Key Concepts:
Classes of Matter:
Pure substances: Elements (e.g., O2) and Compounds (e.g., H2O).
Mixtures: Homogeneous (e.g., saltwater) vs. Heterogeneous (e.g., oil and water).
Properties of Matter:
Physical properties: Measured without changing substance identity (e.g., boiling point, color).
Chemical properties: Observable during a reaction (e.g., flammability, rusting).
Changes:
Physical change: Alter appearance, not composition (e.g., melting ice).
Chemical change: Create new substances (e.g., burning wood).
Intensive vs. Extensive Properties:
Intensive: Independent of amount (e.g., density, boiling point).
Extensive: Dependent on amount (e.g., mass, volume).
Examples:
Identify if mixing vinegar and baking soda is a physical or chemical change.
Determine if density is an intensive or extensive property.
Key Concepts:
Scientific Notation:
Convert 0.000560 to 5.6×10−4
Write 7.89×103 as 7890
SI Units:
Mass: kilogram (kg)
Length: meter (m)
Temperature: Kelvin (K)
Volume: cubic meter (m³), derived from length.
SI Prefixes:
Memorize relationships like 1 Gm=109 m,1 nm=10−9
Examples:
Convert 8.38 kilograms to grams.
Rewrite 0.045 in millimeters.
Key Concepts:
States: Solid (fixed shape), Liquid (flows), Gas (expands).
Physical Changes: Melting, freezing, evaporation.
Examples:
Identify the state change when water boils (liquid to gas).
Key Concepts:
Counting: 0.00450 has 3 significant figures.
Rules for Operations:
Addition/Subtraction: Round to least decimal places.
Multiplication/Division: Round to least significant figures.
Examples:
Add 12.1+3.45=15.6 (rounded to 1 decimal place).
Multiply 4.56×0.020=0.091 (3 significant figures).
Key Concepts:
Conversion Formulas:
Celsius to Kelvin:K=C+273.15
Fahrenheit to Celsius:C=5/9(F−32)
Examples:
Convert 25∘C to Kelvin.
Convert 98.6∘F to Celsius.
Key Concepts:
Protons, Neutrons, Electrons:
Proton: +1, in nucleus.
Neutron: 0, in nucleus.
Electron: −1, orbits nucleus.
Examples:
For Carbon-12 (612C): Protons = 6, Neutrons = 6, Electrons = 6.
For an ion Mg2+: Protons = 12, Neutrons = 12, Electrons = 10.
Key Concepts:
Molar Mass: Sum of atomic masses (e.g., H2O: 2(1.01)+16.00=18.02 g/mol
Mole Concept:
1 mole=6.022×1023 particles
Examples:
Find moles in 36.04 g of water.
Convert 3.01×1023 molecules of CO2 to moles.
Key Concepts:
Identify isotopes by mass spectra.
Use peaks to infer molecular mass.
Examples:
Interpret a spectrum to find the mass of Cl2.
Key Concepts:
Relationships:
λν=c (Speed of light, c=3.00×108 m/s
E=hν (Planck's constant, h=6.626×10−34 J.
Examples:
Calculate the energy of a photon with ν=5×1014 Hz
Relate UV light to higher energy compared to visible light.
Key Concepts:
Quantum Numbers:
Principal (n): Energy level.
Angular (l): Shape.
Magnetic (mll): Orientation.
Spin (mss): +1/2,−1/2
Orbital Shapes:
s: Spherical, p: Dumbbell, d: Clover.
Examples:
Assign quantum numbers to an electron in 2p32p3.
Sketch 3d3d orbital shape.
Key Concepts:
Effective Nuclear Charge (Zeff):
Increases across a period (more protons).
Remains relatively constant down a group (shielding effect).
Atomic Radius:
Decreases across a period (higher Zeff).
Increases down a group (additional electron shells).
Ionization Energy (IE):
First IE increases across a period (stronger attraction to nucleus).
Decreases down a group (electrons further from nucleus).
Successive IEs increase dramatically after removing valence electrons.
