ScienceCBSE Class 9 · Chapter 9

Atomic Foundations of Matter Class 9 Notes, Formulae and Numericals

Updated 15 October 2026 · Fact-checked

Atomic Foundations of Matter in Class 9 covers the two laws of chemical combination (conservation of mass and constant proportions), Dalton's atomic theory, how atoms bond by sharing electrons (covalent) or transferring them (ionic), naming compounds, writing formulae by criss-crossing valencies, properties of ionic and covalent compounds, and calculating molecular and formula unit masses.

What Atomic Foundations of Matter is about

This is Chapter 9 of the NCERT book *Exploration* and part of CBSE's Unit II, Matter: Its Nature and Behaviour. It follows straight on from Chapter 8: now that you know how electrons are arranged, you can see why atoms join and in what ratios.

What the chapter covers:

Law of Conservation of Mass (Lavoisier, 1789): mass is neither created nor destroyed in a chemical reaction (Activities 9.1 to 9.3, including the vinegar-baking soda balloon test and sodium sulphate with barium chloride).
Law of Constant (Definite) Proportions (Proust): a compound always has its elements in a fixed ratio by mass (water H : O = 1 : 8).
Dalton's atomic theory: postulates that explain both laws.
Molecules and bonding: covalent bonds by sharing (H $_{2}$ , Cl $_{2}$ , O $_{2}$ , HCl, H $_{2}$ O) and ionic bonds by transfer (Na $^{+}$ and Cl $^{-}$ in NaCl); cations, anions and polyatomic ions.
Naming and formulae: prefixes for covalent compounds; criss-cross method for both types.
Properties: ionic compounds dissolve in water, conduct when dissolved or molten, and have high melting points; covalent compounds generally do the opposite.
Molecular mass and formula unit mass.

Expect law-verification numericals, electron-dot diagrams, writing formulae and names, comparing ionic and covalent compounds, and mass calculations.

Weightage: This chapter is part of Unit II, Matter - Its Nature and Behaviour, which carries 25 of the 80 theory marks in the school annual exam, shared with Exploring Mixtures and their Separation and Journey Inside the Atom.Source: CBSE Science (086) syllabus, Class IX 2026-27, course structure, Unit II Matter - Its Nature and Behaviour

Key concepts

Law of Conservation of Mass

NCERT section 9.1 Law of Conservation of Mass

Physical change (Activity 9.1): dissolving salt in water on a digital balance: the mass of the solution equals water + salt. Tearing paper does not change its mass either.

Chemical change (Activity 9.2): vinegar and baking soda react and give carbon dioxide gas.

$NaHCO_{3} + CH_{3} COOH \to CH_{3} COONa + H_{2} O + CO_{2}$

Set-up 1 (open): baking soda is tipped into the flask; the gas escapes and the reading falls.
Set-up 2 (closed): the baking soda is held in a balloon tied to the flask's mouth; when mixed, the balloon inflates and the reading stays the same.

Activity 9.3: sodium sulphate and barium chloride solutions give a white precipitate of barium sulphate; no gas forms, so it can be done in open flasks, and the total mass is unchanged.

$Na_{2} SO_{4} + BaCl_{2} \to BaSO_{4} + 2 NaCl$

Law of Conservation of Mass (Antoine Lavoisier, 1789, the 'Father of Modern Chemistry'): mass can neither be created nor destroyed in a chemical reaction; total mass of reactants = total mass of products.

A small difference of ±1 in the last digit of a digital balance is within experimental error.

Worked example

In a flask closed with a balloon, 8.4 g of baking soda reacts with 6.0 g of acetic acid. The products are sodium acetate 8.2 g, water 1.8 g and carbon dioxide 4.4 g. Show that these data obey the law of conservation of mass. What would the balance show if the flask were open?

Try it first, then show the solutionSolution

Mass of reactants $= 8.4 + 6.0 = 14.4$ g.
Mass of products $= 8.2 + 1.8 + 4.4 = 14.4$ g: equal, so the law is obeyed.
In an open flask the 4.4 g of carbon dioxide escapes, so the reading would fall by about 4.4 g, although no mass is destroyed.

Answer: 14.4 g = 14.4 g, so the law holds; an open flask would read about 4.4 g less because the gas escapes.

Law of Constant Proportions

NCERT section 9.2 Law of Constant Proportions

Joseph Proust found that a compound always contains the same elements in the same fixed ratio by mass, however it is made and wherever it comes from. This is the Law of Constant Proportions (Law of Definite Proportions, or Proust's Law).

Pure water from a river, borewell or ocean always has hydrogen : oxygen = 1 : 8 by mass: 9 g of water gives 1 g H and 8 g O.
Carbon dioxide: C : O = 12 : 32 = 3 : 8.
Sodium chloride: Na : Cl = 23 : 35.5.
Cinnabar (*hingula*), a red pigment used in ancient India and elsewhere, gives mercury and sulphur in about 86.2% : 13.8% by mass.
Proust showed copper carbonate always has the same proportions of copper, carbon and oxygen.

The law applies to compounds, not mixtures (a mixture can have any proportions). If two students get copper oxide with Cu : O as 4 : 1 and 8 : 2, both obey the law, because 8 : 2 = 4 : 1.

Using the ratio: mass of B needed = mass of A × $\frac{ratio part of B}{ratio part of A}$ .

Worked example

(i) How much hydrogen and oxygen are in 4.5 g of pure water? (ii) Magnesium oxide has Mg : O = 3 : 2 by mass. How much oxygen combines with 6 g of magnesium, and what mass of oxide forms?

Try it first, then show the solutionSolution

(i) H : O = 1 : 8, total 9 parts. Hydrogen $= \frac{1}{9} \times 4.5 = 0.5$ g; oxygen $= \frac{8}{9} \times 4.5 = 4$ g.
(ii) Oxygen $= 6 \times \frac{2}{3} = 4$ g.
Magnesium oxide $= 6 + 4 = 10$ g (conservation of mass).

Answer: (i) 0.5 g H and 4 g O (ii) 4 g oxygen; 10 g MgO.

Dalton's atomic theory

NCERT section 9.3 Dalton’s Atomic Theory

In 1808, John Dalton explained the two laws with a set of postulates (basic assumptions accepted without formal proof):
1. All matter is made of very tiny particles called atoms, which take part in chemical reactions.
2. Atoms are indivisible and cannot be created or destroyed in a chemical reaction.
3. Atoms of a given element are identical in mass and chemical properties.
4. Atoms of different elements have different masses and chemical properties.
5. Atoms combine in the ratio of small whole numbers to form compounds.
6. The relative number and kinds of atoms are constant in a given compound.

