ✦ Answer: (a) Al(NO₃)₃, (b) (NH₄)₂SO₄, (c) Mg(OH)₂. Rule: any time a polyatomic ion is taken more than once, bracket it.
Q10MEDIUM· Verification
In an experiment, 1.375 g of cupric oxide on reduction yields 1.097 g of copper. In another, 1.179 g of copper, on dissolving in nitric acid and then heating, gave 1.476 g of cupric oxide. Show that the law of constant proportions is obeyed.
Show solution
Step 1 — In experiment 1, find Cu : O ratio. Mass of O = 1.375 − 1.097 = 0.278 g. Ratio Cu : O = 1.097 : 0.278 = 1.097 / 0.278 ≈ 3.946 : 1. Simplify: ~ 4 : 1 (round). Step 2 — In experiment 2, find Cu : O ratio. Mass of O in 1.476 g of CuO = 1.476 − 1.179 = 0.297 g. Ratio Cu : O = 1.179 : 0.297 ≈ 3.97 : 1. Simplify: ~ 4 : 1. Step 3 — Compare. Both experiments give Cu : O ≈ 4 : 1 (matching the known atomic masses 63.5 : 16 ≈ 3.97 : 1). ✦ Answer: Both experiments give the same Cu : O mass ratio (~ 4 : 1) — verifying the law of constant proportions. CuO has a fixed composition regardless of how it's made.
Q11HARD· Mole concept
12 g of magnesium reacts with oxygen to form magnesium oxide. (a) How many moles of Mg is this? (b) How many moles of MgO are formed? (c) What is the mass of MgO produced? (M(Mg) = 24, M(O) = 16)
Show solution
Step 1 — Write balanced equation. 2 Mg + O₂ → 2 MgO. Step 2 (a) — Moles of Mg. n = 12 / 24 = 0.5 mol. Step 3 (b) — Moles of MgO. Stoichiometry: 2 mol Mg → 2 mol MgO (1 : 1 ratio). So 0.5 mol Mg → 0.5 mol MgO. Step 4 (c) — Mass of MgO. M(MgO) = 24 + 16 = 40 g/mol. Mass = n × M = 0.5 × 40 = 20 g. ✦ Answer: (a) 0.5 mol Mg, (b) 0.5 mol MgO, (c) 20 g MgO. Sanity check via conservation of mass: O used = 20 − 12 = 8 g. Moles of O = 8/16 = 0.5 mol O atoms = 0.25 mol O₂. From 2 Mg + O₂, ratio Mg:O₂ = 2:1 = 0.5:0.25 ✓.
Q12HARD· Mole concept
Find the number of (a) molecules and (b) atoms in 1.7 g of NH₃. (M(NH₃) = 17 g/mol, Nₐ = 6.022 × 10²³)
Show solution
Step 1 — Moles of NH₃. n = m/M = 1.7 / 17 = 0.1 mol. Step 2 (a) — Molecules. N = n × Nₐ = 0.1 × 6.022 × 10²³ = 6.022 × 10²². Step 3 (b) — Total atoms. Each NH₃ has 4 atoms (1 N + 3 H). Atoms = 4 × 6.022 × 10²² = 2.4088 × 10²³. ✦ Answer: (a) 6.022 × 10²² molecules; (b) 2.41 × 10²³ atoms. Lesson: 'atoms' is always 'molecules × atomicity'. Read the question carefully — 'how many atoms' vs 'how many molecules' is a common trap.
Q13HARD· Formula by composition
An oxide of nitrogen contains 30.4 % nitrogen and 69.6 % oxygen by mass. Find its empirical formula. (N = 14, O = 16)
Show solution
Step 1 — Take 100 g of the compound for convenience. Mass of N = 30.4 g. Mass of O = 69.6 g. Step 2 — Convert to moles. n(N) = 30.4 / 14 = 2.17 mol. n(O) = 69.6 / 16 = 4.35 mol. Step 3 — Find simplest ratio. Divide both by the smaller (2.17): N : O = 2.17/2.17 : 4.35/2.17 = 1 : 2.005 ≈ 1 : 2. Step 4 — Write empirical formula. N₁O₂ = NO₂. ✦ Answer: Empirical formula = NO₂ (nitrogen dioxide). Class 11 note: empirical formula = simplest whole-number ratio. The molecular formula could be a multiple (e.g., N₂O₄ has the same empirical formula NO₂).
Q14HARD· Real-world
(a) Calculate the molar mass of ZnO. (b) A mineral sample of zinc oxide has mass 100 g. How many zinc atoms does it contain? (c) Comment on whether this is a small or large number. (Zn = 65, O = 16)
Show solution
Step 1 (a) — Molar mass. M(ZnO) = 65 + 16 = 81 g/mol. Step 2 (b) — Moles in 100 g. n = 100 / 81 ≈ 1.235 mol. Step 3 — Each ZnO formula unit has 1 Zn atom. Moles of Zn atoms = 1.235. Number of Zn atoms = 1.235 × 6.022 × 10²³ ≈ 7.44 × 10²³. Step 4 (c) — Comment. Comparison: world population ~ 8 × 10⁹. The number of Zn atoms in 100 g of ZnO is roughly 10¹⁴ times the world population — utterly enormous. ✦ Answer: (a) 81 g/mol, (b) 7.44 × 10²³ Zn atoms, (c) Enormous — it would take 100,000 years to count them at 1 per second × 100 billion humans counting in parallel.
Q15HARD· Multi-step
Convert 23 g of sodium into (a) moles of Na, (b) moles of Na⁺ ions, (c) number of Na⁺ ions, (d) total charge in coulombs (charge per ion = 1.6 × 10⁻¹⁹ C). (M(Na) = 23 g/mol, Nₐ = 6.022 × 10²³)
Show solution
Step 1 (a) — Moles of Na atoms. n = 23 / 23 = 1 mol. Step 2 (b) — Each Na atom becomes 1 Na⁺ ion on oxidation. So moles of Na⁺ ions = 1 mol. Step 3 (c) — Number of Na⁺ ions. N = 1 × Nₐ = 6.022 × 10²³. Step 4 (d) — Total charge. Q = N × charge per ion = 6.022 × 10²³ × 1.6 × 10⁻¹⁹ = 9.6352 × 10⁴ C ≈ 96,500 C. ✦ Answer: (a) 1 mol, (b) 1 mol, (c) 6.022 × 10²³ ions, (d) ≈ 96,500 C. This '96,500 C' has a name: it's the Faraday constant (charge of 1 mole of electrons). You'll meet it again in Class 12 electrochemistry.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • Dalton: matter = indivisible atoms. Three postulates have since been refined (divisible atoms, isotopes, nuclear reactions).
  • Conservation of mass: m(reactants) = m(products) in any chemical reaction.
  • Constant proportions: a pure compound has fixed % composition regardless of source.
  • 1 u = (1/12) mass of C-12 atom = 1.66 × 10⁻²⁷ kg.
  • Molecular mass = sum of atomic masses × subscripts. Same formula for ionic 'formula unit mass'.
  • Valency cross-multiplication: write formula, simplify to lowest whole numbers, bracket polyatomic ions taken > 1 time.
  • Mole concept: 1 mole = Nₐ = 6.022 × 10²³ particles. Mass = atomic/molecular mass in grams.
  • Master formula: n = m/M = N/Nₐ. Use it for every mole conversion.

