ACS First Year Chemistry Comprehensive Review Notes

Atomic Structure and Fundamental Theory

  • Definition of an Atom: An atom represents the smallest unit of matter that retains chemical and physical properties (Question 1).
  • Nuclear Structure discovery: Ernest Rutherford is credited with discovering that the nucleus is extremely small compared to the overall volume of the atom (Question 47).
  • Atomic Composition and Isotopes:
    • Protons define the element's identity (the atomic number).
    • Isotopes of a single element have the same number of protons but different mass numbers due to varying numbers of neutrons.
    • Example for $_{7}^{15}\text{N}^{2+}$ (Question 16, 88):
      • Atomic Number: 77 (7 protons).
      • Mass Number: 1515
      • Neutron Count: 157=815 - 7 = 8 neutrons.
      • Electron Count: For a charge of +2+2, the atom has 72=57 - 2 = 5 electrons.
  • Periodic Trends:
    • Atomic Radius: Generally decreases across a row as the atomic number increases because the addition of more protons increases the effective nuclear charge, pulling electrons closer to the nucleus (Question 36).
    • Ionization Energy (IE):
      • In the alkaline-earth group, atoms with the smallest radii possess the highest ionization energies (Question 66).
      • Elements like nitrogen and oxygen generally have high first ionization energies (Question 37).
      • Successive Ionization Energies: Significant jumps in IE indicate the removal of core electrons. For an unknown element with IE values of 590kJ/mol590\,kJ/mol, 1145kJ/mol1145\,kJ/mol, 2412kJ/mol2412\,kJ/mol, 9474kJ/mol9474\,kJ/mol, and 12326kJ/mol12326\,kJ/mol, the large jump after the third electron suggests the most common ion is X3+X^{3+} (Question 28).
      • Second Ionization Energy: Sodium (Na) has a very high second IE because its second electron is removed from a stable, filled inner shell (Question 63).
    • Electronegativity: Oxygen is characterized by having a higher electronegativity compared to sodium, chlorine, xenon, or sulfur (Question 32).
  • Electronic Configuration:
    • Silicon (Si): In its ground state, silicon has 44 valence electrons and 22 unpaired electrons (Question 67).
    • Orbitals: There are exactly 55 d-orbitals in the fourth energy level (n=4n = 4) (Question 45).

Chemical Bonding and Molecular Geometry

  • Molecular Shapes (VSEPR Theory):
    • Methane (CH4CH_4): Tetrahedral geometry (Question 6).
    • Carbon Dioxide (CO2CO_2): Linear geometry (Question 33).
    • Water (H2OH_2O): Bent, angular, or V-shaped (Question 65).
    • Ammonia (NH3NH_3): Trigonal pyramid (Question 86).
    • Aluminum Chloride (AlCl3AlCl_3): Planar triangular (Question 82).
  • Polarity and Intermolecular Forces:
    • Polar Molecules: Ammonia (NH3NH_3) contains a permanent dipole (Question 15). Compounds like CCl4CCl_4, CO2CO_2, and Cl2Cl_2 are non-polar.
    • Hydrogen Bonding: This explains why methanol (CH3OHCH_3OH) has a higher boiling point than dimethyl ether (CH3OCH3CH_3OCH_3) (Question 80).
    • Ionic Character: The bond between potassium (K) and chlorine (Cl) exhibits high ionic character due to the large electronegativity difference (Question 74).
  • Lewis Structures: The hydroxide ion (OHOH^-) must show 88 electrons in its Lewis structure (11 from Hydrogen, 66 from Oxygen, and 11 from the negative charge) (Question 53).
  • Flame Tests: Sodium chloride (NaClNaCl) produces a characteristic yellowish-orange flame when placed in a Bunsen burner (Question 38).

States of Matter and Gas Laws

  • Kinetic Molecular Theory (KMT):
    • Average kinetic energy of gas molecules is determined solely by temperature. Different gases in identical conditions (same TT and PP) have the same average kinetic energy (Question 5, 68).
    • Real gases most closely resemble ideal gases under conditions of low pressure and high temperature (Question 14).
  • Gas Law Calculations:
    • Graham's Law of Effusion: Velocity is inversely proportional to the square root of molar mass (v1/Mv ∝ 1/\sqrt{M}). If oxygen (M=32M = 32) travels at 1000m/sec1000\,m/sec, hydrogen (M=2M = 2) travels at 4000m/sec4000\,m/sec because its mass is 1616 times smaller (16=4\sqrt{16} = 4) (Question 4).
    • Combined Gas Law: Used to find new volume when PP and TT change (P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}). For a sample at 300.0mL300.0\,mL, 1.00atm1.00\,atm, and 27.0C27.0^{\circ}C (300K300\,K), increasing temperature to 327C327^{\circ}C (600K600\,K) and decreasing pressure to 0.500atm0.500\,atm results in a volume of 1200mL1200\,mL (Question 12).
    • Boyle’s Law: (P1V1=P2V2P_1V_1 = P_2V_2). A 2.0L2.0\,L gas sample at 1.00atm1.00\,atm compressed to 500.0mL500.0\,mL requires a pressure of 4.0atm4.0\,atm (Question 87).
    • Charles’s Law: (V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}). A 5.0L5.0\,L gas at 50.0C50.0^{\circ}C (323K323\,K) heated to 100.0C100.0^{\circ}C (373K373\,K) at constant pressure expands to 5.8L5.8\,L (Question 51).
    • Dalton’s Law of Partial Pressures: The partial pressure of a dry gas collected over water is the total pressure minus the water vapor pressure (Pgas=PtotalPH2OP_{\text{gas}} = P_{\text{total}} - P_{\text{H}_2\text{O}}). At 600.0mmHg600.0\,mmHg and 40C40^{\circ}C (vapor pressure =55.3mmHg= 55.3\,mmHg), the gas pressure is 544.7mmHg544.7\,mmHg (Question 54).
  • Phase Changes: During boiling, the temperature of a liquid remains constant while the potential energy increases (Question 8).

