Mr. Kalix A.P. Chemistry - Overall Formula Sheet:
Chapter#1:
1cm3 = 1mL
1dm3 = 1L
2cups = 1pt
1in = 2.54cm
1kg = 2.2lbs
2pt = 1 qt
1lbs = 453g
1mi = 5280ft
4 qt = 1gal
3ft = 1yd
1mi = 1760yd
Exa . . Peta . . Tera . . Giga . . Mega . . Kilo Hecto Deca BASE deci centi milli . . micro . . nano . . pico . . fempto . . atto
% 𝐸𝑟𝑟𝑜𝑟 = !
𝑡𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒 − 𝑒𝑥𝑝𝑒𝑟𝑖𝑚𝑒𝑛𝑡𝑎𝑙 𝑣𝑎𝑙𝑢𝑒
𝑥100
𝑡𝑟𝑢𝑒 𝑣𝑎𝑙𝑢𝑒
𝐶 = ! 𝐹 − 32
𝐷𝑒𝑛𝑠𝑖𝑡𝑦 = !
𝐹 = ! 𝐶 + 32
K = C + 273
𝑚𝑎𝑠𝑠(𝑔)
𝑣𝑜𝑙𝑢𝑚𝑒(𝑐𝑚! 𝑜𝑟 𝑚𝐿)
Other things to memorize:
o Significant Figures Rules
1. All non-zero numbers are significant.
2. Imbedded zeros are significant (trapped zeros).
3. Leading zeros are never significant.
4. Trailing zeros are significant only if the decimal point is specified.
5. Zeros following a decimal in scientific notation are significant. All digits in scientific
notation count.
6. When an exact number appears in a calculation, it does not effect the
numbers significant digits.
Add or Subtract: the answer has the same number of decimal places as the least precise
measurement used in the calculation.
Multiply and Divide: the answer has the number of digits equal to the smallest number of
significant figures total.
•
1.
2.
3.
4.
Hg2
2+
o Scientific Notation Rules:
Write down the significant digits in a number and leave out any insignificant zeros
Place a decimal after the 1st significant digit
Place a x10 after the number
Count number of times the decimal has moved and place that superscript number after
step 3. If the original number is larger than one the superscript is a positive number. If
the original number is smaller than one the superscript is a negative number.
o Polyatomic Ions:
Dimercury(I) NH4+ Ammonium
SO42- Sulfate
HPO42- Hydrogen
Phosphate
ClO- Hypochlorite
MnO4- Permanganate
HSO4- Bisulfate
H2PO4- Dihydrogen
Phosphate
ClO2- Chlorite
Cr2O72- Dichromate
NO2-
Nitrite
NO3-
Nitrate
SO32- Sulfite
OH- Hydroxide
SCN- Thiocyanate
CN- Cyanide
CO32- Carbonate
PO43- Phosphate
HCO3 - Bicarbonate
ClO3- Chlorate
CrO42- Chromate
ClO4- Perchlorate
O22Peroxide
C2H3O2- Acetate
C2O42- Oxalate
Chapter#2:
Other things to memorize:
o Naming Ionic Bonding Rules
Ionic Bonding:
• metal and non-metal (2 different atoms)
• metal and a polyatomic ion (3 or more different atoms)
• find oxidation numbers and crisscross charges
• if you need more than one polyatomic ion add ( ) and a subscript telling how many are
needed.
Ex) Ca(OH)2
• cation (+) charged ion (written 1st) and then anion (-) charged ion (written 2nd)
• Naming a binary compound: name of cation, then name of anion and change ending to “ide”
• Naming a polyatomic compound: name of cation, then name of polyatomic
(does not usually end in “ide”)
o 2 exceptions
1. Polyatomic ions that end in “ide”
2. Ammonium and a single element second
• reduce subscript outside parenthesis to lowest whole numbers (reduce)
•
Ex) Binary compounds (2 elements)
NaCl
sodium chloride
MgCl2
magnesium chloride
Al2O3
aluminum oxide
Ex) Tertiary compounds (3 elements)
CaSO4
calcium sulfate
K3PO4
potassium phosphate
Be(OH)2
beryllium hydroxide
NH4Br
ammonium bromide
Do not follow the “ide” rule
Roman numerals: some transition elements have multiple oxidation numbers
which is shown by using roman numerals.
Ex) Fe2O3
Iron (III) Oxide
FeO
Iron (II) Oxide
Old multiple oxidation naming system:
• Higher charge ends in “ic”
• Lower charge ends in “ous”
ex) ferric oxide
ex) ferrous oxide
o Naming Covalent Bonding Rules
Covalent Bonding:
• 2 non-metals
• use non-metal naming prefixes
o (mono-, di-, tri-, tetra-, pena-, hexa-, hepta-, octa-, nona-, deca-)
• mono- never leads
• ends with “ide”
NO
nitrogen monoxide
CO2
carbon dioxide
N2 O2
dinitrogen dioxide
o Oxyanion Rules: polyatomics with different numbers of oxygen
•
Note: Usually a halogen group issue
hypo-ite
1 oxygen
ClO1hypochlorite
1-ite
2 oxygen’s
ClO2
chlorite
1-ate
3 oxygen’s
ClO3
chlorate
per-ate
4 oxygen’s
ClO41perchlorate
o Naming Acids:
Acids: hydrogen will be in the lead
•
•
if only 2 elements (no O)
hydro- + 2nd elements name + change “ide” ending to “ic” + acid
Ex) HCl
hydrochloric acid
•
•
if 3 or more elements (O)
polyatomic name + change “ite” ending to “ous” + acid
“ate” ending to “ic”
Ex) HNO3
nitric acid
HNO2
nitrous acid
o Non-metal Prefixes:
! (mono-, di-, tri-, tetra-, penta-, hexa-, hepta-, octa-, nona-, deca-)
o Variable Valence Cations:
Cobalt (II) and (III)
Copper (I) and (II)
Iron (II) and (III)
Lead (II) and (IV)
Mercury (II) Hg2+
Mercury (I) Hg22+
Tin (II) and (IV)
✤Silver (1+)
✤Zinc (2+)
✤Cadmium (2+)
✤Form only one type of ion and Roman numerals are not used.
Chapter#3:
Percent Composition: the percent of the total mass of the compound that is due to that component
Empirical Formula: the smallest whole-number ratio of atoms in a compound
-find % composition
-each % represents the mass in g of that element out of 100g total
- convert gram amounts to moles
-divide all by the smallest mole number to find a ratio
-if ratio is not whole numbers, multiply each number by an integer to get whole numbers,
rounding can only occur when the subscripts are within 0.1 of the nearest whole number.
