Group One:
3.1 The Mole
A. Avogadro's number
1. 6.022 x 1023 units = 1 mole
B. Measuring moles
1. An element's atomic mass expressed in grams contains 1 mole of atoms of that element
3.2 Molar Mass
A. Molar Mass (Gram molecular weight)
1. The mass in grams of one mole of a compound
2. The sum of the masses of the component atoms in a compound
3.3 Percent Composition of Compounds
A. Calculating any percentage
1. "The part, divided by the whole, multiplied by 100"
B. Percentage Composition
1. Calculate the percent of each element in the total mass of the
compound:
(#atoms of the element)x(atomic mass of element)/ (molar mass of the compound) x 100
3.4 Determining the Formula of a Compound
A. Determining the empirical formula
1. Determine the percentage of each element in your compound
2. Treat % as grams, and convert grams of each element to moles of each element
3. Find the smallest whole number ratio of atoms
4. If the ratio is not all whole number, multiply each by an integer so that all elements are in whole number ratio
B. Determining the molecular formula
1. Find the empirical formula mass
2. Divide the known molecular mass by the empirical formula mass,
deriving a whole number, n
3. Multiply the empirical formula by n to derive the molecular formula
***Know how to balance chemical equations***
3.5 Stoichiometric Calculations: Amounts of Reactants and Products
A. Balance the chemical equation
B. Convert grams of reactant or product to moles
C. Compare moles of the known to moles of the desired substance
1. A ratio derived from the coefficients in the balanced equation
D. Convert from moles back to grams if required
3.6 Calculations Involving a Limiting Reactant
A. Concept of limiting reactant (limiting reagent):
1. The limiting reactant controls the amount of product that can form
B. Solving limiting reactant problems
1. Convert grams of reactants to moles
2. Use stoichiometric ratios to determine the limiting reactant
3. Solve as before, beginning the stoichiometric calculation with the
grams of the limiting reactant
C. Calculating Percent Yield
Actual Yield x 100% = percent yield
Theoretical Yield
● Accomplishments of:
○ Rutherford's Metal Foil Experiment
■ ***!!!!!Discovered the nucleus!!!!!***
■ Most alpha particles pass straight through thin metal foil
■ Some particles were greatly deflected
a. Could not have been deflected by electrons or single protons
b. Must have been deflected by a positively charged object of substantial mass
c. Supported concept of a small, central, positive nucleus where most of the atom's mass was concentrated
***!!!!!Disproved Thomson's "plum pudding" model!!!!!***
○ Wolfgang Pauli:
■ Pauli Exclusion Principle:
a. No two electrons in an atom can have identical quantum numbers
○ Antoine Lavoisier:
■ The role of oxygen in combustion
■ Law of conservation of mass
a. The mass of the products equal the mass of the reactants
■ First modern chemistry textbook
Group Two:
Chapter 5
*Kinetic Molecular theory
-The particles are so small compared with the distance between them that the volume of the individual particles can be assumed to be zero.
-The particles are in constant motion. Collisions of the particles with the walls of the container cause pressure.
Assume the particles exert no forces on each other.
-KE(avg)=(3/2)RT ~R=8.3145J/mol*K
- The average kinetic energy of a collection of the gas particles is assumed to be directly proportional to the Kelvin temperature of the gas
*Gas Laws
-Boyle’s Law: “Ideal gases” P1V1 = P2V2 (constant temperature)
The product of pressure times volume is a constant, provided the temperature remains the same.
PV = k P is inversely related to V
-Charles Law: The volume of a gas increases linearly with temperature provided the pressure remains constant.
V = bT V1/T1 = V2/T2 *Temperature must be measured in Kelvin
-Avagadro’s Law: For a gas at constant temperature and pressure, the volume is directly proportional to the # of moles, n.
V = an V1/n1 = V2/n2
-Dalton’s Law of Partial Pressures: For a mixture of gases in a container, the total pressure exerted is the sum of the pressures each gas would exert if it were alone.
Ptotal = P1 + P2 + P3 P1 = (n1RT) / V
Mole Fraction: x1 = n1/ntotal = P1/Ptotal
*Relationships between molar mass and velocity
● As molar mass increases, velocity decreases.
