Mole Concept Class 10 ICSE — Chemistry Chapter 5

Mole concept class 10 ICSE chemistry begins in Chapter 5 of the Class 10 Chemistry textbook with a review of the gas laws. This CISCE-aligned chapter then builds the tools chemists use to count atoms and molecules: relative atomic mass, the mole, Avogadro’s number, molar volume, and stoichiometric calculations. Every section below follows the textbook’s own order and numbering. By the end you will read a balanced equation as masses and volumes.

Download Mole Concept and Stoichimetry Class 10 ICSE Chemistry Chapter 5 PDF

Download the Mole Concept and Stoichimetry Class 10 ICSE Chemistry chapter PDF to revise offline or print for practice. It is the full chapter as printed in the book — 32 pages, 16 sections, 148 questions.

What is inside Chapter 5

32 pages · 16 sections · 31 worked examples · 148 exercise questions

• EXERCISE-5A: 20 questions
• ANSWERS: 3 questions
• EXERCISE-5B: 29 questions
• ANSWERS: 3 questions
• EXERCISE-5C: 23 questions
• ANSWERS: 3 questions
• EXERCISE-5D: 13 questions
• MISCELLANEOUS EXERCISES: 53 questions
• ANSWERS: 1 questions

5.1 INTRODUCTION

Section 5.1 recalls the gas laws that the rest of the chapter builds on. Boyle’s law states that the volume of a given mass of dry gas is inversely proportional to its pressure at constant temperature, so \mathrm{P_1V_1} = \mathrm{P_2V_2}. Charles’s law states that volume is directly proportional to absolute temperature at constant pressure, so \frac{\mathrm{V_1}}{\mathrm{T_1}} = \frac{\mathrm{V_2}}{\mathrm{T_2}}. Combining the two gives the gas equation \frac{\mathrm{P_1V_1}}{\mathrm{T_1}} = \frac{\mathrm{P_2V_2}}{\mathrm{T_2}}.

Gas volumes change with conditions, so the chapter fixes standard temperature and pressure (STP) at 0^\circ\mathrm{C} or 273\ \mathrm{K}, and 1 atmosphere or 760\ \mathrm{mm} of mercury. The Kelvin scale starts from absolute zero, 0\ \mathrm{K} = -273^\circ\mathrm{C}. Convert Celsius to Kelvin before using the gas equation: 27^\circ\mathrm{C} + 273 = 300\ \mathrm{K}.

5.2 GAY-LUSSAC’S LAW OF COMBINING VOLUMES

Gay-Lussac’s law of combining volumes states that when gases react, their volumes bear a simple ratio to one another and to the volume of the gaseous product, provided all volumes are measured at the same temperature and pressure. The law applies only to gases. The volumes of solids and liquids are counted as zero. Gay-Lussac’s law is the first numerical tool in this ICSE mole concept chapter.

The chapter illustrates the law with three reactions. \mathrm{H_2} + \mathrm{Cl_2} \rightarrow 2\mathrm{HCl} gives volumes 1 : 1 : 2. \mathrm{N_2} + 3\mathrm{H_2} \rightarrow 2\mathrm{NH_3} gives 1 : 3 : 2. 2\mathrm{CO} + \mathrm{O_2} \rightarrow 2\mathrm{CO_2} gives 2 : 1 : 2. When one reactant is used up completely, it is called the limiting reagent.

Worked pattern with original numbers: burn 150 mL of ethene. The balanced equation \mathrm{C_2H_4} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 2\mathrm{H_2O} reads as 1 volume : 3 volumes : 2 volumes : 2 volumes. So 150 mL of ethene needs 3 \times 150 = 450\ \mathrm{mL} of oxygen and produces 2 \times 150 = 300\ \mathrm{mL} of carbon dioxide.

5.3 AVOGADRO’S LAW

Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain the same number of molecules. That is why, under identical conditions, one litre of hydrogen holds the same number of molecules as one litre of oxygen. Larger volume means more molecules.

Avogadro also separated the idea of an atom from a molecule. An atom is the smallest particle of an element that can take part in a chemical reaction but may or may not exist alone. A molecule is the smallest particle of an element or compound that can exist by itself. The number of atoms in a molecule of an element is its atomicity.

