General Chemistry: The Complete Guide With Worked Calculations

General chemistry is the foundation course that establishes the quantitative framework for understanding matter and its transformations. It covers eight major topics that build on each other: atomic structure and the mole concept, stoichiometry of chemical reactions, behavior of gases, solutions and concentration, thermochemistry, kinetics, equilibrium, and acid-base chemistry. Every topic shares a single unifying tool — the mole — which converts between the macroscopic world (grams, liters, molarity) and the microscopic world (atoms, molecules, ions). Most general chemistry problems reduce to a three-step procedure: convert given quantities to moles, apply the relevant relationship (stoichiometric ratio, gas law, equilibrium expression), and convert back to the requested quantity. The course is challenging not because individual topics are conceptually difficult but because problem types are dense and require fluent unit conversion across all eight topics. This pillar guide presents each topic with worked examples and the common error patterns that catch students.

The mole is a count of 6.022 × 10^23 entities (Avogadro's number, NA), defined to give a numerical equivalence between atomic mass units (amu) and grams. One mole of any element has mass equal to the element's atomic mass in grams. Carbon-12 has atomic mass 12.011 amu and one mole has mass 12.011 g. The molar mass of a compound is the sum of the atomic masses of all atoms. Worked Example. How many moles are in 50.0 g of CaCO3? Molar mass CaCO3 = 40.08 + 12.011 + 3(16.00) = 100.09 g/mol Moles = 50.0 g / 100.09 g/mol = 0.4995 mol Number of CaCO3 formula units = 0.4995 mol × 6.022 × 10^23 = 3.008 × 10^23 units Number of oxygen atoms = 3.008 × 10^23 × 3 = 9.025 × 10^23 atoms The mole concept enables every other quantitative topic. Stoichiometry uses moles to track reaction ratios. Gas laws use the molar volume at STP (22.414 L/mol). Solutions use moles per liter (molarity). Thermochemistry tabulates enthalpies per mole. The investment in fluency with mole-gram and mole-particle conversions pays dividends across the rest of the course.

Stoichiometry uses balanced equation coefficients as mole ratios to predict product yields and identify limiting reactants. The standard procedure: convert all given quantities to moles, apply the mole ratio from the balanced equation, identify the limiting reactant, compute theoretical yield in moles, then convert to the requested unit. Worked Example. 50.0 g of N2 reacts with 12.0 g of H2 to form NH3: N2 + 3H2 → 2NH3 Moles N2 = 50.0 / 28.02 = 1.785 mol Moles H2 = 12.0 / 2.016 = 5.952 mol From the balanced equation, 1 mol N2 needs 3 mol H2. Moles H2 needed for 1.785 mol N2 = 1.785 × 3 = 5.355 mol Moles H2 available = 5.952 mol > 5.355 needed Limiting reactant: N2 (we have excess H2). Theoretical yield NH3 = 1.785 mol N2 × (2 mol NH3 / 1 mol N2) = 3.570 mol NH3 Mass NH3 = 3.570 × 17.03 = 60.8 g (theoretical) If actual yield is 48.0 g NH3, percent yield = (48.0 / 60.8) × 100% = 78.9%. Common errors: using gram ratios instead of mole ratios (the balanced equation gives MOLE ratios; gram ratios only work after converting); identifying the limiting reactant by mass instead of by mole comparison; forgetting that excess reactant remains unreacted at the end.

The ideal gas law combines Boyle's, Charles's, Avogadro's, and Gay-Lussac's laws into a single relationship: PV = nRT. P is pressure, V is volume, n is moles, T is temperature in Kelvin, and R is the gas constant (0.0821 L·atm/(mol·K) when P is in atm and V in L). Worked Example. What volume does 2.50 mol of O2 occupy at 27 °C and 1.50 atm? T = 27 + 273 = 300 K (must be Kelvin) V = nRT / P = (2.50 × 0.0821 × 300) / 1.50 = 61.6 / 1.50 = 41.0 L Dalton's law of partial pressures: in a gas mixture, the total pressure equals the sum of partial pressures: P_total = P_A + P_B + ..., where each component's partial pressure follows PV = nRT for that component. The mole fraction relates partial pressure to total: P_A = X_A × P_total, where X_A = n_A / n_total. Graham's law of effusion: the rate at which a gas effuses through a small opening is inversely proportional to the square root of its molar mass. Lighter gases effuse faster. Rate(A)/Rate(B) = √(M_B / M_A). Real-gas deviations matter at high pressure (molecules close together; intermolecular forces matter) and low temperature (kinetic energy comparable to attractive forces). The van der Waals equation corrects for these: (P + a(n/V)²)(V − nb) = nRT. For intro problems, ideal gas behavior is assumed unless explicitly told otherwise. Common error: using °C instead of Kelvin. The gas law requires absolute temperature; T must always be in Kelvin.