How it explains the laws:

Atoms are only rearranged in a reaction, not made or destroyed (postulate 2) → mass is conserved.
A compound always has the same number and kind of atoms, each with a fixed mass (postulates 3 and 6) → fixed ratio by mass.

What we now know (Chapter 8): atoms are divisible (electrons, protons, neutrons), and atoms of one element can differ in mass (isotopes). The theory is still the foundation of chemistry.

Worked example

Which postulates of Dalton's theory explain (i) the law of conservation of mass (ii) the law of constant proportions? Name one postulate that later turned out to be incorrect.

Try it first, then show the solutionSolution

(i) Atoms cannot be created or destroyed in a reaction; they only rearrange, so total mass stays the same.
(ii) Atoms of an element have a fixed mass, and a compound always has the same relative number and kind of atoms, so its mass ratio is fixed.
Incorrect later: atoms are indivisible (they contain subatomic particles), and atoms of an element are identical in mass (isotopes differ).

Answer: (i) Atoms are neither created nor destroyed (ii) fixed atom masses and fixed atom ratios; 'indivisible' and 'identical mass' were later revised.

Molecules and covalent bonding

NCERT section 9.4 How Atoms Combine?

A molecule is an electrically neutral group of two or more atoms that can exist independently and shows all the properties of the substance. Noble gases such as helium exist as single atoms because they are already stable.

Atoms bond to reach a stable outer shell (an octet, or a duplet for hydrogen); the bonded arrangement has lower energy than the separate atoms. The force holding atoms together is a chemical bond.

Covalent bond: formed by sharing a pair of electrons; each shared pair attracts both nuclei.

**H $_{2}$ : each H has 1 electron and needs 1 more: they share one pair: single bond**, H–H.
**Cl $_{2}$ :** each Cl (2, 8, 7) needs 1: one shared pair: Cl–Cl.
**O $_{2}$ : each O (2, 6) needs 2: two shared pairs: double bond**, O=O.
**N $_{2}$ : each N (2, 5) needs 3: three shared pairs: triple bond**, N≡N.
HCl: H and Cl share one pair: H–Cl.
**H $_{2}$ O:** O needs 2 and each H needs 1, so O shares one pair with each of two H atoms.
Others: CH $_{4}$ (C shares four pairs with four H), NH $_{3}$ , CO $_{2}$ (O=C=O), H $_{2}$ S.

Elements made of molecules: H $_{2}$ , O $_{2}$ , N $_{2}$ , Cl $_{2}$ . So 'O' means one oxygen atom; 'O $_{2}$ ' means one oxygen molecule.

Worked example

Explain the bonding in a nitrogen molecule (Z = 7) and in ammonia, NH $_{3}$ .

Try it first, then show the solutionSolution

N is 2, 5: it needs 3 electrons for an octet.
N $_{2}$ : two N atoms share three pairs of electrons, a triple bond N≡N; each N now has 8.
NH $_{3}$ : N shares one pair with each of three H atoms; N gets 8 electrons and each H gets 2.

Answer: N $_{2}$ : triple covalent bond; NH $_{3}$ : three single N–H covalent bonds.

Naming covalent compounds

NCERT section 9.4 How Atoms Combine?

For covalent compounds of two non-metals:
1. Name the first element as it is; the second ends in -ide.
2. Use prefixes for numbers of atoms: mono (1), di (2), tri (3), tetra (4), penta (5), hexa (6).
3. Mono is usually left out for the first element but used for the second.
4. If the prefix ends in 'o' or 'a' and the element starts with a vowel, drop that vowel: monoxide, pentoxide, tetroxide. Keep 'i' (dioxide, trioxide).
5. When hydrogen comes first, no prefix: H $_{2}$ S is hydrogen sulphide.

Formula	Name
CO	carbon monoxide
CO $_{2}$	carbon dioxide
CS $_{2}$	carbon disulfide
PCl $_{3}$	phosphorus trichloride
SF $_{6}$	sulphur hexafluoride
N $_{2}$ O $_{4}$	dinitrogen tetroxide
N $_{2}$ O $_{5}$	dinitrogen pentoxide
SO $_{2}$	sulphur dioxide

Some have common names only: H $_{2}$ O is water, NH $_{3}$ is ammonia.

Worked example

Name: (i) NO $_{2}$ (ii) P $_{2}$ O $_{5}$ (iii) CCl $_{4}$ . Write the formula of (iv) sulphur trioxide (v) dinitrogen monoxide.

Try it first, then show the solutionSolution

(i) One N, two O: nitrogen dioxide.
(ii) Two P, five O: diphosphorus pentoxide.
(iii) One C, four Cl: carbon tetrachloride.
(iv) SO $_{3}$ . (v) N $_{2}$ O.

Answer: Nitrogen dioxide; diphosphorus pentoxide; carbon tetrachloride; SO $_{3}$ ; N $_{2}$ O.

Ionic bonding and ions

NCERT section 9.4 How Atoms Combine?

Atoms with fewer than 4 valence electrons (usually metals) tend to lose them; atoms with more than 4 (non-metals) tend to gain electrons.

Sodium (2, 8, 1) loses 1 electron → **Na $^{+}$ (11 protons, 10 electrons): a cation** (positive ion).
Chlorine (2, 8, 7) gains 1 electron → **Cl $^{-}$ (17 protons, 18 electrons): an anion** (negative ion), called chloride.
The electrostatic attraction between oppositely charged ions is an ionic bond. Na gives its electron to Cl, forming Na $^{+}$ Cl $^{-}$ .

More: Mg → Mg $^{2 +}$ (loses 2); O → O $^{2 -}$ oxide (gains 2); S → S $^{2 -}$ sulphide; Al → Al $^{3 +}$ . In magnesium chloride, one Mg $^{2 +}$ joins two Cl $^{-}$ : MgCl $_{2}$ .

Ionic compounds do not form separate molecules. They form 3-D crystals: in NaCl each Na $^{+}$ is surrounded by six Cl $^{-}$ and each Cl $^{-}$ by six Na $^{+}$ , in a repeating pattern (crystal lattice).

Naming: cation first, then anion; simple anions end in -ide (sodium chloride, calcium oxide, magnesium sulphide).

Polyatomic ions (groups of atoms with an overall charge): hydroxide OH $^{-}$ , nitrate NO $_{3}^{-}$ , hydrogencarbonate HCO $_{3}^{-}$ , carbonate CO $_{3}^{2 -}$ , sulphate SO $_{4}^{2 -}$ , ammonium NH $_{4}^{+}$ . Iron and copper show two valencies: ferrous Fe $^{2 +}$ , ferric Fe $^{3 +}$ , cuprous Cu $^{+}$ , cupric Cu $^{2 +}$ .

