ScienceCBSE Class 9 · Atomic Foundations of Matter

Laws of Chemical Combination and Dalton's Atomic Theory (Class 9)

Updated 17 October 2026 · Fact-checked

In a chemical reaction the total mass of the reactants equals the total mass of the products (Lavoisier's law of conservation of mass), and a pure compound always contains the same elements in the same fixed ratio by mass (Proust's law of constant proportions). Dalton's atomic theory of 1808 explains both laws: atoms are only rearranged, never created or destroyed, and every atom of an element has the same mass, so a compound has a fixed ratio by mass.

The samples have different masses, but hydrogen to oxygen is always 1 : 8 by mass, which is Proust's law of constant proportions.

Illustration drawn for QuestionBuddy.

Understand Laws of Chemical Combination and Dalton's Theory

Law of conservation of mass. When vinegar meets baking soda in an open flask, the balance reading falls, because carbon dioxide escapes. Tie a balloon over the mouth so that the gas stays inside, and the reading does not change. Lavoisier's statement is therefore about the whole system: the total mass of everything you start with equals the total mass of everything you end up with, provided nothing is allowed to leave or enter. A precipitation reaction such as sodium sulphate with barium chloride gives a solid but no gas, so even an open flask shows no change.

Law of constant proportions. Proust found that a given compound is always made of the same elements joined in the same ratio by mass, whatever its source. Pure water from a river or from the sea has hydrogen : oxygen = 1 : 8 by mass, so 9 g of water holds 1 g of hydrogen and 8 g of oxygen. Carbon dioxide has C : O = 12 : 32 = 3 : 8, magnesium oxide has Mg : O = 24 : 16 = 3 : 2, and sodium chloride has Na : Cl = 23 : 35.5. Two samples can have different total masses and still obey the law, as long as the ratio reduces to the same numbers (8 : 2 is the same ratio as 4 : 1). The law belongs to compounds. A mixture has no fixed ratio, since you can mix its parts in any amounts.

Using a ratio. If the ratio of A to B is a : b, then for a known mass of A, the mass of B is (mass of A) × b / a, and the mass of the compound is the sum. Always check by adding the parts back and comparing with the total.

Dalton's atomic theory (1808). The six postulates: (1) matter is made of tiny atoms that take part in chemical reactions; (2) atoms cannot be divided, created or destroyed in a chemical reaction; (3) atoms of one element are identical in mass and chemical properties; (4) atoms of different elements differ in mass and properties; (5) atoms combine in ratios of small whole numbers to form compounds; (6) the relative numbers and kinds of atoms in a compound are constant. Postulate 2 explains conservation of mass: atoms are only rearranged, so the total number of atoms, and with it the total mass, stays the same. Postulates 3 and 6 explain constant proportions: if every carbon atom has the same mass, every oxygen atom has the same mass, and CO2 always has one C for two O, the mass ratio cannot change. Check with numbers: (1 × 12) : (2 × 16) = 12 : 32 = 3 : 8.

What we know now. Atoms are divisible into electrons, protons and neutrons, and atoms of one element can have slightly different masses (isotopes). So postulates 2 and 3 need modification, but the idea that atoms are rearranged in a reaction, and not destroyed, is still the foundation of the subject.

Laws of Chemical Combination and Dalton's Theory: formulas and key facts

Conservation of mass: $T o t a l ma sso f r e a c t an t s = t o t a l ma sso f p r o d u c t s (ina c l ose d sy s t e m)$
Constant proportions: some ratios by mass: $W a t er H : O = 1 : 8∣ C O 2 C : O = 3 : 8∣ M g O M g : O = 3 : 2∣ N a C l N a : C l = 23 : 35.5$
Mass of B for a given mass of A: $M a sso f B = ma sso f A \times (r a t i o p a r t o f B \div r a t i o p a r t o f A)$
Mass ratio from atoms: $M a ssr a t i o = (n u mb er o f a t o m s \times a t o mi c ma sso f X) : (n u mb er o f a t o m s \times a t o mi c ma sso f Y)$
Limiting reactant check: Compare available masses with the ratio; the reactant that runs out first decides how much compound forms

How to solve Laws of Chemical Combination and Dalton's Theory questions

Use this method for constant-proportion and conservation-of-mass numericals.