To find the Molecular Formula from the empirical formula
-divide total mass by empirical formula to find the multiplication factor
-take the multiplication factor and multiply each element in molecule by the multiplication factor
6.022x1023
Atomic Mass (g)
Mole (mol)
atoms
22.42
L
Volume of gas
Percent Yield =
Actual (lab)
•100 =
Expected (math)
Hints for balancing equations:
Find the whole number coefficients, which give the same amount of each element on each side of the equation.
This follows Lavoisier’s Law of Constant Mass says matter cannot be created or destroyed.
• Identify each element involved in the reaction.
• All molecules must have a net zero charge.
• Single metal atoms have no oxidation numbers.
• Watch out for diatomic molecules: H2, O2, N2, Cl2, Br2, I2, F2
• You can only change coefficients, not the formulas subscripts.
• Determine the net charge for each side of the equation: must be balanced in the final equation.
• Metal + Non-metal or Metal + Polyatomic = Ionic Bond
• 2 Non-metals = Covalent Bond
1. If an element appears in pure form, leave it to the end.
2. If a polyatomic travels both sides of the equation, balanced it first as a packaged group.
3. Leave complex materials for later. (Elements that are only in one molecule on one side but in
multiple molecules on the other side)
4. Final step, make sure coefficients are the smallest whole numbers.
Remember: Always check your work!! Make sure that the same number of each type of atom and the same total
charge are on each side of the equation.
quantity
of given
Basic Sequence
convert to
moles
moles of
given
quantity
of unknown
convert to
desired unit
find molar ratio
moles of
unknown
Diatomic
Molecules
H2
O2
N2
Cl2
Br2
I2
F2
Chapter#4:
Strong Acids:
HCl, HBr, HI
H2SO4 (HSO4- = weak acid)
HNO3
HClO4
Strong Bases:
Group I hydroxides (LiOH, NaOH, KOH…)
Group II hydroxides (Ca(OH)2, Sr(OH)2, Ba(OH)2…)
Molarity =
moles of solute
Liters of solution
To Dilute Stock Solutions:
1. find the number of moles of material in final solution
2. find the amount of volume of stock solution L = mol
M
(M)(L) = mol
Simple Rules for the Solubility of Salts in H2O:
1. Most nitrate (NO3-) salts are soluble
2. Most salts containing the alkali metal ions (Li+, Na+, K+, Rb+, Cs+,) and the ammonium ion (NH4+)
are soluble
3. Most Cl, Br, I “-ide” salts are soluble. Notable exceptions are salts containing the ions Ag+, Cu+,
Pb2+, and Hg22+.
4. Most sulfate salts are soluble. Notable exceptions are BaSO4, PbSO4, Hg2SO4, and CaSO4
5. Most hydrated salts are only slightly soluble. The important soluble hydroxides are NaOH and
KOH. The compounds Ba(OH)2, Sr(OH)2, and Ca(OH)2 are marginally soluble.
6. Most sulfide (S2-), carbonate (CO32-), chromate (CrO42-), and phosphate (PO43-) salts are only
slightly soluble.
Note: slightly soluble = insoluble
marginally soluble = soluble
Solving Stoichiometry Problems for Reactions in Solution:
1. Identify the species present in the combined solution, and determine what reaction occurs.
2. Write the balanced net ionic equation for the reaction.
3. Calculate the moles of reactant.
4. Determine which reactant is limiting.
5. Calculate the moles of product or products as required.
6. Convert to unit required.
Rules for Assigning Oxidation States (Numbers):
Oxidation State of……
Summary
Ex)
An atom in a element is 0
Element = 0
Na(s), O2(g), Hg(l)
A monatomic ion is the same as
Monatomic ion = charge of ion
Na+, Clits charge
Fluorine is -1 in its compounds
Fluorine = -1
HF, PF3
Oxygen is usually 2 in its compounds Oxygen = 2
H2O, CO2
Exception: Peroxides (O2-2) in which
Peroxide (O2-2) = -1
H2 O2
Oxygen is 1
Hydrogen is +1 in covalent compounds Hydrogen = +1
H2O, HCl
Half – Reaction Method for Balancing Equations for Redox in Acids and Bases:
1. Write separate equations for oxidation and reduction ½ reactions
2. For each ½ reaction:
a. balance all the elements except H and O
b. balance O using H2O
c. balance H using H+
d. balance the charge using e3. If needed, multiply 1 or both balanced ½ reactions by an integer to equalize the number of etransferred in the 2 ½ reactions
4. Add ½ reactions and cancel identical species.
5. Check that the elements and charges are balanced
For Basic Solutions, need to add more steps:
6. To both sides of the equation, add a number of OH- ions that is equal to the number of H+ ions
(same # to both sides to keep it equal)
Step 5 is skipped
until the end
7. Form H2O on the side containing both H+ and OH- ions and eliminate the number of H2O
molecules that appear on both sides of the equation
5. NOW check that the elements and charges are balanced
Learn these types of decomposition reactions:
1. Metallic carbonates, when heated form metallic oxides and CO2(g).
Ex. CaCO3(s) " CaO(s) + CO2(g)
2. Most metallic hydroxides, when heated, decompose into metallic oxides and water.
Ex. Ca(OH)2(s) " CaO(s) + H2O(g)
3. Metallic chlorates, when heated, decompose into metallic chlorides and oxygen.
Ex. 2 KClO3(s) " 2 KCl(s) + 3 O2(g)
4. Some acids, when heated, decompose into nonmetallic oxides and water.
Ex. H2SO4 " H2O(l) + SO3(g)
5. Some oxides, when heated, decompose.
Ex. 2 HgO(s) " 2 Hg(l) + O2(g)
Chapter#5:
•
•
1atm = 760 mm Hg or 760 Torr or 101,325 Pa or 29.92 in Hg or 14.7 lb/in2
Force Newton N
Pr essure =
=
=
= Pascal(Pa)
Area
meter 2 m 2
•
Boyles Law: P1V1 = P2V2
•
Charles Law:
•
P P
Gay-Lussac’s Law: 1 = 2 with constant V
T1 T2
€
€
with constant T
V1 V2
=
with constant P
T1 T2
The Combined Gas Law: combines Boyles, Charles, and Gay-Lussac’s Laws
€
V V
• Avogadro’s Law: 1 = 2 with constant T and P
n1 n 2
•
•
•
P1 • V1 P2 • V2
=
T1
T2
€
Ideal Gas Law: Combination of Boyles, Charles, Gay-Lussac’s and Avogadro’s Laws.
