Short Answer Questions — Chapter 7: Atoms and Molecules
Dalton's Atomic Theory (1803):
1. All matter is composed of extremely small, indivisible particles called atoms. 2. Atoms of the same element are identical in mass and properties. Atoms of different elements differ. 3. Atoms cannot be created, destroyed, or converted from one element to another in chemical reactions. 4. Compounds are formed when atoms of different elements combine in fixed, simple whole-number ratios. 5. Chemical reactions involve the rearrangement of atoms.
Limitations (where Dalton was wrong):
1. Atom is divisible: Atoms contain subatomic particles — protons, neutrons, electrons. 2. Atoms of same element can differ: Isotopes — atoms of the same element with different masses (e.g., C-12 and C-14). 3. Atoms of different elements can have same mass: Isobars (e.g., Ar-40 and Ca-40). 4. Matter is not always made of atoms: Free electrons, plasma, etc.
Mole: A mole is the amount of a substance that contains as many particles (atoms, molecules, or ions) as there are atoms in exactly 12 grams of carbon-12.
Avogadro's Number (Nₐ):
Nₐ = 6.022 × 10²³ mol⁻¹
This means:
- 1 mole of any substance = 6.022 × 10²³ particles
- 1 mole of atoms = 6.022 × 10²³ atoms
- 1 mole of molecules = 6.022 × 10²³ molecules
- 1 mole of ions = 6.022 × 10²³ ions
Significance: The mole connects the macroscopic world (grams, litres) to the microscopic world (individual atoms and molecules). It allows chemists to count particles by weighing.
Examples:
- 1 mole of H₂O = 18 g = 6.022 × 10²³ molecules of water
- 1 mole of NaCl = 58.5 g = 6.022 × 10²³ formula units of NaCl
- 1 mole of Fe = 56 g = 6.022 × 10²³ atoms of iron
Atomic Mass: The atomic mass of an element is the mass of one atom of that element relative to 1/12 the mass of a carbon-12 atom.
Unit: Unified atomic mass unit (u) or amu. 1 u = 1/12 mass of C-12 = 1.66 × 10⁻²⁷ kg
| Element | Atomic Mass |
|---|---|
| Hydrogen (H) | 1 u |
| Carbon (C) | 12 u |
| Oxygen (O) | 16 u |
| Nitrogen (N) | 14 u |
| Sodium (Na) | 23 u |
| Iron (Fe) | 56 u |
Molecular Mass: The molecular mass of a compound is the sum of the atomic masses of all atoms in one molecule.
Examples:
- H₂O: 2(1) + 16 = 18 u
- CO₂: 12 + 2(16) = 44 u
- H₂SO₄: 2(1) + 32 + 4(16) = 2 + 32 + 64 = 98 u
- NaCl: 23 + 35.5 = 58.5 u
| Property | Empirical Formula | Molecular Formula |
|---|---|---|
| Definition | Simplest whole-number ratio of atoms | Actual number of each type of atom in one molecule |
| Example (glucose) | CH₂O | C₆H₁₂O₆ |
| Example (benzene) | CH | C₆H₆ |
| Example (water) | H₂O | H₂O (same in this case) |
| Relationship | Molecular = n × Empirical | where n = 1, 2, 3, ... |
How to find n: n = Molecular mass / Empirical formula mass
Example: Glucose
- Empirical formula: CH₂O; empirical mass = 12 + 2 + 16 = 30
- Molecular mass = 180
- n = 180/30 = 6
- Molecular formula = 6 × CH₂O = C₆H₁₂O₆
1. Law of Conservation of Mass (Lavoisier, 1789): In a chemical reaction, the total mass of reactants equals the total mass of products. Matter is neither created nor destroyed.
Example: 2H₂ + O₂ → 2H₂O; 4 + 32 = 36 g reactants; 36 g products ✓
2. Law of Definite Proportions (Proust, 1799): A compound always contains the same elements in the same fixed ratio by mass, regardless of its source.
Example: Water (H₂O) always has H:O = 1:8 by mass (whether from rain or distillation)
3. Law of Multiple Proportions (Dalton, 1803): When two elements combine to form more than one compound, the masses of one element that combine with a fixed mass of the other are in simple whole-number ratios.
Example: CO and CO₂ — oxygen combining with 12 g carbon: CO = 16 g O; CO₂ = 32 g O → ratio = 1:2
4. Gay-Lussac's Law of Gaseous Volumes: Gases react in simple whole-number ratios by volume at same temperature and pressure.
Example: H₂ + Cl₂ → 2HCl: 1 volume + 1 volume → 2 volumes
5. Avogadro's Law: Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules.
Percentage composition gives the percentage by mass of each element in a compound.
% of element = (mass of element in 1 mol / molar mass of compound) × 100
Example: Calculate percentage composition of water (H₂O)
Molar mass of H₂O = 2(1) + 16 = 18 g/mol
% H = (2/18) × 100 = 11.1%
% O = (16/18) × 100 = 88.9%
Check: 11.1 + 88.9 = 100% ✓
Example: Percentage composition of CO₂
Molar mass of CO₂ = 12 + 32 = 44 g/mol
% C = (12/44) × 100 = 27.3%
% O = (32/44) × 100 = 72.7%
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