Molecules made of one kind of atom are homoatomic, like \mathrm{P_4} and \mathrm{O_3}. Molecules made of different atoms are heteroatomic, like \mathrm{HCl} and \mathrm{NH_3}.

5.4 RELATIVE ATOMIC MASS (ATOMIC WEIGHT)

Relative atomic mass is the number of times one atom of an element is heavier than one-twelfth the mass of one carbon-12 atom.

\text{Relative atomic mass} = \frac{\text{mass of one atom of the element}}{\frac{1}{12}\ \text{of the mass of one C-12 atom}}

The atomic mass unit (a.m.u.) is defined as one-twelfth the mass of a C-12 atom. Most atomic masses are fractional because natural elements are mixtures of isotopes, so the value is a weighted average. Chlorine mixes isotopes of mass 35 and 37 in the ratio 3 : 1, giving \frac{35 \times 3 + 37 \times 1}{4} = 35.5.

Original check: bromine has two isotopes of mass 79 and 81 in equal amounts, so its relative atomic mass is \frac{79 + 81}{2} = 80.

5.5 RELATIVE MOLECULAR MASS (MOLECULAR WEIGHT)

Relative molecular mass is how many times one molecule of a substance is heavier than one-twelfth the mass of a carbon-12 atom. It equals the sum of the relative atomic masses of all atoms in the molecule. For sulphuric acid: \mathrm{H_2SO_4} = 2 \times 1 + 32 + 4 \times 16 = 98 a.m.u.

Original calculation: calcium carbonate, \mathrm{CaCO_3} = 40 + 12 + 3 \times 16 = 100 a.m.u. Hydrated salts include water of crystallisation in the total. Copper sulphate crystals, \mathrm{CuSO_4 \cdot 5H_2O}, give 63.5 + 32 + 64 + 90 = 249.5 a.m.u.

5.6 GRAM ATOMIC MASS

Gram atomic mass is the atomic mass of an element expressed in grams. A quantity weighing that mass is called one gram atom of the element. Oxygen has atomic mass 16 a.m.u., so one gram atom of oxygen is 16 g. One gram atom of sodium is 23 g, and one gram atom of helium is 4 g.

Original check: 46 g of sodium contains \frac{46}{23} = 2 gram atoms.

A common error in this chapter is mixing gram atomic mass with gram molecular mass:

• Gram atomic mass of oxygen is 16 g — one mole of O atoms.
• Gram molecular mass of oxygen is 32 g — one mole of O₂ molecules.
• Using the wrong value changes every mole calculation.

5.7 GRAM MOLECULAR MASS

Gram molecular mass is the molecular mass of a substance expressed in grams, also called molar mass. A sample with that mass is one gram molecule. Water has molecular mass 18 a.m.u., so its gram molecular mass is 18 g. Ammonia has molecular mass 17, so one gram molecule of ammonia is 17 g.

Contrast this with section 5.6. Gram atomic mass belongs to one type of atom, so atomic oxygen is 16 g. Gram molecular mass belongs to the whole molecule, so oxygen gas is 32 g. Match the unit to the particle you are counting.

5.8 MOLE CONCEPT

The mole concept rests on one fixed number. A mole is a collection of 6.022 \times 10^{23} particles, just as a dozen is a collection of 12 objects. This number is Avogadro’s number, N_\mathrm{A}, the number of atoms in exactly 12 g of carbon-12.

One mole of atoms contains 6.022 \times 10^{23} atoms and weighs the gram atomic mass, so one mole of oxygen atoms is 16 g. One mole of molecules weighs the gram molecular mass, so one mole of oxygen gas is 32 g. For gases, one mole occupies the molar volume of 22.4 litres at STP.

1\ \text{mole} = 6.022 \times 10^{23}\ \text{particles} = \text{gram atomic or molecular mass} = 22.4\ \text{litres at STP (for a gas)}

Original check: carbon dioxide has molar mass 44 g, so 0.5 mole of \mathrm{CO_2} weighs 0.5 \times 44 = 22 g. The mole idea carries through the rest of the Class 10 Chemistry book, so it pays to master it here.

5.8.2 Mole of Molecules

This short section applies the mole to molecules. One mole contains 6.022 \times 10^{23} molecules and equals the gram molecular mass of the substance. One mole of water holds 6.022 \times 10^{23} molecules and weighs 18 g. The chapter’s table shows the same pattern for sulphur dioxide (64 g), ammonia (17 g) and calcium chloride (111 g).

