Topics Covered on the Final
Chemical Reactions & Equations
Predict products of double replacement reactions
Write full and net ionic equations
Mole Conversions
Atoms ↔ Moles using Avogadro’s number 6.022 \times 10^{23}
grams ↔ Moles using molar mass
Percent composition
Empirical and Molecular Formulas
Calculate from percent composition
Determine molecular formula using molar mass
Hydrates
Determine the formula and name based on mass loss
Use mole ratios
Stoichiometry
Identify limiting reactants
Theoretical yield and excess reactant calculations
Percent yield formula: \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
Gas Laws
Combined and Ideal Gas Law: PV = nRT
\frac{V1}{T1} = \frac{V2}{T2}
STP conditions (1 atm, 273.15 K)
Solutions
Molarity: M = \frac{\text{mol solute}}{\text{L solution}}
Dilution formula: M1V1 = M2V2
Acids & Bases
Identify acid/base, conjugate acid/base pairs
pH/pOH: \text{pH} = -\log[H^+], \text{pOH} = -\log[OH^-], \text{pH} + \text{pOH} = 14
Nomenclature
Know names of common acids (HF, HNO₂, H₂CO₃, etc.)
Redox Reactions
Assign oxidation numbers
Identify oxidation and reduction
Intermolecular Forces
Identify dispersion, dipole-dipole, and hydrogen bonding
Ion-dipole interactions (especially in solutions like CaCl₂ in water)
Thermochemistry
Review energy transfer, endo/exothermic, q = mc\Delta T, etc.
Important Formulas
Avogadro’s Number: 6.022 \times 10^{23} \text{ particles/mol}
Molar Mass: Sum of atomic masses (periodic table)
Empirical Formula Steps:
Convert % to grams
Convert grams to moles
Divide all by smallest number of moles
Multiply to get whole numbers if needed
Ideal Gas Law: PV = nRT, where R = 0.0821 L·atm/mol·K
Percent Composition: % \text{Element} = \left( \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \right) \times 100
Definitions to Know
Limiting Reactant: The reactant that runs out first
Empirical Formula: Simplest ratio of elements
Molecular Formula: Actual number of atoms
Hydrate: Compound with water molecules bound
Oxidation: Loss of electrons
Reduction: Gain of electrons
Strong vs. Weak Acids/Bases: Complete vs. partial ionization
Intermolecular Forces: Forces between molecules
Key Thermochemistry Formulas
Heat (q) Formula: q = mc\Delta T
q = heat (J or kJ)
m = mass (g)
c = specific heat capacity (J/g·°C)
Water: 4.18 J/g·°C
\Delta T = change in temperature (°C)
Phase Change Heat: q = n\Delta H
n = moles
\Delta H = enthalpy change (kJ/mol)
Use for phase changes like melting (\Delta H{\text{fusion}}) or boiling (\Delta H{\text{vaporization}})
Calorimetry: q{\text{lost}} = q{\text{gained}}
Use to calculate heat transfer between substances in a system (e.g., metal in water)
Enthalpy of Reaction: \Delta H = \sum \Delta Hf^\circ \text{(products)} - \sum \Delta Hf^\circ \text{(reactants)}
Use standard enthalpies of formation \Delta H_f^\circ (kJ/mol)
Hess’s Law:
Add chemical equations and their enthalpy changes to find overall \Delta H
If a reaction is reversed, change the sign of \Delta H
If a reaction is multiplied, multiply \Delta H by the same factor
Important Thermochemistry Definitions
Thermochemistry: Study of energy changes in chemical reactions.
System: The part of the universe you’re focusing on (e.g., the chemical reaction).
Surroundings: Everything outside the system.
Endothermic: Absorbs heat (positive \Delta H); feels cold.
Exothermic: Releases heat (negative \Delta H); feels hot.
Enthalpy (H): Heat content of a system at constant pressure.
Specific Heat Capacity (c): Amount of energy needed to raise 1 g of a substance by 1°C.
Calorimeter: Device used to measure heat changes.
