Some Basic Concepts in Chemistry

Chemistry begins with understanding how matter is quantified. The mole concept is the bridge between the atomic world and the laboratory scale. One mole of any substance contains exactly 6.022 × 10²³ entities (Avogadro's number, Nₐ), and its mass in grams equals its molar mass. The atomic mass unit (1 amu = 1.66054 × 10⁻²⁴ g) provides the scale for individual atoms, while molar mass (g/mol) scales this up to measurable quantities. At STP (0°C, 1 atm), one mole of any ideal gas occupies 22.4 L — the molar volume.

Percentage composition of an element in a compound is calculated as: % = (n × atomic mass / molar mass) × 100. The empirical formula gives the simplest whole-number ratio of atoms, while the molecular formula gives the actual count; they are related by: molecular formula = n × empirical formula, where n = molar mass / empirical formula mass.

Stoichiometry uses balanced equation coefficients as mole ratios to interconvert between moles, mass, volume, and number of particles. When reactants are not in stoichiometric proportion, the limiting reagent — the one consumed first — determines the theoretical yield.

Solved Example 1: Calculate the number of molecules in 36 g of water (H₂O, molar mass = 18 g/mol). Moles = mass / molar mass = 36 g ÷ 18 g/mol = 2 mol Number of molecules = 2 mol × 6.022 × 10²³ mol⁻¹ = 1.2044 × 10²⁴ molecules Dimensional analysis: g ÷ (g/mol) = mol; mol × (entities/mol) = entities ✓

Solved Example 2: A compound contains 40% C, 6.67% H, and 53.33% O by mass. Find its empirical formula. C : H : O = $\frac{40}{12}$ : 6.$\frac{67}{1}$ : 53.$\frac{33}{16}$ = 3.33 : 6.67 : 3.33 Dividing by the smallest (3.33): 1 : 2 : 1 Empirical formula = CH₂O

Solved Example 3: 5.3 g of Na₂CO₃ reacts with excess HCl. Calculate the volume of CO₂ at STP. Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂ Moles of Na₂CO₃ = 5.3 g ÷ 106 g/mol = 0.05 mol From stoichiometry: moles of CO₂ = 0.05 mol (1:1 ratio) Volume at STP = 0.05 mol × 22.4 L/mol = 1.12 L Dimensional analysis: mol × (L/mol) = L ✓

Concentration terms express solute amount relative to solvent or solution. Molarity (M = mol/L) depends on volume and hence temperature. Molality (m = mol/kg solvent) depends on mass and is temperature-independent. Mole fraction (x = nᵢ/$n_{total}$) is dimensionless. Normality (N = equivalents/L) depends on the reaction (n-factor). ppm = mg/kg = mg/L (for dilute aqueous solutions).

Interconversion formulas (derivation from definitions): M = (1000 × d × mass%) / (molar mass × 100), where d = density in g/mL. m = (1000 × mass%) / (molar mass × (100 − mass%)). Dimensional analysis for molarity: M = mol/L → (g/mL × unitless) / (g/mol) = mol/mL, then ×1000 mL/L = mol/L ✓

The key testable concept is mole-based interconversions and limiting reagent identification in stoichiometric calculations.