PV = nRT
R = combined proportionality constant or the universal gas constant
€
0.08206( L)( atm)
R=
(K )(mol)
NOT an ideal gas when: large molecules (attraction), low temperatures, high pressure, or
anything that forces a gas together.
Standard Temperature and Pressure: common reference conditions for the properties of
€
gases. OOC and 1atm
•
mass
RT
nRT
DRT
DRT
( mass)RT
molar
mass
P=
and P =
and P =
and P =
and molar mass =
V
V
V ( molar mass)
molar mass
P
•
Partial Pressure: the P that a particular gas would exert if it were alone in the container.
•
Ptotal = P1 + P2 + P3 + … =
€
n1RT n 2 RT n 3 RT (n1 + n 2 + n 3 )RT n total RT
+
+
=
=
V
V
V
V
V
The Kinetic Molecular Theory of Gases (KMT):
• A model that attempts to explain the properties of an ideal gas.
€
• This model is based on speculations about behavior of individual gas particles.
1. The particles are so small compared with the distances between them that the V of
individual particles can be assumed to be negligible (zero).
2. Particles are in constant motion. The collisions of the particles with the walls of the
container are the cause of the P exerted by gas.
3. The particles are assumed to exert no forces on each other. They are assumed neither
to attract nor to repel each other.
4. The average kinetic energy of a collection of gas particles is assumed to be directly
proportional to the K temp. of the gas.
1
mV 2
2
Energy due to the motion of an object. Dependent on mass of the object and the square of its
velocity.
KEavg = average kinetic energy
€
3
J
KEavg = RT
R = universal gas constant in Joules R = 8.3145
2
mol • K
T = Kelvin
Kinetic Energy:
•
3kT
=
m
J
R = 8.3145
mol • K
3RT
€ M = molar mass in Kg
M
Kg • m 2
Joule = J =
s2
•
Root
€ Mean Square Velocity (υ rms ): (υ rms )=
•
R = 0.08206
•
€
rate1 rate of effusion for gas1
=
=
Graham’s Law of Effusion:
rate2 rate of effusion
€ for gas2
€
• (M1 and M2 represent molar masses of the gas)
€
L • atm
K • mol
Van der Waals Equation:
€ 2
#n&
nRT
Pobs =
− a% (
V − nb $ V '
V correction
€
•
and is rearranged to give
P correction
due to attraction
(
" n %2 +
*Pobs + a$ ' - x (V − nb) = nRT
#V & ,
)
correct P
P ideal
€
M2
M1
corrected V
V ideal
" atm • L2 %
" L %
b =$
a =$
'
2 '
# mol &
# mol &
a and b values come from experimental observations.
€
€
Chapter#6:
•
Work = w = -PΔV
•
Energy = ΔE = q + w and by combining these equations we get ΔE = q + (-PΔV)
Temperature: A property that reflects the random motions of the particles in a substance.
Heat(q): involves the transfer of energy between 2 objects
State Function: a property that is independent of the pathway.
J=
€
kg • m 2
s2
1L•atm = 101.3J
760torr = 1atm
Yes
No
Energy
Work
Enthalpy Heat
Entropy
1st Law of Thermodynamics: energy can be converted from 1 form to another but cannot be
created or destroyed
Symbol Unit Exothermic
Endothermic
Energy
E
(J)
-E
+E
Heat
q
(J)
-q
+q
Work
w
(J)
(expansion)
-w
(compression)
+w
H
(kJ)
-H
+H
Enthalpy
•
ΔH = ΔE + PΔV
•
q = mCΔT
•
Constant Volume ΔE = q
m= mass(g)
! J $
C = specific heat capacity # O
&
" C•g%
Constant Pressure ΔH = q
ΔT = Tf – Ti
3 Ways to Express Heat Capacity:
" J %
1. Heat Capacity(C) or (s) in $ o '
# C&
" J %" J %
2. Specific Heat of Capacity: heat capacity per gram of substance $ o
'$
'
# C • g& # K • g&
"
%" J %
J
€
3. Molar Heat of Capacity: heat capacity per mole of substance $ o
'$
'
# C • mol & # K • mol &
€
Extensive Property: depends directly on the amount of substance
Intensive Property: not related to the amount of substance
€
Hess’s Law:
• Enthalpy changes are state functions.
• It does not matter if ΔH for a reaction is calculated in one step or a series of steps.
2 characteristics of ΔH for a reaction:
1. If a reaction is reversed, the sign of ΔH is also reversed.
2. The magnitude of ΔH is directly proportional to the quantities of reactants and products in a
reaction. If the coefficients in a balanced reaction are multiplied by an integer, the value of ΔH
is multiplied by the same integer.
Hints for Hess’s Law:
• Work backward from the required reaction, using reactants and products to decide how to
manipulate the other given reactions at your disposal.
• Reverse any reactions as needed
• Multiply reactions to give correct numbers of reactants and products.
Key concepts when doing enthalpy calculations:
• when a reaction is reversed, the magnitude of ΔH remains the same, but its sign changes
• when the balanced equation for a reaction is multiplied or divided by an integer, the value of
ΔH for that reaction must be multiplied or divided by the same integer
• the change in enthalpy for a given reaction can be calculated from the enthalpies of
formation of the reactants and products
o
ΔH reaction
= ∑ n product ΔH of ( products) − ∑ n reac tan tsΔH of (reac tan ts)
ΔHfo sets parameters at 25oC, 1atm, 1mol, 1M, and pure materials
€
Chapter#7:
Planck’s Constant (h) = 6.626x10-34 J⋇ s
!
𝑣=!
ΔE = hv
ΔE =
!
Broglie’s Equation = 𝜆 = !" (!"#$%&'()
Speed of light (C) = 3.00x108
!!
!
!
!
Joules (J) =
!"⋇!!
!!
# 1
1 &(
E n = −2.178x10−18 J%%
−
(
$ n final 2 n initial 2 '
n = principle quantum energy level
o Quantum Numbers: €
There are 4 quantum numbers and Pauli Exclusion Principle states that no two electrons can have
the same set of 4 quantum numbers.
