ICSE Class 9 Chemistry Solutions — Chapter 1
These ICSE Class 9 Chemistry solutions explain Chapter 1, The Language of Chemistry, section by section. This chapter builds the vocabulary you will use all year: symbols, valency, radicals, chemical formulae, balancing equations, and molecular mass. By the end you can write a formula from two radicals, balance a skeleton equation, and find percentage composition. Worked patterns here use fresh numbers so you can practise each method.
Download The Language of Chemistry Class 9 ICSE Chemistry Chapter 1 PDF
Download the The Language of Chemistry Class 9 ICSE Chemistry chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 19 pages, 18 sections, 19 questions.
What is inside Chapter 1
19 pages · 18 sections · 14 worked examples · 19 exercise questions
| Section | Title |
|---|---|
| 1.1 | INTRODUCTION |
| 1.2 | CHEMICAL SYMBOLS |
| 1.3 | FORMULA |
| 1.4 | VALENCY |
| 1.5 | RADICALS |
| 1.6 | WRITING CHEMICAL FORMULAE |
| 1.7 | NAMING CERTAIN COMPOUNDS |
| 1.8 | TO CALCULATE THE VALENCY FROM THE FORMULA |
| 1.9 | CHEMICAL EQUATION |
| 1.9.1 | Balanced equation |
| 1.9.2 | How to balance a chemical equation |
| 1.9.3 | Information conveyed by a balanced chemical equation |
| 1.9.4 | Limitations of a chemical equation |
| 1.9.5 | A chemical equation can be made more informative by writing the following additional information to it. |
| 1.10 | RELATIVE ATOMIC MASS (ATOMIC WEIGHT) |
| 1.11 | RELATIVE MOLECULAR MASS (MOLECULAR WEIGHT) |
| 1.12 | PERCENTAGE COMPOSITION |
| 1.13 | EMPIRICAL FORMULA OF A COMPOUND |
- EXERCISE 1(C): 17 questions
- ANSWERS: 2 questions
1.1 INTRODUCTION
The chapter opens by defining the basic units of matter. An element is a pure substance that cannot be broken into simpler substances. Atoms are the smallest particles of an element and carry all its properties. Molecules are sets of identical atoms bound together.
Molecules are classified by atom count. Monoatomic molecules have one atom, such as helium and neon. Diatomic molecules have two, like hydrogen \mathrm{H_2}, oxygen \mathrm{O_2}, and nitrogen \mathrm{N_2}. Phosphorus forms tetratomic molecules \mathrm{P_4}, and sulphur forms octatomic molecules \mathrm{S_8}.
Symbols have a long history. Alchemists drew pictograms, such as a crescent for silver. Dalton used circles with a dot for hydrogen. Berzelius proposed using the element’s initial letter in capitals, and that system became the basis of the IUPAC system. Some symbols come from Latin names: sodium is Na (natrium), potassium is K (kalium), iron is Fe (ferrum), and gold is Au (aurum).
One mnemonic groups the Latin symbols into trios — Na–K–Fe for common metals, Cu–Ag–Au for precious metals, and Hg–Pb–Sn for heavy metals — which makes the list easier to recall. The CISCE syllabus for this chapter follows the IUPAC standard, and you can verify it on the official CISCE website.
1.2 CHEMICAL SYMBOLS
A symbol is the short form that stands for the atom of a specific element. Each element gets a symbol, usually the first letter of its English or Latin name, written as a capital. Sulphur is S, hydrogen is H, and carbon is C.
When several elements share a first letter, a second small letter is added. Carbon is C, but cobalt is Co, copper is Cu, calcium is Ca, cadmium is Cd, chromium is Cr, and chlorine is Cl. The capital letter always comes first, and the second letter is always small.
One trap matters here. Co means the element cobalt, while CO means the compound carbon monoxide. The capital and small letters change the meaning completely.
A symbol carries three pieces of information. It represents the element’s name, one atom of that element, and a quantity equal to its atomic mass. For nitrogen, the symbol N means the element nitrogen, one nitrogen atom, and 14 parts by weight of nitrogen.
1.3 FORMULA
A formula is the symbolic representation of a molecule, written with the symbols of its constituent atoms. It is also called the molecular formula or chemical formula. Element molecules may contain one or more atoms. Hydrogen, oxygen, nitrogen, chlorine, bromine, and iodine form diatomic molecules: \mathrm{H_2}, \mathrm{O_2}, \mathrm{N_2}, \mathrm{Cl_2}, \mathrm{Br_2}, \mathrm{I_2}.