● Inversely proportional relationship
● Velocity of a gas is dependent on molar mass and temperature
● Velocity=sqrt(3RT/M) R=8.3145 j/k*mol
Group Three:
Chapter 7 & 8
Periodic Trends
● Trends exist because a proton is added as you move left to right on the periodic table, and a layer of electrons is added as you move top to bottom on the periodic table
● Ionization energy increases bottom to top, left to right
● Electronegativity increases bottom to top, left to right
● Atomic radius decreases bottom to top, left to right
● As wavelength increases frequency decreases
● When energy decreases wavelength increases
● Energy increases frequency increases
● Equation: wavelength(frequency)=speed
Ionic Bonding
● Electrons are being transferred
● Metals and nonmetals
● Paired ions are more stable and have less energy
● Equation: E=(2.31x10^-19J nm)(Q1Q2/r)
● Q= the charge
● R= distance in nanometers
Covalent Bonds
● Electrons shared
● Nonpolar covalent equally sharing electrons
● Polar covalent unequally sharing electrons
● Dipole moment means that it’s polar
Group Four:
Steric 0 lone pairs 1 lone pair 2 lone pairs 3 lone pairs 4 lone pairs
Hybridization: a mixing of the native orbitals on a given atom to form special atomic orbitals for bonding (makes all bonds the same energy)
● dependent upon the vsper model shape
● steric #1: sp
● steric #2:Sp^2
● Steric #3: sp^3
● Steric #4: sp^3d
● Steric #4: sp^3d^2
Sigma and pi bonds:
Sigma: Initial bond
Pi: Additional bonds
Ex.
Single bond: 1 Sigma bond
Double bond: 1 sigma bond and 1 pi bond
Triple bond: 1 sigma and 2 pi bonds
2 double bonds in 1 molecule: 2 sigma and 2 pi bonds
Group Five: Intermolecular Forces, Boiling Point, Melting Point
Intermolecular Forces: Forces that occur within molecules.
● Dipole-dipole attraction: Molecules with polar bonds act in an electric field if they have a center of negative and positive charges. Molecules with dipole moments attract electrostatically by lining up so that the positive and negative ends are close to each other. They strengthen (+-) interactions and weakens (++) and (--) interactions. They ar 1% strong as covalent or ionic bonds and they become weaker as the distance between the dipoles increases.
● Hydrogen Bonding: The strongest intermolecular force. This occurs because hydrogen is bound to highly electronegative atoms, such as nitrogen, fluorine, and oxygen. There are two reasons that account for the strength. One is the high polarity of the bond and the other is the closeness of the dipoles.
● London-dispersion Forces: This force exists in noble gas atoms and non-polar atoms. They induce a momentary dipole in a neighboring atom.
Boiling Point: The temperatures at which intermolecular forces break between the atoms of a molecule. Liquid water reaches its boiling point at 100 degrees Celsius and the temperature then remains constant as the added energy is used to vaporize the liquid. When liquid is completely changed to vapor, the temperature again begins to rise. These are physical changes, as the intermolecular forces are broken between the bonds but no chemical bonds are broken.
Melting Point: At melting point, added energy is used to break up the ice structure by breaking the hydrogen bonds and increasing the potential energy of water molecules. Molecules become so energetic that they break loose from their lattice positions and change the state of matter from solid to liquid. When the solid melts, the enthalpy change that occurs at the melting point is called heat of fusion.
Group Six:
Metallic Bonding
Metallic bonding is the electrostatic attractive forces between the de-localized electrons, called conduction electrons, gathered in an "electron sea", and the positively charged metal ions.
-Electron Sea Model: valence electrons circulate freely among the metal cations.
-Band Model: electrons are assumed to occupy molecular orbitals.
Triple Point Graphs
-The change from liquid to gas (vapor) is called vaporization or evaporation.
-Condensation is the reverse of vaporization.
-Equilibrium vapor pressure: the pressure that occurs over a liquid or solid in a closed system when the rate of evaporation equals the rate of condensation.....
● Liquids whose components have high intermolecular forces have relatively low vapor pressures.
● Normal boiling point: the temperature at which the vapor pressure of a liquid equals one atmosphere.
● Normal melting point: the temperature at which a solid and its liquid have the same vapor pressure (at 1amt external pressure)
-Phase Diagram
● Shows what state exists at a given temp and pressure in a closed system
● Triple point: temp at which all 3 phases exist simultaneously
● Critical point: defined by the critical temp and pressure
● Critical temp: the temp above which the vapor cannot be liquefied no matter the applied pressure
● Critical pressure: the pressure required to produce liquefaction at the critical temp
Covalent Network
C, Si, SiO2 = covalent network when they are by themselves
Exercise: Chapter 10 #71
Group Seven:
Chapter 11
Solution Composition
● Molarity: moles of solute
● Liters of solution
○ If you have 2.0M CaCl2 (aq) solution, 1 liter will contain 2 moles of Ca, but 4 of Chlorine.
