ChemistryICSE Class 9 · Chapter 1

The Language of Chemistry ICSE Class 9: Symbols, Valency, Formulae and Equations

Updated 15 October 2026 · Fact-checked

The language of chemistry is the set of symbols, formulae and equations chemists use. Valency is the combining capacity of an element or radical; formulae are written by exchanging valencies. Equations are balanced by hit and trial, and relative molecular mass is the sum of the relative atomic masses in a formula.

What The Language of Chemistry is about

Chemistry has its own shorthand. A symbol stands for one atom of an element, a formula for one molecule (or formula unit) of a compound, and a chemical equation for a reaction. To write formulae you need valency, the combining capacity of an element or a radical.

The CISCE 2028 syllabus (Class IX, topic 1) covers: symbols of common elements; valency in terms of hydrogen combination and valence electrons, with mono-, di-, tri- and tetravalent elements; radicals, their formulae and valencies; the existence of radicals in the atmosphere, their effects on living things and how to reduce them; names and formulae of compounds; chemical equations of the four reactant-product patterns and balancing by hit and trial; and relative atomic and molecular masses (hydrogen or carbon-12 standard), with calculations of relative molecular mass and percentage composition.

Weightage: CISCE publishes no chapter weightage for ICSE Chemistry. The Language of Chemistry is topic 1 of the Class IX part of the 2028 syllabus; the Class IX paper (80 marks, 2 hours) is set and marked by the school.Source: CISCE ICSE Chemistry (Science Paper 2) syllabus, examination year 2028 (Class IX)

Key concepts

Symbols and valency

A symbol is the short form of the name of an element; it also stands for one atom of it. Many come from English names (H, O, C, N, Cl), some from Latin: sodium Na (natrium), potassium K (kalium), iron Fe (ferrum), copper Cu (cuprum), silver Ag (argentum), lead Pb (plumbum), mercury Hg (hydrargyrum).

Valency is the combining capacity of an element. It can be seen in two ways:

Hydrogen combination: the number of hydrogen atoms that one atom combines with or displaces. In HCl chlorine has valency 1; in H₂O oxygen 2; in NH₃ nitrogen 3; in CH₄ carbon 4.
Valence electrons: the number of electrons an atom loses, gains or shares to complete its outer shell. Metals lose electrons (Na loses 1, Mg 2, Al 3); non-metals gain or share them (Cl gains 1, O 2, N 3).

Valency	Examples
Monovalent (1)	H, Na, K, Ag, Cl, Br, I
Divalent (2)	Mg, Ca, Zn, Ba, O, S
Trivalent (3)	Al, N (in NH₃), P (in PH₃)
Tetravalent (4)	C, Si

Variable valency: some elements show more than one valency. The lower one takes the ending -ous, the higher -ic: iron(II) or ferrous Fe²⁺ and iron(III) or ferric Fe³⁺; copper(I) or cuprous Cu⁺ and copper(II) or cupric Cu²⁺.

Worked example

Using hydrogen combination, find the valency of phosphorus, silicon and sulphur in (a) PH₃, (b) SiH₄, (c) H₂S.

Try it first, then show the solutionSolution

Count the hydrogen atoms one atom of the element holds.
(a) P holds 3 H, so valency 3. (b) Si holds 4 H, so valency 4. (c) S holds 2 H, so valency 2.

Answer: P = 3, Si = 4, S = 2.

Radicals and their valencies

A radical is an atom or a group of atoms that carries a charge and behaves as a single unit in reactions. Basic radicals are positive (mostly metal ions and NH₄⁺); acid radicals are negative (from acids). The charge is the valency.

