AFNS Chemistry Chapter 01# Basic Concepts
TOPIC 1: Matter and Its Classification
What is Chemistry?
Chemistry is the science that studies the composition, structure, properties, and reactions of matter. For AFNS candidates, chemistry explains how medicines work in the body, how drugs are prepared, and how biological molecules behave.
Classification of Matter
| Type | Definition | Examples |
| Element | Cannot be broken down by chemical means | Gold (Au), Iron (Fe), Oxygen (O₂), Carbon (C) |
| Compound | Two or more elements chemically combined in fixed ratio | Water (H₂O), NaCl, Glucose (C₆H₁₂O₆), CO₂ |
| Homogeneous mixture | Uniform composition throughout (one phase) | Saltwater, air, blood plasma, alloys |
| Heterogeneous mixture | Non-uniform — components visible or in separate phases | Sand + water, oil + water, blood (cells in plasma) |
Physical vs Chemical Changes
| Change Type | Definition | Examples |
| Physical change | Form/state changes but composition does NOT change | Melting ice, boiling water, dissolving salt, cutting paper |
| Chemical change | NEW substance(s) with DIFFERENT properties are formed | Burning, rusting, digestion, photosynthesis, neutralisation |
TOPIC 2: Atomic Mass, Molecular Mass and Molar Mass
What is Atomic Mass Unit (amu)
1 amu = 1/12 the mass of one carbon-12 atom = 1.66 × 10⁻²⁴ g. All relative atomic masses are compared to this standard.
- H = 1 | C = 12 | N = 14 | O = 16 | Na = 23 | Mg = 24 | Al = 27 | S = 32 | Cl = 35.5 | K = 39 | Ca = 40 | Fe = 56 | Cu = 64
Molecular Mass Calculations
| 📌 WORKED EXAMPLE Q: Calculate the molecular mass of glucose (C₆H₁₂O₆). A: C: 6×12=72 | H: 12×1=12 | O: 6×16=96 Molecular mass = 72+12+96 = 180 g/mol ⭐ Key molar masses to memorise: H₂O=18 | CO₂=44 | NaCl=58.5 | HCl=36.5 | NaOH=40 H₂SO₄=98 | NH₃=17 | CaCO₃=100 | HNO₃=63 | MgO=40 |
TOPIC 3: The Mole Concept — Most Important Topic in AFNS Chemistry
What is a Mole?
The MOLE is the SI unit for amount of substance. Just as a ‘dozen’ = 12, one MOLE = 6.022 × 10²³ particles. It allows chemists to count atoms/molecules by weighing them.
| KEY FACT | Avogadro’s Number (Nₐ) = 6.022 × 10²³ particles/mol ⭐ • 1 mol of ANY element = 6.022 × 10²³ atoms • 1 mol of ANY compound = 6.022 × 10²³ molecules • 1 mol of ANY gas at STP = 22.4 litres (Molar Volume) Named after Amedeo Avogadro (Italian scientist, 1776–1856) |
The Mole — Three Key Relationships
| To Find | Formula | Example |
| Moles (n) | n = mass(g) ÷ molar mass(g/mol) n = particles ÷ Nₐ n = volume(L) ÷ 22.4 (at STP) | n of H₂O from 36g: 36÷18 = 2 mol |
| Mass | mass = n × molar mass | Mass of 3 mol NaCl: 3×58.5 = 175.5 g |
| Particles | N = n × 6.022×10²³ | Molecules in 2 mol CO₂: 2×6.022×10²³ = 1.204×10²⁴ |
| Volume at STP | V = n × 22.4 L | Volume of 3 mol O₂: 3×22.4 = 67.2 L |
| 📌 WORKED EXAMPLE Q: How many molecules are in 9 g of water (H₂O)? A: Molar mass H₂O = 18 g/mol Moles = 9÷18 = 0.5 mol Molecules = 0.5 × 6.022×10²³ = 3.011×10²³ molecules ⭐ |
| 📌 WORKED EXAMPLE Q: Calculate the number of atoms in 0.5 mol of Na₂SO₄. A: Na₂SO₄ has 7 atoms per formula unit (2 Na + 1 S + 4 O) Formula units = 0.5 × 6.022×10²³ = 3.011×10²³ Total atoms = 3.011×10²³ × 7 = 2.108×10²⁴ atoms |
TOPIC 4: Empirical and Molecular Formulae
Empirical Formula — Simplest Ratio
The EMPIRICAL FORMULA gives the simplest whole-number ratio of atoms in a compound. It does NOT show actual numbers in a molecule.
