Mole Concept and Stoichiometry

Q: How many moles and how many molecules are present in 9 g of water?

Moles = (9)/(18) = 0.5 mol; molecules = 0.5 × 6.022 × 1023 = 3.011 × 1023.

Q: Find the number of atoms in 3.4 g of ammonia.

Moles of NH3 = (3.4)/(17) = 0.2 mol; molecules = 0.2 × 6.022 × 1023 = 1.2044 × 1023; each has 4 atoms, so atoms = 4.8176 × 1023.

ICSE Class 10 · Chemistry · Chapter 5

What this chapter covers

Why chemists count particles by weighing
Gay-Lussac's law of combining volumes
Avogadro's law
Atomicity of hydrogen, oxygen, nitrogen and chlorine
The mole and Avogadro's number
Mole and atomic mass: gram atomic mass and gram atom
Mole and molecular mass: gram molecular mass and gram molecule
Molar volume of a gas at STP
Calculations relating mole to mass, volume and Avogadro's number
Vapour density and relative molecular mass
Empirical formula and molecular formula
Calculations based on chemical equations

Key definitions

Stoichiometry is the branch of chemistry that deals with the quantitative relationships between the masses, volumes and numbers of particles of the reactants and products in a chemical reaction.

Gay-Lussac's law of combining volumes is the law that when gases react together, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided all the volumes are measured at the same temperature and pressure.

Avogadro's law is the law that equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules.

Atomicity is the number of atoms present in one molecule of an element.

A mole is the amount of a substance that contains as many elementary particles (atoms, molecules or ions) as there are atoms in exactly 12 g of the carbon-12 isotope.

Avogadro's number is the number of particles present in one mole of any substance, equal to 6.022 × 10²³.

Gram atomic mass is the relative atomic mass of an element expressed in grams, and it is the mass of one mole of atoms of that element.

A gram atom is the quantity of an element whose mass in grams is numerically equal to its relative atomic mass, so one gram atom of any element is one mole of its atoms.

Gram molecular mass is the relative molecular mass of a substance expressed in grams, and it is the mass of one mole of molecules of that substance.

A gram molecule is the quantity of a substance whose mass in grams is numerically equal to its relative molecular mass, so one gram molecule of any substance is one mole of its molecules.

Molar mass is the mass of one mole of a substance, expressed in grams per mole (g mol^-1).

STP (standard temperature and pressure) is the set of conditions 0 °C (273 K) and 1 atmosphere (760 mm of mercury) used for comparing volumes of gases.

Key points

Stoichiometry deals with the quantitative relations between reactants and products in a reaction.
Gay-Lussac's law: reacting gases and gaseous products are in a simple whole-number ratio by volume, at the same temperature and pressure; the ratio equals the ratio of coefficients in the balanced equation.
Avogadro's law: equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Using the formation of HCl, NH₃ and NO with Avogadro's law shows that hydrogen, chlorine, nitrogen and oxygen are diatomic.
One mole of any substance contains Avogadro's number, 6.022 × 10²³, of particles.
Atomic mass refers to one atom; gram atomic mass is the mass of one mole of atoms. Molecular mass refers to one molecule; gram molecular mass is the mass of one mole of molecules.
One mole of any gas occupies 22.4 L at STP (0 °C, 760 mm Hg), so the mass of 22.4 L of any gas at STP equals its gram molecular mass.
Moles = mass ÷ gram molecular mass = particles ÷ 6.022 × 10²³ = volume at STP in litres ÷ 22.4.
Vapour density compares the mass of a gas with that of an equal volume of hydrogen; relative molecular mass = 2 × vapour density.
The empirical formula gives the simplest whole-number atom ratio; the molecular formula is n times it, with n = molecular mass ÷ empirical formula mass.
In equation calculations, convert the given quantity to moles, apply the mole ratio from the balanced equation, then convert to the mass or volume required.

Questions answered

State Gay-Lussac's law of combining volumes.

When gases react together, they do so in volumes which bear a simple whole-number ratio to one another and to the volumes of the gaseous products, provided all volumes are measured at the same temperature and pressure.

State Avogadro's law.

Equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules.

Distinguish between atomic mass and gram atomic mass, taking oxygen as the example.

Atomic mass is a number dealing with one atom: one oxygen atom has a mass of 16 u. Gram atomic mass is the mass of one mole of atoms: 16 g of oxygen atoms contains 6.022 × 10²³ atoms.

