Mole Concept & Stoichiometry

The mole concept is the foundation of quantitative chemistry. JEE tests molar mass calculations, stoichiometric relationships, limiting reagent identification, percent yield, empirical/molecular formula determination, and volumetric analysis.

Mole and Avogadro's Number: One mole contains 6.022 x $10^{23}$ entities (atoms, molecules, ions, electrons). Molar mass = mass of 1 mole in grams = numerically equal to atomic/molecular mass in amu. Number of moles n = mass(g)/molar mass = number of particles/$N_{A}$ = volume(L at STP)/22.4 (for gases). At STP (0 C, 1 atm): 1 mole of any ideal gas occupies 22.4 L. At 25 C, 1 atm: molar volume = 24.5 L.

Atomic and Molecular Mass: Atomic mass unit (amu or u) = $\frac{1}{12}$ of the mass of C-12 atom = 1.66 x $10^{-24}$ g. Average atomic mass accounts for isotopic abundances: $M_{avg}$ = sum(fraction_i x mass_i). Molecular mass = sum of atomic masses of all atoms in the formula. Formula mass is used for ionic compounds (no discrete molecules).

Equivalent Weight: Equivalent weight = molar mass / n-factor. For acids: n-factor = basicity (number of replaceable H+). For bases: n-factor = acidity (number of replaceable OH-). For salts in double decomposition: n-factor = total positive charge. For oxidising/reducing agents: n-factor = change in oxidation state per formula unit. Number of equivalents = mass/equivalent weight = n x n-factor. Law of equivalence: equivalents of reactant 1 = equivalents of reactant 2 (at stoichiometric point).

Percent Composition: Mass % of element = (number of atoms x atomic mass / molecular mass) x 100. Used to find empirical formula: convert mass% to moles, divide by smallest, get whole number ratio.

Empirical and Molecular Formula: Empirical formula = simplest whole number ratio of atoms. Molecular formula = actual number of atoms = n x empirical formula. n = molecular mass / empirical formula mass. Molecular formula can be determined from empirical formula + molar mass (from vapour density, colligative properties, or mass spectrometry). Vapour density = molar mass / 2 (relative to H₂).

Stoichiometry: Balanced equation gives mole ratios. Steps: (1) Write balanced equation. (2) Convert given quantity to moles. (3) Use mole ratio to find moles of desired species. (4) Convert to required units. For gases at STP: use 22.4 L/mol. For solutions: moles = M x V(L).

Limiting Reagent: The reactant that is completely consumed first, determining the maximum product. Method: calculate moles of product from each reactant separately — the one giving the least product is the limiting reagent. The other reactant is in excess.

Percent Yield: Yield% = (actual yield / theoretical yield) x 100. Theoretical yield = maximum possible from stoichiometry with limiting reagent. Actual yield is always <= theoretical yield (due to side reactions, losses, incomplete reaction, equilibrium).

Volumetric Analysis (Titration): At equivalence point: meq of acid = meq of base, or M₁V₁n1 = M₂V₂n2. For acid-base: N₁V₁ = N₂V₂ (normality x volume). For redox: equivalents of oxidant = equivalents of reductant. Back titration: add known excess reagent, titrate the unreacted excess. Double titration: mixture of two substances titrated successively (e.g., NaOH + Na₂CO₃ with HCl using phenolphthalein then methyl orange).

Concentration Terms: Molarity (M) = mol/L solution. Normality (N) = equivalents/L = M x n-factor. Molality (m) = mol/kg solvent. Mass fraction = mass solute/mass solution. ppm = mg/kg. Dilution: M₁V₁ = M₂V₂.

Laws of Chemical Combination: Law of conservation of mass (Lavoisier). Law of definite proportions (Proust). Law of multiple proportions (Dalton). Gay-Lussac's law of gaseous volumes. Avogadro's law: equal volumes of gases at same T, P contain equal number of molecules.

The key problem-solving concept is systematic application of mole ratios from balanced equations combined with limiting reagent analysis.