Valency	Basic radicals	Acid radicals
1	Na⁺, K⁺, Ag⁺, Cu⁺, NH₄⁺ (ammonium)	Cl⁻, Br⁻, I⁻, OH⁻ (hydroxide), NO₃⁻ (nitrate), NO₂⁻ (nitrite), HCO₃⁻ (bicarbonate), HSO₄⁻ (bisulphate), MnO₄⁻ (permanganate), AlO₂⁻ (aluminate)
2	Mg²⁺, Ca²⁺, Zn²⁺, Fe²⁺, Cu²⁺, Pb²⁺, Ba²⁺	O²⁻, S²⁻, SO₄²⁻ (sulphate), SO₃²⁻ (sulphite), CO₃²⁻ (carbonate), Cr₂O₇²⁻ (dichromate), ZnO₂²⁻ (zincate), PbO₂²⁻ (plumbite)
3	Al³⁺, Fe³⁺	N³⁻ (nitride), PO₄³⁻ (phosphate)

Reactivity of radicals: radicals do not exist on their own in a compound's formula; they always pair up so that the total positive and negative valencies are equal. A group radical such as SO₄²⁻ stays together through most reactions (barium chloride with sodium sulphate moves the whole SO₄ group).

Worked example

Name the radicals and give the valency of each in (a) NaHCO₃, (b) (NH₄)₂SO₄, (c) KMnO₄.

Try it first, then show the solutionSolution

(a) Na⁺ (1) and HCO₃⁻ bicarbonate (1).
(b) NH₄⁺ ammonium (1) and SO₄²⁻ sulphate (2); two ammonium balance one sulphate.
(c) K⁺ (1) and MnO₄⁻ permanganate (1).

Answer: As above.

Writing formulae by exchanging valencies

Steps (the criss-cross method):
1. Write the basic radical first, then the acid radical.
2. Write each valency above its radical.
3. Exchange the valencies and write them as subscripts.
4. Simplify to the lowest ratio, and put a group radical in brackets if it is taken more than once. Do not write the subscript 1.

Examples:

Aluminium oxide: Al 3, O 2 → Al₂O₃.
Calcium hydroxide: Ca 2, OH 1 → Ca(OH)₂.
Magnesium oxide: Mg 2, O 2 → Mg₂O₂ → simplify → MgO.
Ammonium phosphate: NH₄ 1, PO₄ 3 → (NH₄)₃PO₄.
Iron(III) sulphate: Fe 3, SO₄ 2 → Fe₂(SO₄)₃.

Names: a binary compound ends in -ide (sodium chloride, calcium oxide); a compound of an oxygen-containing radical takes the radical's name (sodium sulphate, potassium nitrate).

Worked example

Write the formulae of (a) calcium nitrate, (b) aluminium sulphate, (c) sodium zincate, (d) copper(II) carbonate.

Try it first, then show the solutionSolution

(a) Ca 2, NO₃ 1 → Ca(NO₃)₂.
(b) Al 3, SO₄ 2 → Al₂(SO₄)₃.
(c) Na 1, ZnO₂ 2 → Na₂ZnO₂.
(d) Cu 2, CO₃ 2 → Cu₂(CO₃)₂ → simplify → CuCO₃.

Answer: Ca(NO₃)₂, Al₂(SO₄)₃, Na₂ZnO₂, CuCO₃.

Radicals in the atmosphere (free radicals)

In the air there is a second kind of 'radical': a free radical, an atom or group with an unpaired electron, such as the hydroxyl radical (·OH), a chlorine atom (Cl·) or nitric oxide (NO). It forms when a covalent bond breaks evenly, for example when ultraviolet light splits a molecule high in the atmosphere, or in hot flames and vehicle engines.

Why they matter: a free radical is very reactive because it 'wants' a partner for its odd electron, so it attacks whatever molecule it meets.

Effects on living organisms:

They damage cell membranes, proteins and DNA, which is linked to ageing, lung irritation and some cancers.
Chlorine radicals from CFCs break down the ozone layer, letting more harmful UV reach the Earth.
Radicals from exhaust gases help form photochemical smog, which irritates the eyes and lungs and damages crops.

Reduction or elimination:

Cut emissions: well-tuned engines and catalytic converters, cleaner fuels, less burning of waste and crop stubble.
Phase out CFCs and other ozone-depleting chemicals.
Avoid tobacco smoke.
In the body, antioxidants (vitamins C and E in fresh fruit and vegetables) mop up radicals.

