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ICSE Class 10 Chemistry Question 2 of 37

Mole Concept and Stoichiometry — Question 14

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Question 13

What volume of air (containing 20% O2 by volume) will be required to burn completely 10 cm3 each of methane and acetylene?

CH4 + 2O2 ⟶ CO2 + 2H2O

2C2H2 + 5O2 ⟶ 4CO2 + 2H2O

Answer

CH4+2O2⟶CO2+2H2O1 vol.:2 vol.⟶1 vol.2C2H2+5O2⟶4CO2+2H2O2 vol.:5 vol.⟶4 vol.\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1\text{ vol.} \\ 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 & \longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} \\ \end{matrix}

[By Gay Lussac's law]

1 Vol. CH4 requires 2 Vol. of O2

∴ 10 cm3 CH4 will require 2 x 10

= 20 cm3 of O2

Given, air contains 20% O2 by volume.

Let xx volume of air contain 20 cm3 of O2

⇒20100×x=20⇒x=10020×20⇒x=100 cm3\Rightarrow \dfrac{20}{100} \times x = 20 \\[1em] \Rightarrow x = \dfrac{100}{20} \times 20 \\[1em] \Rightarrow x = 100 \text{ cm}^3

∴ 20 cm3 O2 is present in 100 cm3 of air.

Similarly, 2 Vol C2H2 requires 5 Vol. of oxygen

∴ 10 cm3 C2H2 will require 52\dfrac{5}{2} x 10

= 25 cm3 of oxygen

Given, air contains 20% O2 by volume

Let xx volume of air contain 25 cm3 of O2

⇒20100×x=25⇒x=10020×25⇒x=125 cm3\Rightarrow \dfrac{20}{100} \times x = 25 \\[1em] \Rightarrow x = \dfrac{100}{20} \times 25 \\[1em] \Rightarrow x = 125 \text{ cm}^3

∴ 25 cm3 O2 is present in 125 cm3 of air.

Hence, total volume of air required is 100 + 125 = 225 cm3

Chapter Overview: Mole Concept and Stoichiometry

The Mole Concept is the foundation of quantitative chemistry. A mole is the amount of substance that contains 6.022 × 1023 particles (Avogadro's number). The molar mass of a substance is numerically equal to its relative molecular mass expressed in grams. This chapter teaches students to calculate the number of moles, number of particles, and mass of substances using inter-conversion formulae. Stoichiometry involves using balanced chemical equations to determine the quantities of reactants and products. The Gay-Lussac's law and Avogadro's law relate volumes of gases in reactions. At STP (Standard Temperature and Pressure: 0°C, 1 atm), one mole of any gas occupies 22.4 litres (molar volume). Students must master percentage composition calculations, empirical and molecular formula determination, and solving problems based on balanced equations involving mass-mass, mass-volume, and volume-volume relationships. This chapter carries the highest weightage in the chemistry paper and demands strong numerical problem-solving skills.

Key Formulas

Formula / Concept Expression
Number of Molesn = Given mass / Molar mass
Number of ParticlesN = n × NA (where NA = 6.022 × 1023)
Moles from Volume (gas at STP)n = Volume at STP / 22400 mL
Vapour DensityMolecular Mass = 2 × Vapour Density
% Composition% of element = (Mass of element in 1 mole / Molar mass) × 100
Empirical FormulaSimplest whole number ratio of atoms in a compound
Molecular Formulan × Empirical Formula (where n = Molecular mass / Empirical formula mass)
Gay-Lussac's LawGases react in simple whole number ratios by volume
Avogadro's LawEqual volumes of gases at same T and P contain equal number of molecules
Molar Volume at STP22.4 L or 22400 mL for one mole of any gas

Must-Know Concepts

  • Relative Atomic Mass (RAM) of H = 1, C = 12, N = 14, O = 16, S = 32, Cl = 35.5, Na = 23, Ca = 40
  • Always balance the equation FIRST before doing stoichiometric calculations
  • Mole ratios from balanced equations: 2H2 + O2 → 2H2O means 2 mol H2 reacts with 1 mol O2
  • STP conditions: 0°C (273 K) and 1 atmosphere (760 mm Hg)
  • For empirical formula: convert % to mass → divide by atomic mass → find simplest ratio
  • Volume-volume problems use the mole ratio directly (from balanced equation coefficients)
  • 1 mole of water = 18 g = 6.022 × 1023 molecules = 3 × 6.022 × 1023 atoms

Important Diagrams to Practice

  • Mole concept inter-conversion triangle (mass, moles, particles, volume)
  • Flowchart for empirical and molecular formula determination
  • Stoichiometry problem-solving steps diagram

Common Mistakes

  • Using unbalanced equations for stoichiometric calculations
  • Confusing atoms and molecules (1 mole of H2 = 2 moles of H atoms)
  • Using 22.4 L for liquids or solids (molar volume applies only to gases at STP)
  • Not converting % to grams when finding empirical formula (assume 100 g sample)
  • Forgetting to multiply empirical formula by n to get molecular formula
  • Using wrong units: molar mass is in g/mol, not g

Scoring Tips

  • Show all steps clearly: balanced equation → mole ratio → calculation → answer with units
  • Double-check atomic masses before starting calculations
  • For percentage composition questions, always verify that all percentages add up to 100%
  • Practice at least 20 numericals from each type (mass-mass, mass-volume, volume-volume)
  • Memorise molar masses of common compounds: H2O = 18, CO2 = 44, NaCl = 58.5, CaCO3 = 100

Frequently Asked Questions

Why is the mole concept important in chemistry?

Atoms and molecules are too small to count individually. The mole provides a bridge between the atomic scale and the macroscopic scale, allowing us to weigh out specific numbers of atoms or molecules for reactions.

Can the empirical formula and molecular formula be the same?

Yes. When n = 1 (molecular mass equals empirical formula mass), both are identical. For example, CH2O is both the empirical formula of formaldehyde and can be the molecular formula (molecular mass = 30).

What happens to gas volume if temperature is not at STP?

At non-STP conditions, the molar volume is not 22.4 L. You would need to use the ideal gas equation (PV = nRT) or convert the given conditions to STP first. However, ICSE problems typically assume STP unless stated otherwise.