Heat of Fusion: Heat required to melt 1 mol of a substance at its melting point.
Heat of Vaporization: Heat required to vaporize 1 mol of a substance at its boiling point.
Standard Enthalpy of Formation (\Delta H_f^\circ): Change in enthalpy when 1 mole of a compound forms from its elements in their standard states.
Important Definitions for Reactions
Reactants: Substances present before a chemical reaction.
Products: Substances formed by a chemical reaction.
Chemical Equation: A representation of a chemical reaction using symbols and formulas.
Example: 2H2 + O2 \rightarrow 2H_2O
Law of Conservation of Mass: Matter is neither created nor destroyed in a chemical reaction. Equations must be balanced.
Types of Chemical Reactions:
Synthesis (Combination): A + B \rightarrow AB
Decomposition: AB \rightarrow A + B
Single Replacement: A + BC \rightarrow AC + B
Double Replacement: AB + CD \rightarrow AD + CB
Combustion: Hydrocarbon + O2 \rightarrow CO2 + H_2O
Aqueous (aq): Dissolved in water
Precipitate: A solid that forms from a solution in a chemical reaction
Spectator Ions: Ions that do not change during the reaction (appear on both sides)
Net Ionic Equation: A simplified chemical equation that only shows species that actually change
Formulas & Procedures for Reactions
Balancing Chemical Equations: Adjust coefficients to have the same number of each atom on both sides.
Mole Ratio (from balanced equations): Used in stoichiometry to relate amounts of reactants and products.
Writing Ionic & Net Ionic Equations:
Write full balanced molecular equation.
Split all aqueous compounds into ions (full ionic equation).
Cancel spectator ions → Net ionic equation
Example:
\text{NaOH (aq) + HCl (aq)} \rightarrow \text{NaCl (aq) + H}_2\text{O (l)}
Full Ionic: \text{Na}^+ + \text{OH}^- + \text{H}^+ + \text{Cl}^- \rightarrow \text{Na}^+ + \text{Cl}^- + \text{H}_2\text{O}
Net Ionic: \text{OH}^- + \text{H}^+ \rightarrow \text{H}_2\text{O}
Solubility Rules (for predicting products):
Nitrates (NO3^⁻), alkali metals, and ammonium (NH4^⁺) are always soluble
Use solubility rules to determine if a precipitate will form
Key Definitions for Moles
Mole (mol): A unit that represents 6.022 \times 10^{23} particles (Avogadro’s number), which could be atoms, molecules, ions, or formula units.
Avogadro’s Number: 1 mole = 6.022 \times 10^{23} particles
Molar Mass: The mass of one mole of a substance (g/mol), equal to the sum of the atomic masses from the periodic table.
Representative Particles:
Atoms (elements like Na or He)
Molecules (covalent compounds like H₂O)
Formula Units (ionic compounds like NaCl)
Conversion Map for Moles
particles (atoms, molecules)
↑ ↓
(\times 6.022\times10^{23}) (\div 6.022\times10^{23})
mass (g) ↔ moles ↔ particles
↑ ↓
(\div mtext{molar mass}) (\times mtext{molar mass})
g/mol
Core Mole Conversion Formulas
Mass ↔ Moles:
\text{moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
\text{mass} = \text{moles} \times \text{molar mass}
Particles ↔ Moles:
\text{moles} = \frac{\text{particles}}{6.022 \times 10^{23}}
\text{particles} = \text{moles} \times 6.022 \times 10^{23}
Volume (at STP) ↔ Moles (for gases):
\text{moles} = \frac{\text{volume (L)}}{22.4}
\text{volume} = \text{moles} \times 22.4
(Only valid at Standard Temperature and Pressure: 0°C and 1 atm)
Moles ↔ Atoms in a Compound:
Multiply moles of compound by the number of atoms of an element per formula unit.
Example: 2 mol of H₂O → 2 \times 2 = 4 mol H atoms
Helpful Tips
Always label your units and cancel them as you go.
For compounds, calculate the total molar mass by adding up all atoms.