Worked example

Why is a chlorine atom in the upper atmosphere far more harmful than a chloride ion in sea water?

Try it first, then show the solutionSolution

A chlorine atom has an unpaired electron: it is a free radical and very reactive.
It attacks ozone molecules again and again.
A chloride ion has a complete octet and is stable.

Answer: The atom is a reactive free radical that destroys ozone; the ion is stable.

Chemical equations and balancing by hit and trial

A chemical equation shows a reaction with symbols and formulae, reactants on the left and products on the right. A balanced equation has the same number of atoms of each element on both sides (law of conservation of mass).

Hit and trial method:
1. Write the correct formulae (never change a subscript to balance).
2. Start with the element in the largest formula; leave H and O (and free elements) to the end.
3. Put coefficients in front; recount; repeat until every element balances.
4. Reduce the coefficients to the lowest whole numbers.

The four syllabus patterns:

One reactant, two or three products: $2 K C l O_{3} \to 2 K C l + 3 O_{2}$ and $2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}$
Two reactants, one product: $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$
Two reactants, two products: $3 F e + 4 H_{2} O ⇌ F e_{3} O_{4} + 4 H_{2}$
Two reactants, three or four products: $C a C O_{3} + 2 H C l \to C a C l_{2} + H_{2} O + C O_{2}$ and $2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$

Worked example

Balance by hit and trial: $C u + H N O_{3} \to C u (N O_{3})_{2} + N O_{2} + H_{2} O$ (concentrated acid).

Try it first, then show the solutionSolution

Cu is already 1 = 1.
Try 4HNO₃: H = 4 on the left, so 2H₂O on the right.
N = 4 on the left; Cu(NO₃)₂ takes 2, so 2NO₂ for the other 2.
O check: left 12; right 6 + 4 + 2 = 12. Balanced.

Answer: $C u + 4 H N O_{3} \to C u (N O_{3})_{2} + 2 N O_{2} + 2 H_{2} O$

Relative atomic mass, relative molecular mass and percentage composition

Relative atomic mass (atomic weight) of an element is the number of times one atom of it is heavier than 1/12 of the mass of one carbon-12 atom (the older standard was one hydrogen atom). It has no unit: O = 16, Na = 23, Cl = 35.5.

Relative molecular mass is the number of times one molecule is heavier than 1/12 of a carbon-12 atom. Find it by adding the relative atomic masses of all the atoms in the formula.

$M_{r} (H_{2} S O_{4}) = 2 (1) + 32 + 4 (16) = 98$

Percentage composition = $\frac{mass of the element in one formula}{M _{r}} \times 100$

For a hydrate, the water of crystallisation counts too: $C u S O_{4} \cdot 5 H_{2} O$ has $M_{r} = 63.5 + 32 + 64 + 5 (18) = 249.5$ .

Worked example

Calculate the percentage of nitrogen in ammonium sulphate, (NH₄)₂SO₄. (N = 14, H = 1, S = 32, O = 16)

Try it first, then show the solutionSolution

$M_{r} = 2 (14 + 4) + 32 + 4 (16) = 36 + 32 + 64 = 132$
Mass of N = 2 × 14 = 28.
% N = $\frac{28}{132} \times 100 = 21.21%$

Answer: 21.21%

Chemistry in real life: The Language of Chemistry

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.

Percentage of nitrogen in fertilisers

Suppose a farmer can buy either urea, CO(NH₂)₂, or ammonium nitrate, NH₄NO₃, at the same price per kilogram and wants the most nitrogen for the money.

Calculate the percentage of nitrogen in each and advise the farmer. (C = 12, O = 16, N = 14, H = 1)

Try it first, then show the solutionSolution

Urea: $M_{r} = 12 + 16 + 2 (14 + 2) = 60$ ; % N = $\frac{28}{60} \times 100 = 46.67%$ .
Ammonium nitrate: $M_{r} = 14 + 4 + 14 + 48 = 80$ ; % N = $\frac{28}{80} \times 100 = 35%$ .
Urea gives more nitrogen per kilogram.