• As n increases the orbital become larger and higher in energy
• Each orbital can hold two electrons maximum and must have opposite spins
• Erwin Schrödinger developed the wave equations describing the probabilities of where
the electrons are located in the atom. These equations require 3 integers, called quantum
numbers, which describe each electron.
Name
Principle Quantum Number
Angular Momentum
Quantum Number
Symbol
n
ℓ
Magnetic Quantum Number
mℓ
Electron Spin Quantum
Number
ms
Property of the Orbital
Related to size and energy of the orbital
Related to the shape of the orbital
0 = s = spherical
1 = p = dumbbell shaped
2 = d = clover leaf shaped
3=f
4=g
Related to the position of the orbital in
space relative to other orbitals
s=1
0
p=3
-1 0 +1
d=5
-2 -1 0 +1 +2
f=7
-3 -2 -1 0 +1 +2 +3
Related to the spin of the electron,
which can be only one of two values
Range of Values
Integers > 0 (1, 2, 3, …)
Integers from 0 to n – 1
Integers from +ℓ to 0 to -ℓ
(2ℓ + 1)
+½ or -½
Chapter#8 & 9:
Coulomb’s Law: can determine energy interaction between a pair of ions
"QQ %
E = (2.31x10-19 J#nm) $ 1 2 '
r in nm
# r &
E = units in J
ClNa+
r = distance between the ion centers in nm
Q1 & Q2 = numerical ion charges
€
Electronegativity: the ability of an atom in a molecule to attract shared electrons to itself.
• Generally increases across a period and decreases down a group
Isoelectronic Ions: ions containing the same number of electrons
Enthalpy Change (ΔH) = sum of the energies required to break old bonds (+) plus the sum of
the energies released in the formation of new bonds(-)
ΔH = ΣD (bonds broken) - ΣD (bonds formed)
(energy required)
(energy released)
Localized Electron (LE) has 3 parts:
1. Description of the valence electrons arrangement in the molecule using Lewis structures
2. Prediction of the geometry of the molecule using the valence shell electron pair repulsion
model (VSEPR)
3. Description of the type of atomic orbitals used by the atoms to share electrons or hold
lone pairs
Resonance: sometimes more than 1 valid Lewis structure is possible for a given molecule.
Formal Charge: the difference between the number of valence electrons on the free atom and
the number of valence electrons assigned to the atom in the molecule.
1. number of valence electrons on free neutral atom
2. number of valence electrons belonging to the atom in a molecule
3. compare these two numbers
Formal charge = # of valence e- on free atom – # of valence e- assigned to atom in molec.
Chapter#10:
Intramolecular Bonding: sharing electrons within the molecule.
Intermolecular Forces: weak interactions that occur between molecules.
London Dispersion Forces (AKA – van der Waals dispersion forces): the forces existing
among noble gas atoms and nonpolar molecules that involve an accidental dipole that induces a
momentary dipole in a neighbor.
Dipole-Dipole Attraction: molecules can attract each other electrostatically by lining up so the
(+) and (-) ends are close to each other.
Hydrogen Bonding: strong dipole – dipole forces when hydrogen is bound to a highly
electronegative atom, such as N, O, or F.
These forces are responsible for:
Surface Tension: The resistance of a liquid to an increase in it surface area.
Capillary Action: the spontaneous rising of a liquid in a narrow tube.
• Cohesive Forces: intermolecular forces among the molecules on the liquid
• Adhesive Forces: forces between the liquid molecule and their container.
Viscosity: measure of a liquid’s resistance to flow.
London Forces
Non-polar or
Noble Gas
Accidental
Dipole
Dipole – Dipole
Polar
H – Bonding
H- to N, O, or F
Electrostatic
Attraction
Strong Dipole
Ionic Forces
True + and –
charge
Stealing of
electrons causes
actual + and ions
Silicon Bonding
Si – O
Si – O
tetrahedral
structures
Bragg Equation:
nλ = 2dsinθ
n = integer
or
d=
nλ
2sin θ
λ = wavelength of X-rays
d = distance between atoms
θ = angle of incidence and reflection
•
•
•
Ionic Solids: (salt) – a solid containing cations and anions that dissolves in water to give a
solution containing the separated ions which are mobile and thus free to conduct electrical
current.
Molecular Solids: a solid composed of neutral molecules at the lattice points (no ions)
3 Types of Atomic Solids: a solid that contains atoms at the lattice points.
o Metallic Solids: a special type of delocalized non - directional covalent bonding
occurs.
o Network Solids: atoms bond to each other with strong directional covalent bonds
that lead to giant molecules or networks of atoms. Ex) Diamond and graphite
o Group 18 Solids: noble gas elements are attracted to each other with London
Dispersion Forces.
"P
% ΔH " 1 1 %
vap
Clausius Clapeyron Equation: ln$$ vap,T1 '' =
$ − '
R # T2 T1 &
# Pvap,T 2 &
• Sublimation: solid to the gaseous state without passing through the liquid state.
• Deposition: gas to the solid state without passing through the liquid state.
• Melting: solid to a €
liquid
• Freezing: liquid to a solid
• Vaporization/Evaporation/Boiling: liquid to a gas
• Condensation: gas to a liquid
• Heat of Fusion or Enthalpy of Fusion(ΔHfus): The enthalpy change that occurs at the melting
point when a solid melts
• Heat of Vaporization or Enthalpy of Vaporization (ΔHvap): energy required to vaporize 1mole
of a liquid at 1 atm.
To calculate energy to raise the temperature of material use: q = mCΔT
"
%
To calculate energy to change state use:
HFusion $ kJ '
gram
mol
# mol &
•
•
kJ
The greater the ΔH the longer the horizontal line will be on a heating curve during a phase change.
The greater the specific heat capacity (C), the less steep the slope of the line will be as a material
€
rises in temperature.
Phase Diagram: a convenient way of representing the phase of a substance as a function of
temperature and pressure.
Triple Point: all three state of a material exist at the same time.
Critical Temperature: temperature above which the vapor cannot be liquefied no matter what
pressure is applied.
Critical Pressure: Pressure required to produce liquefication at the critical temperature.
Critical Point: critical temperature and pressure together.
•
If the solid/liquid boundary line has a (-) slope also tells us that its solid is less dense than its
liquid state.
Chapter#11:
Solute: substance being dissolved
Solvent: the dissolving medium (liquid present in largest amount if 2 liquids)
Solution: when components of a mixture are uniformly intermingled and is homogeneous.