Be extra careful with diatomic elements like H₂, O₂, N₂, etc.
Key Definitions for Formulas
Empirical Formula: The simplest whole-number ratio of atoms in a compound.
Example: CH₂O is the empirical formula of glucose (C₆H₁₂O₆).
Molecular Formula: The actual number of atoms of each element in a molecule. It’s a multiple of the empirical formula.
Percent Composition: The percentage by mass of each element in a compound.
Formulas & Procedures for Formulas
Percent Composition
% \text{Element} = \left( \frac{\text{mass of element in 1 mol of compound}}{\text{molar mass of compound}} \right) \times 100
Empirical Formula from Percent Composition Steps:
Assume 100 g of compound (so % becomes g).
Convert grams to moles for each element: \text{moles} = \frac{\text{grams}}{\text{atomic mass}}
Divide each by the smallest number of moles.
Multiply to get whole numbers, if needed: If you get something like 1.5, multiply all by 2. If you get 2.33, multiply all by 3, etc.
Molecular Formula from Empirical Formula
Formula: \text{Molecular Formula} = (\text{Empirical Formula}) \times n
Where: n = \frac{\text{Molar Mass of Compound}}{\text{Molar Mass of Empirical Formula}}
Example Problem for Formulas
Given: 40.00% C, 6.71% H, and 53.29% O
Assume 100 g → 40.00 g C, 6.71 g H, 53.29 g O
Convert to moles:
C: \frac{40.00}{12.01} = 3.33 mol
H: \frac{6.71}{1.008} = 6.66 mol
O: \frac{53.29}{16.00} = 3.33 mol
Divide by smallest (3.33):
C: 1, H: 2, O: 1 → Empirical formula = CH₂O
If the molar mass is 180 g/mol, then: Molar mass of CH₂O = 30 g/mol
\frac{180}{30} = 6 → Molecular formula = C₆H₁₂O₆
Key Definitions for Hydrate
Hydrate: A compound that includes water molecules chemically bound within its crystal structure.
Example: \text{CuSO}4 \cdot 5\text{H}2\text{O} (copper(II) sulfate pentahydrate)
Anhydrate: The compound left behind after the water is removed (usually by heating).
Example: CuSO₄ is the anhydrate of CuSO₄·5H₂O.
Water of Hydration: The water molecules in a hydrate’s structure.
Formulas & Calculations for Hydrate
Mass of Water Lost
\text{Mass of Water} = \text{Mass of Hydrate} - \text{Mass of Anhydrate}
Moles of Water and Anhydrate
\text{Moles of H}_2\text{O} = \frac{\text{Mass of Water Lost}}{18.02 \, \text{g/mol}}
\text{Moles of Anhydrate} = \frac{\text{Mass of Anhydrate}}{\text{Molar Mass of Anhydrate}}
Mole Ratio of Water to Salt
\text{Ratio} = \frac{\text{Moles of H}_2\text{O}}{\text{Moles of Anhydrate}} \Rightarrow \text{Round to nearest whole number}
This gives the formula of the hydrate: \text{Salt} \cdot x\text{H}_2\text{O}
Example Problem of Hydrate
Given: Mass of hydrate = 5.061 g
Mass of anhydrate = 2.472 g → Mass of water = 5.061 - 2.472 = 2.589 g
Moles of water:
\frac{2.589 \, \text{g}}{18.02 \, \text{g/mol}} \approx 0.144 \, \text{mol}
Moles of MgSO₄:
\frac{2.472 \, \text{g}}{120.37 \, \text{g/mol}} \approx 0.0205 \, \text{mol}
Ratio:
\frac{0.144}{0.0205} \approx 7 \Rightarrow \text{Formula} = \text{MgSO}4 \cdot 7\text{H}2\text{O}
→ Name: Magnesium sulfate heptahydrate
Key Definitions for Stoichiometry
Stoichiometry: The calculation of reactant and product quantities in chemical reactions using balanced equations.
Mole Ratio: The ratio of moles of one substance to another, derived from the coefficients in a balanced chemical equation.