Worked example

A species has 17 protons, 18 neutrons and 18 electrons. Find its atomic and mass numbers, say whether it is neutral, a cation or an anion, write its configuration and name it.

Try it first, then show the solutionSolution

Z = 17, A = 17 + 18 = 35.
17 positive and 18 negative charges: net −1, so an anion.
Configuration of 18 electrons: 2, 8, 8.
Z = 17 is chlorine: it is the chloride ion, Cl $^{-}$ .

Answer: Z 17, A 35; an anion, 2, 8, 8; chloride ion Cl $^{-}$ .

Writing chemical formulae (criss-cross method)

NCERT section 9.5 Writing Chemical Formulae

Covalent compounds:
1. Write the symbols.
2. Write each element's valency below it.
3. Cross over the valencies as subscripts (do not write 1).

H (1), Cl (1) → HCl; H (1), S (2) → **H $_{2}$ S; C (4), Cl (1) → CCl $_{4}$ **.

Ionic compounds:
1. Cation first, then anion.
2. Write the charge numbers below.
3. Cross over the numbers.
4. Divide by a common factor to get the simplest ratio.
5. Put brackets round a polyatomic ion if more than one is needed.

Ions	Cross over	Formula
Ca $^{2 +}$ , Cl $^{-}$	Ca $_{1}$ Cl $_{2}$	CaCl $_{2}$
Al $^{3 +}$ , O $^{2 -}$	Al $_{2}$ O $_{3}$	Al $_{2}$ O $_{3}$
Mg $^{2 +}$ , O $^{2 -}$	Mg $_{2}$ O $_{2}$ → simplify	MgO
Mg $^{2 +}$ , OH $^{-}$	Mg(OH) $_{2}$	Mg(OH) $_{2}$
Ca $^{2 +}$ , CO $_{3}^{2 -}$	simplify	CaCO $_{3}$
Al $^{3 +}$ , SO $_{4}^{2 -}$	Al $_{2}$ (SO $_{4}$ ) $_{3}$	Al $_{2}$ (SO $_{4}$ ) $_{3}$

Charges are not shown in the final formula; the total positive and negative charges balance, so the compound is neutral. Al(OH) $_{3}$ is correct; 'AlOH $_{3}$ ' is wrong.

Worked example

Write formulae for (i) ferric oxide (ii) aluminium nitrate (iii) ammonium sulphate (iv) the compound of magnesium and nitrogen (N $^{3 -}$ ).

Try it first, then show the solutionSolution

(i) Fe $^{3 +}$ , O $^{2 -}$ → Fe $_{2}$ O $_{3}$ .
(ii) Al $^{3 +}$ , NO $_{3}^{-}$ → Al(NO $_{3}$ ) $_{3}$ .
(iii) NH $_{4}^{+}$ , SO $_{4}^{2 -}$ → (NH $_{4}$ ) $_{2}$ SO $_{4}$ .
(iv) Mg $^{2 +}$ , N $^{3 -}$ → Mg $_{3}$ N $_{2}$ .

Answer: Fe $_{2}$ O $_{3}$ ; Al(NO $_{3}$ ) $_{3}$ ; (NH $_{4}$ ) $_{2}$ SO $_{4}$ ; Mg $_{3}$ N $_{2}$ .

Properties of ionic and covalent compounds

NCERT section 9.6 Properties of the Ionic and the Covalent Compounds

Activity 9.4: test camphor, sodium chloride, copper sulphate, sugar and naphthalene for solubility in water, kerosene and petrol, and for conductivity with a 9 V battery, bulb and two electrodes.

Property	Ionic (NaCl, CuSO $_{4}$ )	Covalent (camphor, naphthalene, sugar)
Made of	ions in a crystal	molecules
Solubility	generally soluble in water, insoluble in kerosene and petrol	mostly insoluble in water but soluble in kerosene/petrol (sugar is an exception: dissolves in water)
Conductivity as solid	no (ions fixed in place)	no
In water solution	conducts (ions free to move)	does not (no ions; sugar solution does not conduct)
Molten	conducts	does not
Melting and boiling points	high (strong attraction between ions)	usually low

Safety: use a low-voltage battery; do not touch the electrodes; keep kerosene and petrol away from flames.

This answers the opening question: dissolved salt conducts because it gives free ions; sugar does not because it stays as molecules.

Worked example

A white solid does not conduct electricity as a solid but its water solution lights the bulb. It melts at about 800 °C. What type of compound is it? Explain each observation.

Try it first, then show the solutionSolution

Ions in a solid crystal are held in fixed positions, so they cannot carry current.
In water the ions separate and move freely, so the solution conducts.
Strong attraction between oppositely charged ions needs a lot of energy to break, giving a high melting point.

Answer: An ionic compound (like sodium chloride).

Molecular mass and formula unit mass

NCERT section 9.7 Molecular Mass of Covalent Compounds

Masses of atoms are given in unified atomic mass units (u). Common values: H 1, C 12, N 14, O 16, Na 23, Mg 24, Al 27, P 31, S 32, Cl 35.5, K 39, Ca 40.

Molecular mass (covalent compounds) = sum of the atomic masses of all atoms in one molecule.

H $_{2}$ O $= 2 (1) + 16 = 18$ u
CO $_{2}$ $= 12 + 2 (16) = 44$ u
HNO $_{3}$ $= 1 + 14 + 3 (16) = 63$ u
CH $_{4}$ $= 12 + 4 (1) = 16$ u

Formula unit mass (ionic compounds): ionic compounds have no molecules, so we use the formula unit, the simplest whole-number ratio of ions.

Na $_{2}$ O $= 2 (23) + 16 = 62$ u
Ca(NO $_{3}$ ) $_{2}$ $= 40 + 2 [14 + 3 (16)] = 164$ u
KCl $= 39 + 35.5 = 74.5$ u
Mg(OH) $_{2}$ $= 24 + 2 (16 + 1) = 58$ u

Method: list atoms and how many of each (multiply out brackets), multiply by atomic masses, add.

Worked example

Find the formula unit mass of (i) potassium sulphate K $_{2}$ SO $_{4}$ (ii) calcium phosphate Ca $_{3}$ (PO $_{4}$ ) $_{2}$ , and the molecular mass of (iii) sucrose C $_{12}$ H $_{22}$ O $_{11}$ .