€
€
€
€
Ways to describe solution composition:
moles of solute
Molarity(M) =
Liter of solution
mass of solute
Mass% (% Comp.) =
x100
mass of solution
number of moles of given component
Mole Fraction(x) =
total number of moles of solution
for a 2 – component solution: mole fraction of Component A =
Molality(m) =
na
na + nb
moles of solute
Kg of solvent
€
Normality (N): the number of equivalents per liter of solution.
Acid/Base Equivalent is defined so that 1 equivalent of acid reacts with exactly 1 equivalent of
base (mass of acid or base that can furnish or accept 1 mole of protons (H+ ions)).
Redox
Equivalent of reducing agent will react with exactly 1 equivalent of oxidizing agent
(quantity of oxidizing or reducing agent that can accept or furnish 1 mole of
electrons).
Henry’s Law: relationship between gas pressure and the concentration of dissolved gas.
C = concentration of dissolved gas (mol/L)
C = kP
k = constant characteristic of a particular solution
(mol/L*atm)
P = partial pressure of gaseous solute above the
solution. (atm)
Psoln = observed vapor pressure of the solution
xsolvent = mole fraction of solvent
POsolvent = vapor pressure of the pure solvent
Raoult’s Law: Psoln = xsolvent POsolvent
For liquid – liquid solutions where both components are volatile, Raoult’s Law must be
modified:
Ptotal = PA + Pb = xA POA + xB POB
Boiling Point Elevation:
ΔT = Kbmsolute
ΔT = difference between boiling points of solution and pure solvent.
" O C • Kg %
Kb = Characteristic of solvent. $
'
# mol &
Called Molal Boiling Point Elevation Constant
moles of solute
msolute = molality of the solute in solution Molality(m) =
Kg of solvent
€
Freezing Point Depression:
ΔT = Kfmsolute
€
ΔT = difference between the freezing point of the pure solvent and that of the
solutions.
Kf = characteristic of solvent.
Called Molal Freezing Point Depression Constant
Osmotic Pressure:
Π = MRT
Π = osmotic pressure (atm)
M = molarity of solution
" 0.08206L • atm %
R = Gas law constant $
'
#
&
mol • K
T = temperature (K)
The relationship of moles of solute dissolved and moles of particles in solution is expressed
using the Van’t Hoff Factor (i):
€ in solution
moles of particles
i=
moles of solution dissolved
€
Chapter#12:
k = rate constant
[ ] = concentration of material
n = order of reactant (can be 1 of 3 types; integer, fraction, or zero)
Rate Law = k[ ]
n
" L(n−1) %
'' , where n is the overall reaction order.
The units for k will be $$
(n−1)
# mol • s &
Rate = k[A]
€
1
1
[ A]
2
t1 =
1
[A]0
versus t
1
k[A]0
Slope = k
[ A]
= kt +
Rate = k[A]2 or Rate = k[A][B]
Summary of the Kinetics for Reactions of the Type aA
Products that are Zero, First, or Second Order in [A]
Order
First
Second
Rate = k
ln[A] = -kt + ln[A]0
Zero
Rate Law:
[A] = -kt + [A]0
0.693
k
ln [A] versus t
2
t1 =
[A] versus t€
€
[A]0
2k
Slope = -k
2
t1 =
Slope = -k
" [A] %
ln$ 0 ' = kt
# [A]t &
ln[A]0 − ln[A]t =€
kt
Integrated Rate Law:
Plot Needed to Give a
Straight Line:
Relationship of Rate Constant
to the Slope of Straight Line:
Half-Life:
€
€
€
€
−E
a
k = Ae−E a / RT when natural log is taken you get ln k =
Arrhenius Equation:
+ ln A
RT
"k % E " 1 1 %
" rate2 % E a " 1 1 %
Used if only the rate constants determined at 2 different temps. ln$ 2 ' = a $ − ' or ln$
'=
$ − '
# k1 & R # T1 T2 &
# rate1 & R # T1 T2 &
€
J
€
k = rate constant, Ea = activation energy, R = 8.314
, T = Kelvin,
A = proportionality constant, e = base of natural logarithm
mol • K
€
[ ]
1
1
1
1
1
1
1
1
[ ]
1
2
3
1
2
3
1
3
[ ]
#
Time
(t)
#
#
#
#
#
#
#
Rate
1.0x10-5
2.0x10-5
3.0x10-5
1.0x10-5
4.0x10-5
9.0x10-5
1.0x10-5
1.0x10-5
•
•
•
•
•
•
Order of Reaction
1 Order Reaction Rate
As [ ] doubles, rate doubles and as [ ] triples, rate triples.
st
2nd Order Reaction Rate
As [ ] doubles, rate becomes 4 times as large [2]2 = 4 and as
[ ] triples, the rate become 9 times as large [3]2 = 9.
Zero Order Reaction Rate
A change in [ ] does not change the rate.
[ ] vs. t
ln [ ] vs. t
1/[ ] vs. t
Not a 0 Order Rxn.
Not a 1st Order Rxn.
A 2nd Order Rxn.
A 0 Order Rxn.
A 1st Order Rxn.
Arrhenius Expression:
Number of collisions with Ea = (total number of collisions) e−E a / RT
• Ea = Activation Energy
" J %
• R = universal gas constant 8.3145$
'
# K • mol
€&
• T = Kelvin
€
€
Ea # 1 &
+ ln(A)
% (€
R $T '
•
ln( k ) = −
•
"k % E " 1 1 %
ln$ 2 ' = a $ − '
# k1 & R # T1 T2 &
Chapter#13:
Proposed Law of Mass Action: general description of the equilibrium condition
• For a reaction of the type jA + kB
lC + mD
• Where A, B, C, D represent chemical species and j, k, l, and m are their coefficients in
the balanced equation.
The Law of Mass Action is represented by the following Equilibrium Expression (K):
l
m
[C ] [ D ]
K=
j
k
[ A] [ B]
[ ] = concentration of the chemical species at equilibrium
K = a constant called the Equilibrium Constant (customarily given without units)
•
If reaction is reversed the new equilibrium expression =
1
(reciprocal)
K
nl
•
If original reaction is multiplied by some factor n equilibrium expression K
n
nm
[C ] [ D]
=
nj
nk
[ A ] [ B]
€
Δn
€
K P = K ( RT ) . (This is used to find K from Kp.)
