Class 9 Atoms and Molecules — the Laws of Chemical Combination Explained
This is the chapter where chemistry stops being descriptive and starts being arithmetic. Two experimental laws lead to Dalton's atomic theory, the theory leads to chemical formulae, formulae lead to molecular mass, and molecular mass leads to the mole. Every numerical you will meet from Class 10 to a competitive exam sits on top of these four steps. Get them right in Class 9 and the next four years become much easier.
Law 1 — the law of conservation of mass
In a chemical reaction, mass can neither be created nor destroyed. The total mass of the reactants equals the total mass of the products, provided nothing escapes.
The word provided nothing escapes matters. If a gas is produced in an open beaker it walks out of the experiment, and the reading on the balance falls. That does not break the law; it means the experiment was not sealed.
Worked example 1. In a closed flask, 5.3 g of sodium carbonate reacts with 6.0 g of ethanoic acid. The products are 8.2 g of sodium ethanoate, 2.2 g of carbon dioxide and 0.9 g of water. Show that this obeys the law of conservation of mass.
Mass of reactants = 5.3 + 6.0 = 11.3 g
Mass of products = 8.2 + 2.2 + 0.9 = 11.3 g
The two totals are equal, so mass is conserved. ✓
Law 2 — the law of constant (definite) proportions
A pure chemical compound always contains the same elements combined in the same fixed proportion by mass, no matter where the sample came from or how it was made. Water from a tap, from rain or made in a laboratory always contains hydrogen and oxygen in the mass ratio 1 : 8.
Worked example 2. In one experiment, 1.375 g of copper(II) oxide was reduced by hydrogen and gave 1.098 g of copper. In a second experiment, 1.179 g of copper(II) oxide gave 0.942 g of copper. Show that the results agree with the law of constant proportions.
Experiment 1
Mass of oxygen = 1.375 − 1.098 = 0.277 g
Cu : O = 1.098 : 0.277 = 3.96 : 1
Experiment 2
Mass of oxygen = 1.179 − 0.942 = 0.237 g
Cu : O = 0.942 : 0.237 = 3.97 : 1
The two ratios agree to two significant figures; the tiny difference is ordinary experimental error in weighing. So the composition of copper(II) oxide is fixed.
Cross-check from atomic masses: in CuO the ratio should be 63.546 : 15.999 = 3.97 : 1 — which is exactly what the experiments found.
An honest note about a third law. Some textbooks (and several ICSE and older syllabuses) also teach the law of multiple proportions — that when two elements form more than one compound, the masses of one element combining with a fixed mass of the other are in a small whole-number ratio, as in CO and CO2. The NCERT Class 9 chapter covers only the first two laws. Learn the two that are in your syllabus first; do not assume a question about CO and CO2 is testing constant proportions, because it is not.
Dalton's atomic theory — and which law each postulate explains
| Postulate | What it explains |
|---|---|
| All matter is made of very small particles called atoms | The basic model |
| Atoms are indivisible and are neither created nor destroyed in a chemical reaction | Law of conservation of mass |
| Atoms of a given element are identical in mass and chemical properties | Why a compound's composition is reproducible |
| Atoms of different elements have different masses and properties | Why elements differ |
| The relative number and kinds of atoms in a given compound are fixed | Law of constant proportions |
| Atoms combine in small whole-number ratios | Why formulae are simple |
Two postulates are now known to be wrong, and examiners do ask for this. Atoms are not indivisible — they contain electrons, protons and neutrons. And atoms of the same element are not always identical in mass, because isotopes exist. Dalton's theory is still the right model for Class 9 arithmetic; it simply is not the final word.
Atoms, molecules, ions and atomicity
- An atom is the smallest particle of an element that takes part in a chemical reaction.
- A molecule is the smallest particle of an element or compound that can exist independently.
- An ion is a charged particle: a positive cation or a negative anion.
- Atomicity is the number of atoms in one molecule.
| Atomicity | Name | Examples |
|---|---|---|
| 1 | Monoatomic | He, Ne, Ar; metals such as Na, Cu |
| 2 | Diatomic | H2, O2, N2, Cl2 |
| 3 | Triatomic | O3 (ozone) |
| 4 | Tetra-atomic | P4 |
| 8 | Polyatomic | S8 |
Writing a chemical formula — the criss-cross method
- Write the symbol of the cation first, then the anion.
- Write the valency (magnitude of the charge) above each.
- Cross the numbers over and write them as subscripts.
- Simplify the subscripts if they share a common factor.
