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: 7 (7 protons).
Mass Number: 15
Neutron Count: 15−7=8 neutrons.
Electron Count: For a charge of +2, the atom has 7−2=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/mol, 1145kJ/mol, 2412kJ/mol, 9474kJ/mol, and 12326kJ/mol, the large jump after the third electron suggests the most common ion is X3+ (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 4 valence electrons and 2 unpaired electrons (Question 67).
Orbitals: There are exactly 5 d-orbitals in the fourth energy level (n=4) (Question 45).
Polar Molecules: Ammonia (NH3) contains a permanent dipole (Question 15). Compounds like CCl4, CO2, and Cl2 are non-polar.
Hydrogen Bonding: This explains why methanol (CH3OH) has a higher boiling point than dimethyl ether (CH3OCH3) (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 (OH−) must show 8 electrons in its Lewis structure (1 from Hydrogen, 6 from Oxygen, and 1 from the negative charge) (Question 53).
Flame Tests: Sodium chloride (NaCl) 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 T and P) 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 (v∝1/M). If oxygen (M=32) travels at 1000m/sec, hydrogen (M=2) travels at 4000m/sec because its mass is 16 times smaller (16=4) (Question 4).
Combined Gas Law: Used to find new volume when P and T change (T1P1V1=T2P2V2). For a sample at 300.0mL, 1.00atm, and 27.0∘C (300K), increasing temperature to 327∘C (600K) and decreasing pressure to 0.500atm results in a volume of 1200mL (Question 12).
Boyle’s Law: (P1V1=P2V2). A 2.0L gas sample at 1.00atm compressed to 500.0mL requires a pressure of 4.0atm (Question 87).
Charles’s Law: (T1V1=T2V2). A 5.0L gas at 50.0∘C (323K) heated to 100.0∘C (373K) at constant pressure expands to 5.8L (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=Ptotal−PH2O). At 600.0mmHg and 40∘C (vapor pressure =55.3mmHg), the gas pressure is 544.7mmHg (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)2): 0.20moles has a mass of approximately 33g (Question 21).
Oxygen gas (O2): 2.0moles has a mass of 64g (2×32g/mol) (Question 27).
Magnesium chloride (MgCl2): Molar mass is approximately 95.3g/mol (Question 44).
Reaction Stoichiometry:
Water Yield: From 2H2+O2→2H2O, 6.00moles of H2 and 6.00moles of O2 produce a maximum of 108g of water (since H2 is limiting, producing 6.00moles of H2O) (Question 17).
Gas Volume to Mass: To produce 44.8L of methane (CH4) at STP (which is 2.0moles) via C+2H2→CH4, one needs 4.0moles of H2, weighing 8.0g (Question 77).
Decomposition: 2KClO3→2KCl+3O2. The decomposition of 4.00mol of KClO3 produces 6.00mol (192g) of oxygen (Question 55).
Haber Process: (N2+3H2→2NH3). Converting 12.0g of H2 (6.0mol) requires 2.0mol of N2 (Question 60).
Empirical Formulas:
A compound with 64g of Oxygen (4mol) and 4g of Hydrogen (4mol) has an empirical formula of HO (Question 75).
If empirical formula is NO2 (46amu) and molecular mass is 85−95amu, the molecular formula is N2O4 (Question 71).
Solutions and Titrations
Concentration Calculations:
Molarity (M): Moles of solute per Liter of solution. For example, 120g of NaOH (3mol) in 600mL (0.6L) results in a 5.0M solution (Question 84).
Volume and Concentration: A 0.40MNH3 solution containing 3.4g (0.2mol) of ammonia has a volume of 500mL (Question 46).
Titrations:
Calculated using balanced equations (nacidMAVA=nbaseMBVB).
Titrating 100.0mL of Ba(OH)2 with 200.0mL of 0.500MHNO3: Since Ba(OH)2 produces two OH−, the concentration is 0.500M (Question 2).
Titrating 50.0mL of HCl with 40.0mL of 0.400MNaOH yields an HCl molarity of 0.320M (Question 50).
In a strong acid/strong base titration, the pH at the end point is approximately 7.0, and salt/water are the primary species present (Question 81).
Solubility and Conductance:
Non-polar molecules like methane (CH4) are not very soluble in water (Question 11).
Strong Electrolytes: Dissociate completely and conduct electricity well (e.g., HNO3, KCl solution) (Question 76, 90).
Weak Electrolytes: Dissociate only partially (Question 9).
Precipitates: Based on solubility guidelines, mixing KBr and Hg(NO3)2 or NH4Cl and Pb(NO3)2 produces a precipitate (Question 39).
Net Ionic Equations: For silver nitrate and sodium sulfide: 2Ag+(aq)+S2−(aq)→Ag2S(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+]=8.26×10−5M, the pH is between 4 and 5 (Question 3).
Distilled water concentration of OH− is 10−7M at neutral pH (Question 25).
A 10−4MNaOH solution has a pH of 10 (Question 62).
Basic substances (pH>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), increasing total pressure shifts equilibrium toward the side with fewer gas moles (CO2 increases, CO decreases) (Question 18).
Concentration: In CH3COOH(aq)+H2O(l)⇌H3O+(aq)+CH3COO−(aq), CH3COOH 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+101kJ⇌CO+2H2) shifts equilibrium to the products, increasing [CO] (Question 59).
Equilibrium Constant (K):
A very high value of K 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 K (Question 85).
Thermochemistry and Nuclear Chemistry
Thermodynamics:
Specific Heat: Energy (Q) absorbed by 10.0g of water heated from 10.0∘C to 40.0∘C is calculated as 10.0×4.2×30.0=1260J or 1.26kJ (Question 34).
Units: Energy as heat is measured in Joules (J) or Kilojoules (kJ) (Question 42).
Gibbs Free Energy: (ΔG=ΔH−TΔS). For a reaction where ΔH=−150kJ/mol, ΔS=+200J/mol⋅K, and T=300K, the value is −210kJ/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 Na2SO4 has an oxidation number of +6 (Question 10).
Single Replacement: Reactions such as 2NaBr+Cl2→2NaCl+Br2 (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 125g remains of a 500g sample (which is two half-lives, 1/4 original) after 8.0 years, the half-life is 4.0 years (Question 78).
Balancing Nuclear Equations: Bombarding a nucleus with an alpha particle (24He) results in transmutations that must conserve mass and atomic numbers (Question 69).
Measurement, Formulas, and Properties
Scientific Notation and Sig Figs:
0.00930m is expressed as 9.30×10−3m (Question 22).
50.0cm is same as 500mm 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)2 is Iron(II) nitrate (Question 20).
Aluminum sulfate is Al2(SO4)3 (Question 57).
Compound formed by lead(IV) and carbonate is Pb(CO3)2 (Question 48).
Mixtures: A solution of sugar and water is a homogeneous mixture, whereas garden soil is heterogeneous (Question 31).