Try it first, then show the solutionSolution

(i) K 2 × 39 = 78; S 32; O 4 × 16 = 64; total $= 174$ u.
(ii) Ca 3 × 40 = 120; P 2 × 31 = 62; O 8 × 16 = 128; total $= 310$ u.
(iii) C 12 × 12 = 144; H 22 × 1 = 22; O 11 × 16 = 176; total $= 342$ u.

Answer: (i) 174 u (ii) 310 u (iii) 342 u.

Science in real life: Atomic Foundations of Matter

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

Suppose you make a fizzy 'volcano' with baking soda and lemon juice on a kitchen scale, and the reading drops by 2 g.

Where did the 2 g go?

Try it first, then show the solutionSolution

The reaction makes carbon dioxide gas, which bubbles out into the air.
If you had trapped the gas with a balloon on a bottle, the reading would not change.

Answer: About 2 g of carbon dioxide escaped; mass was not destroyed.

Memory hook: Missing mass? Check for gas.

Ionic compounds conduct in solution

Suppose you are told never to touch a switch with wet, sweaty hands.

Why does sweat make it more dangerous than pure water would?

Try it first, then show the solutionSolution

Sweat contains dissolved salts such as sodium chloride.
These split into ions that move freely and carry electric current.

Answer: Ions from dissolved salts make sweat a good conductor.

Memory hook: Salty water carries current.

Constant proportions

Suppose a shop sells 'mineral water from a mountain spring' and claims it is 'a different kind of water'.

Is the water itself chemically different from tap water?

Try it first, then show the solutionSolution

Pure water from any source is H $_{2}$ O with H : O = 1 : 8 by mass.
Only the dissolved minerals differ; the water molecules are the same.

Answer: No; the water is identical; only dissolved minerals vary.

Memory hook: Water is water: one to eight.

Formula unit mass

Suppose your grandmother takes an antacid containing sodium hydrogencarbonate, NaHCO $_{3}$ .

Find its formula unit mass and the mass of sodium in 84 g of it (Na 23, H 1, C 12, O 16).

Try it first, then show the solutionSolution

$23 + 1 + 12 + 3 (16) = 84$ u.
In 84 g, sodium is 23 g (same fraction as 23 out of 84).

Answer: 84 u; 23 g of sodium.

Memory hook: Add the atoms, keep the fraction.

Topics in this chapter

Atomic Foundations of Matter: formulas and key facts

Conservation of mass: $T o t a l ma sso f r e a c t an t s = t o t a l ma sso f p r o d u c t s .$
Constant proportions: Water H : O = 1 : 8; CO $_{2}$ C : O = 3 : 8; NaCl Na : Cl = 23 : 35.5; MgO Mg : O = 3 : 2.
Valency rule: 1 to 3 valence electrons: lose them (cation); 5 to 7: gain 8 − n (anion); 4: share.
Bond types: Covalent: sharing (single, double, triple). Ionic: transfer, then attraction between ions.
Covalent prefixes: mono 1, di 2, tri 3, tetra 4, penta 5, hexa 6; second element ends in -ide.
Criss-cross: Swap valencies (or charge numbers) as subscripts; simplify; bracket repeated polyatomic ions.
Common polyatomic ions: OH⁻, NO $_{3}^{-}$ , HCO $_{3}^{-}$ , CO $_{3}^{2 -}$ , SO $_{4}^{2 -}$ , NH $_{4}^{+}$ .
Molecular mass: Sum of atomic masses in one molecule, e.g. H $_{2}$ O = 18 u, CO $_{2}$ = 44 u.
Formula unit mass: Sum of atomic masses in one formula unit, e.g. NaCl = 58.5 u, CaCO $_{3}$ = 100 u.
Key equations: $2 H_{2} + O_{2} \to 2 H_{2} O$ ; $Na_{2} SO_{4} + BaCl_{2} \to BaSO_{4} + 2 NaCl$

Atomic Foundations of Matter: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

Closed-flask test of conservation of mass (Fig. 9.3)

Draw the set-up to show mass is conserved when vinegar reacts with baking soda; links to the CBSE practical.

Draw

A conical flask with vinegar standing on a digital balance.
A balloon containing baking soda tied over the mouth of the flask with thread, hanging to one side.
A second drawing with the balloon lifted, baking soda falling in, and the balloon inflated; the balance reading unchanged.

Label

Digital weighing balance
Conical flask
Vinegar
Balloon with baking soda
Inflated balloon (carbon dioxide)
Same reading before and after

Watch out: Contrast with the open flask, where the reading falls because gas escapes.

Electron-dot (sharing) diagrams of molecules (Figs. 9.6 to 9.10)

Show the formation of H $_{2}$ , Cl $_{2}$ , O $_{2}$ , N $_{2}$ , HCl, H $_{2}$ O, NH $_{3}$ , CH $_{4}$ or CO $_{2}$ .

Draw

Each atom as its symbol with valence electrons as dots (one atom) and crosses (the other).
Overlapping circles round the shared pairs between atoms.
The line structure beside it: H–H, O=O, N≡N, H–O–H.

Label

Shared pair
Single bond
Double bond
Triple bond
Octet / duplet completed

Watch out: Count: every atom except H should end with 8 electrons round it; H ends with 2.

Formation of sodium chloride (Figs. 9.11 to 9.13)

Show the transfer of an electron from sodium to chlorine and the ions formed.

Draw

Sodium atom with shells 2, 8, 1 and chlorine atom with shells 2, 8, 7.
A curved arrow showing the single outer electron of sodium moving to chlorine.
The ions after transfer in square brackets: [Na] $^{+}$ with 2, 8 and [Cl] $^{-}$ with 2, 8, 8.

Label

Sodium atom (2, 8, 1)
Chlorine atom (2, 8, 7)
Electron transfer
Sodium cation Na⁺
Chloride anion Cl⁻
Ionic bond (attraction)

Watch out: Show charges on the ions, and note that the ion keeps the same number of protons.

Conductivity test set-up (Fig. 9.15)

Draw an apparatus to test whether a solution conducts electricity.

Draw

A beaker with the solution and two electrodes pushed through a piece of cardboard resting on top.
Wires from the electrodes to a 9 V battery and a bulb in series.

Label

Beaker with solution
Electrodes
Cardboard
Bulb
9 V battery

Watch out: Mention that the bulb glows for salt solution but not for sugar solution.