Example: 2H2 + O2 \rightarrow 2H2O \Rightarrow \frac{2 \, \text{mol } H2}{1 \, \text{mol } O2}, \frac{2 \, \text{mol } H2O}{2 \, \text{mol } H_2}, \text{etc.}
Limiting Reactant: The reactant that gets used up first, limiting the amount of product formed.
Excess Reactant: The reactant that is not completely used up in the reaction.
Theoretical Yield: The maximum amount of product that can be formed from the given reactants.
Actual Yield: The amount of product actually obtained from the reaction (usually given in a lab).
Percent Yield: \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100
Core Stoichiometry Formulas
Moles to Moles:
\text{mol A} \times \frac{\text{mol B}}{\text{mol A}} = \text{mol B}
Mass to Mass:
Convert grams of A to moles of A
Use mole ratio to find moles of B
Convert moles of B to grams of B
\text{mass A} \xrightarrow{\div \text{molar mass A}} \text{mol A} \xrightarrow{\times \text{mol B/mol A}} \text{mol B} \xrightarrow{\times \text{molar mass B}} \text{mass B}
Limiting Reactant Steps:
Convert both reactants to moles of product.
The one that makes less product is the limiting reactant.
Excess Reactant Left Over:
Start - Used = Left Over
Example of Stoichiometry
Given: 25.0 g N₂ and 25.0 g H₂
Reaction: N2 + 3H2 \rightarrow 2NH_3
Convert grams to moles:
N₂: \frac{25.0}{28.02} \approx 0.893 \, \text{mol}
H₂: \frac{25.0}{2.02} \approx 12.38 \, \text{mol}
Mole ratio:
N₂ needs 3 mol H₂ per 1 mol N₂ → Needs 0.893 \times 3 = 2.679 mol H₂
Available H₂ = 12.38 mol → Excess
N₂ is limiting reactant
Product moles:
0.893 \, \text{mol N}2 \times \frac{2 \, \text{mol NH}3}{1 \, \text{mol N}2} = 1.786 \, \text{mol NH}3
Convert to grams if needed:
1.786 \times 17.03 \approx 30.4 \, \text{g NH}_3
Key Definitions for Gas Laws
Gas Laws: Describe the relationships between pressure (P), volume (V), temperature (T), and amount (n) of a gas.
Standard Temperature and Pressure (STP):
0°C (273.15 K) and 1 atm
At STP, 1 mole of any ideal gas = 22.4 L
Kelvin Temperature: Always convert Celsius to Kelvin: K = °C + 273.15
Core Gas Law Formulas
Boyle’s Law (Pressure-Volume)
P1V1 = P2V2
Inverse relationship: as pressure increases, volume decreases (T constant)
Charles’s Law (Volume-Temperature)
\frac{V1}{T1} = \frac{V2}{T2}
Direct relationship: as temperature increases, volume increases (P constant)
Gay-Lussac’s Law (Pressure-Temperature)
\frac{P1}{T1} = \frac{P2}{T2}
Direct relationship: as temperature increases, pressure increases (V constant)
Combined Gas Law
\frac{P1V1}{T1} = \frac{P2V2}{T2}
Use when P, V, and T all change
Ideal Gas Law
PV = nRT
P = pressure (atm)
V = volume (L)
n = moles
R = 0.0821 L·atm/mol·K
T = temperature (K)
Gas Density & Molar Mass Density: d = \frac{PM}{RT}
Where M is molar mass
Molar mass: M = \frac{dRT}{P}
Avogadro’s Law (Volume-Moles)
\frac{V1}{n1} = \frac{V2}{n2}
More moles = more volume (at same T and P)
Dalton’s Law of Partial Pressures
\text{P}{\text{total}} = P1 + P2 + P3 + \ldots
Quick Tips for Gas Laws
Always use Kelvin for temperature in gas law calculations.
Check units—especially for volume (L) and pressure (atm).
For gas stoichiometry at STP: 1 mol gas = 22.4 L
Key Definitions for Solutions
Solution: A homogeneous mixture of a solute dissolved in a solvent.