Electron Affinity (EA):
Energy change when an atom gains an electron.
Negative EA: Energy is released (e.g., chlorine).
Electronegativity:
Tendency to attract electrons in a bond.
Increases across a period; decreases down a group.
Examples:
Predict which is larger: Na or Na+ (Answer: Na).
Compare the IE of Mg and Al (Answer: Al has a slightly lower IE due to sublevel shielding).
Key Concepts:
Ionization:
Example: Mg→Mg++ e−
Electron Affinity:
Example: Cl+e−→Cl− (ΔE=−X).
Examples:
Write the equation for the second ionization of calcium.
Describe the sign and magnitude of the EA for sulfur.
Key Concepts:
Ionic Compounds: Metal + nonmetal (e.g., NaCl).
Molecular Compounds: Nonmetal + nonmetal (e.g., CO2).
Examples:
Determine if K2O is ionic or molecular (Answer: Ionic).
Key Concepts:
Bond polarity is determined by the electronegativity difference:
Nonpolar: ΔEN<0.5 (e.g., H-H).
Polar: 0.5≤ΔEN≤1.7 (e.g., H-F).
Ionic: ΔEN>1.7 (e.g., NaCl).
Examples:
Rank HCl,H2,NaCl in increasing bond polarity.
Key Concepts:
Ionic Compounds: Use metal name + nonmetal with “-ide” (e.g., magnesium chloride).
Molecular Compounds: Use prefixes (mono-, di-, tri-) (e.g., carbon dioxide).
Examples:
Name SO3 (Answer: Sulfur trioxide).
Write the formula for aluminum nitrate (Answer: Al(NO3)3)
Key Concepts:
Represent valence electrons around atoms.
Include lone pairs, bonds, and formal charges.
Follow the octet rule; consider exceptions (e.g., BF3).
Steps:
Count total valence electrons.
Assign bonds and lone pairs.
Minimize formal charges.
Examples:
Draw the Lewis structure for SO2 and predict its resonance structures.
Key Concepts:
Predict molecular shape based on electron-pair repulsion.
Electron-pair geometry (EPG) vs. molecular geometry (MG).
Bond angles vary with lone pairs (e.g., H2O: bent, NH3: trigonal pyramidal).
Examples:
Predict the geometry of CH4, SF4, and ICl3
Key Concepts:
A molecule’s polarity depends on:
Bond dipoles.
Symmetry of the molecule.
Examples:
Explain why CO2 is nonpolar but H2O is polar.
Key Concepts:
Orbitals overlap to form bonds:
Sigma (σ) bonds: Head-on overlap.
Pi (π) bonds: Side-by-side overlap.
Hybridization:
sp (linear, 180°), sp2 (trigonal planar, 120°), sp3 (tetrahedral, 109.5°).
Examples:
Identify the hybridization of the central atom in NH3
Count the number of σ and π bonds in C2H4
NameFormula | |
Water | H2O |
Ammonia | NH3 |
Methane | CH4 |
Carbon dioxide | CO2 |
Sodium chloride | NaCl |
Calcium carbonate | CaCO3 |
Sulfuric acid | H2SO4 |
Nitric acid | HNO3 |
Glucose | C6H12O6 |
Ethanol | C2H5OH |
Magnesium hydroxide | Mg(OH)2 |
Iron(III) oxide | Fe2O3 |
Aluminum sulfate | Al2(SO4)3 |
Sodium bicarbonate | NaHCO3 |
Phosphoric acid | H3PO4 |
Electron Groups | Electron Geometry | Molecular Geometry | Bond Angles | Examples |