Answer: Urea 46.67%, ammonium nitrate 35%: buy urea.

Memory hook: Same price, compare the percentage of the useful element.

Water in washing soda

Suppose a shopkeeper sells washing soda crystals, Na₂CO₃·10H₂O, by weight, and a customer wonders how much of what she pays for is just water.

Find the percentage of water of crystallisation in washing soda. (Na = 23, C = 12, O = 16, H = 1)

Try it first, then show the solutionSolution

$M_{r} = 2 (23) + 12 + 3 (16) + 10 (18) = 46 + 12 + 48 + 180 = 286$
% water = $\frac{180}{286} \times 100 = 62.94%$

Answer: About 62.9% of the crystals' mass is water.

Memory hook: Ten waters weigh 180: most of a washing soda crystal is water.

Free radicals and health

Suppose a family lives beside a busy highway and a doctor advises them to eat more fresh fruit and vegetables and to keep windows shut at rush hour.

Explain the chemistry behind both pieces of advice.

Try it first, then show the solutionSolution

Vehicle exhaust and smog contain free radicals, which have unpaired electrons and are very reactive.
They damage cells; shutting windows at peak traffic cuts the dose breathed in.
Fruit and vegetables supply antioxidants such as vitamins C and E, which react with free radicals and make them harmless.

Answer: Less exposure to radicals, and antioxidants to neutralise those that get in.

Memory hook: Odd electron, odd behaviour: avoid them and neutralise them.

Topics in this chapter

The Language of Chemistry formula sheet

Relative molecular mass: $M_{r} = \sum (atomic mass \times number of atoms)$; No unit.
Percentage of an element: $% = \frac{mass of element in formula}{M _{r}} \times 100$; Count every atom of the element.
Percentage of water in a hydrate: $% H_{2} O = \frac{n \times 18}{M _{r} of hydrate} \times 100$; n = molecules of water of crystallisation.
Decomposition of potassium chlorate: $2 K C l O_{3} \to 2 K C l + 3 O_{2}$; One reactant, two products.
Decomposition of lead nitrate: $2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}$; One reactant, three products.
Potassium permanganate with HCl: $2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$; Two reactants, four products.

The Language of Chemistry: diagrams to practise

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

Criss-cross layout for a formula

Show the steps for writing the formula of a compound, e.g. aluminium sulphate.

Draw

Write the symbols side by side, basic radical first: Al and SO₄.
Write the valencies above: 3 over Al, 2 over SO₄.
Draw two crossing arrows from each valency to the foot of the other symbol.
Write the result with brackets: Al₂(SO₄)₃.

Label

basic radical
acid radical
valency
exchange
final formula

Watch out: Always show the simplify step if the numbers share a factor.

How the school exam asks The Language of Chemistry

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

Question type	How it is asked	Method
Valency and radicals	Give the valency of an element or radical; name the radicals in a compound; fill a table of radicals.	Learn the radical table by valency; charge = valency.
Writing formulae	Write formulae of 5-10 named compounds, often with group radicals and variable valency.	Criss-cross, brackets for groups, simplify.
Balancing equations	Balance 3-5 skeletal equations; classify by number of reactants and products.	Largest formula first, H and O last, recount every element.
Molecular mass and percentage composition	Find Mr and % of an element or of water of crystallisation; compare two fertilisers.	Write Mr working in a line; multiply atomic mass by atom count.
Free radicals in the atmosphere	Short answers: what a free radical is, one source, one effect, one way to reduce it.	Unpaired electron → very reactive → cell or ozone damage.