1Write the fixed ratio by mass for the compound, reduced to simplest whole numbers (for example H : O = 1 : 8).
2Write down what is given and what is asked, with units (grams).
3Find the value of one ratio part: (given mass) ÷ (its ratio number).
4Multiply that value by the other ratio number to get the second mass.
5If a reactant is in excess, work out which one runs out first and use only that one to decide how much compound forms.
6Check: the two parts must add up to the mass of compound, and in a reaction the total mass before must equal the total mass after.

Laws of Chemical Combination and Dalton's Theory in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Conservation of mass in an open pan

Suppose you place a small steel bowl with 40.0 g of vinegar and a small paper packet holding 6.0 g of baking soda on a kitchen scale, and the scale shows 120.0 g in total with the bowl. You tip the baking soda in and watch it fizz. After the fizzing stops, the scale shows 118.5 g.

(a) How much mass has the scale 'lost'? (b) Where has it gone? (c) What would the scale show if the same bowl were sealed inside a bag so that no gas could leave?

Try it first, then show the solutionSolution

(a) Loss = 120.0 g − 118.5 g = 1.5 g.
(b) The reaction produced carbon dioxide gas, which left the open bowl and spread into the air. The 1.5 g is the mass of that gas.
(c) In a sealed bag the gas stays inside, so the reading stays at 120.0 g (the bag would puff up). Conservation of mass holds for the whole closed system.

Answer: (a) 1.5 g; (b) it left as carbon dioxide gas; (c) 120.0 g, unchanged.

Memory hook: Mass is never lost, it just walks out of an open door.

Fixed ratio in a compound

Suppose a school lab makes magnesium oxide by burning magnesium ribbon. One group burns 3.0 g of ribbon, and another group burns 9.0 g. Both groups collect all of the white powder.

Using Mg : O = 3 : 2 by mass, find the mass of white powder each group should collect and show that both groups get the same ratio of magnesium to oxygen.

Try it first, then show the solutionSolution

Group 1: Mg = 3.0 g, so one ratio part is 1.0 g and O = 2 × 1.0 = 2.0 g. Powder = 3.0 + 2.0 = 5.0 g.
Group 2: Mg = 9.0 g, so one ratio part is 3.0 g and O = 2 × 3.0 = 6.0 g. Powder = 9.0 + 6.0 = 15.0 g.
Ratios: 3.0 : 2.0 = 3 : 2 and 9.0 : 6.0 = 3 : 2. Both are the same, as the law of constant proportions says.

Answer: 5.0 g and 15.0 g; both have Mg : O = 3 : 2.

Memory hook: Bigger heap, same recipe.

Tricks to remember Laws of Chemical Combination and Dalton's Theory

Short memory aids. Say each one aloud once, then use it on a question the same day.

Closed box, same weight
Imagine the reaction happening inside a sealed lunch box on a scale. Nothing can leave, so the reading cannot change. If the reading does change, something escaped or came in.
Use it when: Explaining why a mass appears to increase or decrease in open-air reactions.
1 : 8 and 3 : 8
Water is 'one hydrogen to eight oxygen'. Carbon dioxide is 'three carbon to eight oxygen'. Both use 8 for oxygen, so remember the two together.
Use it when: Quick recall of the two ratios the exam uses most.
Compare mixture with compound by asking 'can I choose the amounts?'
For a mixture you may choose any amounts. For a compound, nature chooses the amounts for you.
Use it when: Questions that ask whether the law of constant proportions applies.
Postulate 2 = mass, postulates 3 and 6 = ratio
Atoms are never destroyed (2) so mass is conserved. Atoms are identical (3) and fixed in number (6) so the ratio is fixed.
Use it when: Questions that ask how Dalton's theory explains each law.

