KAP Chemistry: Atomic Structure, Nuclear Chemistry, and Dimensional Analysis

  • PART 1 — ATOMIC STRUCTURE

    • Atom: smallest particle of an element that retains its properties.
    • Subatomic particles
    • Electron (e⁻): charge −1; mass negligible; located in electron cloud.
    • Proton (p⁺): charge +1; mass ≈ 1.673×10⁻²⁴ g; located in nucleus.
    • Neutron (n⁰): charge 0; mass ≈ 1.675×10⁻²⁴ g; located in nucleus.
    • Mass, charge, and volume
    • Mass: almost all mass in nucleus; nucleus densely packed (high density).
    • Volume: electron cloud accounts for most of atom’s volume.
    • Mass vs charge vs volume: nucleus = mass; electron cloud = volume.
    • Protons and electrons have same magnitude of charge (+1 vs −1) but vastly different masses.
    • Mass of electrons is negligible in atomic mass calculations.

  • PART 1 — Atomic Number, Ions, Isotopes, Mass Number and Average Atomic Mass

    • Atomic number Z: number of protons; defines the element; unique for each element.
    • In a neutral atom: electrons = protons (electrons = Z).
    • Ions: neutral atoms can gain or lose electrons without changing the element.
    • Cation: positively charged (loss of electrons). Usually from metals.
    • Anion: negatively charged (gain of electrons). Usually from nonmetals.
    • Naming: cations retain element name; anions end with -ide (e.g., chloride).
    • Notation: element symbol with charge, e.g., Na⁺, Ca²⁺, Cl⁻.
    • Isotopes: same element (same Z) with different number of neutrons (N).
    • Mass number A = Z + N.
    • Isotope notation
    • Neutral atom: X with mass number: e.g., ${}^{12}\mathrm{C}$ or ${}^{12}{6}\mathrm{C}$; forms also ${}^{A}\mathrm{X}$ or ${}^{A}{Z}\mathrm{X}$.
    • Ion notation: isotope notation followed by charge, e.g., ${}^{23}{11}\mathrm{Na}^{+}$, ${}^{32}{16}\mathrm{S}^{2-}$.
    • APE MAN
    • Atomic number = Protons = Electrons (in neutral atom).
    • Mass number = Protons + Neutrons.
    • Isotopes of carbon (example)
    • Carbon-12: Z = 6; protons = 6; neutrons = 6; mass = 12.
    • Carbon-13: protons = 6; neutrons = 7; mass = 13.
    • Carbon-14: protons = 6; neutrons = 8; mass = 14.
    • Practice concept: electrons have negligible mass; use Z for element identity; A for isotope identity.

  • PART 1 — Isotopes, Neutrons, and Abundances

    • Isotopes differ in neutrons; mass number changes accordingly.
    • Mass on the periodic table is average atomic mass (weighted average of isotopes).
    • 1 amu = ${1.660538 \times 10^{-24}}$ g; carbon-12 defined as exactly 12 amu.
    • Average atomic mass is the weighted average of isotope masses by their natural abundances:
    • ${\bar{M}} = \sumi (mi \times fi)$ where $fi$ is the fractional abundance.
    • Example: Lithium isotopes 6.015 amu (7.5% = 0.075) and 7.016 amu (92.5% = 0.925)
    • ${\bar{M}} = (6.015 \times 0.075) + (7.016 \times 0.925) \approx 6.94\,\text{amu}$ (check against table).
    • Practice data: Mg and Ne isotopes and their abundances.