€ of the
Δn = sum of the coefficients of the gaseous products minus the sum
coefficients of the gaseous reactants
Δn = (l + m) – (j + k)
Reaction Quotient (Q): obtained by applying the Law of Mass Action using INITIAL
concentration instead of equilibrium concentration
Ex) 1N2(g) + 3H2(g)
2NH3(g)
[NH 3 ] 0 2
Expression for reaction quotient is Q =
[N 2 ] 0 [ H 2 ] 0 3
Subscript zeros indicate initial concentrations
€ will shift to reach equilibrium, we compare the
To determine in which direction a system
values of Q and K. 3 possible cases:
1. Q = K system at equilibrium, no shift will occur
2. Q > K the ratio of initial concentration of products to initial concentration of reactants is
to large to reach equilibrium, a net change of products to reactants must occur. System
shifts left, consuming products and forming reactants.
3. Q < K ratio of initial concentration of products to initial concentration of reactants is too
small. System must shift right, consuming reactants and forming products.
Procedure for Solving Equation Problems:
1. Write the balanced equation for the reaction.
2. Write the equilibrium expression using the law of mass action.
3. List the initial concentrations.
4. Calculate Q and determine the direction of the shift to equilibrium.
5. Create an I.C.E. table (Initial, Change, Equilibrium) Define the change needed to reach
equilibrium, and define the equilibrium concentrations by applying the change to the
initial concentrations.
6. Substitute the equilibrium concentration into the equilibrium expression, and solve for the
unknown.
7. Check your calculated equilibrium concentration by making sure they give the correct
value of K.
Le Chåtelier’s Principle: if a change is imposed on a system at equilibrium the position
of the equilibrium will shift in a direction that tends to reduce that change.
• If a component (reactants or products) is added to a reaction system at equilibrium (at
constant T and P or constant T and V) the equilibrium position will shift in the direction that
lowers the concentration of that component. If a component is removed, the opposite effect
occurs.
• Changes in concentration may shift equilibrium position but has no effect on the value of the
equilibrium constant, the new equilibrium concentration satisfy the original equilibrium
constant.
• The value of K changes with the temperature.
• Treat energy (heat) as a reactant (in an endothermic process) or as a product (in an
exothermic process), and predict the direction of shift in the same way using Le Chåtelier’s
Principle. This cannot predict size of change in K, but does predict direction of change.
Chapter#14:
•
•
•
•
•
•
•
•
•
•
•
Arrhenius: Acids produce hydrogen ions (H+) in aqueous solution, while bases produce
hydroxide ions (OH-).
Bronsted – Lowry Model: An acid is a proton (H+) donor and a base is a proton acceptor.
!" H 3O + #$!" A − #$ !" H + #$!" A − #$
K a = Acid Dissociation Cons tan t =
=
[ HA]
[ HA]
Kw = Ka x Kb
Kw = Water Dissociation Constant = [H+] [OH-] Kw = (1.0x10-7) (1.0x10-7) = 1.0x10-14
pH = -log [H+]
pOH = -log[OH-]
14.00 = pH + pOH
[H+] = 10-pH and [OH-] = 10-pOH
Significant Figures for Logarithms: The rule is that the number of decimal places in
the log is equal to the number of significant figures in that original number.
o Ex) [H+] = 1.0x10-9M (2 sig. figs. means 2 digits after the decimal) pH = 9.00
Common Strong Acids: HCl, HNO3, H2SO4, HClO4, HBr, HI
Solving Weak Acid Equilibrium Problem:
1. List the major species in the solution.
2. Choose the species that can produce H+, and write balanced equation for the reactions
producing H+
3. Using the values of the equilibrium constants for the reactions you have written, decide
which equilibrium will dominate in producing H+.
4. Write the equilibrium expression for the dominate equilibrium.
5. List the initial concentration of the species participating in the dominant equilibrium.
6. Define the change needed to achieve equilibrium (x)
7. Write the equilibrium concentration in terms of x.
8. Substitute the equilibrium concentration into the equilibrium expression.
9. Solve the x for the “easy” way, that is, assume [HA]0 – x ≈ [HA]0
10. Use the 5% rule to verify if approximation is valid.
11. Calculate [H+] and pH.
•
Strong Bases: LiOH, NaOH, KOH, RbOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2
1.
2.
3.
Acid Used
Strong
Weak
Strong
Base Used
Strong
Strong
Weak
Salt Type
Neutral
Basic
Acidic
Salt Solution
Neutral
Basic
Acidic
pH
7.00
> 7.00
< 7.00
Ideas and generalizations :
1. Memorize the following strong acids: HCl, HBr, HI, HNO3, HClO4, and H2SO4
2. Memorize the following strong bases: LiOH, NaOH, KOH, RbOH, Ca(OH)2, Sr(OH)2,
and Ba(OH)2
3. All weak acids have a Ka value less than 1 but greater than Kw. All weak bases have a Kb
value less than 1 but greater than Kw.
4. All conjugate bases of weak acids are weak bases. All have a Kb value less than 1 but
greater than Kw.
5. All conjugate acids of weak bases are weak acids. All have a Ka value less than 1 but
greater than Kw.
6. Alkali metal ions (Li+, Na+, K+, Rb+, Cs+) and heavier alkaline earth metal ions (Ca2+, Sr2+,
Ba2) have no acidic or basic properties in water.
7. All conjugate bases of strong acids (Cl-, Br-, I-, NO3-, ClO4-, HSO4-) have no basic
properties in water (Kb << Kw) and only HSO4- has any acidic properties in water.
Structure: There are 2 main factors that determine whether a molecule containing X – H bond
will behave as a Bronsted – Lowry acid:
1. Strength of the bond: as H – X bond strength decreases, acid strength increases
2. Polarity of the bond: greater the electronegativity, stronger the acid of neighboring atoms.
Oxyacids: acid strength increases with an increase in the # of oxygen atoms attached to
the central atom.
Lewis Acids: an electron pair acceptor
Lewis Base: an electron pair donor
• A Lewis acid has an empty atomic orbital that it can use to accept (share) an electron pair
from a molecule that has a lone pair of electrons (Lewis base).
Chapter#15
Common Ion: ion produced by both acid and salt
Common Ion Effect: the shift in equilibrium position that occurs because of the addition of an
ion already involved in the equilibrium reaction.
Equivalence Point: Point at which equal quantities of [H+] and [OH-] exist.
o Defined by the reactions stoichiometry
End Point: Change in color of an indicator.
Pay attention to special points:
1. Buffered solutions are simply solutions of weak acids or bases containing a common ion.
2. When a strong acid or base is added to a buffered solution, deal with stoichiometry first, then
consider the equilibrium calculations.