Solute: The substance being dissolved (usually present in smaller amount).
Solvent: The substance doing the dissolving (usually water in aqueous solutions).
Aqueous (aq): A substance dissolved in water.
Molarity (M): A measure of concentration: M = \frac{\text{mol of solute}}{\text{L of solution}}
Dilution: The process of adding solvent to a solution to decrease its concentration.
Saturated Solution: Contains the maximum amount of solute that can dissolve at a given temperature.
Supersaturated Solution: Contains more solute than normally possible at that temperature (unstable).
Unsaturated Solution: More solute can still dissolve in the solvent.
Core Formulas for Solution
Molarity (Concentration)
M = \frac{n}{V} = \frac{\text{mol of solute}}{\text{L of solution}}
Convert grams to moles if needed: \text{mol} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}
Dilution Equation
M1V1 = M2V2
M1, V1: concentration and volume of stock solution
M2, V2: concentration and volume of diluted solution
Percent by Mass
% \text{mass} = \left( \frac{\text{mass of solute}}{\text{mass of solution}} \right) \times 100
Percent by Volume
% \text{volume} = \left( \frac{\text{volume of solute}}{\text{volume of solution}} \right) \times 100
Molality (less common unless specified)
m = \frac{\text{mol of solute}}{\text{kg of solvent}}
Other Useful Concepts for Solutions
Solubility: How much solute dissolves in a given amount of solvent at a specific temperature.
“Like dissolves like”: Polar solvents dissolve polar solutes; nonpolar dissolves nonpolar.
Ion-Dipole Interactions: When ionic compounds dissolve in polar solvents like water.
Example Problem (Molarity) for Solutions:
Q: What is the molarity of a solution containing 45 g of NaOH in 1.059 L of solution?
Find moles of NaOH:
\text{mol} = \frac{45 \, \text{g}}{39.997 \, \text{g/mol}} \approx 1.125 \, \text{mol}
Use molarity formula:
M = \frac{1.125}{1.059} \approx 1.06 \, \text{M}
Key Definitions for Acids and Bases
Acid:
Arrhenius: Produces H⁺ (or H₃O⁺) in water
Brønsted-Lowry: Proton (H⁺) donor
Base:
Arrhenius: Produces OH⁻ in water
Brønsted-Lowry: Proton (H⁺) acceptor
Conjugate Acid: Formed when a base gains a proton (H⁺)
Conjugate Base: Formed when an acid loses a proton (H⁺)
Strong Acids/Bases: Completely ionize in solution
Example acids: HCl, HNO₃, H₂SO₄
Example bases: NaOH, KOH, Ba(OH)₂
Weak Acids/Bases: Partially ionize in solution
Example: HC₂H₃O₂ (acetic acid), NH₃ (ammonia)
Core Formulas for Acids and Bases
pH and pOH
\text{pH} = -\log[H^+]
\text{pOH} = -\log[OH^-]
pH + pOH Relationship
pH + pOH = 14
Ion Concentration from pH or pOH
[H^+] = 10^{-\text{pH}}
[OH^-] = 10^{-\text{pOH}}
Kw — Ion Product of Water
K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \quad (\text{at 25°C})
Neutralization Reaction for Acids and Bases
Acid + Base → Salt + Water
Example: HCl + NaOH → NaCl + H₂O
Naming acids
Acid Formula Name Type
No oxygen (HX) “Hydro- + -ic acid”
HCl → hydrochloric acid
With oxygen ending in -ate “-ic acid”
HNO₃ → nitric acid
With oxygen ending in -ite “-ous acid”
HNO₂ → nitrous acid
Example Problem: pH Calculation
Q: What is the pH of a solution with [H^+] = 1.995 \times 10^{-7}\, \text{M}?
pH = -log(1.995 \times 10^{-7}) ≈ 6.70 → Slightly acidic
Basic Definitions for Nomenclature
Element: A substance made up of atoms that all have the same number of protons.
