Mole Concept & Stoichiometry Miscellaneous Exercises — Question 35
Back to all questionsSamples of the gases O2, N2, CO2 and CO under the same conditions of temperature and pressure contain the same number of molecules x. The molecules of oxygen occupy V litres and have a mass of 8 g under the same conditions of temperature and pressure
What is the volume occupied by:
(a) x molecules of N2
(b) 3x molecules of CO
(c) What is the mass of CO2 in grams?
(d) In answering the above questions, which law have you used?
By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.
∴ If gases under the same conditions have same number of molecules then they must have the same volume.
(i) So, X molecules of N2 occupy 1V litres.
(ii) 3X molecules of CO will occupies 3V litres.
(iii) Molar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g
1 mole of O2 weighs 32 g and occupy 22.4 lit. volume
∴ 8 g of O2 will occupy x 8 = 5.6 lit. vol. = Molar volume
If 8 g of O2 occupies 5.6 lit. vol.
And 1 mole of CO2 occupy 22.4 lit and weighs = 44 g at s.t.p.
∴ 5.6 lit. vol. of CO2 will weigh = x 5.6 = 11 g
Hence, mass of CO2 = 11 g
(iv) Avogadro's Law is used above.
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.