Subscripts show the number of each atom in one molecule. In \mathrm{H_2SO_4}, there are two hydrogen atoms, one sulphur atom, and four oxygen atoms. The coefficient in front, as in 2\mathrm{H_2O}, counts whole molecules, not atoms.
A molecular formula carries quantitative meaning. It shows the molecule and its molecular mass, the number of each atom present, and the mass ratio of the elements. Consider methane, \mathrm{CH_4}. It has one carbon atom (mass 12) and four hydrogen atoms (mass 1 each), giving a molecular mass of 12 + 4 = 16.
1.4 VALENCY
Valency is the combining capacity of an atom or a radical. It is measured against hydrogen, whose valency is taken as one. The modern definition is the number of electrons an atom loses, gains, or shares during a reaction.
Metals with one, two, or three valence electrons lose them to form cations. Sodium loses one electron to become \mathrm{Na^+} (univalent). Magnesium loses two to become \mathrm{Mg^{2+}} (divalent). Aluminium loses three to become \mathrm{Al^{3+}} (trivalent). Non-metals with five, six, or seven valence electrons gain electrons to form anions. Chlorine gains one to give \mathrm{Cl^-}, oxygen gains two to give \mathrm{O^{2-}}, and nitrogen gains three to give \mathrm{N^{3-}}.
Follow the same pattern with fresh elements. Potassium (\mathrm{K}, 2,8,8,1) loses one electron to form \mathrm{K^+}, so it is univalent. Calcium (\mathrm{Ca}, 2,8,8,2) loses two to form \mathrm{Ca^{2+}}, so it is divalent.
Some elements show variable valency, losing electrons from the penultimate shell too. The lower valency takes the suffix ous, and the higher takes ic. Ferrous [katex display="true"]\mathrm{Fe(II)}[/katex] and ferric [katex display="true"]\mathrm{Fe(III)}[/katex], cuprous [katex display="true"]\mathrm{Cu(I)}[/katex] and cupric [katex display="true"]\mathrm{Cu(II)}[/katex] are the common pairs. Modern names use Roman numerals in brackets.
1.5 RADICALS
A radical is an atom or a group of atoms that behaves as a single charged unit. A molecule of a compound splits into two parts called radicals. Potassium chloride gives the potassium radical and the chloride radical. A radical is simple when it is a single atom, like \mathrm{Na^+} or \mathrm{Mg^{2+}}. It is a compound radical when it is a polyatomic group, like sulphate (\mathrm{SO_4^{2-}}).
Radicals are named by their source. A base contributes the basic radical, which carries a positive charge and is called a cation. An acid contributes the acid radical, which carries a negative charge and is called an anion. Common basic radicals include \mathrm{NH_4^+}, \mathrm{Ca^{2+}}, and \mathrm{Al^{3+}}. Common acid radicals include \mathrm{Cl^-}, \mathrm{CO_3^{2-}}, and \mathrm{PO_4^{3-}}.
Radicals combine according to their valency. In calcium nitrate \mathrm{Ca(NO_3)_2}, the basic radical \mathrm{Ca^{2+}} has valency 2, and the acid radical \mathrm{NO_3^-} has valency 1, so two nitrate units pair with one calcium unit.
1.6 WRITING CHEMICAL FORMULAE
The criss-cross method writes a formula in four steps. Write the symbols side by side, basic radical first. Write each valency above its symbol. Divide both valencies by their highest common factor. Interchange the valencies and place them as subscripts. If a polyatomic radical has valency above 1, enclose it in brackets.
Step 1: Write zinc and phosphate side by side: \mathrm{Zn} \; \mathrm{PO_4}.
Step 2: Write valencies above each: \mathrm{Zn^{2+}} \; \mathrm{PO_4^{3-}}.
Step 3: The valencies 2 and 3 have no common factor, so interchange them directly.
\mathrm{Zn^{2+}} \; \mathrm{PO_4^{3-}} \rightarrow \mathrm{Zn_3(PO_4)_2}
Final formula: \mathrm{Zn_3(PO_4)_2}, with the bracket around the phosphate radical.
Two common errors to avoid. Never drop the bracket when a polyatomic radical has valency above 1. Always divide by the HCF first when the valencies share a factor. Calcium oxide is \mathrm{CaO}, the simplified form, not \mathrm{Ca_2O_2}.
