6/13/2009 10:13:00 PM
[2/4/09]
did lab
[2/5/09]
Mole-volume hypothesis
 Avogadro hypothesized that equal volumes of a gas at the same
temperature and pressure contain equal number of particles
o If PV = nRT (and P,V,T, are the same then n is the same)
 STP – 273 K, 1 atm, 1 mol
o (1 atm)(V) = 1 mol (.0821)(273)
 V = 22.4 L
o At STP  1 mol (NA particles) of any gas occupies a volume
of 22.4 liters
 22.4 molar liters
o At STP  volume of gas = (# mol) x (22.4 L/mol)
 Ex. how many moles of H2 are in 0.2 L?
 8.92 x 10-3 mol H2
o At STP  molar mass = density x molar volume
 (g/mol) = (g/l) (l/mol)
The density of a compound containing C and O is
1.94 g/l at STP what is the molar mass?
 44 g/mol (1.94*22.4)
All the ways to convert moles
 Moles to particles
o Multiply by NA
 Particles to moles
o Divide by NA
 Mole to mass




o Multiply by molar mass
Mass to mole
o Divide by molar mass
Mole to volume of gas (Must be STP)
o Multiply by 22.4 mol/l
Volume of gas to mole (Must be STP)
o Divide by 22.4 mol/l
Q: what volume of SO2 will result from the complete burning (S + O2 = SO2)
of pure sulfur in 8 L of Oxygen gas?
 8 L / 22.4 L/mol = .36 mol O2
 .36 mol S O2 * 22.4 = 8L
Q: How many liters of N2 will react with 36 grams of H2 at STP, given N2 +
3H2  2NH3?
 36 g / 2 g/ mol = 18 mol H2 x (1/3) = 6 mol N2 * 22.4 L/mol =
134.4 L
 Q: How many liters of carbon dioxide at STP are produced by
combustion of 342 grams of octane (C8H18) according to the
equation 2C8H18 + 25O2  16CO2 +18H2O?
o 342/114 = 3 mol x (16/2) = 24 mol x 22.4 L/mol = 537.6 L
Q: How many liters of oxygen gas at 755 mm Hg and 305 K form when 294
g of KClO3 completely react in 2KClO3 (s)  2KCl (s) + 3O2 (g)?
 294 / 122.5 = 2.4 mol (3/2) = 3.6 mol O2
 755/760 = .993 atm
 (.993)V = (3.6)(.0821)(305)
o V = 90.78 L
[2/6/09]
Co (g) + 2H2 (g)  CH2OH (g)
 What volume of H2 (in L) at 748 mmHg and 86 C is required to
produce 25.8 g CH3OH?
o 25.8/32 = .80625 mol x (2/1) = 1.6125
o 748/760 = .984 mmHg
o 359 K
o PV = nRT
o (.984)(v) =(1.6125)(.0821)(359)

V= 47.9
Mixture of gases
 Air – 78% nitrogen, 21% oxygen, .9% argon, .03% CO2
o Percent by volume
o Each gas exerts partial pressure
 Partial pressure – pressure due to any individual component in a
gas mixture
 Daltons law of partial pressure
o Total pressure - sum of all partial pressures
 Ptotal = Pa + Pb + Pc + …]
[2/9/09]
 continued
 PA= XAPtotal
o XA = mol fraction of gas A
 XA = Moles of A / total # of moles of gas
o PA = Partial pressure
 Ex. .75 mol nitrogen, .2 mol hydrogen, .005 mol fluorine. Total
pressure 2.5 atm
o .955 total mol
 Partial pressure of
 Nitrogen – 1.88 atm
 (.75/.955)*2.5
 Hydrogen – .52 atm
 (.2/.955)*2.5
 Fluorine – .013 atm
 (.005/.955)*2.5
Diffusion/Effusion
 Diffusion
o Tendency of molecules to move toward areas of how
concentration until the concentration
 Effusion
o Gas escapes through a tiny hole in a container
o * lower molar mass effuse faster
 Graham’s law of effusion
o The rate of effusion is inversely proportional to the square
root of gas’s molar mass
 RateA / rateB = √(molar mass B / molar mass A)
o Ex. How much faster does Helium effuse than nitrogen?
 Rate He / Rate N2 = √(28/4) ≈ 2.7
 Rate He is 2.7 times the rate of N2
o Rate is anything over time
[2/10/09]
When doing calculations remember diatomic atoms
Partial pressure
 Pressure that one atom or molecule would exert if it alone was in a
container
o In problems you may need to use boyle’s law PV = PV ( to
find new pressure of something)
 Total pressure = added partial pressures
A certain gas effusion 4 times as fast as oxygen gas. What is the molar mass
of this gas?
 Rate ? / rate O2 = 4 = √32 / x
o X=2