2 | Linear | Linear | 180∘ | CO2, BeCl2 |
3 | Trigonal planar | Trigonal planar | 120∘ | BF3, SO3 |
3 | Trigonal planar | Bent (1 lone pair) | <120∘ | SO2 |
4 | Tetrahedral | Tetrahedral | 109.5∘ | CH4, NH4+ |
4 | Tetrahedral | Trigonal pyramidal (1 lone pair) | <109.5∘ | NH3 |
4 | Tetrahedral | Bent (2 lone pairs) | <109.5∘ | H2O |
5 | Trigonal bipyramidal | Trigonal bipyramidal | 90∘, 120∘ | PCl5 |
5 | Trigonal bipyramidal | Seesaw (1 lone pair) | <90∘, <120∘ | SF4 |
5 | Trigonal bipyramidal | T-shaped (2 lone pairs) | <90∘ | ClF3 |
5 | Trigonal bipyramidal | Linear (3 lone pairs) | 180∘ | XeF2 |
6 | Octahedral | Octahedral | 90∘ | SF6 |
6 | Octahedral | Square pyramidal (1 lone pair) | <90∘ | BrF5 |
6 | Octahedral | Square planar (2 lone pairs) | 90∘ | XeF4 |
NameFormula | |
Water | H2OH2O |
Ammonia | NH3NH3 |
Methane | CH4CH4 |
Carbon dioxide | CO2CO2 |
Sodium chloride | NaClNaCl |
Calcium carbonate | CaCO3CaCO3 |
Sulfuric acid | H2SO4H2SO4 |
Nitric acid | HNO3HNO3 |
Glucose | C6H12O6C6H12O6 |
Ethanol | C2H5OHC2H5OH |
Magnesium hydroxide | Mg(OH)2Mg(OH)2 |
Iron(III) oxide | Fe2O3Fe2O3 |
Aluminum sulfate | Al2(SO4)3Al2(SO4)3 |
Sodium bicarbonate | NaHCO3NaHCO3 |
Phosphoric acid | H3PO4H3PO4 |
Electron GroupsElectron GeometryMolecular GeometryBond AnglesExamples | ||||
2 | Linear | Linear | 180∘180∘ | CO2CO2, BeCl2BeCl2 |
3 | Trigonal planar | Trigonal planar | 120∘120∘ | BF3BF3, SO3SO3 |
3 | Trigonal planar | Bent (1 lone pair) | <120∘<120∘ | SO2SO2 |
4 | Tetrahedral | Tetrahedral | 109.5∘109.5∘ | CH4CH4, NH4+NH4+ |
4 | Tetrahedral | Trigonal pyramidal (1 lone pair) | <109.5∘<109.5∘ | NH3NH3 |
4 | Tetrahedral | Bent (2 lone pairs) | <109.5∘<109.5∘ | H2OH2O |
5 | Trigonal bipyramidal | Trigonal bipyramidal | 90∘90∘, 120∘120∘ | PCl5PCl5 |
5 | Trigonal bipyramidal | Seesaw (1 lone pair) | <90∘<90∘, <120∘<120∘ | SF4SF4 |
5 | Trigonal bipyramidal | T-shaped (2 lone pairs) | <90∘<90∘ | ClF3ClF3 |
5 | Trigonal bipyramidal | Linear (3 lone pairs) | 180∘180∘ | XeF2XeF2 |
6 | Octahedral | Octahedral | 90∘90∘ | SF6SF6 |
6 | Octahedral | Square pyramidal (1 lone pair) | <90∘<90∘ | BrF5BrF5 |
6 | Octahedral | Square planar (2 lone pairs) | 90∘90∘ | XeF4XeF4 |
Hybridization | Electron Groups | Geometry | Bond Angles | Examples |
sp | 2 | Linear | 180∘ | CO2, C2H2 |
sp2 | 3 | Trigonal planar | 120∘ | BF3, C2H4 |
sp3 | 4 | Tetrahedral | 109.5∘ | CH4, NH3 |
sp3d | 5 | Trigonal bipyramidal | 90∘,120∘ | PCl5, SF4 |
sp3d2 | 6 | Octahedral | 90∘ | SF6, XeF4 |
Reading: Section 7.2
Objective:
Balance chemical equations for reactions while adhering to the law of conservation of mass.
Key Concepts:
Steps for Balancing Equations:
Write the unbalanced equation.
Count the number of atoms of each element on both sides.