# [ A− ] &
(
Henderson – Hasselbalch Equation: pH = pKa + log %%
(
HA
[
]
$
'
=
" [ Base] %
pKa + log $
'
# [ Acid ] &
Titration Set-up Steps:
€
€
1. There are 4 types of problems in a Strong
Acid/Strong Base
Titration:
a. 0.00mL titrant:
• Only HCl in solution and 100% dissociation due to being a strong acid.
• Molarity is [H+]
• pH = -log[H+]
b. Any amount of titrant up until just before equivalence point:
• Need a before and after stoich table with conversions to mmol.
• Only HCl left in solution and 100% dissociation due to being a strong acid.
mmol [ H + ] left
+
• Molarity is [ H ] =
volume total ( mL)
+
• pH = -log[H ]
c. At equivalence point:
€
• [H+] = [OH-] and water is formed giving a pH = 7
d. Any point after equivalence point:
• Excess strong base and 100% ionization due to being a strong base.
• Need a before and after stoich table with conversions to mmol.
mmol [OH − ] excess
−
• Molarity is [OH ] =
volume total ( mL)
• pOH = -log[OH ]
• pH = 14 – pOH
•
€
Titration of a strong base with strong acid uses similar math. [OH-] before equivalence point and
[H+] in excess.
2. There are 5 types of problems in a weak acid/strong base titration:
a. 0.00mL titrant:
• Weak acid problem
• ICE table to solve for [H+]
• pH = -log[H+]
b. Any amount of titrant except ½ way to equivalence point and equivalency point.
• Before and after stoich table.
mmol
• M=
volume total ( mL)
• Need pKa = -log Ka
" [ Base] %
• Now Henderson – Hasselbalch problem
pH = pKa + log $
'
€
# [ Acid ] &
c. At ½ way to the equivalence point:
• Before and after stoich table.
mmol
• M=
volume total ( mL)
• pH = pKa
• Optimal Buffer
€
€
d. Equivalency Point:
• All acid in container is now equal to all base in container but there is a weak base salt and so
is a weak base problem.
• Before and after stoich problem.
K
• In need of K b = w
Ka
• ICE table to solve for [OH-]
!" H + #$ = K w
• pOH = -log[OH-]
OR
!"OH − #$
€
•
pH = 14 – pOH
pH = -log [H+]
e. Any Amount Past Equivalence Point:
• Strong base is in excess and 100% ionization occurs.
• Before and after stoich table to solve for [OH-]
mmol [OH − ] excess
−
• Molarity is [OH ] =
volume total ( mL)
!" H + #$ = K w
• pOH = -log[OH-]
OR
!"OH − #$
•
€ = 14 – pOH
pH
pH = -log [H+]
Ksp (Solubility Product Constant) of the Solubility Product:
Chapter#16:
•
•
•
•
•
•
•
•
•
•
••
€
•
The First Law of Thermodynamics: mass-energy cannot be created or destroyed.
Second Law of Thermodynamics: All processes occur spontaneously in the direction that
increases the total entropy (disorder) of the universe (system plus surroundings).
Entropy(S): Entropy is a measure of the disorder of the system
Gibb Free Energy: The change in free energy of a system at constant temperature and
pressure is found with ΔGsys = ΔHsys - TΔSsys
Standard entropy of reaction, ΔSorxn: ΔSorxn = ∑nSoproducts - ∑nSoreactants
ΔHosys = ΔHorxn = ∑nΔHof (products) - ∑nΔHof (reactants)
State Functions are not pathway dependent, they depend only on the beginning and end point
values; ΔE, ΔH, ΔG, ΔS, and Pressure
Reaction Spontaneity and the Signs of ΔHo, ΔSo, and ΔGo:
ΔGosys = ΔHosys - TΔSosys
Description
ΔHo ΔSo -TΔSo
ΔGo
+
Spontaneous at all T
+
-
+
+
Non-spontaneous at all T
+
+
-
+ or -
-
-
+
+ or -
Spontaneous at higher T
Non-spontaneous at lower T
Spontaneous at lower T
Non-spontaneous at higher T
So there are 3 ways to measure and solve for ΔG:
o
ΔH reaction
= ∑ n product ΔH of ( products) − ∑ n reac tan tsΔH of (reac tan ts)
1.
o
ΔSreaction
= ∑ n product ΔS of ( products) − ∑ n reac tan tsΔS of (reac tan ts)
and ΔG = ΔH - TΔS
2. Hess’s Law: total is sum of the smaller parts
o
ΔGreaction
= ∑ n product ΔG of ( products) − ∑ n reac tan tsΔG of (reac tan ts)
3.
€
Free Energy and Equilibrium:
The relationship between free energy and the reaction quotient can be derived from the ideal gas
€
law and the definition of Q:
ΔG = ΔGo + RT lnQ
ΔGo = -RT lnK
At equilibrium, ΔG =0
and
K = e-(ΔG/RT)
Symbol
Unit
Energy
E
(J)
-E
+E
Heat
q
(J)
-q
+q
Work
w
(J)
-w
+w
(Heat Flow)
Enthalpy
H
(kJ)
-H
+H
(Disorder)
Entropy
S
"J%
$ '
#K &
+S
-S
(Spontaneous)
-G
+G
+E
-E
Free Energy
G
(kJ)
Exothermic
Endothermic
€
Cell Potential
E
V
"J%
$ '
#C&
At equilibrium S and G = 0
€
Chapter#17:
•
•
•
•
•
€
Oxidation is the loss of electrons (the increase in oxidation number)
Reduction is the gain of electrons (the decrease in oxidation number)
Anode: electrode compartment in which oxidation occurs. Loss of e-, (-) charged, s
Cathode: electrode compartment in which reduction occurs. Gain of e-, (+) charged, aq
work ( J ) joules
Vvolt =
ch arge(C ) coulomb
Line Notation: used to describe electrochemical cells
• Anode component on left/cathode on right.
• Separated by double vertical lines (indicating the salt bridge).
• Phase difference (boundary) is indicated by a single vertical line.
• Substances constituting the anode are listed far left/ cathode far right.
• If no electrode (all in solution), a non-reacting (inert) conductor must be used, usually Pt.
aq
s
Complete Description includes 4 items:
1. Cell potential (always +) for galvanic cell, where E Ocell = E Ocathode + E Oanode and the balanced cell
reaction.