Solutions are homogeneous mixtures of solute (the dissolved substance) and solvent. Concentration can be expressed several ways: Molarity (M): moles of solute per liter of solution. Most common. Molality (m): moles of solute per kg of solvent. Used for colligative properties because it is temperature-independent. Mole fraction (X): moles of component / total moles. Used in gas mixtures and Raoult's law. Mass percent (%): mass of solute / total mass × 100%. Parts per million (ppm): mass solute / mass solution × 10^6. Used for trace concentrations. Dilution Calculation: M_1 V_1 = M_2 V_2, where 1 is initial and 2 is diluted. Moles of solute do not change in dilution; only volume increases. Worked Example. How much 12.0 M HCl is needed to prepare 500. mL of 2.00 M HCl? M_1 V_1 = M_2 V_2 (12.0)(V_1) = (2.00)(500. mL) V_1 = (2.00 × 500.) / 12.0 = 83.3 mL of 12.0 M stock Dilute by adding 83.3 mL of stock to enough water to make 500. mL final volume (NOT 500 − 83.3 = 416.7 mL of water; volumes do not always add for concentrated solutions). Colligative properties depend on number of solute particles, not their identity. Boiling point elevation: ΔT_b = K_b × m × i, where i is the van't Hoff factor (number of particles per formula unit; 1 for non-electrolytes, 2 for NaCl, 3 for CaCl2). Freezing point depression: ΔT_f = K_f × m × i. Osmotic pressure: π = MRT × i. These predict how much salt depresses freezing point of water (road salt application) or how much sugar elevates boiling point of cooking water.

Thermochemistry quantifies energy changes in chemical reactions. The first law of thermodynamics: ΔU = q + w (energy is conserved; energy is exchanged as heat q or work w). At constant pressure, the heat exchanged is the enthalpy change ΔH. Exothermic reactions release heat (ΔH < 0); endothermic reactions absorb heat (ΔH > 0). Calorimetry measures heat exchanged: q = m × c × ΔT, where m is mass, c is specific heat, ΔT is temperature change. Water has c = 4.184 J/(g·°C). Hess's Law: the enthalpy change for a reaction is the sum of enthalpy changes for individual steps. Used to calculate ΔH for reactions difficult to measure directly by combining reactions whose ΔH values are known. Worked Example. Compute ΔH for: 2C + O2 → 2CO using: C + O2 → CO2, ΔH_1 = −393 kJ 2CO + O2 → 2CO2, ΔH_2 = −566 kJ Multiply step 1 by 2: 2C + 2O2 → 2CO2, ΔH = 2(−393) = −786 kJ Reverse step 2: 2CO2 → 2CO + O2, ΔH = +566 kJ (sign flips when reaction reversed) Add: 2C + 2O2 + 2CO2 → 2CO2 + 2CO + O2 → simplify: 2C + O2 → 2CO ΔH = −786 + 566 = −220 kJ Standard enthalpy of formation (ΔHf°) is the enthalpy change to form 1 mole of compound from elements in standard state. ΔH°rxn = Σ ΔHf°(products) − Σ ΔHf°(reactants), each multiplied by the coefficient. The second law connects entropy to spontaneity: ΔS_universe > 0 for spontaneous processes. The Gibbs free energy ties enthalpy and entropy: ΔG = ΔH − TΔS. ΔG < 0: spontaneous; ΔG > 0: non-spontaneous; ΔG = 0: equilibrium. Temperature dependence determines whether endothermic-but-entropically-favored reactions become spontaneous at high T.

Kinetics describes how fast a reaction proceeds; thermodynamics describes how far. A reaction can be highly exothermic (favored thermodynamically) but slow (kinetically inhibited). Rate Law: rate = k[A]^m [B]^n, where m and n are reaction orders (determined experimentally, NOT from coefficients). Order is the exponent; total order = m + n. Integrated rate laws (for first-order): ln[A] = ln[A]_0 − kt Half-life t_1/2 = 0.693 / k (independent of [A]_0 for first-order; this is the characteristic property of first-order kinetics, including radioactive decay). For zero-order: [A] = [A]_0 − kt; t_1/2 = [A]_0 / (2k). For second-order: 1/[A] = 1/[A]_0 + kt; t_1/2 = 1/(k[A]_0). The Arrhenius equation describes temperature dependence of rate constants: k = A × e^(−Ea/RT), where Ea is activation energy. A 10 °C temperature increase typically doubles the rate (the "rule of 10"). Catalysts lower the activation energy by providing alternate pathways; they speed both forward and reverse reactions equally and do not change the equilibrium position. Reaction mechanisms describe the elementary step-by-step path. The rate-determining step is the slowest step; the overall rate law is determined by it. Experimentally measured order may not match the stoichiometry of the overall reaction — only the rate-determining step's mechanism predicts the observed rate law. Common error: using the overall balanced equation to predict reaction order. Order is determined experimentally from initial-rate data or from integrated rate-law plots, not from stoichiometric coefficients.