Ionic compounds split into ions. One mole of \mathrm{CaCl_2} gives 6.022 \times 10^{23} calcium ions and twice that number of chloride ions.

5.9 APPLICATIONS OF AVOGADRO’S LAW

Section 5.9 lists five applications of Avogadro’s law. It explains Gay-Lussac’s law, because equal volumes contain equal numbers of molecules, so volume ratios become molecule ratios. It determines the atomicity of gases. It determines the molecular formula of a gaseous compound. It links molecular mass to vapour density, and it fixes the molar volume of a gas at 22.4 litres at STP.

The vapour density relation is worth memorising: relative molecular mass = 2 × vapour density. Chlorine has vapour density 35.5, so its molecular mass is 2 \times 35.5 = 71, the mass of \mathrm{Cl_2}.

Original pattern: a cylinder holds 50 g of hydrogen and 400 g of gas Z under the same conditions. Gas Z has vapour density \frac{400}{50} = 8, so its molecular mass is 2 \times 8 = 16.

5.10 Percentage composition

Percentage composition is the percentage by weight of each element present in a compound.

\text{Percentage of an element} = \frac{\text{total weight of the element in one molecule}}{\text{gram molecular weight of the compound}} \times 100

Original example: find the percentage of oxygen in potassium nitrate, \mathrm{KNO_3}. Its molecular mass is 39 + 14 + 48 = 101. Oxygen contributes 48, so \frac{48}{101} \times 100 = 47.5\%.

Hydrated salts use the same formula for water. Washing soda, \mathrm{Na_2CO_3 \cdot 10H_2O}, has molecular mass 286, of which 180 is water. The percentage of water is \frac{180}{286} \times 100 = 62.9\%.

5.11 EMPIRICAL FORMULA OF A COMPOUND

The empirical formula is the simplest whole-number ratio of atoms of each element in one molecule. Hydrogen peroxide has the actual formula \mathrm{H_2O_2} but the empirical formula HO, a 1 : 1 ratio. Glucose \mathrm{C_6H_{12}O_6} has the empirical formula \mathrm{CH_2O}, a 1 : 2 : 1 ratio.

Empirical formula mass is the sum of the atomic masses in the empirical formula. For HO, that is 1 + 16 = 17. Section 5.12 shows how to find this ratio from percentage composition.

5.12 DETERMINATION OF EMPIRICAL FORMULA

Finding an empirical formula from percentage composition takes five steps. First, if one element’s percentage is missing, subtract the given percentages from 100. Second, divide each percentage by its atomic mass. Third, if water of crystallisation is present, divide its percentage by 18. Fourth, divide every result by the smallest one. Fifth, write each symbol with the whole-number ratio as its subscript.

Worked pattern with original numbers: a compound contains C = 52.2%, H = 13.0% and O = 34.8%. The atomic ratios are \frac{52.2}{12} = 4.35, \frac{13.0}{1} = 13.0 and \frac{34.8}{16} = 2.175. Dividing by 2.175 gives C : H : O = 2 : 6 : 1, so the empirical formula is \mathrm{C_2H_6O}.

Hydrated compounds follow the same steps with water treated as a component. The book’s worked example turns given percentages of Mg, S, O and water into \mathrm{MgSO_4 \cdot 7H_2O}.

5.13 DETERMINATION OF MOLECULAR FORMULA

The molecular formula gives the actual number of atoms of each element in one molecule. To find it, calculate the empirical formula mass, then divide the molecular weight by it to get n, and multiply the empirical formula by n.

\text{Molecular formula} = \text{empirical formula} \times n, \quad \text{where} \quad n = \frac{\text{molecular weight}}{\text{empirical formula weight}}

The book’s hydrogen peroxide case shows the idea. Empirical formula HO has mass 17, the molecular weight is 34, so n = \frac{34}{17} = 2 and the molecular formula is \mathrm{H_2O_2}.

Original pattern: a hydrocarbon has vapour density 15, so its molecular weight is 2 \times 15 = 30. If its empirical formula is \mathrm{CH_3}, mass 15, then n = \frac{30}{15} = 2 and the molecular formula is \mathrm{C_2H_6}.