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Natural chlorine is a mixture of two isotopes — Cl-35 (75 %) and Cl-37 (25 %). The 'atomic mass' on the periodic table is the weighted average of these isotopes: 0.75 × 35 + 0.25 × 37 ≈ 35.5 u.

H was easy to start with but caused inconsistencies. O was more precise but had 3 natural isotopes complicating things. C-12 is monoisotopic when chosen as a specific isotope, gives a clean integer atomic mass, and is universally available — making it the most stable reference.

Numerically identical. Molecular mass has units of atomic mass units (u) and refers to one molecule. Molar mass has units of g/mol and refers to one mole (Nₐ molecules). Different units, same number.

Only partially. CH₃OH (methanol) and H₂CO (formaldehyde) both have the same atoms but different arrangements (different structural formulas). Class 9 deals with empirical/molecular formulas; structure comes later (Class 11–12).

Because the subscript applies to whatever is immediately to its left. Ca(NO₃)₂ means 1 Ca and 2 NO₃ units (total 2 N and 6 O). Without brackets — CaNO₃₂ — you'd read it as 1 N and 32 O, which is meaningless.

Because chemical reactions happen at the level of individual particles, not by mass. To stoichiometry-balance a reaction, what matters is COUNTING the atoms — and the mole gives us a count we can derive from a mass we can weigh.
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Last reviewed on 18 May 2026. Written and reviewed by subject-matter experts — read about our process.
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