Tricks to remember The Language of Chemistry

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

Latin names: 'No King Feeds Cute Silver Plum Hogs'
Na (natrium), K (kalium), Fe (ferrum), Cu (cuprum), Ag (argentum), Pb (plumbum), Hg (hydrargyrum).
Use it when: Symbols from Latin names.
-ous is lower, -ic is higher
'-ous' sounds like 'less': ferrous Fe²⁺, cuprous Cu⁺. '-ic' is the higher: ferric Fe³⁺, cupric Cu²⁺.
Use it when: Variable valency names.
Criss-cross, bracket, cut down
Exchange the valencies, bracket any group taken twice, divide by the common factor.
Use it when: Writing formulae.
HO last
When balancing, leave Hydrogen and Oxygen to the end; free elements (O₂, Cl₂) are easiest to fix last.
Use it when: Hit and trial balancing.

The Language of Chemistry: exam-level solved questions

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

Example 1: Valency of a radical · 1 mark · MCQExam level

The valency of the dichromate radical is:
(a) 1
(b) 2
(c) 3
(d) 7

Try it first, then show the solutionSolution

Dichromate is Cr₂O₇²⁻; its charge is 2-.

Answer: (b)

Example 2: Formulae from valency · 3 marks · Short answerExam level

Element X has valency 3 and forms an oxide. Write the formula of (a) its oxide, (b) its sulphate, (c) its nitrate.

Try it first, then show the solutionSolution

(a) X 3, O 2 → X₂O₃.
(b) X 3, SO₄ 2 → X₂(SO₄)₃.
(c) X 3, NO₃ 1 → X(NO₃)₃.

Answer: X₂O₃, X₂(SO₄)₃, X(NO₃)₃.

Example 3: Variable valency · 4 marks · Short answerHarder exam level

Write the formulae of (a) ferrous sulphate, (b) ferric chloride, (c) cuprous oxide, (d) cupric hydroxide, and give the valency of the metal in each.

Try it first, then show the solutionSolution

(a) Fe²⁺ + SO₄²⁻ → FeSO₄ (Fe = 2).
(b) Fe³⁺ + Cl⁻ → FeCl₃ (Fe = 3).
(c) Cu⁺ + O²⁻ → Cu₂O (Cu = 1).
(d) Cu²⁺ + OH⁻ → Cu(OH)₂ (Cu = 2).

Answer: FeSO₄, FeCl₃, Cu₂O, Cu(OH)₂.

Example 4: Balancing: one reactant, three products · 2 marks · Short answerHarder exam level

Balance $P b (N O_{3})_{2} \to P b O + N O_{2} + O_{2}$ and state the reactant-product pattern.

Try it first, then show the solutionSolution

Pb and N: try 2Pb(NO₃)₂ → 2PbO + 4NO₂.
O: left 12; right 2 + 8 = 10, so add O₂ (2 more O).
$2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}$ ; one reactant, three products.

Answer: $2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}$

Example 5: Balancing: two reactants, four products · 3 marks · Short answerHarder exam level

Balance $K M n O_{4} + H C l \to K C l + M n C l_{2} + H_{2} O + C l_{2}$ .

Try it first, then show the solutionSolution

Start with 2KMnO₄ (8 O): gives 2KCl, 2MnCl₂ and 8H₂O.
8H₂O needs 16 H, so 16HCl.
Cl: left 16; right 2 + 4 = 6 in salts, so 10 Cl left over = 5Cl₂.
Check: K 2, Mn 2, O 8, H 16, Cl 16 on both sides.

Answer: $2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$

Example 6: Percentage of water of crystallisation · 3 marks · Short answerExam level

Calculate the percentage of water of crystallisation in blue vitriol, CuSO₄·5H₂O. (Cu = 63.5, S = 32, O = 16, H = 1)

Try it first, then show the solutionSolution

$M_{r} = 63.5 + 32 + 4 (16) + 5 (18) = 249.5$
Mass of water = 5 × 18 = 90.
% water = $\frac{90}{249.5} \times 100 = 36.07%$

Answer: 36.07%

Example 7: Comparing compounds · 3 marks · Short answerHOTS / competency

Which contains the greater percentage of oxygen: calcium carbonate, CaCO₃, or iron(III) oxide, Fe₂O₃? Show working. (Ca = 40, C = 12, O = 16, Fe = 56)

Try it first, then show the solutionSolution

CaCO₃: $M_{r} = 100$ ; % O = 48/100 × 100 = 48%.
Fe₂O₃: $M_{r} = 112 + 48 = 160$ ; % O = 48/160 × 100 = 30%.
Both hold 48 parts of oxygen, but in a smaller total for CaCO₃.