1.7 NAMING CERTAIN COMPOUNDS
Five rules cover most compound names. Rule one: for a metal and a non-metal, write the metal first and add the suffix ide to the non-metal. Calcium and nitrogen form calcium nitride, \mathrm{Ca_3N_2}. Rule two: for two non-metals, add prefixes tri, tetra, or penta. Phosphorus trichloride is \mathrm{PCl_3}, and phosphorus pentachloride is \mathrm{PCl_5}.
Rule three: for two elements and oxygen, the name depends on the oxygen count. Hypo and ite name fewer oxygen atoms, ate names the middle count, and per names more. Sodium hypochlorite is \mathrm{NaClO}, sodium chlorite is \mathrm{NaClO_2}, sodium chlorate is \mathrm{NaClO_3}, and sodium perchlorate is \mathrm{NaClO_4}.
Rule four covers acids. Binary acids take hydro and ic, so \mathrm{HCl} is hydrochloric acid. Oxy-acids take ic or ous from the second element, so \mathrm{H_2SO_4} is sulphuric acid and \mathrm{H_2SO_3} is sulphurous acid. Rule five: some names are trivial, like ammonia for \mathrm{NH_3} and water for \mathrm{H_2O}. Apply rule one to \mathrm{Mg_3N_2}: magnesium is the metal, and nitrogen takes the ide suffix, giving magnesium nitride.
1.8 TO CALCULATE THE VALENCY FROM THE FORMULA
You can find an element’s valency from its formula using the fixed valencies of hydrogen (1), oxygen (2), and chlorine (1). The steps are: write the formula, interchange the subscripts as superscripts, multiply both by 2 because oxygen’s valency is 2, and read the result as the valency.
Step 1: Write the formula: \mathrm{SO_3}.
Step 2: Interchange subscripts as superscripts: \mathrm{S^3O^1}.
Step 3: Multiply both by 2: \mathrm{S^6O^2}.
Answer: Sulphur has valency 6 in \mathrm{SO_3}.
Why does this work? Oxygen’s valency of 2 is the anchor. The subscript of oxygen tells how many oxygen atoms bind to one atom of the element, and the exchange converts that ratio into the element’s combining capacity.
1.9 CHEMICAL EQUATION
A chemical equation is the symbolic representation of a reaction using the symbols and formulae of the substances involved. A word equation names the substances in words, while a chemical equation uses formulae. Burning magnesium is written as a word equation — magnesium plus oxygen gives magnesium oxide — and as a chemical equation: 2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}.
Four steps go into writing an equation. Write the reactants on the left with plus signs between them. Write the products on the right with plus signs. Balance the equation so atom counts match. Represent all substances in their molecular forms.
A skeleton equation shows a chemical change but is not balanced. An example is \mathrm{KNO_3} \rightarrow \mathrm{KNO_2} + \mathrm{O_2}, where the oxygen atoms do not match on the two sides.
Reactions follow four patterns. One reactant gives two or more products, as in \mathrm{CaCO_3} \rightarrow \mathrm{CaO} + \mathrm{CO_2}. Two reactants give one product, as in \mathrm{NH_3} + \mathrm{HCl} \rightarrow \mathrm{NH_4Cl}. Two give two, as in \mathrm{AgNO_3} + \mathrm{NaCl} \rightarrow \mathrm{AgCl} + \mathrm{NaNO_3}. Two give three or more, as in \mathrm{Cu} + 2\mathrm{H_2SO_4} \rightarrow \mathrm{CuSO_4} + \mathrm{SO_2} + 2\mathrm{H_2O}.
1.9.1 Balanced equation
A balanced equation has the same number of atoms of each element on both sides. Ammonia formation, \mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3}, is already balanced. Two nitrogen atoms and six hydrogen atoms appear on each side.
An unbalanced equation violates the law of conservation of matter, which states that matter is neither created nor destroyed in a reaction. An unbalanced equation would imply that atoms appear or vanish, which cannot happen.
1.9.2 How to balance a chemical equation
The hit and trial method balances an equation by counting atoms and adjusting coefficients. Count how many times each element appears on each side. Balance the element with the least frequency first. When two elements have equal frequency, balance the metal first.