Compound: A substance formed when two or more elements chemically bond.
Molecule: The smallest unit of a compound that retains its chemical properties.
Ions: Atoms or molecules that have gained or lost electrons, resulting in a charged species.
Ionic Compounds Nomenclature
Ionic compounds are made from metal cations (positive ions) and nonmetal anions (negative ions).
Naming Ionic Compounds: Name the cation (metal) first, followed by the anion (nonmetal).
For monatomic cations (e.g., Na⁺), use the element’s name.
For monatomic anions (e.g., Cl⁻), use the element’s root and add “-ide.”
If the metal can have multiple oxidation states (transition metals), include the oxidation state in Roman numerals in parentheses (e.g., Fe²⁺ → Iron(II)).
Example: NaCl is Sodium Chloride.
Formula for Ionic Compounds: The formula reflects the balance of charges to form a neutral compound. For Na⁺ and Cl⁻, one of each is needed for neutrality, giving the formula NaCl.
Covalent (Molecular) Compounds Nomenclature
Covalent compounds form when two nonmetals share electrons.
Naming Molecular Compounds:
The first element is named first, with its full element name.
The second element is named as if it were an anion, ending in “-ide.”
Prefixes (mono-, di-, tri-, etc.) are used to indicate the number of atoms of each element.
Examples:
CO₂ = Carbon Dioxide.
N₂O₄ = Dinitrogen Tetroxide.
Formula for Molecular Compounds: The formula indicates the number of atoms of each element.
Acids Nomenclature
Acids are compounds that release hydrogen ions (H⁺) when dissolved in water.
Naming Acids:
Binary Acids (two elements, often hydrogen + a halogen): Start with “hydro-” and end with “-ic acid.”
Example: HCl = Hydrochloric Acid.
Oxyacids (hydrogen + a polyatomic ion with oxygen): The name depends on the polyatomic ion.
If the ion ends in “-ate”, the acid name ends in “-ic acid.”
If the ion ends in “-ite”, the acid name ends in “-ous acid.”
Example: H₂SO₄ (sulphate ion) = Sulfuric Acid, H₂SO₃ (sulfite ion) = Sulfurous Acid.
Polyatomic Ions
Polyatomic ions are ions made up of more than one atom.
Common Polyatomic Ions:
Ammonium: NH₄⁺
Hydroxide: OH⁻
Sulfate: SO₄²⁻
Nitrate: NO₃⁻
Phosphate: PO₄³⁻
Oxidation States Nomenclature
Oxidation states (or numbers) represent the charge on an atom in a molecule or ion.
Some elements have fixed oxidation states (e.g., alkali metals always have an oxidation state of +1), while others (like transition metals) can have multiple oxidation states.
Percent Composition and Empirical Formula Nomenclature
Percent Composition: The percentage by mass of each element in a compound.
\text{Percent Composition} = \left( \frac{\text{Mass of element in compound}}{\text{Total mass of compound}} \right) \times 100
Empirical Formula: The simplest whole number ratio of atoms in a compound.
Find the ratio of moles of each element in the compound and simplify to the smallest whole numbers.
Molar Mass Nomenclature
The molar mass is the mass of one mole of a substance (in g/mol).
Add up the atomic masses of all atoms in the formula.
Stoichiometry Nomenclature
Stoichiometry is the calculation of reactants and products in chemical reactions.
Use the mole ratio from the balanced chemical equation to find quantities of reactants or products.
Basic Definitions for Redox Reactions
Redox Reactions: A redox (reduction-oxidation) reaction involves the transfer of electrons between two species.
One species undergoes oxidation (loses electrons), and the other undergoes reduction (gains electrons).
Oxidation: The process of losing electrons.
Reduction: The process of gaining electrons.
Oxidizing Agent (Oxidant): The substance that gains electrons and gets reduced.
Reducing Agent (Reductant): The substance that loses electrons and gets oxidized.
Oxidation States (Numbers) for Redox Reactions
Oxidation State: A measure of the degree of oxidation of an atom in a compound.
The oxidation state of free