5.14 CHEMICAL EQUATIONS

A balanced equation is a quantitative account of a reaction. It gives the molecular proportions, the relative masses, and the volumes of gaseous substances at STP.

For 2\mathrm{KClO_3} \xrightarrow{\mathrm{MnO_2},\ \Delta} 2\mathrm{KCl} + 3\mathrm{O_2}, the equation says two molecules of potassium chlorate give two molecules of potassium chloride and three molecules of oxygen. In masses, 245 g of potassium chlorate gives 149 g of KCl and 96 g of oxygen. In volume, that oxygen occupies 3 \times 22.4 = 67.2 litres at STP.

The book’s problem method has four steps: write the balanced equation, write the molecular weight below each formula, keep the same units throughout, and write 22.4 L × the number of molecules below each gaseous substance.

5.14.2 Problems based on mass-volume relationship

This section applies that method to problems mixing mass and volume. Worked pattern with original numbers: how much sodium chloride is needed to form 0.5 g of silver chloride?

\mathrm{NaCl} + \mathrm{AgNO_3} \rightarrow \mathrm{AgCl} + \mathrm{NaNO_3}

The masses are 58.5 for NaCl, 143.5 for AgCl and 85 for sodium nitrate. Since 143.5 g of silver chloride comes from 58.5 g of NaCl, the required mass is \frac{58.5 \times 0.5}{143.5} = 0.204 g of NaCl. The sodium nitrate formed is \frac{85 \times 0.204}{58.5} = 0.296 g.

The common error is mixing units. Compare grams with grams and litres with litres. Convert any gas volume to STP through 22.4 litres before using it in a proportion. Stoichiometry returns later in the Electrolysis chapter of the same book, where equations give the masses of products.

What to practise from Mole Concept Class 10 ICSE

Mole Concept Class 10 ICSE uses five exercise blocks. Each block tests a defined range of sections.

• EXERCISE-5A — 20 questions on Gay-Lussac’s law and Avogadro’s law.
• EXERCISE-5B — 29 questions on relative atomic mass, relative molecular mass and the mole concept.
• EXERCISE-5C — 23 questions on percentage composition, empirical formula and molecular formula.
• EXERCISE-5D — 13 questions on calculations based on chemical equations.
• MISCELLANEOUS EXERCISES — 53 mixed questions covering the whole chapter.

The mapping shows which block tests which section. Exercise 5A covers sections 5.2 and 5.3. Exercise 5B covers sections 5.4 to 5.9. Exercise 5C covers sections 5.10 to 5.13. Exercise 5D covers section 5.14, and the miscellaneous set mixes everything. Check the latest syllabus on the official CISCE website.

For other subjects, browse the Class 10 ICSE books collection.

Mole Concept Class 10 ICSE frequently asked questions

What is the mole concept in class 10 ICSE chemistry?

The mole concept class 10 ICSE chapter defines a mole as a collection of 6.022 \times 10^{23} particles, called Avogadro’s number. One mole of atoms weighs the gram atomic mass, one mole of molecules weighs the gram molecular mass, and one mole of any gas occupies 22.4 litres at STP.

How do I apply Gay-Lussac’s law to gas volume problems?

Write the balanced equation and read the coefficients as volume ratios. In \mathrm{C_2H_4} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 2\mathrm{H_2O}, one volume of ethene needs three volumes of oxygen. Multiply the given volume by that ratio, keeping temperature and pressure constant.

What is the difference between gram atomic mass and gram molecular mass?

Gram atomic mass is the atomic mass in grams and belongs to one type of atom, so atomic oxygen is 16 g. Gram molecular mass is the molecular mass in grams and belongs to the whole molecule, so oxygen gas is 32 g. Pick the value that matches the particle in your calculation.

How do I find the molecular formula from the percentage composition?

First find the empirical formula by dividing each element’s percentage by its atomic mass and simplifying to whole numbers. Then divide the molecular weight by the empirical formula mass to get n, and multiply the empirical formula by n.

Reference: ICSE Class 10 Chemistry textbook material, chapter 5: Mole Concept and Stoichiometry — CISCE-aligned.

More from this book

• Chapter 2: Chemical Bonding
• Chapter 8: Study of Compounds Hydrogen Chloride
• Chapter 9: Study of Compounds Ammonia
• Chapter 10: Study of Compounds Nitric Acid
• Chapter 11: Sulphuric Acid