Stoichiometry and Molar Relationships

  • Formula Weights and Molar Mass:
    • Calcium nitrate (Ca(NO3)2Ca(NO_3)_2): 0.20moles0.20\,moles has a mass of approximately 33g33\,g (Question 21).
    • Oxygen gas (O2O_2): 2.0moles2.0\,moles has a mass of 64g64\,g (2×32g/mol2 \times 32\,g/mol) (Question 27).
    • Magnesium chloride (MgCl2MgCl_2): Molar mass is approximately 95.3g/mol95.3\,g/mol (Question 44).
  • Reaction Stoichiometry:
    • Water Yield: From 2H2+O22H2O2H_2 + O_2 \rightarrow 2H_2O, 6.00moles6.00\,moles of H2H_2 and 6.00moles6.00\,moles of O2O_2 produce a maximum of 108g108\,g of water (since H2H_2 is limiting, producing 6.00moles6.00\,moles of H2OH_2O) (Question 17).
    • Gas Volume to Mass: To produce 44.8L44.8\,L of methane (CH4CH_4) at STP (which is 2.0moles2.0\,moles) via C+2H2CH4C + 2H_2 \rightarrow CH_4, one needs 4.0moles4.0\,moles of H2H_2, weighing 8.0g8.0\,g (Question 77).
    • Decomposition: 2KClO32KCl+3O22KClO_3 \rightarrow 2KCl + 3O_2. The decomposition of 4.00mol4.00\,mol of KClO3KClO_3 produces 6.00mol6.00\,mol (192g192\,g) of oxygen (Question 55).
    • Haber Process: (N2+3H22NH3N_2 + 3H_2 \rightarrow 2NH_3). Converting 12.0g12.0\,g of H2H_2 (6.0mol6.0\,mol) requires 2.0mol2.0\,mol of N2N_2 (Question 60).
  • Empirical Formulas:
    • A compound with 64g64\,g of Oxygen (4mol4\,mol) and 4g4\,g of Hydrogen (4mol4\,mol) has an empirical formula of HOHO (Question 75).
    • If empirical formula is NO2NO_2 (46amu46\,amu) and molecular mass is 8595amu85-95\,amu, the molecular formula is N2O4N_2O_4 (Question 71).

Solutions and Titrations

  • Concentration Calculations:
    • Molarity (MM): Moles of solute per Liter of solution. For example, 120g120\,g of NaOHNaOH (3mol3\,mol) in 600mL600\,mL (0.6L0.6\,L) results in a 5.0M5.0\,M solution (Question 84).
    • Volume and Concentration: A 0.40MNH30.40\,M \, NH_3 solution containing 3.4g3.4\,g (0.2mol0.2\,mol) of ammonia has a volume of 500mL500\,mL (Question 46).
  • Titrations:
    • Calculated using balanced equations (nacidMAVA=nbaseMBVBn_{\text{acid}}M_AV_A = n_{\text{base}}M_BV_B).
    • Titrating 100.0mL100.0\,mL of Ba(OH)2Ba(OH)_2 with 200.0mL200.0\,mL of 0.500MHNO30.500\,M \, HNO_3: Since Ba(OH)2Ba(OH)_2 produces two OHOH^-, the concentration is 0.500M0.500\,M (Question 2).
    • Titrating 50.0mL50.0\,mL of HClHCl with 40.0mL40.0\,mL of 0.400MNaOH0.400\,M \, NaOH yields an HClHCl molarity of 0.320M0.320\,M (Question 50).
    • In a strong acid/strong base titration, the pH at the end point is approximately 7.07.0, and salt/water are the primary species present (Question 81).
  • Solubility and Conductance:
    • Non-polar molecules like methane (CH4CH_4) are not very soluble in water (Question 11).
    • Strong Electrolytes: Dissociate completely and conduct electricity well (e.g., HNO3HNO_3, KClKCl solution) (Question 76, 90).
    • Weak Electrolytes: Dissociate only partially (Question 9).
    • Precipitates: Based on solubility guidelines, mixing KBrKBr and Hg(NO3)2Hg(NO_3)_2 or NH4ClNH_4Cl and Pb(NO3)2Pb(NO_3)_2 produces a precipitate (Question 39).
    • Net Ionic Equations: For silver nitrate and sodium sulfide: 2Ag+(aq)+S2(aq)Ag2S(s)2Ag^+(aq) + S^{2-}(aq) \rightarrow Ag_2S(s) (Question 40).
  • Colligative Properties: Solutes like sugar in distilled water depress the freezing point and elevate the boiling point (Question 61).