- Put a polyatomic ion in brackets if its subscript is more than 1.
| Ions | Criss-cross gives | Final formula |
|---|---|---|
| Al3+, O2− | Al2O3 | Al2O3 |
| Mg2+, O2− | Mg2O2 | MgO (divide by 2) |
| Ca2+, PO43− | Ca3(PO4)2 | Ca3(PO4)2 |
| NH4+, SO42− | (NH4)2SO4 | (NH4)2SO4 |
Step 4 is the one students skip. Mg2O2 is marked wrong; MgO is right.
Molecular mass and formula unit mass
Molecular mass is the sum of the atomic masses of all the atoms in one molecule, expressed in unified atomic mass units (u). Formula unit mass is exactly the same sum, but used for ionic compounds, which have no separate molecules — only a repeating lattice. Sodium chloride has a formula unit mass, not a molecular mass.
Worked example 3. Calculate the molecular mass of H2SO4 and the formula unit masses of NaCl and CaCO3. (H = 1.008, O = 15.999, S = 32.06, Na = 22.990, Cl = 35.45, Ca = 40.078, C = 12.011)
H2SO4: H 2 × 1.008 = 2.016; S 1 × 32.06 = 32.06;
O 4 × 15.999 = 63.996
Total = 2.016 + 32.06 + 63.996 = 98.07 u
NaCl: 22.990 + 35.45 = 58.44 u
CaCO3: 40.078 + 12.011 + (3 × 15.999 = 47.997) = 100.09 u
The mole — counting by weighing
Atoms are far too small to count one by one, so chemists count them in packets. One mole is 6.022 × 1023 particles — the Avogadro constant. The neat part is that the mass of one mole in grams is numerically the same as the atomic or molecular mass in u.
Number of particles, N = n × 6.022 × 1023
So one mole of H2SO4 has a mass of 98.07 g, and one mole of NaCl has a mass of 58.44 g. A small honesty point: 12 g of pure carbon-12 is exactly one mole by definition, but ordinary carbon is a mixture of isotopes with an average atomic mass of 12.011, so 12 g of ordinary carbon is 12 ÷ 12.011 = 0.999 mol — near enough to 1 mol for Class 9 work, and worth knowing why the two numbers differ.
Worked example 4. How many molecules are present in 22 g of carbon dioxide? How many atoms in total? (C = 12.011, O = 15.999)
M(CO2) = 12.011 + (2 × 15.999 = 31.998) = 44.009 g/mol
n = 22 ÷ 44.009 = 0.4999 mol
Number of molecules = 0.4999 × 6.022 × 1023 = 3.01 × 1023 molecules
Each CO2 molecule has 3 atoms (1 C + 2 O), so
Number of atoms = 3 × 3.01 × 1023 =
9.03 × 1023 atoms
Sanity check: 22 g is almost exactly half of 44 g, so the answer should be about half of 6.022 × 1023 — and 3.01 × 1023 is exactly that.
Mistakes that cost marks
- Forgetting to simplify after the criss-cross. Mg2O2 must be reduced to MgO.
- Saying mass "disappeared" in an open beaker. The escaping gas took it. The law applies to a closed system.
- Calling 58.44 u the molecular mass of NaCl. Ionic compounds have a formula unit mass. It is a one-word answer worth a full mark.
- Confusing one mole of atoms with one mole of molecules. One mole of oxygen atoms is 16.00 g; one mole of oxygen molecules, O2, is 32.00 g. Read whether the question says atoms or molecules.
- Mixing up the units. Atomic and molecular masses are in u; molar masses are in g/mol. Same number, different unit, different meaning.
- Applying constant proportions across two different compounds. CO and CO2 are different compounds, so their compositions are allowed to differ.
- Treating 6.022 × 1023 as "the mole". The mole is the unit; 6.022 × 1023 is how many particles are in it.
Where this chapter is examined
| Class / exam | How it appears |
|---|---|
| Class 9 (CBSE and ICSE) | Statement of the laws, numerical verification, Dalton's postulates, formula writing, molecular/formula unit mass, simple mole calculations |
| Class 10 | Balancing equations and the mass relationships in chemical reactions rest directly on conservation of mass |
| Class 11 | The same mole concept, extended to empirical formulae, limiting reagents and solution concentration |
| Competitive papers | Mole and Avogadro-number questions of exactly the type in Worked Example 4 |
Qualitatively, this is the highest-leverage chapter in Class 9 chemistry: not because it carries the most questions, but because almost every later numerical chapter assumes you can do it quickly and without thinking.
Practise the mass-to-mole step until it is automatic. The Mass ↔ Mole calculator converts between grams, moles and number of particles for any formula, so you can set yourself twenty conversions and check every one — the exact skill Worked Example 4 uses.
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