How the school exam asks Atomic Foundations of Matter

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Laws of chemical combination	Verify conservation of mass from data; find masses from a fixed ratio; explain open versus closed flask; why the law applies to compounds, not mixtures.	Add reactant and product masses; use ratio × mass; mention escaping gas.
Dalton's theory	State postulates; which postulate explains which law; limitations in light of Chapter 8.	Link 'atoms not created or destroyed' to conservation and 'fixed number and kinds' to constant proportions.
Bonding diagrams	Show formation of covalent molecules and ionic compounds; predict bond type from configurations.	Write configurations, count electrons needed, then share (non-metal + non-metal) or transfer (metal + non-metal).
Names and formulae	Name covalent compounds with prefixes; write formulae from names or ion pairs; complete grids of cations and anions.	Criss-cross, simplify, bracket; check total charge is zero.
Properties	Solubility and conductivity tables; why salt solution conducts but sugar does not; molten state predictions.	Ions present and free to move → conducts; strong ionic attraction → high melting point.
Mass calculations	Molecular or formula unit masses for fertilisers, acids, antacids, salts.	Expand brackets, multiply counts by atomic masses, add, give units (u).

Tricks to remember Atomic Foundations of Matter

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

Closed box, same weight
If nothing escapes, nothing changes on the balance.
Use it when: Explaining conservation-of-mass experiments.
Water is 1 : 8
One gram of hydrogen, eight of oxygen: 'one for H, eight for O'.
Use it when: Constant proportions problems.
Metals give, non-metals take
Metals become cations (+), non-metals become anions (−); two non-metals share.
Use it when: Predicting ionic or covalent bonding.
Cross, cancel, bracket
Cross the charges, cancel common factors, bracket a repeated polyatomic ion.
Use it when: Writing formulae.
Salt sparks, sugar sleeps
Salt solution lights the bulb (ions); sugar solution does not (molecules).
Use it when: Conductivity questions.

Atomic Foundations of Matter: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Formula of aluminium oxide · 1 mark · MCQExam level

The formula of aluminium oxide is
(a) AlO
(b) Al $_{2}$ O $_{3}$
(c) Al $_{3}$ O $_{2}$
(d) AlO $_{3}$

Try it first, then show the solutionSolution

Al $^{3 +}$ and O $^{2 -}$ : cross over 3 and 2.
Al $_{2}$ O $_{3}$ : $2 \times 3 = 6$ positive and $3 \times 2 = 6$ negative charges.

Answer: (b)

Example 2: Balanced charges · 1 mark · MCQExam level

Which combination gives a total charge of 3+ and 3−?
(a) 1 Al $^{3 +}$ and 1 PO $_{4}^{3 -}$
(b) 3 Na $^{+}$ and 1 SO $_{4}^{2 -}$
(c) 1 Fe $^{3 +}$ and 1 O $^{2 -}$
(d) 3 K $^{+}$ and 3 CO $_{3}^{2 -}$

Try it first, then show the solutionSolution

(a) +3, −3 (b) +3, −2 (c) +3, −2 (d) +3, −6.
Only (a) balances, giving AlPO $_{4}$ .

Answer: (a)

Example 3: Sugar solution (assertion-reason) · 1 mark · Assertion-reasonExam level

Assertion (A): A sugar solution does not conduct electricity.
Reason (R): Sugar is a covalent compound that stays as molecules in water and gives no ions.
(a) A and R are both true, and R explains A correctly.
(b) A and R are both true, but R does not explain A.
(c) A is true, R is false.
(d) A is false, R is true.

Try it first, then show the solutionSolution

In Activity 9.4 the bulb does not glow with sugar solution: A is true.
Current in a solution needs free ions; sugar gives none, so R explains A.

Answer: (a)

Example 4: Naming covalent compounds · 2 marks · Very short answerExam level

Name (i) SF $_{6}$ (ii) PCl $_{3}$ and write the formulae of (iii) carbon disulfide (iv) dinitrogen pentoxide.

Try it first, then show the solutionSolution

(i) Sulphur hexafluoride (ii) phosphorus trichloride.
(iii) CS $_{2}$ (iv) N $_{2}$ O $_{5}$ .

Answer: Sulphur hexafluoride; phosphorus trichloride; CS $_{2}$ ; N $_{2}$ O $_{5}$ .

Example 5: Hydrogen and oxygen to water · 3 marks · Short answerHarder exam level

In a sealed vessel, 6 g of hydrogen and 48 g of oxygen combine fully. Find the mass of water formed and check it against the 1 : 8 ratio. What happens if 10 g of hydrogen is used with the same 48 g of oxygen?

Try it first, then show the solutionSolution

Water $= 6 + 48 = 54$ g (conservation of mass).
Ratio $6 : 48 = 1 : 8$ : agrees with constant proportions.
48 g of oxygen needs only 6 g of hydrogen, so again 54 g of water forms and 4 g of hydrogen is left unreacted.

Answer: 54 g; yes; 54 g water with 4 g hydrogen left over.

Example 6: Magnesium gaining mass · 3 marks · Short answerHarder exam level

A student heats 2.4 g of magnesium ribbon in an open crucible and finds 4.0 g of white ash. Does this break the law of conservation of mass? Explain, and say how you would show that the law holds.

Try it first, then show the solutionSolution

Magnesium combines with oxygen from the air: $2 Mg + O_{2} \to 2 MgO$ .
The extra $4.0 - 2.4 = 1.6$ g is the oxygen taken from the air, which was not weighed at the start.
Mass of magnesium + oxygen = mass of oxide, so the law holds; to show it, heat the magnesium in a closed container and weigh everything before and after.

Answer: No; 1.6 g of oxygen from the air joined the magnesium. Weighing a closed system shows no change.

Example 7: Bonding in A and B · 3 marks · Short answerHarder exam level

Element A is 2, 4 and element B is 2, 8, 7. (i) Which tends to share electrons rather than gain or lose them? (ii) Will they form ionic or covalent bonds? (iii) Predict the formula and name the compound.

Try it first, then show the solutionSolution

(i) A (carbon) has 4 valence electrons: losing or gaining 4 is hard, so it shares.
(ii) B (chlorine) is a non-metal needing 1 electron; two non-metals share electrons: covalent.
(iii) A needs 4 and B needs 1, so one A shares with four B: AB $_{4}$ , carbon tetrachloride (CCl $_{4}$ ).

Answer: (i) A (ii) covalent (iii) AB $_{4}$ , carbon tetrachloride.

Example 8: Formula unit masses · 3 marks · Short answerHarder exam level

Find the masses of (i) sulphuric acid H $_{2}$ SO $_{4}$ (molecular) (ii) calcium carbonate CaCO $_{3}$ (iii) magnesium nitrate Mg(NO $_{3}$ ) $_{2}$ . (H 1, C 12, N 14, O 16, Mg 24, S 32, Ca 40)

Try it first, then show the solutionSolution

(i) $2 (1) + 32 + 4 (16) = 98$ u.
(ii) $40 + 12 + 3 (16) = 100$ u.
(iii) $24 + 2 [14 + 3 (16)] = 24 + 124 = 148$ u.