Using Avogadro's law and the reaction H₂ + Cl₂ → 2HCl, explain why a hydrogen molecule must contain at least two atoms.

1 volume of hydrogen gives 2 volumes of hydrogen chloride, so 1 molecule of hydrogen gives 2 molecules of hydrogen chloride. Each of these contains at least one hydrogen atom, and atoms cannot be split, so the one hydrogen molecule must contain at least 2 atoms.

Why is the mass of 22.4 litres of any gas at STP equal to its gram molecular mass?

By Avogadro's law, 22.4 litres of any gas at STP contains one mole of molecules, 6.022 × 10²³ of them; the mass of one mole of a substance is its gram molecular mass.

How many moles and how many molecules are present in 9 g of water?

Moles = (9)/(18) = 0.5 mol; molecules = 0.5 × 6.022 × 10²³ = 3.011 × 10²³.

Find the volume at STP occupied by 3.2 g of oxygen gas.

Moles = (3.2)/(32) = 0.1 mol; volume = 0.1 × 22.4 = 2.24 L.

Find the number of atoms in 3.4 g of ammonia.

Moles of NH₃ = (3.4)/(17) = 0.2 mol; molecules = 0.2 × 6.022 × 10²³ = 1.2044 × 10²³; each has 4 atoms, so atoms = 4.8176 × 10²³.

Past board exam questions

ICSE 2027 sample paper · 1 mark

An organic compound X containing carbon and hydrogen only has Vapour Density (V.D.) 14. The molecular formula of X is _____________. (At. wt. of C=12, H=1)

CH
C₂H₄
C₂H₂
CH₂

Show answer

C₂H₄

ICSE 2027 sample paper · 1 mark

The volume occupied by 0.2 moles of a gas at S.T.P. is:

22.4 litres
2.24 litres
44.8 litres
4.48 litres

Show answer

4.48 litres

ICSE 2027 sample paper · 2 marks

20 grams of a gas M₂P has lesser number of moles than 20 grams of the gas MP. Justify the above statement using mathematical calculations. [At.wt. M=14, P=16]

Show answer

Molar mass of M₂P = 2 × 14 + 16 = 44 g; moles in 20 g = 20/44 = 0.45 mol Molar mass of MP = 14 + 16 = 30 g; moles in 20 g = 20/30 = 0.67 mol 0.45 < 0.67, therefore 20 g of M₂P has a lesser number of moles than 20 g of MP.

ICSE 2027 sample paper · 3 marks

Solid ammonium dichromate decomposes as under: (NH₄)₂Cr₂O₇ → N₂ + Cr₂O₃ + 4H₂O If 126 g of ammonium dichromate decomposes, calculate: (a) the number of moles of ammonium dichromate that undergoes decomposition. (b) the mass of chromic oxide formed at the same time. (c) the volume of nitrogen gas evolved at STP. [At wt: N = 14, Cr = 52, O = 16, H = 1]

Show answer

Molar mass of (NH₄)₂Cr₂O₇ = 2(14 + 4) + 2(52) + 7(16) = 36 + 104 + 112 = 252 g (a) Moles = 126/252 = 0.5 mol (b) Molar mass of Cr₂O₃ = 104 + 48 = 152 g. 1 mol of dichromate gives 1 mol Cr₂O₃, so mass = 0.5 × 152 = 76 g (c) 1 mol of dichromate gives 1 mol N₂ = 22.4 L at STP, so volume = 0.5 × 22.4 = 11.2 L

ICSE 2026 sample paper · 1 mark

The volume occupied by 2 moles of a gas at STP is:

22.4L
2.24L
44.8L
4.48L

Show answer

44.8L

ICSE 2026 board paper · 1 mark

What volume of carbon dioxide is produced at STP when 5 litres of propane is burnt completely according to the equation given below? C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O

10 litres
15 litres
20 litres
25 litres

Show answer

15 litres

ICSE 2026 sample paper · 1 mark

The ratio between the volumes occupied by 22 grams of carbon dioxide and 10 grams of hydrogen gas is:

2.2 :1
1: 2.2
1:10
10:1

Show answer

1:10

ICSE 2026 sample paper · 2 marks

What will be the mass of carbon dioxide that will contain the same number of molecules as present in 3.2g of oxygen gas? [At. Wt: O=16, C=12]

Show answer

Moles of O₂ = 3.2/32 = 0.1 mol (6.022 × 10²² molecules). The same number of CO₂ molecules = 0.1 mol of CO₂. Mass of CO₂ = 0.1 × 44 = 4.4 g