Laws of Chemical Combination and Dalton's Theory: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Fixed ratio in two samples of water · 3 marks · Short answerBoard level

Pure water contains hydrogen and oxygen in the ratio 1 : 8 by mass.
(a) A sample of 18 g of water is split into its elements. Find the masses of hydrogen and oxygen. (1)
(b) A second sample of water has 6.4 g of oxygen. Find the mass of hydrogen and the mass of this sample. (1)
(c) A student finds 3 g of hydrogen combined with 30 g of oxygen in a sample of liquid. Can it be pure water? Give a reason. (1)

Try it first, then show the solutionSolution

(a) Total parts = 1 + 8 = 9. One part = 18 g ÷ 9 = 2 g. Hydrogen = 2 g, oxygen = 8 × 2 = 16 g. Check: 2 + 16 = 18 g. (1)
(b) One part = 6.4 g ÷ 8 = 0.8 g, so hydrogen = 0.8 g. Sample = 0.8 + 6.4 = 7.2 g. (1)
(c) The ratio is 3 : 30 = 1 : 10, which is not 1 : 8. Pure water always has H : O = 1 : 8, so this liquid is not pure water (it may be a different compound, a mixture, or impure). (1)

Answer: (a) H = 2 g, O = 16 g; (b) H = 0.8 g, sample = 7.2 g; (c) no, the ratio is 1 : 10, not 1 : 8.

Example 2: Conservation of mass in a precipitation reaction · 3 marks · Short answerBoard level

20.8 g of barium chloride solution reacts completely with 14.2 g of sodium sulphate in water (both amounts are of the dissolved solids). The reaction gives 23.3 g of a white precipitate of barium sulphate and a solution of sodium chloride.
(a) Find the mass of sodium chloride formed. (1)
(b) Explain why this reaction can be done in an open flask. (1)
(c) Name the law and state it. (1)

Try it first, then show the solutionSolution

(a) Mass of reactants = 20.8 + 14.2 = 35.0 g. Mass of products = 23.3 g + mass of NaCl. So NaCl = 35.0 − 23.3 = 11.7 g. (1)
(b) No gas is produced, so nothing escapes from the flask and the total mass cannot change. (1)
(c) Law of conservation of mass (Lavoisier): mass can neither be created nor destroyed in a chemical reaction, so the total mass of reactants equals the total mass of products. (1)

Answer: (a) 11.7 g; (b) no gas escapes; (c) law of conservation of mass.

Example 3: Magnesium burning: how much oxygen is used? · 5 marks · Long answerHarder board level

Magnesium and oxygen combine in the ratio Mg : O = 3 : 2 by mass to give magnesium oxide.
(a) How much oxygen combines with 12 g of magnesium, and what mass of oxide forms? (1)
(b) A student heats 6 g of magnesium in a closed container holding 6 g of oxygen. Which reactant is left over, and how much? (2)
(c) What mass of oxide forms in (b)? (1)
(d) How much magnesium and oxygen are present in 15 g of magnesium oxide? (1)

Try it first, then show the solutionSolution

(a) One ratio part = 12 g ÷ 3 = 4 g. Oxygen = 2 × 4 = 8 g. Oxide = 12 + 8 = 20 g. (1)
(b) 6 g of magnesium needs oxygen = 6 × 2/3 = 4 g. Only 4 g of the 6 g of oxygen is needed, so magnesium is used up completely and oxygen is in excess. (1)
Oxygen left over = 6 − 4 = 2 g. (1)
(c) Oxide = 6 + 4 = 10 g (the leftover 2 g of oxygen does not join in). Check: total mass of everything = 6 + 6 = 12 g = 10 g oxide + 2 g unused oxygen. (1)
(d) Total parts = 3 + 2 = 5. One part = 15 ÷ 5 = 3 g. Magnesium = 9 g and oxygen = 6 g. (1)

Answer: (a) O = 8 g, oxide = 20 g; (b) 2 g of oxygen is left over; (c) 10 g; (d) Mg = 9 g, O = 6 g.