Answer: (i) 98 u (ii) 100 u (iii) 148 u.

Example 9: Cation-anion grid · 5 marks · Long answerHarder exam level

Write formulae of compounds formed by K $^{+}$ , Fe $^{3 +}$ and Mg $^{2 +}$ with Cl $^{-}$ , CO $_{3}^{2 -}$ and PO $_{4}^{3 -}$ .

Try it first, then show the solutionSolution

K $^{+}$ : KCl, K $_{2}$ CO $_{3}$ , K $_{3}$ PO $_{4}$ .
Fe $^{3 +}$ : FeCl $_{3}$ , Fe $_{2}$ (CO $_{3}$ ) $_{3}$ , FePO $_{4}$ (3 and 3 cancel).
Mg $^{2 +}$ : MgCl $_{2}$ , MgCO $_{3}$ (2 and 2 cancel), Mg $_{3}$ (PO $_{4}$ ) $_{2}$ .
Check each: total positive charge = total negative charge.

Answer: KCl, K $_{2}$ CO $_{3}$ , K $_{3}$ PO $_{4}$ ; FeCl $_{3}$ , Fe $_{2}$ (CO $_{3}$ ) $_{3}$ , FePO $_{4}$ ; MgCl $_{2}$ , MgCO $_{3}$ , Mg $_{3}$ (PO $_{4}$ ) $_{2}$ .

Example 10: Metal E and oxygen · 4 marks · Case-basedHOTS / competency

Metal E has three electrons in its M shell and burns in oxygen to form a hard white compound with a very high melting point. Identify E and predict (i) the formula (ii) the bond type (iii) whether the molten compound conducts electricity. Calculate the formula unit mass (Al 27, O 16).

Try it first, then show the solutionSolution

Three electrons in the M shell: 2, 8, 3, so E is aluminium.
(i) Al $^{3 +}$ and O $^{2 -}$ : cross over to Al $_{2}$ O $_{3}$ .
(ii) Metal and non-metal by electron transfer: ionic (which fits the very high melting point). (iii) Molten, its ions are free to move, so it conducts.
Formula unit mass $= 2 (27) + 3 (16) = 54 + 48 = 102$ u.

Answer: Aluminium; Al $_{2}$ O $_{3}$ ; ionic; conducts when molten; 102 u.

Example 11: Testing samples of lime · 3 marks · Short answerHOTS / competency

Calcium oxide has Ca : O = 40 : 16 by mass. Three samples are analysed: P (2.8 g) contains 2.0 g of calcium; Q (5.6 g) contains 4.0 g; R (3.0 g) contains 1.8 g. Which samples are pure calcium oxide? Suggest a reason for the odd one.

Try it first, then show the solutionSolution

Pure CaO: calcium fraction $= \frac{40}{56} \approx 0.714$ .
P: $\frac{2.0}{2.8} \approx 0.714$ ; Q: $\frac{4.0}{5.6} \approx 0.714$ : both fit the law of constant proportions.
R: $\frac{1.8}{3.0} = 0.6$ : too little calcium, so R is not pure CaO; it may be mixed with water, chalk (CaCO $_{3}$ ) or another impurity, or mis-weighed.

Answer: P and Q are pure CaO; R has the wrong ratio, so it is impure or wrongly weighed.

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 2 marks.

Fertiliser shop

A farmer reads fertiliser labels at a shop: urea, ammonium nitrate (NH $_{4}$ NO $_{3}$ ), ammonium sulphate ((NH $_{4}$ ) $_{2}$ SO $_{4}$ ) and a phosphate fertiliser made from phosphoric acid. The shopkeeper says ammonium nitrate dissolves easily in water and its solution conducts electricity.

(i) · 1 mark
Name the two ions in ammonium sulphate and write their formulae.
Answer
Answer: Ammonium NH $_{4}^{+}$ and sulphate SO $_{4}^{2 -}$
Both are polyatomic ions; two NH $_{4}^{+}$ balance one SO $_{4}^{2 -}$ .
(ii) · 1 mark
Find the formula unit mass of ammonium sulphate (H 1, N 14, O 16, S 32).
Answer
Answer: 132 u
N 2 × 14 = 28, H 8 × 1 = 8, S 32, O 4 × 16 = 64; total 132 u.
(iii) · 2 marks
Explain why its solution conducts electricity, and why its formula has no charges shown.
Answer
Answer: Free ions; charges balance
It is ionic; in water NH $_{4}^{+}$ and NO $_{3}^{-}$ ions separate and move, carrying current (1 mark). One +1 and one −1 charge cancel, so the formula unit is neutral and charges are not written (1 mark).

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Write the formulae of (i) calcium chloride (ii) aluminium sulphate. (2 marks)

Step	Marks
Ca $^{2 +}$ and Cl $^{-}$ criss-crossed: CaCl $_{2}$	1
Al $^{3 +}$ and SO $_{4}^{2 -}$ criss-crossed with brackets: Al $_{2}$ (SO $_{4}$ ) $_{3}$	1

12 g of carbon burns in oxygen to give 44 g of carbon dioxide. (i) How much oxygen was used? (ii) How much carbon dioxide forms from 3.6 g of carbon? (iii) Which law did you use in each part? (3 marks)

Step	Marks
(i) Oxygen $= 44 - 12 = 32$ g (conservation of mass)	1
(ii) CO $_{2}$ $= \frac{44}{12} \times 3.6 = 13.2$ g	1
(iii) Part (ii) uses the law of constant proportions (fixed C : CO $_{2}$ ratio)	1

Element P has two electrons in its third shell and element Q has seven electrons in its second shell. (i) Identify P and Q (ii) state what ions they form (iii) name the bond and write the formula of their compound (iv) show the electron transfer (v) state two properties of the compound. (5 marks)

Step	Marks
P is 2, 8, 2: magnesium; Q is 2, 7: fluorine	1
P loses 2 electrons: Mg $^{2 +}$ ; Q gains 1: F $^{-}$	1
Ionic bond; formula MgF $_{2}$	1
Diagram: one Mg atom transferring one electron to each of two F atoms	1
Properties: high melting point; conducts electricity when molten	1