2. The direction of electron flow, obtained by inspecting the ½ reactions and using the direction that
gives a (+) E Ocell.
3. Designation of anode and cathode.
4. Nature of each electrode and the ions present in each compartment. A chemically inert conductor
(usually graphite) is required if none of the substances participating in the ½ reaction is a
conducting solid.
•
E=
−W (work)
q(ch arge)
or
W = −q E
February 27, 2012
€
•
q = quantity of charge in coulombs transferred
• The charge on 1 mole of electrons is a constant called the
Coulombs of ch arge
Faraday (F) = 96,485 coulombs of
mole of electrons
• q = nF
n = moles of e
F = charge per mole of e•
ΔG = -qE max = -nFE max
•
= E Ocell -
0.0591
log(Q)
n
•
log K =
n EO
0.0591
€
10.0C " 30min 60sec % 1mole e − 1mol Cu 63.546g Cu
•$
•
•
•
= 5.94gCu
'•
sec # 1
1min & 96,485C 2mole e −
1mol Cu
Ex)
€
€
€
€
€
€
€
Chapter#18:
1
−1
antiproton
H
1
1
1
0
0
−1
4
2
0
+1
H
n
e
He
e
2 00γ
proton
neutron
electron (β particle)
α particle
Positron
Gamma Ray
Integrated 1st order rate =
Half Life =
•
Understand and be able to perform α and β decay, positron emission, γ rays, and electron
capture.
•
Understand how the zone of stability determines which of the above is needed to allow an
atom to become stable.
Above the line of stability means the
atom has too many neutrons and will
undergo β decay.
Below the line of stability means the
atom has to many protons and will do 1
of 3 things:
1. positron emission
2. electron capture
3. α decay (for heavy nuclides only)
Chapter#22:
Rules for Naming Alkanes:
1. Parent Chain: longest continuous chain of C-C
2. Identify shorter chains that make up branches and locations.
3. Number parent chains C atoms from end to end, starting with end that is closer to a branch.
4. Count number of C-C in branch chain, find root word for that number and add suffix “-yl” to
root word.
5. Put parts together in this order: # of branch carbon, name of branch, name of parent chain.
• Hyphen is written between the number and the substituent name.
• Substituent’s are listed in alphabetical order
Organic Chemistry Prefixes
Prefixes Di, Tri, … are used to indicate multiple, identical
• 1 = Meth• 6 = Hex• 2 = Eth• 7 = HeptStructural Isomers: when 2 molecules have the same atoms
• 3 = Prop• 8 = Octbut different bonds
• 4 = But• 9 = NonConformation: A structure that has been spun in space.
• 5 = Pent• 10 = DecCyclic Alkanes: C atoms that form rings
• Root name is proceeded by the prefix cyclo• Ring is numbered to yield smallest substituent number possible.
Alkenes: hydrocarbons that contain at least 1 C C double bond.
general formula CnH2n
1. Root ends in –ene
2. In alkenes containing more than 3 C atoms, location of double bonds is indicated by the
lowest numbered C atom involved in the double bond.
• sp2 hybridized (trigonal planar)
• 1σ and 1π bond
• prevents rotation
• Cis – Trans Isomerism:
! Identical substituents on same side of double bond are Cis! Identical substituents on opposite sides of double bond are TransAlkynes: unsaturated hydrocarbons containing at least 1 C
• 1σ and 2π bonds
• sp hybridized (linear)
• root ends in –yne
C bond
Aromatic Hydrocarbons: a class of cyclic unsaturated hydrocarbons
• Benzene C6H6
• Planar
resonance
2
• sp hybridized
structures
O
• 120 Bond angles
• C – C and C – H σ bonds
• π bonds delocalization is usually indicated by circle inside the ring
• Undergoes substitution reactions in which H atoms are replaced by other atoms .
Nomenclature: if more than 1 substituent is present, number’s are used to indicate substituents
postition.
Ortho (o) – for 2 adjacent substituents
Meta (m) – for 2 substituents with 1 C between them
Para (p) – for 2 substituents opposite each other
Name
Methyl
Formula
-CH3
Ethyl
-CH2CH3
Propyl
-CH2CH2CH3
Isopropyl
-CH(CH3)2
Butyl
-CH2CH2CH2CH3
sec-Butyl
-CH(CH3)(CH2CH3)
Isobutyl
-CH2-CH(CH3)2
tert-Butyl
-C(CH3)3
•
•
•
Structure
Primary alcohols can be oxidized into aldehydes, which can then be oxidized into carboxylic
acids.
Secondary alcohols can be oxidized into ketones.
Tertiary alcohols do not undergo oxidation.
Functional Groups Flow Chart
Class
Functional General
Group Formula
Halohydrogens –X
R–X
(F, Cl, Br, I)
Naming Rules
*Suffix “o”
*Seen as a substitute.
*Same priority as an alkyl group.
*Floro-, Chloro-, Bromo-, Iodo-
Example
CH3I
iodomethane
3-bromo-1-propene
Alcohols
–OH
R – OH
*Suffix “ol”
CH3OH
*Primary, secondary, and tertiary
alcohols are based # of C’s attached
to the C – OH group.
Ethers
–O–
R – O – R’
*Name the chains listed on either side CH3 – O – CH3
of the oxygen in alphabetical order dimethylether
and add ether to the end.
CH3 – CH2 – O – CH3
ethylmethylether
Aldehydes
O
–C–H
O
*Suffix “al”
*Always on the 1st C, or C #1.
CH2O
methanol
methanal
R–C–H
2-methyl-butanal
Ketones
Carboxylic
Acids
O
O
–C–
R–C–R
O
O
– C – OH
Esters
O
–C–O–
Amines
-NH2
Naming Priority:
*Suffix “one”
*Never on 1st C
*Give ketone lowest possible #.
CH3 – CH2 – CO – CH2 – CH3
3-pentanone (IUPAC) or
diethylketone
*suffix “oic acid”
CH3COOH or HC2H3O2
ethanoic or acetic acid
R – C – OH
O
*Suffix “oate”
*The R group bound to esters O – R
R – C – O – R’ is a substituent and not part of the
parent.
R – NH2
*As substituent – prefix “amino”
*As parent – suffix “amine”
*Primary, secondary, and tertiary
amines are based # of C’s attached
to the NHn structure. For every C
added, one less H atom exists.
Functional Groups
Multiple Bonds
Halogens and Alkyl Groups
methylethanoate
1-aminobutane (IUPAC)
butylamine