Topic 1 is the quantitative foundation of chemistry. Students master the mole concept and apply it to balanced equations, reacting mass calculations, limiting reagent problems, and solution stoichiometry.
One mole = 6.022 × 10²³ particles (Avogadro\'s number). Molar mass (g/mol) = relative atomic/molecular mass in grams. n = m/M (moles = mass/molar mass). For gases at STP (0°C, 100 kPa): molar volume = 22.7 dm³/mol. Concentration c = n/V (mol/dm³).
Empirical formula: simplest whole-number ratio of atoms (found from percentage composition). Steps: mass → moles → divide by smallest → round to whole numbers. Molecular formula = n × empirical formula, where n = molar mass / empirical formula mass.
Balanced equations show mole ratios. Steps: (1) Write balanced equation. (2) Convert given mass to moles. (3) Use mole ratio to find moles of target. (4) Convert to required units. Limiting reagent: the reactant that runs out first and determines the maximum product. Excess reagent: left over after reaction.
Theoretical yield: maximum product from stoichiometry. Actual yield: what is obtained experimentally. Percentage yield = (actual/theoretical) × 100%. Atom economy = (mass of desired product / total mass of products) × 100% — a green chemistry measure.
Calculate the moles of each reactant, then divide each by its coefficient in the balanced equation. The reactant with the smallest value is the limiting reagent — it runs out first and determines how much product forms. The other reactant(s) are in excess.
Book a Trial + Diagnostic session. Get a personalized Learning Path with clear milestones, tutor match, and a plan recommendation — all within 24 hours.
Book Trial + Diagnostic →