Where students lose marks in Atomic Foundations of Matter

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

Mistake: Claiming mass is lost when a gas forms
The balance reading falls in an open flask.
Wrong working
The mass decreased, so mass is not conserved in this reaction.
Right working
The carbon dioxide escaped; in a closed system the mass stays the same.
Do this: Always account for gases.
Usually costs: 1 to 2 marks.
Mistake: Not simplifying ionic formulae
Criss-crossing is applied blindly.
Wrong working
Magnesium oxide: Mg $_{2}$ O $_{2}$ ; calcium carbonate: Ca $_{2}$ (CO $_{3}$ ) $_{2}$ .
Right working
MgO and CaCO $_{3}$ (divide by the common factor 2).
Do this: Always reduce to the simplest ratio.
Usually costs: 1 mark each.
Mistake: Missing brackets on polyatomic ions
Brackets seem optional.
Wrong working
Aluminium hydroxide: AlOH $_{3}$ .
Right working
Al(OH) $_{3}$ : three OH $^{-}$ groups.
Do this: Bracket any polyatomic ion that appears more than once.
Usually costs: 1 mark.
Mistake: Calling NaCl a molecule and using 'molecular mass'
All compounds are thought to be molecules.
Wrong working
The molecular mass of NaCl is 58.5 u.
Right working
NaCl is ionic and forms a crystal; its formula unit mass is 58.5 u.
Do this: Ionic → formula unit mass; covalent → molecular mass.
Usually costs: The term mark.
Mistake: Wrong covalent prefixes
Mono and vowel rules are forgotten.
Wrong working
CO = monocarbon monooxide; N $_{2}$ O $_{4}$ = dinitrogen tetraoxide; H $_{2}$ S = dihydrogen sulphide.
Right working
Carbon monoxide; dinitrogen tetroxide; hydrogen sulphide.
Do this: No mono on the first element; drop o/a before a vowel; no prefix for hydrogen first.
Usually costs: 1 mark each.

NCERT exercise guide: Atomic Foundations of Matter

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Revise, Reflect, Refine

What it tests: Ions from elements with given electrons, diatomic molecules, designing an ionic compound with balanced charges, true/false on molecules, formulae from names and ion pairs, identifying Cl⁻, formula unit masses of fertilisers and antacids, compounds of metals with nitrogen, sulphur and oxygen, a cation-anion grid, conservation-of-mass data, a species with 11 p and 10 e, reactivity and bonding of 2, 8, 5 and 2, 8, 7 elements, assertion-reason on copper sulphate, and electrons and neutrons in ions.

How to approach it: Write configurations, decide lose, gain or share, then criss-cross; for masses, expand brackets and add; for ions, electrons = protons − charge.

Like the ion-counting question in Revise, Reflect, Refine

Find the electrons and neutrons in $^{24} Mg^{2 +}$ (12 protons), $^{32} S^{2 -}$ (16 protons) and $^{40} Ca^{2 +}$ (20 protons).

Show the solutionSolution

Mg $^{2 +}$ : electrons $= 12 - 2 = 10$ ; neutrons $= 24 - 12 = 12$ .
S $^{2 -}$ : electrons $= 16 + 2 = 18$ ; neutrons $= 32 - 16 = 16$ .
Ca $^{2 +}$ : electrons $= 20 - 2 = 18$ ; neutrons $= 40 - 20 = 20$ .

Answer: Mg $^{2 +}$ 10 e, 12 n; S $^{2 -}$ 18 e, 16 n; Ca $^{2 +}$ 18 e, 20 n.

Think It Over

What it tests: Whether water from different sources is chemically the same, the difference between O and O $_{2}$ , and why salt solution conducts but sugar solution does not.

How to approach it: Use the law of constant proportions, the idea of a molecule, and ions versus molecules.

Like the opening Think It Over questions

Rainwater and purified seawater are both analysed. Will they have the same composition? Explain.

Show the solutionSolution

After purification, both are pure water, H $_{2}$ O.
By the law of constant proportions, pure water always has H : O = 1 : 8 by mass.
Differences in taste come from dissolved substances, which purification removes.

Answer: Yes; pure water from any source has hydrogen and oxygen in the fixed ratio 1 : 8.

Let us explore

What it tests: The Journey Beyond: showing water's fixed composition from different sources, bar graphs of electrons in atoms and ions, a cation-anion card game, and a molecule-building simulation.

How to approach it: Use the ion table for card values; check each 'compound' has zero total charge.