Example 4: Burning carbon with limited oxygen, and the atom ratio · 5 marks · Long answerHarder board level

Carbon and oxygen combine as C : O = 3 : 8 by mass to form carbon dioxide (atomic masses: C = 12 u, O = 16 u).
(a) 6 g of carbon is burnt in 20 g of oxygen. Find the mass of carbon dioxide formed and the mass of oxygen left. (2)
(b) 6 g of carbon is burnt in only 10 g of oxygen. Find the mass of carbon dioxide formed and the mass of carbon left. (2)
(c) Use the ratio 3 : 8 and the atomic masses to show that carbon dioxide has 1 carbon atom for 2 oxygen atoms. (1)

Try it first, then show the solutionSolution

(a) Oxygen needed for 6 g carbon = 6 × 8/3 = 16 g. Oxygen available is 20 g, so oxygen is in excess. CO2 = 6 + 16 = 22 g. Oxygen left = 20 − 16 = 4 g. (2)
(b) Carbon that can burn in 10 g oxygen = 10 × 3/8 = 3.75 g. Oxygen is the limiting reactant. CO2 = 3.75 + 10 = 13.75 g. Carbon left = 6 − 3.75 = 2.25 g. Check: 13.75 + 2.25 = 16 g = 6 + 10. (2)
(c) Number of atoms ∝ mass ÷ atomic mass. Carbon: 3 ÷ 12 = 0.25. Oxygen: 8 ÷ 16 = 0.5. Ratio = 0.25 : 0.5 = 1 : 2, so one carbon atom for two oxygen atoms (CO2). (1)

Answer: (a) 22 g CO2, 4 g O left; (b) 13.75 g CO2, 2.25 g C left; (c) C : O atoms = 1 : 2.

Example 5: Do these samples obey the law? · 4 marks · Case-basedHOTS / competency

A student analyses three samples of a compound P made only of elements A and B and records the data below.

Sample 1: total 10 g, element A 6 g.
Sample 2: total 25 g, element A 15 g.
Sample 3: total 8 g, element A 5 g.

(a) Find the ratio A : B by mass in each sample. (1)
(b) Which samples are consistent with the law of constant proportions? (1)
(c) The student concludes 'the law fails because sample 3 is different'. Give two possible reasons for the data that do not require the law to be wrong. (1)
(d) Using sample 1, find the mass of B in 40 g of P. (1)

Try it first, then show the solutionSolution

(a) Sample 1: B = 10 − 6 = 4 g, A : B = 6 : 4 = 3 : 2. Sample 2: B = 25 − 15 = 10 g, A : B = 15 : 10 = 3 : 2. Sample 3: B = 8 − 5 = 3 g, A : B = 5 : 3. (1)
(b) Samples 1 and 2 (both 3 : 2). Sample 3 (5 : 3 = 1.67 : 1, against 1.5 : 1) does not match. (1)
(c) Sample 3 may not be pure P (it may be a mixture or have an impurity), or it may be a different compound of A and B, or there was an error in the measurement. (1)
(d) A is 3 parts of 5, so B is 2 parts of 5: B = 40 × 2/5 = 16 g. Check: A = 24 g, 24 + 16 = 40 g. (1)

Answer: (a) 3 : 2, 3 : 2, 5 : 3; (b) samples 1 and 2; (c) impure sample or a different compound or a measurement error; (d) 16 g.

Where students lose marks in Laws of Chemical Combination and Dalton's Theory

The slips examiners see most often, and what to write instead.