Adjust coefficients to balance each element (start with elements appearing in a single reactant and product).
Check to ensure coefficients are in the lowest ratio.
Verify that all atoms balance and total charge (if applicable) is conserved.
Combustion Reactions: Hydrocarbon + O₂ → CO₂ + H₂O
Example:
C3H8+O2→CO2+H2O
Balanced: C3H8+5O2→3CO2+4H2O
Practice Problems:
Ch. 7: 20, 25a-b, 27
Reading: Sections 7.3 – 7.4
Objective:
Use stoichiometric calculations to determine reactant consumption and product formation.
Identify the limiting reactant and calculate leftover reagents.
Key Concepts:
Mole Ratio: Derived from the coefficients in a balanced equation.
Limiting Reactant: The reactant that determines the maximum amount of product formed.
Steps for Limiting Reactant Problems:
Balance the chemical equation.
Convert all reactants to moles.
Use the mole ratio to calculate the theoretical product yield for each reactant.
The reactant producing the least product is the limiting reactant.
Calculate the excess amount of other reactants.
Example:
2H2+O2→2H2O
Given: 3 moles of H₂ and 2 moles of O₂.
H₂: 3 moles H2×2 moles H2O2 moles H2=3 moles H2O3 moles H2×2 moles H22 moles H2O=3 moles H2O
O₂: 2 moles O2×2 moles H2O1 mole O2=4 moles H2O2 moles O2×1 mole O22 moles H2O=4 moles H2O
Limiting Reactant: H₂; produces 3 moles of water.
Practice Problems:
Ch. 7: 5, 35, 41, 49, 51
Reading: Sections 7.4 – 7.7
Objective:
Determine the efficiency of a reaction (percent yield).
Derive formulas based on mass and molar composition.
Key Concepts:
Percent Yield: %Yield=(Actual Yield/Theoretical Yield)×100%
Empirical Formula Steps:
Convert % composition to grams (assume 100g sample).
Convert grams to moles using molar mass.
Divide by the smallest mole value to find whole-number ratios.
Example:
Compound with 40.0% C, 6.7% H, and 53.3% O.
C: 40.0/ 12.01=3.33, H: 6.7/1.008=6.65, O: 53.3/16.00=3.33
Mole Ratio: C = 1, H = 2, O = 1 → Empirical Formula = CH₂O
Practice Problems:
Ch. 7: 94, 97, 3, 71, 78
Reading: Sections 8.1 – 8.3
Objective:
Calculate concentrations of solutions and determine outcomes of dilutions.
Key Concepts:
Molarity Formula: M=moles of soluteliters of solutionM=liters of solutionmoles of solute
Dilution Formula: M1V1=M2V2
Example:
Calculate the molarity of a solution with 0.5 moles of NaCl in 2.0 L of solution.
M=0.52.0=0.25 M NaCl
Practice Problems:
Ch. 8: 13, 15, 19a-b, 29, 32
Reading: Sections 8.3 – 8.5
Objective:
Understand acid-base properties and equations.
Key Concepts:
Strong Acids and Bases: Completely dissociate in water.
Weak Acids and Bases: Partially dissociate.
Neutralization Reaction:
Acid+Base→Salt+WaterAcid+Base→Salt+Water
Practice Problems:
Ch. 8: 1, 51, 53, 57, 62
Reading: Sections 8.4, 8.6
Key Concepts:
Use solubility rules (Table 8.4) to predict precipitation.
Identify spectator ions in net ionic equations.
Example:
BaCl2(aq)+Na2SO4(aq)→BaSO4(s)+2NaCl(aq)
Net Ionic:
Ba2+(aq)+SO42−(aq)→BaSO4(s)
Practice Problems:
Ch. 8: 67, 69, 73
Reading: Section 8.7
Objective:
Assign oxidation states and identify redox components.
Key Concepts:
Rules for Oxidation States:
Elements in elemental form = 0.
Group 1 metals = +1; Group 2 metals = +2.
Oxygen = -2 (except in peroxides = -1).
