QCC Notes
CLASS 11 · CHEMISTRY JEE MAIN × NEET हिंदी
§ 1.5NCERT Class 11 · Chemistry · Chapter 1

Laws of Chemical Combinations: All 5 Laws, Simply Explained

Before anyone had seen an atom, chemists weighed reactions and found rules — five of them. Each law is a fingerprint of atomic behaviour, and together they forced Dalton to invent the atom. Statements, examples, exam traps, and a live combination ledger.

01

What are the Laws of Chemical Combinations? — Complete Theory

Between 1789 and 1811 — long before X-ray crystals or mass spectrometers — chemists did one thing superbly: they weighed. Reactants in, products out, balances swinging. From those weights emerged five laws of such regularity that matter itself seemed to be keeping accounts. Dalton read the ledgers and concluded: matter must be atomic. Every law below is a macroscopic echo of microscopic discreteness — keep that thread as you read.

Law 1 · Lavoisier, 1789

Law of Conservation of Mass

Matter can neither be created nor destroyed in a chemical reaction — total mass of reactants equals total mass of products.

12 g of carbon burned in 32 g of oxygen yields exactly 44 g of carbon dioxide — nothing vanishes, nothing appears, atoms merely rearrange. The law belongs to closed systems: an open beaker lets escaping CO₂ carry mass away, making a reaction look “lossy” when it isn’t.

Law 2 · Proust, 1799

Law of Definite Proportions (Constant Composition)

A pure compound always contains the same elements combined in the same fixed proportion by mass — whatever its source or sample size.

River water, tap water, rain, or water synthesised in a lab from exploding gases — every sample is hydrogen : oxygen = 1 : 8 by mass, and carbon dioxide is always C : O = 3 : 8. Compounds, unlike mixtures, are recipes with no chef’s discretion. This is the law that percentage composition (§ 1.9) turns into a working tool.

Law 3 · Dalton, 1803

Law of Multiple Proportions

When two elements 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.

Carbon and oxygen make CO (1.33 g C per 1 g O) and CO₂ (2.67 g C per 1 g O) — ratio 1 : 2. Sulfur oxides: 1 : 1.5 → 2 : 3. Water vs hydrogen peroxide: 1 : 2. And the nitrogen oxides run the full 1 : 2 : 3 : 4 : 5 ladder. Why whole numbers? Because atoms combine as whole, indivisible units — this law is the atom peeking through the balance.

Law 4 · Gay-Lussac, 1808

Law of Gaseous Volumes (Gay-Lussac’s Law)

Gases react in volumes that bear a simple whole-number ratio to one another and to the volumes of gaseous products — all measured at the same temperature and pressure.

1 volume H2 + 1 volume Cl2 → 2 volumes HCl (1 : 1 : 2); 1 volume N2 + 3 volumes H2 → 2 volumes NH3 (1 : 3 : 2). Volumes behave like mole counts because, at equal T and P, equal volumes hold equal numbers of molecules — the bridge is § 1.10’s gas-volume shortcut. The catch: gases only — 100 mL of liquid water tells you nothing about 100 mL of steam’s chemistry.

Law 5 · Avogadro, 1811

Avogadro’s Law

Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.

One sentence that fixed two problems at once: it explained Gay-Lussac’s simple volume ratios, and it forced the world to distinguish atoms from molecules — hydrogen gas travels as H2, not lone H atoms. Its gift to you is the 22.4 L molar volume at STP (§ 1.8), the “÷ 22.4” door on every stoichiometry bridge.