| Compound | Molecular Formula | Empirical Formula |
| Glucose | C₆H₁₂O₆ | CH₂O |
| Benzene | C₆H₆ | CH |
| Hydrogen peroxide | H₂O₂ | HO |
| Acetic acid | C₂H₄O₂ | CH₂O |
| Water | H₂O | H₂O (already simplest) |
Molecular Formula from Empirical Formula
Molecular formula = (Empirical formula) × n, where n = Molar mass ÷ Empirical formula mass
| 📌 WORKED EXAMPLE Q: A compound has empirical formula CH₂O and molar mass = 60 g/mol. Find molecular formula. A: EF mass of CH₂O = 12+2+16 = 30 g/mol n = 60÷30 = 2 Molecular formula = (CH₂O)₂ = C₂H₄O₂ (this is acetic acid / ethanoic acid) |
Finding Empirical Formula from % Composition
Step 1: Assume 100 g sample → % becomes grams directly Step 2: Divide each mass by atomic mass → moles of each element Step 3: Divide all by the SMALLEST mole value → ratio Step 4: Round to nearest whole number → empirical formula
| 📌 WORKED EXAMPLE Q: A compound contains 40% C, 6.67% H, 53.33% O. Find empirical formula. A: In 100 g: C=40g, H=6.67g, O=53.33g Moles: C=40/12=3.33 | H=6.67/1=6.67 | O=53.33/16=3.33 Divide by smallest (3.33): C=1 | H=2 | O=1 Empirical formula = CH₂O ⭐ (shared by glucose, acetic acid, formaldehyde) |
TOPIC 5: Chemical Equations and Stoichiometry
Balancing Chemical Equations
LAW OF CONSERVATION OF MASS: atoms are neither created nor destroyed. Atoms on left side = atoms on right side. Only change COEFFICIENTS — never subscripts.
| 📌 WORKED EXAMPLE Q: Balance: Fe + O₂ → Fe₂O₃ A: Start with Fe₂O₃ (most complex molecule): 4Fe + 3O₂ → 2Fe₂O₃ Check: Fe: 4=4 ✓ | O: 6=6 ✓ |
Mole Ratios in Stoichiometry
| 📌 WORKED EXAMPLE Q: In 2H₂ + O₂ → 2H₂O. How many grams of water from 4 g H₂? A: Moles H₂ = 4÷2 = 2 mol Ratio H₂:H₂O = 2:2 = 1:1 → 2 mol H₂ gives 2 mol H₂O Mass H₂O = 2×18 = 36 g ⭐ |
Limiting Reagent and Percentage Yield
LIMITING REAGENT = reactant completely consumed first; determines max product formed.
% Yield = (Actual yield ÷ Theoretical yield) × 100%
| KEY FACT | Finding limiting reagent: 1. Calculate moles of each reactant 2. Divide by stoichiometric coefficient from balanced equation 3. Reactant with SMALLER value = LIMITING REAGENT 4. Use limiting reagent to calculate theoretical yield Example: 2H₂ + O₂ → 2H₂O | Given: 4 mol H₂ + 1 mol O₂ H₂: 4÷2=2 | O₂: 1÷1=1 → O₂ is LIMITING (smaller value) Max H₂O = 1×2 = 2 mol |