Answer: Calcium carbonate (48% against 30%).

Example 8: Identify the radical · 3 marks · Short answerHOTS / competency

A compound of sodium has the formula Na₃Y, and the same radical Y forms an aluminium salt. (a) What is the valency of Y? (b) Write the formula of the aluminium salt. (c) If Y contains one phosphorus and four oxygen atoms, name Y.

Try it first, then show the solutionSolution

(a) Three Na⁺ (1 each) balance one Y, so Y has valency 3.
(b) Al 3, Y 3 → Al₃Y₃ → AlY.
(c) PO₄³⁻, phosphate (aluminium phosphate, AlPO₄).

Answer: Valency 3; AlY; phosphate.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 5-mark question split into 1 + 2 + 2 marks.

Fertiliser label check

Suppose a gardening shop sells a bag labelled 'Ammonium sulphate, 21% nitrogen' and another labelled 'Calcium nitrate, 17% nitrogen'. A Class 9 student wants to test the labels with calculation. (N = 14, H = 1, S = 32, O = 16, Ca = 40)

(i) · 1 mark
Write the formulae of both compounds.
Answer
Answer: (NH₄)₂SO₄ and Ca(NO₃)₂.
(ii) · 2 marks
Calculate the relative molecular mass of each.
Answer
Answer: (NH₄)₂SO₄ = 132; Ca(NO₃)₂ = 40 + 2(14 + 48) = 164.
(iii) · 2 marks
Calculate the percentage of nitrogen in each and say whether the labels are fair.
Answer
Answer: 28/132 × 100 = 21.21%; 28/164 × 100 = 17.07%. Both labels match the pure compounds.

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 (a) ammonium carbonate, (b) potassium dichromate. (2 marks) (2 marks)

Step	Marks
(NH₄)₂CO₃ (brackets around NH₄)	1
K₂Cr₂O₇	1

Balance: $F e + H_{2} O \to F e_{3} O_{4} + H_{2}$ and state why the equation must be balanced. (3 marks) (3 marks)

Step	Marks
3Fe on the left to match Fe₃O₄	1
4H₂O for 4 O, then 4H₂: $3 F e + 4 H_{2} O ⇌ F e_{3} O_{4} + 4 H_{2}$	1
Mass is conserved: atoms are neither created nor destroyed	1

Calculate the percentage of each element in calcium phosphate, Ca₃(PO₄)₂. (Ca = 40, P = 31, O = 16) (4 marks) (4 marks)

Step	Marks
$M_{r} = 3 (40) + 2 (31) + 8 (16) = 310$	1
% Ca = 120/310 × 100 = 38.71%	1
% P = 62/310 × 100 = 20%	1
% O = 128/310 × 100 = 41.29% (total 100%)	1