● Mass Percent: (Mass of solute/Mass of solution) x 100%
● Mole Fraction: Mole Fraction of Component A = XA= nA/nA+nB
● Molality: moles of solute/ kilograms of solvent
Solution Conductivity
● “LIKE” DISSOLVES “LIKE”
○ Polar molecules and ionic compounds are dissolved in polar solvents.
○ Nonpolar molecules are dissolved in nonpolar solvents.
● DH1 + DH2 + DH3 ==> DHtotal (D means delta!!!!)
○ DH1 = Seperating Solute (+)
○ DH2 = Seperating Solvent (+)
○ DH3 = Mixing (-)
Factors Affecting Solubility
Structure Effects
Nonpolar – Hydrophobic (water-fearing)
Polar – Hydrophilic (water-loving)
Pressure Effects
Amount of gas dissolved is directly proportional to the pressure of the gas.
Temperature Effects
Solids
Increase in Temp = increase of speed of dissolving
Increase in Temp = increase in solubility
Gases
Increase in Temp = Decrease in solubility (But not always)
Group Eight:
Chapter 11
Freezing point of Depression:When a solute is dissolved in a solvent,the freezing point of the solution is lower than that of the pure solvent.
What mass of ethylene glycol (C2H6O2, molar mass = 62.1 g/mol), the main component of antifreeze, must be added to 10.0 L of water to produce a solution for use in a car’s radiator that freezes at –10.0°F (-23.3°C)? Assume the density of water is exactly 1 g/ml.**Answer 7.76 kg
Boiling Point Elevation: A non-volatile solute elevates the boiling point of the solvent (depends on the concentration of solute)
A solution was prepared by dissolving 18.00 grams of glucose in 150.0 grams of water. The resulting solution was found to have a boiling point of 100.34°C. Calculate the molar mass of glucose. Glucose is a molecular solid that is present as individual molecules in solution. ***Answer : 180 g/mol
● Use T=KbMsolute
● The freezing point solution <Freezing point solvent
● A concentration ↑, freezing point ↓
Vapor pressure- Raoult’s Law: nonvolatile solutes lower the vapor pressure of a solvent
● Psolution=XsolventPsolvent X=mole fraction of solvent P=vapor pressure of solvent
● Ions separate when dissolved, meaning lowered vapor pressure
● Non-idea solutions Ptotal=Pa+Pb=XaP a+ XbPb Pa and Pb=partial pressure of molecules
Group Nine:
Chapter 12
1. Differential Rate Law
a. Expresses how the rate of a reaction depends on the concentration of the reactants
b. Rate=k[reactant]^n
c. n=coefficient in front of the reactant
d. k=rate constant that must be found through experiments
1. Integrated Rate Law
a. Expresses how the concentration of a reactant or product in a reaction depends on the time
b. Different rate laws for 0, 1st, and 2nd order reaction
i. 0 order
1. [A] = -kt + [A]0
ii. 1st order
1. ln[A] = -kt + ln[A] 0
iii. 2nd order
1. 1/[A] = kt + 1/[A] 0
1. Reaction Mechanisms
a. A series of elementary steps that must satisfy 2 requirements
i. The sum of the elementary steps must give the overall balanced equation for the reaction
ii. The mechanism must agree with the experimentally determined rate law
a. Molecularity
i.The # of species that must collide to produce the reaction indicated by that step
1. Unimolecular step – a reaction involving one molecule
2. Bimolecular step – a reaction involving the collisions of 2 species
3. Termolecular step – a reaction involving the collisions of 3 species
a. Rate determining step
i. The slowest step in a reaction mechanism that determines the rate of the overall reaction; the second of the 2 requirements that the mechanism must satisfy
1. Collision Theory
a. Molecules must collide in order to reaction
i. They must collide with sufficient energy
ii. They must collide with correct orientation
a. This is why reaction mechanisms make more sense than a reaction just occurring with all of the molecules colliding with the correct force and orientation to produce all of the products
2. Reaction Profiles
a.
a. “delta”H = Energy of products – energy of reactants
b. Activation Energy is decreased with catalysts, but the “delta”H is not
Group Ten:
Chapter 13 – Chemical Equilibrium
Chemical Equilibrium:
The rate of the forward reaction equals the rate of reverse reaction and the concentration of products and reactants remain unchanged.
H2O + CO --> H2 + CO2