Balance the combustion of propane, \mathrm{C_3H_8} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O}.
| Element | Reactants | Products |
|---|---|---|
| C | 3 | 1 |
| H | 8 | 2 |
| O | 2 | 3 |
Carbon appears once on each side, so balance it first with 3 \mathrm{CO_2}. Hydrogen appears once on each side, so balance it with 4 \mathrm{H_2O}. Oxygen now has 2 on the left and 10 on the right, so write 5\mathrm{O_2}. The balanced equation is \mathrm{C_3H_8} + 5\mathrm{O_2} \rightarrow 3\mathrm{CO_2} + 4\mathrm{H_2O}.
The partial equation method is a second approach. It splits a complex reaction into steps, balances each step, and adds them. The book shows this with the liberation of iodine from potassium iodide using hydrogen peroxide. For more Class 9 ICSE Chemistry resources, see the Class 9 ICSE Chemistry hub.
1.9.3 Information conveyed by a balanced chemical equation
A balanced equation carries six pieces of information. It states the actual result of the change, the reactants and products, the number of molecules of each substance, the molecular composition of each species, the relative masses involved, and it proves the law of conservation of mass.
Consider 2\mathrm{Mg} + \mathrm{O_2} \rightarrow 2\mathrm{MgO}. Two molecules of magnesium react with one molecule of oxygen to give two molecules of magnesium oxide. The masses work out as 2(24) + 32 = 2(40), so 48 + 32 = 80. The reactant mass equals the product mass.
A useful gas rule follows. At STP (0°C and 760 mm pressure), 22.4 litres of any gas weighs its molar mass in grams. So 22.4 litres of hydrogen weighs 2 g, and 22.4 litres of ammonia weighs 17 g.
1.9.4 Limitations of a chemical equation
A chemical equation leaves out seven details. It does not state the physical state of the reactants or products, the time taken for the reaction, whether heat is given out or absorbed, the concentrations, the rate of reaction, whether the reaction is reversible, or whether it reaches completion.
The equation is a shorthand, not a full description. The missing details are added separately, as section 1.9.5 shows.
1.9.5 A chemical equation can be made more informative by writing the following additional information to it.
Four additions make an equation more informative. Conditions such as temperature, pressure, and catalyst go above or below the arrow. For example, sulphur dioxide reacts with oxygen over a vanadium pentoxide catalyst: 2\mathrm{SO_2} + \mathrm{O_2} \xrightarrow{\mathrm{V_2O_5}} 2\mathrm{SO_3}.
State symbols show the physical state: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous. Hydrogen burning in oxygen becomes 2\mathrm{H_2}(g) + \mathrm{O_2}(g) \rightarrow 2\mathrm{H_2O}(l).
A heat term on the product side shows heat evolution, while a negative heat term shows absorption. Methane burning gives \mathrm{CH_4}(g) + 2\mathrm{O_2}(g) \rightarrow \mathrm{CO_2}(g) + 2\mathrm{H_2O}(g) + \text{Heat}.
Concentration of acids can also be written. Magnesium with dilute sulphuric acid gives \mathrm{Mg}(s) + \mathrm{H_2SO_4}(dil) \rightarrow \mathrm{MgSO_4}(aq) + \mathrm{H_2}(g).
1.10 RELATIVE ATOMIC MASS (ATOMIC WEIGHT)
Atoms are too small to weigh directly in grams. A hydrogen atom has a mass of about 1.66 \times 10^{-24} g, which is inconvenient. Chemists instead compare atom masses to a standard. Relative atomic mass (RAM) is the number of times one atom of an element is heavier than \frac{1}{12} of the mass of a carbon-12 atom.
Hydrogen was the first standard, but carbon-12 replaced it in 1961 because the change least disturbed existing values. One atomic mass unit (amu) equals \frac{1}{12} the mass of a carbon-12 atom, or 1.6605 \times 10^{-24} g.
On this scale, hydrogen has RAM 1, oxygen has 16, and helium has 4. This ICSE Class 9 Chemistry chapter builds every later mass calculation on this standard.
1.11 RELATIVE MOLECULAR MASS (MOLECULAR WEIGHT)
Relative molecular mass (RMM) tells how many times one molecule is heavier than \frac{1}{12} of a carbon-12 atom. It is found by adding the atomic masses of all atoms in the molecule.
Calculate the RMM of calcium carbonate, \mathrm{CaCO_3}.
| Element | Atoms | Atomic mass | Total |
|---|---|---|---|
| Ca | 1 | 40 | 40 |
| C | 1 | 12 | 12 |
| O | 3 | 16 | 48 |
Adding the column gives 40 + 12 + 48 = 100. The RMM of calcium carbonate is 100 amu.