Where students lose marks in The Language of Chemistry

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

Mistake: Right valency, wrong formula
CISCE examiners' comments on the ICSE Class 10 paper, 2024: asked to identify an anion (carbonate) from a test, some candidates gave the correct valency but the wrong formula. The same slip begins in Class 9.
Wrong working
Sulphate: SO₃²⁻; nitrate: NO₂⁻.
Right working
Sulphate SO₄²⁻ (sulphite is SO₃²⁻); nitrate NO₃⁻ (nitrite is NO₂⁻).
Do this: -ate has one more oxygen than -ite.
Usually costs: 1 mark each.
Mistake: Missing brackets on group radicals
Without brackets the subscript applies only to the last atom.
Wrong working
Calcium hydroxide: CaOH₂; ammonium sulphate: NH₄₂SO₄.
Right working
Ca(OH)₂; (NH₄)₂SO₄.
Do this: Bracket any group radical taken more than once.
Usually costs: 1 mark.
Mistake: Not simplifying
Criss-cross applied blindly to equal valencies.
Wrong working
Calcium oxide: Ca₂O₂.
Right working
CaO.
Do this: Divide by the common factor at the end.
Usually costs: 1 mark.
Mistake: Balancing by changing subscripts
Changing a subscript changes the substance.
Wrong working
H₂ + O₂ → H₂O₂ (now hydrogen peroxide, not water).
Right working
$2 H_{2} + O_{2} \to 2 H_{2} O$
Do this: Only coefficients may change.
Usually costs: 1 mark.
Mistake: Forgetting the atom count in percentage
Students use one atomic mass when the formula has two or more atoms of the element.
Wrong working
% N in (NH₄)₂SO₄ = 14/132 × 100 = 10.6%.
Right working
28/132 × 100 = 21.21%.
Do this: Mass of element = atomic mass × number of atoms in the formula.
Usually costs: 1-2 marks.

Practice questions on The Language of Chemistry

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
Which pair has the same valency?
(a) Na and Mg
(b) Al and N (in NH₃)
(c) C and O
(d) K and Ca
Hint
Count H atoms combined or electrons lost.
Answer
(b) Both 3.
Q2 · 1 mark · MCQExam level
The formula of the plumbite radical is:
(a) PbO₃²⁻
(b) PbO₂⁻
(c) PbO₂²⁻
(d) Pb²⁺
Hint
Like zincate.
Answer
(c)
Q3 · 2 marks · Short answerExam level
Write the formulae: (a) magnesium nitride, (b) sodium aluminate, (c) calcium bicarbonate, (d) silver nitrate.
Hint
N³⁻; AlO₂⁻; HCO₃⁻.
Answer
(a) Mg₃N₂ (b) NaAlO₂ (c) Ca(HCO₃)₂ (d) AgNO₃
Q4 · 3 marks · Short answerHarder exam level
Balance: (a) $A l + O_{2} \to A l_{2} O_{3}$ (b) $M n O_{2} + H C l \to M n C l_{2} + H_{2} O + C l_{2}$ (c) $C + H_{2} S O_{4} \to C O_{2} + S O_{2} + H_{2} O$
Hint
H and O last.
Answer
(a) $4 A l + 3 O_{2} \to 2 A l_{2} O_{3}$ (b) $M n O_{2} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}$ (c) $C + 2 H_{2} S O_{4} \to C O_{2} + 2 S O_{2} + 2 H_{2} O$
Q5 · 2 marks · Short answerExam level
Calculate the relative molecular mass of (a) sugar, C₁₂H₂₂O₁₁, (b) potassium dichromate, K₂Cr₂O₇. (C = 12, H = 1, O = 16, K = 39, Cr = 52)
Hint
Add atomic masses.
Answer
(a) 144 + 22 + 176 = 342 (b) 78 + 104 + 112 = 294
Q6 · 2 marks · Short answerHarder exam level
Calculate the percentage of water of crystallisation in Epsom salt, MgSO₄·7H₂O. (Mg = 24, S = 32, O = 16, H = 1)
Hint
Mr of hydrate first.
Answer
Mr = 24 + 32 + 64 + 126 = 246; % water = 126/246 × 100 = 51.22%.
Q7 · 3 marks · Short answerHarder exam level
Element M forms an oxide MO and a chloride. (a) What is the valency of M? (b) Write the formula of its chloride and its phosphate. (c) Is M more likely a metal or a non-metal? Give a reason.
Hint
O is 2.
Answer
(a) 2 (b) MCl₂; M₃(PO₄)₂ (c) A metal: it forms a basic radical M²⁺ by losing two electrons.
Q8 · 3 marks · Short answerHarder exam level
Explain, with one example each, what is meant by (a) a monovalent acid radical, (b) a divalent basic radical, (c) a trivalent acid radical.
Hint
Sign and size of the charge.
Answer
(a) Negative group or ion of charge 1, e.g. NO₃⁻. (b) Positive ion of charge 2, e.g. Ca²⁺. (c) Negative ion of charge 3, e.g. PO₄³⁻.
Q9 · 4 marks · Long answerHarder exam level
Name one source of free radicals in city air, one harmful effect on people and one way of reducing them. Why does a free radical react so readily?
Hint
Unpaired electron.
Answer
Source: vehicle exhaust (or smoke, UV splitting molecules). Effect: damage to cells and DNA, lung irritation (or ozone loss). Reduction: catalytic converters and cleaner fuels (or less burning, antioxidant-rich diet). It has an unpaired electron, so it readily attacks other molecules to pair it.
Q10 · 3 marks · Short answerHOTS / competency
A compound contains 40% calcium, 12% carbon and 48% oxygen. Show, using Mr, that it is calcium carbonate and not calcium bicarbonate. (Ca = 40, C = 12, O = 16, H = 1)
Hint
Find % Ca in both.
Answer
CaCO₃: Mr 100, Ca 40%, C 12%, O 48%: matches. Ca(HCO₃)₂: Mr 162, Ca = 40/162 × 100 = 24.69%: does not match.
Q11 · 3 marks · Short answerHOTS / competency
Two oxides of copper are CuO and Cu₂O. Calculate the percentage of copper in each and say which oxide gives more copper per kilogram. (Cu = 63.5, O = 16)
Hint
Mr 79.5 and 143.
Answer
CuO: 63.5/79.5 × 100 = 79.87%. Cu₂O: 127/143 × 100 = 88.81%. Cu₂O gives more copper.