  • PART 2 — NUCLEAR CHEMISTRY

    • Nuclear chemistry: study of the nucleus and its changes; may form new elements.
    • Radioactivity: discovered by Becquerel; Curie contributions; background context (not required to memorize).
    • Why decay occurs: unstable isotopes; decay to achieve a more stable neutron:proton ratio.
    • Nuclear notation for particles
    • Alpha (α): $^{4}_{2}\mathrm{He}$ (helium nucleus).
    • Beta (β): $^{0}_{-1}e$ (high-speed electron formed in the nucleus).
    • Neutron: $^{1}_{0}n$.
    • Proton: $^{1}_{1}p$.
    • Examples of decay
    • Alpha decay: $^{238}{92}\mathrm{U} \rightarrow {}^{234}{90}\mathrm{Th} + {}^{4}_{2}\mathrm{He}$
    • Beta decay: $^{210}{82}\mathrm{Pb} \rightarrow {}^{210}{83}\mathrm{Bi} + {}^{0}_{-1}\mathrm{e}$
    • Transmutation: identity of element changes when protons/neutrons change; occurs in alpha and beta decay; gamma emission is energy, not a particle, and not shown as a particle in the equation.
    • Comparison: Alpha vs Beta vs Gamma
    • Alpha: largest mass, +2 charge, least penetrating; stopped by paper.
    • Beta: smaller than alpha, −1 charge, moderate penetration; stopped by aluminum/wood/glass.
    • Gamma: no mass or charge; most penetrating; requires lead/large shielding.
    • Half-life: time for half the sample to decay; isotope-dependent; constant (first-order process).
    • Nuclear reactions
    • Fission: heavy nucleus splits into smaller nuclei; can be chain reactions (U-235, Pu-239).
    • Fusion: small nuclei fuse to form larger nucleus; releases more energy per mass than fission; powers stars; difficult to achieve and control.
    • Energy release: $E = mc^{2}$; small mass conversion yields huge energy (e.g., 1 g converts to energy equivalent to ~7×10^8 L of gasoline).
    • Practice prompts
    • Identify fission vs fusion vs chemical reaction in provided reactions; chemical reactions conserve elements, nuclei unchanged; nuclear reactions form new elements.
    • Example: $^{1}{1}\mathrm{H} + {}^{3}{1}\mathrm{H} \rightarrow {}^{4}{2}\mathrm{He} + {}^{1}{0}\mathrm{n}$
    • Nuclear chemistry applications
    • Fission power: used in nuclear power plants; safety concerns and radioactive waste.
    • Fusion power: difficult to achieve; produces many neutrons; safety and containment challenges; potential for higher energy output; used in weapons research.
    • Radiation in medicine: imaging (X-ray, CT, PET); diagnostic isotopes (e.g., I-123);
      radiation therapy for cancer.
    • Carbon dating: C-14 dating relies on the constant ratio in living organisms; after death, C-14 decays; age inferred by remaining C-14.

  • PART 3 — DIMENSIONAL ANALYSIS

    • Dimensional Analysis (factor-label method): convert units using conversion factors; a conversion factor equals 1 (numerator and denominator represent equal quantities).
    • Conversion factors
    • Write as fractions: e.g., $2.44\;\$/dozen = 1$ dozen / $2.44$ or $1$ figuratively equals the other quantity.
    • Examples of common factors:
      • 60 s = 1 min
      • 1 \$ = 100¢
      • 1 L = 1000 mL
      • 1 km = 10^3 m
      • 1 mi = 1.609 km
      • 1 gal = 3.785 L
    • Using conversion factors
    • Align units so that units cancel from numerator and denominator, leaving desired unit.
    • Problem example: 3480 s to minutes using $1\text{ min} = 60\text{ s}$ → 58 min.
    • Three major pieces in any dimensional analysis problem
    • Given amount and unit (start here)
    • Unknown amount and unit (target)
    • Conversion factor(s) connecting the two units
    • Practice problems (illustrative, not exhaustive here)
    • Mass in g from moles: Na: 2.38 mol Na × (22.99 g/mol) = 54.76 g Na (example format)
    • Gas volume and moles at STP: 15.9 L with 22.4 L per mole → 0.710 mol
    • Metric conversions (do not move the decimal; use base units)
    • 587,000 m to km: 587 km
    • 4.56 L to mL: 4.56×10^3 mL
    • Complex unit conversions
    • Involve units like g/mL, kg/L, etc.
    • Example: 65 mi/hr to m/s; 15 mi/gal to km/L; 0.78 g/mL to lbs/gal
    • Useful conversion factors
    • A chart is provided with many factors; you do not need to memorize them all; consult the chart as needed.

  • NOTE: Values, constants, and example numbers are taken from the provided notes. Use the defined relationships (Z, A, N, isotopes, half-life, E=mc^2, etc.) for recall and quick problem solving.