Table 1 — The five laws at a glance (exam revision grid)
Law · YearOne-line statementSignature exampleScope caveat
Conservation · 1789Σm(reactants) = Σm(products)12 g C + 32 g O₂ → 44 g CO₂Closed systems; chemical change only
Definite proportions · 1799Fixed mass ratio per compoundH : O = 1 : 8 in all waterCompounds, not mixtures; isotopes nudge it
Multiple proportions · 1803Fixed A → m(B) in whole ratiosN-oxides: 1 : 2 : 3 : 4 : 5Needs two or more compounds
Gay-Lussac · 1808Gas volumes in whole ratios1 : 3 : 2 for N₂ + H₂ → NH₃Gases only; same T & P
Avogadro · 1811Equal V ⇔ equal molecules22.4 L per mol at STPGases only; molecules, not atoms

Read the five laws in sequence and the plot writes itself: masses conserved (something persistent is rearranging), compositions fixed (something with fixed counts is being built), multiple proportions whole-numbered (those counts are integers), gas volumes simple (the counts are visible through volume), and Avogadro hands everyone the counting unit. Five ledgers, one conclusion: matter is atomic. That is exactly the inference Dalton formalised in § 1.6.

02

Visualising the Five Laws

Two decades, five laws — the timeline shows the story arc, and the ledger below lets you verify the laws with your own arithmetic.

The five laws on one timeline — each a step toward the atomic conclusion FIG. 1 — TWO DECADES THAT DISCOVERED THE ATOM Lavoisier, 1789 — conservation of mass 1789 LAVOISIER mass conserved Dalton, 1803 — multiple proportions 1803 DALTON whole-number ratios Avogadro, 1811 — molecules 1811 AVOGADRO equal V = equal molecules Proust, 1799 — definite proportions 1799 PROUST fixed composition Gay-Lussac, 1808 — gaseous volumes 1808 GAY-LUSSAC simple volume ratios Each law tightened the noose: if compositions are fixed and ratios whole-numbered, matter must be built from discrete, indivisible units. five ledgers → one conclusion
FIG. 1 — 1789 to 1811: conservation fixed the bookkeeping, definite proportions fixed the recipe, multiple proportions exposed whole numbers, Gay-Lussac found the counts in gas volumes, and Avogadro explained them with molecules. Dalton’s 1803 theory (red dots) sits at the pivot — built from the laws, explaining them back.
Try it live

Combination Ledger

Compoundg of O per 1 g NRatio (normalised)

Fill any three masses — the fourth solves itself, because Σ reactants = Σ products. Leave exactly one blank.

Reactant AReactant BProduct CProduct D
g
g
g
g

Ledger masses follow exam convention (N = 14, O = 16, C = 12, S = 32, H = 1). Balance pane defaults to the classic: 10 g CaCO₃ → 5.6 g CaO + 4.4 g CO₂.

03

Solved Examples (Step-by-Step)

Statement → data → arithmetic → verdict. Write these same steps on your rough sheet — that is what earns full method marks.

EXAMPLE 01Foundation · Conservation

Missing mass, found by bookkeeping

24.5 g of KClO3 is heated until decomposition is complete (2KClO3 → 2KCl + 3O2). The residue of KCl weighs 14.9 g. What mass of oxygen escaped, and does the experiment obey conservation of mass?

  1. LedgerReactants = 24.5 g (closed balance). Products = KCl + O2.
  2. Subtractm(O2) = 24.5 − 14.9 = 9.6 g
  3. Mole checkn(KClO3) = 24.5 ÷ 122.5 = 0.2 mol → O2 = 0.3 mol × 32 = 9.6 g ✓

9.6 g O₂ — 14.9 + 9.6 = 24.5 g, conserved ✓

EXAMPLE 02JEE Main · Definite proportions

The fixed 1 : 8 recipe meets a shortage

2.0 g of hydrogen is sparked with 16.3 g of oxygen. Water always contains H : O = 1 : 8 by mass. Find the mass of water formed and the mass of any leftover gas.

  1. Recipe1 g H needs 8 g O → 2 g H needs 2 × 8 = 16 g O.
  2. CompareAvailable O = 16.3 g > 16 g needed → H2 fully consumed, O2 excess.
  3. Formm(H2O) = 2 + 16 = 18 g
  4. Leftoverm(O2) left = 16.3 − 16 = 0.3 g — the LR logic of § 1.10, wearing this law’s clothes.

