AP Course

AP Chemistry

Updated for the AP Chemistry framework effective fall 2024, with particle models, quantitative reasoning, labs, quizzes, and practice problems.

Build general chemistry fluency through atomic structure, bonding, reactions, kinetics, equilibrium, acids and bases, thermodynamics, and electrochemistry.

Lessons

Moles and Molar MassMass Spectra of ElementsElemental Composition of Pure SubstancesComposition of MixturesAtomic Structure and Electron ConfigurationPhotoelectron SpectroscopyPeriodic TrendsValence Electrons and Ionic CompoundsTypes of Chemical BondsIntramolecular Force and Potential EnergyStructure of Ionic SolidsStructure of Metals and AlloysLewis DiagramsResonance and Formal ChargeVSEPR and HybridizationIntermolecular and Interparticle ForcesProperties of SolidsSolids, Liquids, and GasesIdeal Gas LawKinetic Molecular TheoryDeviation from Ideal Gas LawSolutions and MixturesRepresentations of SolutionsSeparation of Solutions and MixturesSolubilitySpectroscopy and the Electromagnetic SpectrumProperties of PhotonsBeer-Lambert LawIntroduction for ReactionsNet Ionic EquationsRepresentations of ReactionsPhysical and Chemical ChangesStoichiometryIntroduction to TitrationTypes of Chemical ReactionsIntroduction to Acid-Base ReactionsOxidation-Reduction (Redox) ReactionsReaction RatesIntroduction to Rate LawConcentration Changes over TimeElementary ReactionsCollision ModelReaction Energy ProfileIntroduction to Reaction MechanismsReaction Mechanism and Rate LawPre-Equilibrium ApproximationMultistep Reaction Energy ProfileCatalysisEndothermic and Exothermic ProcessesEnergy DiagramsHeat Transfer and Thermal EquilibriumHeat Capacity and CalorimetryEnergy of Phase ChangesIntroduction to Enthalpy of ReactionBond EnthalpiesEnthalpy of FormationHess’s LawIntroduction to EquilibriumDirection of Reversible ReactionsReaction Quotient and Equilibrium ConstantCalculating the Equilibrium ConstantMagnitude of the Equilibrium ConstantProperties of the Equilibrium ConstantCalculating Equilibrium ConcentrationsRepresentations of EquilibriumIntroduction to Le Châtelier’s PrincipleReaction Quotient and Le Châtelier’s PrincipleIntroduction to Solubility EquilibriaCommon-Ion EffectIntroduction to Acids and BasespH and pOH of Strong Acids and BasesWeak Acid and Base EquilibriaAcid-Base Reactions and BuffersAcid-Base TitrationsMolecular Structure of Acids and BasespH and pK_aProperties of BuffersHenderson-Hasselbalch EquationBuffer CapacitypH and SolubilityIntroduction to EntropyAbsolute Entropy and Entropy ChangeGibbs Free Energy and Thermodynamic FavorabilityThermodynamic and Kinetic ControlFree Energy and EquilibriumFree Energy of DissolutionCoupled ReactionsGalvanic (Voltaic) and Electrolytic CellsCell Potential and Free EnergyCell Potential Under Nonstandard ConditionsElectrolysis and Faraday’s Law

Quizzes

Practice Problems Equation practice, lab analysis, and FRQ-style reasoning resources can be added here.

AP Chemistry · Unit 1 · Reviewed representative lesson

Moles and Molar Mass

The mole connects a measurable sample to a count of microscopic entities. One mole contains exactly \(6.02214076\times10^{23}\) specified entities, while molar mass tells how many grams correspond to one mole of a substance.

Definition and Units

The amount \(n\) is measured in moles, sample mass \(m\) in grams, and molar mass \(M\) in grams per mole. The entity must be named: atoms, molecules, ions, or formula units are not interchangeable labels.

\[n=\frac{m}{M},\qquad N=nN_A\]

Here \(N\) is the number of specified entities and \(N_A=6.02214076\times10^{23}\,\mathrm{mol^{-1}}\).

Why the Conversion Works

Molar mass is a conversion factor. Dividing grams by grams per mole cancels grams and leaves moles. Multiplying moles by entities per mole then cancels moles and leaves a count of entities.

Worked Example

Question: Approximately how many water molecules are in \(18.0\,\mathrm g\) of \(\mathrm{H_2O}\), using \(M=18.0\,\mathrm{g\,mol^{-1}}\)?

\[n=\frac{18.0\,\mathrm g}{18.0\,\mathrm{g\,mol^{-1}}}=1.00\,\mathrm{mol}\]

\[N=(1.00\,\mathrm{mol})(6.022\times10^{23}\,\mathrm{mol^{-1}})=6.02\times10^{23}\]

Answer: The sample contains approximately \(6.02\times10^{23}\) water molecules.

Common Mistakes

Multiplying by molar mass when the units require division.
Reporting “particles” without naming the chemical entity.
Using an element's atomic mass as the molar mass of an entire compound.

Key Takeaways

Mass-to-mole conversion uses molar mass.
Mole-to-entity conversion uses the exact Avogadro constant.
Unit cancellation reveals whether the conversion direction is correct.

Practice connection

Use the quiz and practice tabs to convert compounds and ionic formula units while tracking significant figures.

Official curriculum reference: College Board AP Chemistry course page. The explanation and worked example are independently written for this study site.