2.1 Dalton’s Atomic Theory and Its Postulates
Definition: Proposed by John Dalton in 1808, this theory explains the nature of matter at the atomic level.
Postulates:
- Matter is made of indivisible atoms.
- All atoms of an element are identical in mass and properties.
- Atoms of different elements have different masses and properties.
- Compounds form when atoms combine in simple whole-number ratios.
- Atoms are neither created nor destroyed in chemical reactions (law of conservation of mass).
Limitations:
- Atoms are divisible into subatomic particles (electrons, protons, neutrons).
- Isotopes of an element have different masses.
- Does not explain molecular bonding or energy changes.
Importance: Provides the foundation for understanding atomic structure and chemical reactions.
2.2 Laws of Stoichiometry
Stoichiometry is based on laws governing mass and proportion in chemical reactions:
- Law of Conservation of Mass: Mass of reactants equals mass of products. Example: In C + O₂ → CO₂, 12 g C + 32 g O₂ = 44 g CO₂.
- Law of Definite Proportions: A compound always contains elements in a fixed mass ratio. Example: Water (H₂O) always has H:O = 1:8 by mass.
- Law of Multiple Proportions: When elements form multiple compounds, the mass ratios of one element to a fixed mass of another are in simple ratios. Example: In CO and CO₂, oxygen masses (16 g and 32 g) per 12 g carbon are in a 1:2 ratio.
- Law of Reciprocal Proportions: When two elements combine with a fixed mass of a third, their mass ratio is the same as when they combine directly. Example: In H₂O and HCl, H combines with O and Cl in ratios consistent with H₂:O and H:Cl.
- Gay-Lussac’s Law of Gaseous Volumes: Gases react in simple whole-number volume ratios at constant temperature and pressure. Example: 2H₂ (2 vol) + O₂ (1 vol) → 2H₂O (2 vol).
2.3 Avogadro’s Law and Deductions
Avogadro’s Law: Equal volumes of all gases, at the same temperature and pressure, contain an equal number of molecules.
Key Points:
- 1 mole of any gas occupies 22.4 L at STP (0°C, 1 atm).
- Avogadro’s number: 6.022 × 10²³ particles per mole.
Deductions:
- Molecular Mass and Vapour Density:
- Vapour density (VD) = Molecular mass / 2.
- Example: CO₂ has molecular mass 44; VD = 44 / 2 = 22.
- Molecular Mass and Volume of Gas:
- 1 mole of gas = 22.4 L at STP = Molecular mass in grams.
- Example: 22.4 L of O₂ at STP weighs 32 g (molecular mass).
- Molecular Mass and Number of Particles:
- 1 mole = 6.022 × 10²³ molecules.
- Example: 18 g of H₂O (1 mole) contains 6.022 × 10²³ molecules.
Numerical Example: Find the volume of 16 g of methane (CH₄) at STP.
- Molecular mass of CH₄ = 12 + 4×1 = 16 g.
- 16 g = 1 mole; 1 mole occupies 22.4 L at STP.
- Answer: Volume = 22.4 L.
2.4 Mole and Its Relations
Mole Concept: A mole is the amount of substance containing Avogadro’s number (6.022 × 10²³) of particles (atoms, molecules, ions).
Relations:
- Mole and Mass: Moles = Mass / Molar mass.
- Mole and Volume: Moles = Volume / 22.4 L (for gases at STP).
- Mole and Number of Particles: Moles = Number of particles / 6.022 × 10²³.
Numerical Example: Calculate moles in 64 g of O₂.
- Molar mass of O₂ = 32 g/mol.
- Moles = 64 / 32 = 2 moles.
- Volume at STP = 2 × 22.4 = 44.8 L.
- Particles = 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules.
2.5 Limiting Reactant and Excess Reactant
Limiting Reactant: The reactant that is completely consumed, limiting the amount of product formed.
Excess Reactant: The reactant that remains after the reaction stops.
Numerical Example: For 2H₂ + O₂ → 2H₂O, 4 g H₂ reacts with 40 g O₂. Find the limiting reactant.
- Moles of H₂ = 4 / 2 = 2 moles.
- Moles of O₂ = 40 / 32 = 1.25 moles.
- From equation: 2 moles H₂ need 1 mole O₂.
- 2 moles H₂ need 1 mole O₂; available O₂ (1.25 moles) is excess.
- Limiting Reactant: H₂; O₂ is excess.
- Moles of H₂O formed = 2 moles (based on H₂).
2.6 Theoretical Yield, Experimental Yield, and % Yield
Theoretical Yield: Maximum amount of product calculated from the limiting reactant.
Experimental Yield: Actual amount of product obtained in a reaction.
Percentage Yield: (% Yield) = (Experimental Yield / Theoretical Yield) × 100%.
Numerical Example: In the above reaction, theoretical yield of H₂O is 36 g (2 moles × 18 g/mol). If 30 g is obtained, calculate % yield.
- % Yield = (30 / 36) × 100% = 83.33%.
2.7 Calculation of Empirical and Molecular Formula from % Composition
Empirical Formula: Simplest whole-number ratio of atoms.
Molecular Formula: Actual number of atoms, n × Empirical Formula.
Steps for Calculation:
- Convert % composition to mass (assume 100 g sample).
- Calculate moles of each element (Mass / Atomic mass).
- Divide by smallest mole value to get ratio.
- Multiply to get whole numbers for empirical formula.
- For molecular formula, calculate n = Molecular mass / Empirical formula mass.
Numerical Example: A compound has 40% C, 6.67% H, 53.33% O, and molecular mass 180. Find empirical and molecular formulas.
- Assume 100 g: C = 40 g, H = 6.67 g, O = 53.33 g.
- Moles: C = 40 / 12 = 3.33; H = 6.67 / 1 = 6.67; O = 53.33 / 16 = 3.33.
- Ratio: 3.33 / 3.33 = 1 (C); 6.67 / 3.33 = 2 (H); 3.33 / 3.33 = 1 (O).
- Empirical Formula: CH₂O (mass = 12 + 2×1 + 16 = 30).
- n = 180 / 30 = 6.
- Molecular Formula: C₆H₁₂O₆.
Summary Table: Key Concepts and Examples
| Concept | Definition | Example |
|---|---|---|
| Dalton’s Theory | Atoms are indivisible, identical for an element | C atoms in CO₂ are identical |
| Law of Conservation | Mass of reactants = Mass of products | C + O₂ → CO₂ (12 g + 32 g = 44 g) |
| Avogadro’s Law | Equal volumes, equal molecules at STP | 22.4 L of O₂ = 1 mole |
| Mole | 6.022 × 10²³ particles | 18 g H₂O = 1 mole |
| Limiting Reactant | Reactant consumed first | H₂ in 2H₂ + O₂ → 2H₂O |
| % Yield | (Experimental / Theoretical) × 100% | 30 g / 36 g = 83.33% |
| Empirical Formula | Simplest atom ratio | CH₂O for glucose |
| Molecular Formula | Actual atom numbers | C₆H₁₂O₆ for glucose |