The two masses differ in scope.
| RAM | RMM | |
|---|---|---|
| Compares | one atom | one molecule |
| Standard | 1/12 of carbon-12 | 1/12 of carbon-12 |
| How found | from the periodic table | sum of atomic masses in the molecule |
| Example | H = 1, O = 16 | \mathrm{CaCO_3} = 100 |
1.12 PERCENTAGE COMPOSITION
Percentage composition is the percentage by weight of each element in a compound.
\text{Percentage of element} = \frac{\text{total mass of the element in one molecule}}{\text{gram molecular mass of the compound}} \times 100
Find the percentage of nitrogen in ammonium nitrate, \mathrm{NH_4NO_3}. The molecular mass is 14 + 4 + 14 + 48 = 80. Nitrogen contributes 28, so \frac{28}{80} \times 100 = 35\%.
Check the other elements. Hydrogen is \frac{4}{80} \times 100 = 5\%, and oxygen is \frac{48}{80} \times 100 = 60\%. The three percentages sum to 100, which confirms the calculation.
1.13 EMPIRICAL FORMULA OF A COMPOUND
The empirical formula gives the simplest whole-number ratio of atoms in a molecule. It is not always the actual molecular formula. Hydrogen peroxide, \mathrm{H_2O_2}, has the empirical formula HO, meaning a 1:1 ratio. Glucose, \mathrm{C_6H_{12}O_6}, has the empirical formula \mathrm{CH_2O}, a 1:2:1 ratio. Ethane, \mathrm{C_2H_6}, has the empirical formula \mathrm{CH_3}.
The empirical formula mass is the sum of the atomic masses in the empirical formula. For HO, that is 1 + 16 = 17. The molecular formula is a whole-number multiple of the empirical formula, and the multiple is found by dividing the molecular mass by the empirical formula mass.
What to practise from ICSE Class 9 Chemistry Solutions, Chapter 1
Work through the exercise blocks in this chapter in order. Each one drills a different skill.
- EXERCISE 1(A): about 13 questions on symbols, valency, and radicals.
- EXERCISE 1(B): 18 items (a to r) on balancing chemical equations.
- EXERCISE 1(C): 17 questions covering formulae, valency, balancing, molecular mass, and percentage composition.
- ANSWERS: 2 entries for the molecular mass and percentage questions.
The questions map to the exam in a clear way. Symbol questions test your memory of the element list. Valency questions test the electron-shell table and the criss-cross method. Balancing questions test the hit and trial method. Molecular mass and percentage questions test the addition and division skills from sections 1.11 and 1.12. For other subjects and chapters, browse the Class 9 ICSE books collection.
The Language of Chemistry Class 9 ICSE frequently asked questions
Why is the symbol for sodium Na and not S in ICSE Class 9 Chemistry?
Because S already stands for sulphur. When two elements share a first letter, the second element takes a two-letter symbol from its Latin name. Sodium’s Latin name is natrium, so its symbol is Na. The same applies to potassium (K, from kalium) and iron (Fe, from ferrum).
How do I use ICSE Class 9 Chemistry solutions to balance equations?
Use the hit and trial method from section 1.9.2. Count atoms on both sides, balance the least frequent element first, and balance the metal first when frequencies tie. The balanced combustion of propane, \mathrm{C_3H_8} + 5\mathrm{O_2} \rightarrow 3\mathrm{CO_2} + 4\mathrm{H_2O}, is a worked example.
What is the valency of an element and how do I find it from its formula?
Valency is the combining capacity of an atom or radical, measured by the electrons it loses, gains, or shares. To find it from a formula, interchange the subscripts as superscripts, multiply by 2 because oxygen has valency 2, and read the result. In \mathrm{SO_3}, sulphur has valency 6.
What is the difference between relative atomic mass and relative molecular mass?
RAM compares one atom to \frac{1}{12} of a carbon-12 atom. RMM compares one molecule to the same standard and is the sum of the RAMs of all atoms in the molecule. Hydrogen has RAM 1, while calcium carbonate has RMM 100.
How do I calculate the percentage composition of a compound?
Divide the total mass of the element in one molecule by the gram molecular mass, then multiply by 100. In ammonium nitrate, \mathrm{NH_4NO_3}, nitrogen gives \frac{28}{80} \times 100 = 35\%.
Reference: ICSE Class 9 Chemistry textbook material, chapter 1: The Language of Chemistry — CISCE-aligned.
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