Mistake: Saying the law of conservation of mass fails when a gas escapes or enters.
Students count only the solid or liquid they can see and forget that gases are matter too.
Wrong working
Reading falls from 120.0 g to 118.5 g, so 1.5 g of mass was destroyed.
Right working
1.5 g of carbon dioxide gas escaped; counting the gas, total mass is unchanged.
Do this: Always ask whether the system is open or closed before judging the law.
Usually costs: 1 to 2 marks in a reasoning question.
Mistake: Using the ratio 1 : 8 as if it were the ratio of atoms.
The ratio 1 : 8 is by mass, but H2O has 2 hydrogen atoms for 1 oxygen atom.
Wrong working
Water has 1 atom of H for 8 atoms of O.
Right working
By mass H : O = 1 : 8; by atoms H : O = 2 : 1 (since 1 ÷ 1 : 8 ÷ 16 = 1 : 0.5).
Do this: Write 'by mass' next to every ratio, and divide by atomic masses to get a ratio of atoms.
Usually costs: 1 to 2 marks.
Mistake: Forgetting to add the two parts back to get the total mass of compound.
The question asks for the mass of compound formed but the student stops at the mass of the second element.
Wrong working
12 g Mg needs 8 g O, so oxide = 8 g.
Right working
Oxide = 12 + 8 = 20 g.
Do this: Finish with a check line: parts add up to the total.
Usually costs: 1 mark.
Mistake: Using all the oxygen supplied when one reactant is in excess.
The student multiplies the whole given mass of both reactants instead of finding which runs out first.
Wrong working
6 g Mg + 6 g O gives 12 g oxide.
Right working
Only 4 g O is needed for 6 g Mg, so oxide = 10 g and 2 g O is left over.
Do this: Work out the amount of the second reactant needed first, and compare.
Usually costs: 2 marks.
Mistake: Stating the law of constant proportions for mixtures.
The word 'substance' is used loosely.
Wrong working
Any two substances always combine in a fixed ratio by mass.
Right working
A pure compound contains its elements in a fixed ratio by mass; a mixture can have any proportions.
Do this: Say 'a chemical compound' every time.
Usually costs: ½ to 1 mark in a definition.

Exam tips

State each law with the key phrase: 'in a chemical reaction' (conservation of mass) and 'a chemical compound' with 'fixed ratio by mass' (constant proportions).
In a numerical, write the ratio, find one ratio part first, then multiply; finish with an addition check.
If two reactants are given, test which one is limiting before you compute the product.
When asked how Dalton's theory explains a law, name the postulate numbers and say in one line what each does.
Mention 'closed system' or 'open flask' whenever an experiment with a gas is described.