18 g H₂O · 0.3 g O₂ left

EXAMPLE 03JEE Main · Multiple proportions

Two oxides of iron, one verdict

Iron forms FeO (77.7% Fe by mass) and Fe2O3 (70.0% Fe). Show that these data illustrate the law of multiple proportions.

  1. Fix AFix iron at 1 g in each oxide.
  2. LedgerFeO: O per 1 g Fe = 22.3/77.7 ≈ 0.287 g · Fe2O3: 30/70 = 0.4286 g
  3. Ratio0.4286 ÷ 0.287 ≈ 1.5 → × 2 → 3 : 2 — simple whole numbers ✓
  4. VerifyFormula check: O per Fe is 1 in FeO and 1.5 in Fe2O3 → Fe2O3 : FeO = 1.5 : 1 = 3 : 2, exactly as computed ✓

O-per-1 g-Fe = 3 : 2 — multiple proportions demonstrated

05

Key Statements & Takeaways

Law card

Eight lines that solve this topic

Σm(reactants) = Σm(products)Conservation — closed systems, chemical change only.
Compound ⇒ fixed % by massDefinite proportions — water H : O = 1 : 8, CO₂ C : O = 3 : 8, always.
fixed A → m(B) in whole ratioMultiple proportions — compute per fixed 1 g, then normalise.
V-ratio = coefficient ratio (gases)Gay-Lussac — same T & P, gases only.
equal V ⇔ equal moleculesAvogadro — molecules, never atoms.
n = V / 22.4 (STP, 1 atm)Avogadro’s gift — the § 1.8 volume door.
N-oxides O-ledger: 1:2:3:4:5Per 1 g N in N₂O, NO, N₂O₃, NO₂, N₂O₅ — the go-to example.
m(missing) = Σ(one side) − Σ(other)Conservation as a solver — find the escaped/collected mass.

Fixed recipes: H : O (water) = 1 : 8  ·  C : O (CO₂) = 3 : 8  ·  CO/CO₂ O-ledger per 1 g C = 1.33 : 2.67 = 1 : 2  ·  STP: 273.15 K, 1 atm → 22.4 L/mol

  1. Five laws, one story — each is a macroscopic rule that only discrete atoms can explain; Dalton read them exactly that way.
  2. Know each law’s boundary — closed systems (conservation), compounds only (definite), gases at same T,P (Gay-Lussac, Avogadro).
  3. The ledger method solves all of them — fix one quantity, tabulate the other, normalise, and check for whole numbers.
  4. Molecules ≠ atoms — the single most-tested word swap in this section, in Avogadro’s law and everywhere it echoes.
06

FAQs

What is the law of conservation of mass?

In every chemical reaction, matter is neither created nor destroyed — the total mass of reactants equals the total mass of products. Burning 12 g of carbon in 32 g of oxygen gives exactly 44 g of carbon dioxide. The law holds for closed systems and chemical changes; nuclear reactions follow the deeper conservation of mass–energy.

What is the law of definite proportions?

A pure compound always contains the same elements combined in the same fixed proportion by mass, whatever its source or sample size. Every sample of water is hydrogen to oxygen in the 1 : 8 mass ratio — drawn from a river, a tap or made in a laboratory.

What is the law of multiple proportions? Give an example.

When two elements form more than one compound, the masses of one element that combine with a fixed mass of the other bear a simple whole-number ratio. Carbon with 1 g of oxygen forms CO (1.33 g C per g O) and CO₂ (2.67 g C per g O) — a 1 : 2 ratio. The nitrogen oxides give the classic 1 : 2 : 3 : 4 : 5 series.

What is Avogadro’s law and why is it important?

Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. It explains why 1 volume of hydrogen combines with 1 volume of chlorine to give 2 volumes of hydrogen chloride, resolves Dalton’s conflict with Gay-Lussac, and leads directly to the 22.4 L molar volume at STP.

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QCC Notes — Class 11 Chemistry

Strictly NCERT-aligned notes for JEE Main & NEET, prepared by QCC Notes (Padho Likho JEE). Content follows the latest NCERT edition and current NTA exam pattern.

Last updated
30 Aug 2026