At equilibrium, forward and reverse reaction rates are equal; concentrations are constant (but not zero). The equilibrium constant K characterizes the position: K = [products]^coeff / [reactants]^coeff. Large K (>>1): products favored. Small K (<<1): reactants favored. For aA + bB ⇌ cC + dD: K = [C]^c [D]^d / ([A]^a [B]^b) Reaction quotient Q has the same form but uses non-equilibrium concentrations. Compare Q to K: Q < K: reaction proceeds forward (more products needed). Q > K: reaction proceeds backward (too many products). Q = K: at equilibrium. Le Chatelier's Principle: a system at equilibrium responds to stress by shifting toward the side that relieves the stress. • Add reactant → shift toward products (forward). • Add product → shift toward reactants (backward). • Increase pressure (decrease volume) → shift toward side with fewer moles of gas. • Increase temperature: endothermic → forward shift; exothermic → backward shift. • Catalyst: no shift (changes only rate). Worked Example. For N2 + 3H2 ⇌ 2NH3 (exothermic), what happens if you add N2? Q momentarily decreases (denominator increases). System responds by shifting forward to consume N2 and produce more NH3 until Q = K again. Result: more NH3, less H2, lower N2 (but still elevated above original). If you raise temperature on this exothermic reaction, system shifts backward (treating heat as a "product" that you are adding). Industrial Haber-Bosch process must balance: high pressure favors NH3 (Le Chatelier favors fewer-moles side); but high temperature actually disfavors NH3 (exothermic) — used anyway because rate at low T is impractically slow. Net practical conditions: ~400 °C, 200 atm, iron catalyst.

Brønsted-Lowry definition: acids are proton (H+) donors; bases are proton acceptors. Lewis definition: acids are electron-pair acceptors; bases are electron-pair donors (broader, includes non-proton-transfer reactions). Water autoionizes: 2H2O ⇌ H3O+ + OH-, with K_w = [H+][OH-] = 1.0 × 10^-14 at 25 °C. pH = -log[H+]; pOH = -log[OH-]; pH + pOH = 14 at 25 °C. Neutral water: pH = 7. Acidic: pH < 7. Basic: pH > 7. Strong acids (HCl, HNO3, H2SO4, HBr, HI, HClO4) ionize completely. [H+] = initial acid concentration. pH = -log(M). Weak acids (HF, CH3COOH, NH4+) partially ionize. Use Ka = [H+][A-]/[HA]. Worked Example. pH of 0.10 M acetic acid (Ka = 1.8 × 10^-5): HA ⇌ H+ + A- Let x = [H+] = [A-] at equilibrium; [HA] = 0.10 - x ≈ 0.10 (small-x approximation). Ka = x² / 0.10 = 1.8 × 10^-5 x² = 1.8 × 10^-6; x = 1.34 × 10^-3 M = [H+] pH = -log(1.34 × 10^-3) = 2.87 Valid the approximation: x = 0.00134, which is ~1.3% of 0.10 (under 5% rule, approximation is valid). Buffer solutions resist pH change. Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). When [A-] = [HA], pH = pKa. Buffers work best when pH is within ±1 of pKa. Titration curves show pH vs added titrant. Strong acid + strong base: equivalence at pH 7. Weak acid + strong base: equivalence at pH > 7 (basic conjugate base). Weak base + strong acid: equivalence at pH < 7. Half-equivalence point of a weak acid titration: pH = pKa. Polyprotic acids (H3PO4, H2CO3, citric) have multiple ionization steps with successive Ka values. Each step is treated as a separate equilibrium; lower Ka values are typically much smaller than the first.

General chemistry covers the densest set of problem types in the undergraduate curriculum, and most students struggle with one or two specific topics — often stoichiometry of limiting reactants, dilution calculations, or weak-acid pH. Snap a photo of any general chemistry problem and ChemistryIQ identifies the topic (mole, stoichiometry, gas law, solutions, thermochemistry, kinetics, equilibrium, or acid-base), sets up the appropriate equation, walks through the unit conversions, and produces the final answer with all intermediate steps. For multi-step problems involving more than one topic (e.g., a reaction yielding a gas at known T and P followed by absorption into a solution to give a known molarity), ChemistryIQ chains the topics in sequence. This content is for educational purposes only and does not constitute chemistry advice.

Six errors recur. First, using gram ratios instead of mole ratios in stoichiometry. The balanced equation gives MOLE ratios; convert to moles before applying. Second, using °C instead of Kelvin in gas law calculations. PV = nRT requires absolute T. Third, using the overall reaction equation to predict reaction order. Order is determined experimentally, not from stoichiometric coefficients (only the rate-determining step's mechanism predicts observed order). Fourth, forgetting to flip the sign of ΔH when reversing a reaction in Hess's law. Reverse the reaction: reverse the sign. Fifth, forgetting the small-x approximation validity check in weak-acid problems. If x exceeds 5% of initial concentration, solve the quadratic formula instead. Sixth, applying Le Chatelier inconsistently — adding inert gas at constant volume does NOT shift equilibrium (partial pressures of reactive species unchanged); adding inert gas at constant pressure DOES shift equilibrium (effectively dilutes).