Practice questions on Laws of Chemical Combination and Dalton's Theory

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
In an open flask, 4.2 g of baking soda is added to excess vinegar, and the balance reading falls by 2.2 g after the fizzing stops. (a) Explain the fall in reading. (b) What would the balance show if the same reaction were done with the baking soda held in a balloon tied over the flask's mouth, and why?
Hint
Ask where the gas goes in each set-up.
Answer
(a) Carbon dioxide gas formed escapes into the air; the 2.2 g is the mass of gas that left (1 mark). (b) The reading would not change, because the gas is trapped in the balloon and the total mass of the closed system stays the same (1 mark).
Q2 · 3 marks · Short answerHarder board level
In sodium chloride, sodium and chlorine are present in the ratio 23 : 35.5 by mass. (a) Find the masses of sodium and chlorine in 11.7 g of sodium chloride. (b) How much chlorine combines with 9.2 g of sodium?
Hint
Total parts = 23 + 35.5 = 58.5.
Answer
(a) One part of the total is 11.7 ÷ 58.5 = 0.2 g. Sodium = 23 × 0.2 = 4.6 g and chlorine = 35.5 × 0.2 = 7.1 g; check 4.6 + 7.1 = 11.7 g (2 marks). (b) 9.2 ÷ 23 = 0.4, so chlorine = 35.5 × 0.4 = 14.2 g (1 mark).
Q3 · 2 marks · Assertion-reasonHarder board level
Assertion (A): When a magnesium ribbon burns in open air, the white ash weighs more than the ribbon, so the law of conservation of mass is violated. Reason (R): The ash contains oxygen taken from the air. Choose: (a) A and R both true, R explains A; (b) A and R both true, R does not explain A; (c) A false, R true; (d) A true, R false.
Hint
Count all the mass at the start, including the air that reacts.
Answer
(c). R is true: the ash is magnesium oxide, which includes oxygen from the air. A is false: the ribbon plus the oxygen it used has the same mass as the ash, so the law is not violated; the apparent gain comes from leaving the oxygen out of the starting mass.
Q4 · 3 marks · Short answerHarder board level
Two of Dalton's six postulates are not fully correct today. (a) Name them and give one piece of modern evidence against each. (b) Which part of Dalton's theory still explains conservation of mass, and why?
Hint
Think of what the atom is made of, and of isotopes.
Answer
(a) Postulate 2, atoms are indivisible: atoms contain electrons, protons and neutrons (1 mark). Postulate 3, all atoms of an element are identical in mass: isotopes of the same element have different masses (1 mark). (b) The idea that atoms are only rearranged in a chemical reaction and are not created or destroyed still holds, so the total number of atoms, and the total mass, stays the same (1 mark).
Q5 · 4 marks · Case-basedHOTS / competency
A student heats three weighed samples of copper with oxygen to make an oxide of copper and records: Sample X: 3.2 g Cu with 0.8 g O. Sample Y: 6.0 g Cu with 1.5 g O. Sample Z: 5.6 g Cu with 1.6 g O. (a) Which two samples agree with the law of constant proportions, and what is their Cu : O ratio? (b) Suggest a reason for Z. (c) Find the mass of oxide formed in X and in Y.
Hint
Divide copper by oxygen in each sample.
Answer
(a) X: 3.2 ÷ 0.8 = 4; Y: 6.0 ÷ 1.5 = 4; Z: 5.6 ÷ 1.6 = 3.5. X and Y agree, ratio Cu : O = 4 : 1 (2 marks). (b) Z may be impure, a mixture of two oxides or an incompletely dried or weighed sample; it cannot be called a pure sample of the same compound (1 mark). (c) X: 3.2 + 0.8 = 4.0 g; Y: 6.0 + 1.5 = 7.5 g (1 mark).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the law of conservation of mass.Show answer
Mass can neither be created nor destroyed in a chemical reaction; total mass of reactants equals total mass of products.
2.Who proposed the law of constant proportions?Show answer
Joseph Proust.
3.How much hydrogen is present in a 9 g sample of pure water?Show answer
1 g, because H : O = 1 : 8 by mass and 1 + 8 = 9.
4.What is the ratio of carbon to oxygen by mass in carbon dioxide?Show answer
3 : 8.
5.In which year did Dalton put forward his atomic theory?Show answer
1808.
6.Which Dalton postulate explains conservation of mass?Show answer
Atoms cannot be created or destroyed in a chemical reaction, only rearranged.

Laws of Chemical Combination and Dalton's Theory: frequently asked questions

Why does the mass seem to increase when magnesium burns in air?

Because the magnesium combines with oxygen from the air. The oxide weighs as much as the magnesium plus the oxygen it used, so the law of conservation of mass is not broken.

Does the law of constant proportions apply to mixtures?

No. It applies to pure compounds. A mixture such as sand and salt can have any amounts of its parts.

If two samples have different total masses, can they still obey the law?

Yes, if the ratio of the elements reduces to the same numbers, such as 4 : 1 and 8 : 2.

Why do we need a closed system for the law of conservation of mass?

Otherwise gases can escape or enter and change the reading, which makes the comparison unfair.

Is Dalton's atomic theory wrong because atoms can be divided?

Not wrong, but incomplete. Atoms are made of smaller particles and isotopes exist, yet atoms are still rearranged in reactions, which is what explains the laws.