Acids, Bases, and pH

  • pH scale:
    • If [H3O+]=8.26×105M[H_3O^+] = 8.26 \times 10^{-5} \, M, the pH is between 44 and 55 (Question 3).
    • Distilled water concentration of OHOH^- is 107M10^{-7} \, M at neutral pH (Question 25).
    • A 104MNaOH10^{-4} \, M \, NaOH solution has a pH of 1010 (Question 62).
    • Basic substances (pH>7.0pH > 7.0) include dish soap (Question 23).
  • Properties of Bases: They taste bitter, feel slippery, and turn red litmus paper blue (Question 13).

Chemical Equilibrium

  • Le Châtelier’s Principle:
    • Pressure: In the system 2CO(g)+O2(g)2CO2(g)2CO(g) + O_2(g) \rightleftharpoons 2CO_2(g), increasing total pressure shifts equilibrium toward the side with fewer gas moles (CO2CO_2 increases, COCO decreases) (Question 18).
    • Concentration: In CH3COOH(aq)+H2O(l)H3O+(aq)+CH3COO(aq)CH_3COOH(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3COO^-(aq), CH3COOHCH_3COOH is the species present in highest concentration because it is a weak acid (Question 7).
    • Temperature: Increasing temperature on an endothermic reaction (e.g., CH3OH+101kJCO+2H2CH_3OH + 101 \, kJ \rightleftharpoons CO + 2H_2) shifts equilibrium to the products, increasing [CO][CO] (Question 59).
  • Equilibrium Constant (KK):
    • A very high value of KK indicates that products are heavily favored (Question 64).
    • Changing the concentration of reactants/products at a constant temperature does not change the value of the equilibrium constant KK (Question 85).

Thermochemistry and Nuclear Chemistry

  • Thermodynamics:
    • Specific Heat: Energy (QQ) absorbed by 10.0g10.0\,g of water heated from 10.0C10.0^{\circ}C to 40.0C40.0^{\circ}C is calculated as 10.0×4.2×30.0=1260J10.0 \times 4.2 \times 30.0 = 1260\,J or 1.26kJ1.26\,kJ (Question 34).
    • Units: Energy as heat is measured in Joules (JJ) or Kilojoules (kJkJ) (Question 42).
    • Gibbs Free Energy: (ΔG=ΔHTΔS\Delta G = \Delta H - T\Delta S). For a reaction where ΔH=150kJ/mol\Delta H = -150\,kJ/mol, ΔS=+200J/molK\Delta S = +200\,J/mol\cdot K, and T=300KT = 300\,K, the value is 210kJ/mol-210\,kJ/mol, making the reaction spontaneous (Question 58).
    • Entropy and Enthalpy: The process of ice freezing is exothermic and results in a decrease in entropy (Question 73).
  • Redox and Reactions:
    • Oxidation Numbers: Sulfur in Na2SO4Na_2SO_4 has an oxidation number of +6+6 (Question 10).
    • Single Replacement: Reactions such as 2NaBr+Cl22NaCl+Br22NaBr + Cl_2 \rightarrow 2NaCl + Br_2 (Question 35).
  • Nuclear Chemistry:
    • Penetrating Power: Gamma radiation has greater penetrating power and velocity compared to alpha and beta radiation (Question 56).
    • Half-life: If 125g125\,g remains of a 500g500\,g sample (which is two half-lives, 1/41/4 original) after 8.08.0 years, the half-life is 4.04.0 years (Question 78).
    • Balancing Nuclear Equations: Bombarding a nucleus with an alpha particle (24He_{2}^{4}He) results in transmutations that must conserve mass and atomic numbers (Question 69).

Measurement, Formulas, and Properties

  • Scientific Notation and Sig Figs:
    • 0.00930m0.00930 \, m is expressed as 9.30×103m9.30 \times 10^{-3} \, m (Question 22).
    • 50.0cm50.0 \, cm is same as 500mm500 \, mm when significant figures are considered (Question 83).
  • Chemical vs. Physical Changes: Boiling and melting are physical changes; rusting and burning are chemical changes (Question 52).
  • Naming Compounds:
    • Fe(NO3)2Fe(NO_3)_2 is Iron(II) nitrate (Question 20).
    • Aluminum sulfate is Al2(SO4)3Al_2(SO_4)_3 (Question 57).
    • Compound formed by lead(IV) and carbonate is Pb(CO3)2Pb(CO_3)_2 (Question 48).
  • Mixtures: A solution of sugar and water is a homogeneous mixture, whereas garden soil is heterogeneous (Question 31).