Chapter 7 — Atoms and Molecules: Long Answer Questions
Dalton's Atomic Theory (1808)
John Dalton proposed the first modern atomic theory based on his studies of chemical reactions.
Postulates:
1. All matter is made of atoms. Atoms are tiny, indivisible particles that cannot be created or destroyed.
2. Atoms of the same element are identical in mass, size, and all properties.
3. Atoms of different elements are different — they have different masses and different properties.
4. Compounds are formed when atoms of different elements combine in simple, fixed, whole-number ratios.
5. In chemical reactions, atoms are neither created nor destroyed — they are rearranged (Law of Conservation of Mass).
Achievements of Dalton's Theory:
- Provided a rational explanation for the Laws of Chemical Combination (see below)
- Led to the concept of atomic masses and the periodic table
- Explained why elements combine in fixed proportions
- Foundation of all subsequent atomic models
Limitations:
1. Atoms are divisible — they contain subatomic particles (proton, neutron, electron) 2. Isotopes — atoms of the same element can have different masses (e.g., ¹²C and ¹⁴C) 3. Isobars — atoms of different elements can have the same mass number 4. Does not explain allotropy — the same element existing in different forms (diamond and graphite) 5. Does not explain the nature of the bond between atoms in a molecule 6. Does not address the electrical nature of matter
Law 1 — Law of Conservation of Mass (Antoine Lavoisier, 1789)
Statement: In a chemical reaction, the total mass of reactants equals the total mass of products. Mass is neither created nor destroyed in a chemical reaction.
Example:
Law 2 — Law of Definite Proportions / Law of Constant Composition (Joseph Proust, 1799)
Statement: A chemical compound always contains the same elements in the same ratio by mass, regardless of its source or method of preparation.
Example: Water (H₂O) always contains hydrogen and oxygen in the mass ratio 1:8, whether it comes from a river, rain, or electrolysis.
Hydrogen : Oxygen = 2 : 16 = 1 : 8 (always)
Law 3 — Law of Multiple Proportions (John Dalton, 1803)
Statement: If two elements A and B form more than one compound, the different masses of B that combine with a fixed mass of A are in a simple whole-number ratio.
Example: Carbon and oxygen form two compounds:
- Carbon monoxide (CO): 12 g C combines with 16 g O → ratio of O = 16
- Carbon dioxide (CO₂): 12 g C combines with 32 g O → ratio of O = 32
Ratio of oxygen masses: 16 : 32 = 1 : 2 (simple whole number ratio) ✓
Law 4 — Gay-Lussac's Law of Combining Volumes (1808)
Statement: When gases react together, the volumes of the gaseous reactants and products (measured at the same temperature and pressure) bear a simple whole-number ratio.
Example: Hydrogen + Oxygen → Water
- 2 L H₂ + 1 L O₂ → 2 L H₂O (steam)
- Ratio = 2 : 1 : 2 (simple whole numbers)
Law 5 — Avogadro's Law (Amedeo Avogadro, 1811)
Statement: Equal volumes of all gases, at the same temperature and pressure, contain equal numbers of molecules.
Significance: At STP (standard temperature and pressure = 0°C, 1 atm):
- 1 mole of any gas occupies 22.4 L (molar volume)
- This volume contains 6.022 × 10²³ molecules
The Mole Concept
A mole is the SI unit of amount of substance. It is defined as the amount of substance that contains as many elementary entities (atoms, molecules, ions, electrons) as there are atoms in exactly 12 grams of carbon-12.
This number is Avogadro's number (Nₐ):
Molar Mass
The molar mass of a substance is the mass of one mole of that substance, numerically equal to its atomic/molecular/formula mass but expressed in grams per mole (g/mol).
Examples:
- Molar mass of H = 1 g/mol
- Molar mass of O = 16 g/mol
- Molar mass of H₂O = 2(1) + 16 = 18 g/mol
- Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Relationships
Worked Examples:
Example 1: Find the number of moles and molecules in 36 g of water.
- M(H₂O) = 18 g/mol
- n = 36/18 = 2 moles
- N = 2 × 6.022 × 10²³ = 1.204 × 10²⁴ molecules
Example 2: Find the mass of 3 moles of CO₂.
- M(CO₂) = 12 + 2(16) = 44 g/mol
- m = n × M = 3 × 44 = 132 g
Example 3: How many atoms are in 4 g of calcium (Ca, atomic mass = 40)?
- n = 4/40 = 0.1 mol
- N = 0.1 × 6.022 × 10²³ = 6.022 × 10²² atoms
Percentage Composition
The percentage by mass of each element in a compound.
Example: Find the percentage composition of water (H₂O).
- M(H₂O) = 18 g/mol
- % H = (2/18) × 100 = 11.1%
- % O = (16/18) × 100 = 88.9%
Empirical Formula
The simplest whole-number ratio of atoms of each element in a compound.
Example: Glucose has molecular formula C₆H₁₂O₆. Ratio C:H:O = 6:12:6 = 1:2:1 Empirical formula = CH₂O
Molecular Formula
The actual number of atoms of each element in one molecule of a compound.
Relationship:
Example:
- Empirical formula of benzene: CH (mass = 13)
- Molecular mass of benzene = 78
- n = 78/13 = 6
- Molecular formula = (CH)₆ = C₆H₆
Finding empirical formula from percentage composition: 1. Write the percentage of each element 2. Divide by atomic mass → get mole ratio 3. Divide by the smallest value → get simplest ratio 4. If not whole numbers, multiply to get whole numbers
Example: A compound contains 40% C, 6.67% H, 53.33% O.
- C: 40/12 = 3.33; H: 6.67/1 = 6.67; O: 53.33/16 = 3.33
- Divide by 3.33: C:H:O = 1 : 2 : 1
- Empirical formula: CH₂O
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