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.Symbol of potassium and its Latin name?Show answer
K, kalium.
2.Valency of the ammonium radical?Show answer
1 (NH₄⁺).
3.Formula of the bicarbonate radical?Show answer
HCO₃⁻.
4.Formula of aluminium oxide?Show answer
Al₂O₃.
5.Standard for relative atomic mass today?Show answer
1/12 of the mass of a carbon-12 atom.
6.Mr of sulphuric acid?Show answer
98.
7.What is a free radical?Show answer
An atom or group with an unpaired electron; very reactive.
8.Ferric is which ion?Show answer
Fe³⁺ (iron(III)).

Best order to study it

Learn the symbols, including the Latin ones.
Learn the radical table by valency (1, 2, 3).
Practise 20 formulae with criss-cross, brackets and simplifying.
Practise balancing in the four reactant-product patterns.
Learn the definitions of relative atomic and molecular mass and do percentage composition sums.
Read about free radicals: source, effect, reduction.

Quick revision checklist

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

Can you give the valency of ten radicals without looking?
Can you write formulae with group radicals correctly bracketed?
Can you balance $K M n O_{4} + H C l$ and $P b (N O_{3})_{2}$ decomposition?
Can you define relative atomic mass on the carbon-12 scale?
Can you find the percentage of water in a hydrate?
Can you explain how free radicals harm living things and how to reduce them?

The Language of Chemistry: frequently asked questions

Is a radical the same as a free radical?

No. In formula writing, a radical is a charged atom or group (SO₄²⁻) that stays together in a compound. A free radical is a neutral (usually) atom or group with an unpaired electron, such as Cl· or ·OH; it is short-lived and very reactive. The 2028 syllabus asks for both.

Why do relative masses have no unit?

They are ratios: how many times heavier an atom or molecule is than 1/12 of a carbon-12 atom. A ratio of two masses has no unit.

Why was carbon-12 chosen instead of hydrogen?

Carbon-12 gives whole-number values for most elements and is easy to measure accurately; on the hydrogen scale many values were awkward. Both are named in the syllabus.

Do I need to balance equations with the ion-electron method?

No. The Class IX syllabus asks only for balancing by the hit and trial method.

Based on: CISCE ICSE 2028 Science (52) Chemistry (Science Paper 2) syllabus, Classes IX and X; CISCE Analysis of Pupil Performance, ICSE (Class 10) Chemistry (Science Paper 2), 2024