Mole Concept and Stoichiometry — Question 2
Back to all questions(a) Number of molecules in 73 g of HCl —
Molecular wt. of any substance contain 6.022 × 1023 molecules.
Mass of 1 mole of HCl is 1 + 35.5 = 36.5 g
36.5 g of HCl contains 6.022 × 1023 molecules
∴ 73 g of HCl contains
= 1.2 × 1024 molecules
(b) Weight of 0.5 mole of O2 —
1 mole of O2 weighs = 2O = 2 x 16 = 32 g
∴ 0.5 moles will weigh = = 16 g
(c) Number of molecules in 1.8 g of H2O —
Molecular wt. of any substance contains 6.022 × 1023 molecules.
Mass of 1 mole of H2O is (2 x 1) + 16 = 2 + 16 = 18 g
18 g of H2O contains 6.022 × 1023 molecules
∴ 1.8 g of H2O contains
= 6.02 × 1022 molecules
(d) Number of moles in 10 g of CaCO3 —
Mass of 1 mole of CaCO3 = 40 + 12 + 3(16) = 52 + 48 = 100 g
100 g of CaCO3 = 1 mole
∴ 10 g of CaCO3 =
= 0.1 mole
(e) Weight of 0.2 mole H2 gas —
1 mole of H2 weighs = 2 g
∴ 0.2 moles will weigh = = 0.4 g
(f) No. of molecules in 3.2 g of SO2 —
Molecular wt. of any substance contain 6 × 1023 molecules.
Mass of 1 mole of SO2 is 32 + 2(16) = 32 + 32 = 64 g
64 g of SO2 contains 6 × 1023 molecules
∴ 3.2 g of SO2 contains
= 3 x 1022 molecules.
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 Moles | n = Given mass / Molar mass |
| Number of Particles | N = n × NA (where NA = 6.022 × 1023) |
| Moles from Volume (gas at STP) | n = Volume at STP / 22400 mL |
| Vapour Density | Molecular Mass = 2 × Vapour Density |
| % Composition | % of element = (Mass of element in 1 mole / Molar mass) × 100 |
| Empirical Formula | Simplest whole number ratio of atoms in a compound |
| Molecular Formula | n × Empirical Formula (where n = Molecular mass / Empirical formula mass) |
| Gay-Lussac's Law | Gases react in simple whole number ratios by volume |
| Avogadro's Law | Equal volumes of gases at same T and P contain equal number of molecules |
| Molar Volume at STP | 22.4 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.