Hydrogen = +1 (except in metal hydrides = -1).
Example:
2H2O2→2H2O+O2
Oxidation: O22−O22− → O20
Reduction: O22−O22− → O2−
Practice Problems:
Ch. 8: 85, 95, 106, 127
This detailed guide follows the learning goals outlined in the roadmap, including key concepts, examples, and practice strategies.
Reading: Sections 6.1 – 6.4
Objective:
Relate particle-level interactions to the physical properties of substances.
Key Concepts:
Intermolecular Forces (IMFs):
London Dispersion Forces (LDFs): Present in all molecules; stronger in larger molecules with greater surface area.
Dipole-Dipole Interactions: Occur between polar molecules.
Hydrogen Bonding: Special dipole-dipole force involving H and N, O, or F.
Ion-Dipole Forces: Between ions and polar molecules; relevant in solutions.
Boiling Points:
Stronger IMFs → Higher boiling points.
Trend: Hydrogen bonding > Dipole-dipole > LDFs.
Example:
Compare boiling points of CH4, CH3OH, and NaCl
CH4CH4: Weak LDFs, low boiling point.
CH3OHCH3OH: H-bonding, higher boiling point.
NaClNaCl: Ionic forces, highest boiling point.
Practice Problems: Ch. 6: 1, 19, 35, 41
Reading: Section 6.5
Objective:
Interpret phase diagrams and heating curves for substances.
Key Concepts:
Phase Diagram: Shows states of matter under varying temperature and pressure.
Critical points: Triple point, critical point, boiling point, freezing point.
Phase changes: Sublimation, deposition, melting, freezing, boiling, condensation.
Heating Curve:
Sloped regions: Temperature changes; q=mCΔT
Flat regions: Phase changes; q=ΔHfusion/vaporization⋅m
Example:
Identify the phase and phase change at 1 atm pressure and 100°C for water: Boiling point → Liquid to gas.
Practice Problems: Ch. 6: 7, 63, 67, 71
Reading: Sections 9.1 – 9.10
Objective:
Apply gas laws and kinetic molecular theory to explain gas behavior.
Key Concepts:
Ideal Gas Law:
PV=nRT
R depends on units of pressure.
Partial Pressures (Dalton’s Law):
Ptotal=∑Pi
Kinetic Molecular Theory:
Average kinetic energy proportional to temperature.
Gas pressure due to collisions of particles with container walls.
Deviations from Ideal Gas Law:
High pressure/low temperature → Real gases deviate.
Example:
Calculate pressure of 1 mole of O2O2 gas at 25°C in a 5.0 L container.
P=nRTV=(1)(0.0821)(298)5.0=4.88 atm
Practice Problems: Ch. 9: 2, 7, 27, 49, 53, 59, 69
Reading: Sections 10.1 – 10.7
Objective:
Understand energy transfer, enthalpy, and calorimetry in chemical and physical processes.
Key Concepts:
First Law of Thermodynamics:
ΔE=q+w
q: Heat; + if absorbed, − if released.
ww: Work; + if done on the system, − if done by the system.
Heat Transfer:
q=mCΔT
Enthalpy (ΔHΔH):
ΔH for a reaction:
ΔHrxn∘=∑(nΔHf∘products)−∑(nΔHf∘reactants)
Hess’s Law: Combine reaction enthalpies to determine overall ΔHΔH.
Examples:
Calorimetry:
If 200 g of water absorbs 8.36 kJ of heat, calculate the temperature change:
q=mCΔT ⟹ ΔT=qmC=8360(200)(4.184)=10.0∘C
Hess’s Law:
Given: A+B→C, ΔH=−100 kJ and C→D,ΔH=−50 kJC→D,ΔH=−50kJ, find A+B→D
ΔH=−100−50=−150 kJ
Practice Problems: Ch. 10: 7, 14, 53, 67, 75, 97
PV=nRT
ΔHrxn=∑(nΔHf∘ products)−∑(nΔHf∘ reactants)
q=mCΔT
ΔE=q+w