Practice questions on Atomic Foundations of Matter

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

Q1 · 1 mark · MCQExam level
The law of constant proportions was given by
(a) Dalton
(b) Lavoisier
(c) Proust
(d) Bohr
Hint
French chemist who studied copper carbonate.
Answer
(c) Proust
Q2 · 1 mark · MCQExam level
A molecule of oxygen contains
(a) a single bond
(b) a double bond
(c) a triple bond
(d) an ionic bond
Hint
Each O needs 2.
Answer
(b) a double bond
Q3 · 1 mark · Assertion-reasonExam level
Assertion (A): Hydrogen burns and oxygen helps burning, yet water puts out fire.
Reason (R): A compound has properties different from those of the elements it is made from.
(a) A and R are both true, and R explains A correctly.
(b) A and R are both true, but R does not explain A.
(c) A is true, R is false.
(d) A is false, R is true.
Hint
Compare the compound with its elements.
Answer
(a). Both are true, and the different properties of a compound explain why water behaves unlike hydrogen and oxygen.
Q4 · 2 marks · Very short answerExam level
Write the formulae of (i) potassium carbonate (ii) calcium hydroxide.
Hint
K⁺ with CO $_{3}^{2 -}$ ; Ca $^{2 +}$ with OH⁻.
Answer
(i) K $_{2}$ CO $_{3}$ (ii) Ca(OH) $_{2}$ .
Q5 · 2 marks · Very short answerHarder exam level
Sodium chloride has Na : Cl = 23 : 35.5 by mass. How much chlorine is needed for 11.5 g of sodium, and how much salt forms?
Hint
Scale the ratio.
Answer
Chlorine $= 11.5 \times \frac{35.5}{23} = 17.75$ g; salt $= 11.5 + 17.75 = 29.25$ g.
Q6 · 3 marks · Short answerHarder exam level
Show the formation of a water molecule by electron-dot diagram and explain why oxygen bonds with two hydrogen atoms.
Hint
Oxygen needs 2, hydrogen needs 1.
Answer
O (2, 6) needs 2 electrons; each H needs 1. O shares one pair with each H: H–O–H with two shared pairs. O completes its octet and each H its duplet, so one O bonds with exactly two H atoms.
Q7 · 3 marks · Short answerHarder exam level
Show how magnesium (Z = 12) and chlorine (Z = 17) form magnesium chloride. Write the formula and the charges on the ions.
Hint
One Mg gives two electrons.
Answer
Mg (2, 8, 2) loses two electrons to become Mg $^{2 +}$ ; each of two Cl atoms (2, 8, 7) gains one electron to become Cl $^{-}$ . The ions attract: MgCl $_{2}$ .
Q8 · 3 marks · Short answerHarder exam level
Write formulae of the compounds formed by (i) calcium and nitrogen (ii) potassium and oxygen (iii) aluminium and sulphur. Find the formula unit mass of the potassium compound (K 39, O 16).
Hint
N $^{3 -}$ , O $^{2 -}$ , S $^{2 -}$ .
Answer
(i) Ca $_{3}$ N $_{2}$ (ii) K $_{2}$ O (iii) Al $_{2}$ S $_{3}$ . K $_{2}$ O $= 2 (39) + 16 = 94$ u.
Q9 · 3 marks · Short answerHarder exam level
Explain, with Activity 9.4, why camphor dissolves in kerosene but not in water, while sodium chloride does the opposite.
Hint
Like dissolves like.
Answer
Sodium chloride is ionic; water separates its ions, so it dissolves in water, not in kerosene. Camphor is covalent with non-polar molecules, which mix with kerosene or petrol (also covalent) but not with water.
Q10 · 5 marks · Long answerHarder exam level
Describe how you would verify the law of conservation of mass using sodium sulphate and barium chloride solutions. Write the balanced equation and state why the flasks need not be closed.
Hint
Activity 9.3.
Answer
Put two flasks with 10 mL of 1% sodium sulphate (A) and 10 mL of 1% barium chloride (B) on a digital balance and note the total. Pour B into A; a white precipitate of barium sulphate forms. Put both flasks back on the balance; the reading is the same. $Na_{2} SO_{4} + BaCl_{2} \to BaSO_{4} + 2 NaCl$ . No gas is produced, so nothing escapes and an open system works; keeping the empty flask on the balance avoids error from solution sticking to its walls.
Q11 · 3 marks · Short answerHOTS / competency
A compound is 40% sulphur and 60% oxygen by mass. A sample contains 20 g of sulphur. How much oxygen must it contain? What is the ratio of S atoms to O atoms if S = 32 u and O = 16 u, and what is the formula?
Hint
Divide each mass by its atomic mass.
Answer
Oxygen $= 20 \times \frac{60}{40} = 30$ g. Atoms: S $\frac{20}{32} = 0.625$ , O $\frac{30}{16} = 1.875$ ; ratio $1 : 3$ , so SO $_{3}$ (sulphur trioxide).
Q12 · 3 marks · Short answerHOTS / competency
Neon (Z = 10) neither gives, takes nor shares electrons, but fluorine (Z = 9) forms F $_{2}$ molecules and NaF with sodium. Explain the difference and the type of bond in each fluorine compound.
Hint
Who already has eight?
Answer
Neon is 2, 8: its octet is complete, so it is stable and exists as single atoms. Fluorine is 2, 7 and needs one electron: in F $_{2}$ two F atoms share one pair (single covalent bond); with sodium, which gives away its one outer electron, F becomes F $^{-}$ and forms the ionic compound NaF.

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.Who proposed the law of conservation of mass, and when?Show answer
Antoine Lavoisier, in 1789.
2.State the law of constant proportions.Show answer
In a compound, the elements are always present in a fixed ratio by mass, whatever its source or method of preparation.
3.What is the ratio by mass of hydrogen to oxygen in water?Show answer
1 : 8.
4.Define a molecule.Show answer
An electrically neutral group of two or more atoms that can exist independently and shows all the properties of the substance.
5.What is a covalent bond?Show answer
A bond formed by sharing a pair of electrons between two atoms.
6.What is an ionic bond?Show answer
The electrostatic attraction between oppositely charged ions formed by transfer of electrons.
7.Name the bond in an oxygen molecule.Show answer
A double covalent bond (O=O).
8.Write the formula of the sulphate ion and the ammonium ion.Show answer
SO $_{4}^{2 -}$ and NH $_{4}^{+}$ .
9.Why does sugar solution not conduct electricity?Show answer
Sugar is covalent and does not form ions in water.
10.Find the molecular mass of methane, CH $_{4}$ .Show answer
$12 + 4 = 16$ u.

Best order to study it

Do or read Activities 9.1 to 9.3 and learn the law of conservation of mass with numericals (9.1).
Learn the law of constant proportions and ratio problems (9.2).
Learn Dalton's postulates and link them to the two laws (9.3).
Practise electron-dot diagrams for covalent molecules and the ionic bond in NaCl (9.4).
Learn the ion tables and practise criss-cross formulae and covalent names (9.5).
Finish with properties (9.6) and molecular and formula unit masses (9.7, 9.8), then all Revise, Reflect, Refine questions.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you explain why the open flask loses mass but the closed one does not?
Can you verify the law of conservation of mass from data?
Can you solve fixed-ratio problems for water, CO $_{2}$ and NaCl?
Can you list Dalton's six postulates?
Can you draw dot diagrams for H $_{2}$ , O $_{2}$ , N $_{2}$ , HCl, H $_{2}$ O and NH $_{3}$ ?
Can you show electron transfer in NaCl and MgO?
Can you name covalent compounds with the right prefixes?
Can you write formulae with polyatomic ions and brackets?
Can you compare ionic and covalent compounds on four properties?
Can you calculate molecular and formula unit masses?

Atomic Foundations of Matter: frequently asked questions

Which practical goes with this chapter?

The CBSE syllabus lists verifying the law of conservation of mass in a chemical reaction, as in Activities 9.2 and 9.3.

What is the difference between O and O$_2$?

O is one oxygen atom; O_2 is one oxygen molecule made of two atoms joined by a double bond. Oxygen gas exists as O_2 molecules.

Do I need to know mole concept?

No. This chapter goes up to molecular mass and formula unit mass in u; the mole is not in the Class 9 2026-27 syllabus.

Which atomic masses should I memorise?

H 1, C 12, N 14, O 16, Na 23, Mg 24, Al 27, S 32, Cl 35.5, K 39, Ca 40 are used most; questions usually give them, but knowing them saves time.

How many marks does this chapter carry?

It is part of Unit II, Matter: Its Nature and Behaviour, which carries 25 of the 80 theory marks together with two other chapters; CBSE does not fix marks per chapter.

Based on: CBSE Science (086) syllabus, Class IX 2026-27; NCERT Exploration, Science Textbook for Grade 9 (First Edition 2026)