H2
Real gases
 At high pressures or low temperatures gases do not behave ideally
o Van Der Waals equation
 [P+a(n/v)2] x [v-nb] = nRT
[2/11/09]
 ex. solve for pressure, 1 mol Cl2 , 5 L , 273 K using
o Van der Waals
 A – 6.49, B - .0562


[P + 6.49(1/5)2] x [5 – 1.0562] = 1(.0821)(273)
P = 4.27 atm
o Ideal
 P(5) = 1(.0821)(273)
 P = 4.48 atm
[2/12/09]
Determine density of CH4. 3.25 L, 5.8 atm, 25.5 C
 (5.8) (3.25) = n (.0821) (298.5)
o .77 mol * 16 g/mol = 12.3 grams
o 3.8 g/l
Calculate molar mass of unknown gas
 Flask
o 143.187 g , empty, 255 ml
 Filled
o 267 torr, 25 C, 143.289 g
 267/760 = .35 atm, 25+273 = 298 K, 143.289-143.187
= .102

PV = nRT
 (.35)(.255) = n(.0821)(298)
 n = .0037
o .102 /.0037 =27.5 g /mol
Chapter 7
6/13/2009 10:13:00 PM
[2/17/09]
Isaac Newton discovered that light traveled in particles
 Later it was discovered that light travels in waves as well
o Amplitude - maximum distance from equilibrium
o Wavelength – distance from one point on a wave to an
identical point on another wave
 Symbol – λ (lamda)
o Crest – top point on a wave
o Trough – bottom point on a wave
o Frequency – number of waves to pass a point in one second

o Measured in hertz
 1 hertz = 1/sec (or s-1)
 Symbol - ν (nu)
o C (speed of light) = λν
Light changes color (pg. 285) as frequency grows greater (this is
only in a small portion of all the frequencies)
o C = λν
 C – 3 x 106 m
 λ – 510 x 10-9 m/s

 ν = 5.88 x 1014 hertz
o if sun is shined on a prison a reflection of all the colors appear
 this happens with a rainbow (water drops reflect the
lights)
Dual nature of lights
o Acts as waves + particles (light is unique that this happens)
 Particles made up of photons (little package of light)
Atomic spectra
o Pass electric current through a gas in a tube




This energizes electrons of atoms
The electrons absorb energy and move to higher energy
levels (rings around nucleus)
The electrons then lose energy + come back to original
position and emit light
[2/18/09]
 Ground state – lowest possible energy state of e- in atom
 E = quantum of energy

o E = hν
 v (nu) – frequency of light emitted
 h = pianck’s constant = 6.63 x 10-34 JS
 Combined with C = λν
  E = hc / λ
 energy = (h) 6.63x 10-34 * (c) frequency /
wavelength λ
o energies unit is J*S
Bohr model
o Named after Neils Bohr
o Said that e- can be found in orbits around the nucleus
o Each orbit has a fixed amount of energy = energy level
Ladder
Atom
Lowest ring
Energy level closest to nucleus
(lowest amount of energy)
Person climbs from ring to ring
e- travel from energy level to
another
To move from energy level to
another e- amount must be
gained or lost contain amount of
energy

Quantum of energy – amount of energy required for an electron to
move from one energy level to another
o Higher energy levels (further from nucleus) are closer
together
o Goes up, needs energy
o Goes down, emit energy
[2/19/09]

44. Calculate the energy of a photon of a electromagnetic radiation
at
o 100.2 MHZ
 E = hv
 6.63 x 10-34 JS * 100,200,000 hz = 6.76 x 10-26 J
o 1070 KHZ
 E = hv



6.63 x 10-34 JS * 1,070,000 = 7.09 x 10-28 J
o 450 nm
 E = hc / λ
 [6.63 x 10-34 JS x (3x108 m/s)] /450 x10-9 m =
4.44 x 10-19 J
Heisenberg uncertainty principle
o Impossible to know both position and velocity of an electron
at the same time
Schrodinger
o Came up with quantum mechanical model
Atom: probability of find an electron within a certain
region of space surrounding the nucleus  represented
as fuzzy cloud
 Cloud is denser where there is a greater
probability of finding electrons
o Atomic orbital’s – regions of space with highest probability of
finding electrons in an energy level
o Energy levels are represented by a principle quantum
numbers: n = 1,2,3,4…



Numbering from nucleus out
Within each energy level are sublevels  corresponds to
orbital’s of different shapes describing where electrons
are likely to be found
 Sublevels are represented by letters  s , p , d , f
 Labeled by angular momentum numbers
o If L is:
 0=s
 1=p


 2=d
 3=f
S: spherical
P: dumbbell shaped
o Every P has three orbital’s one on –
px, pz, py orbital’s
 On that axis
o Every orbital’s holds on two electrons




S – 2 electrons
P – 6 electrons
D – 10 electrons
F – 14 electrons
[2/23/09]
Sublevels
 L – angular momentum
o Angular momentum numbers relate to sublevels which have
different amounts of orbitals with each hold two electrons
L
sublevel
Electrons
0
S
1 orbital - 2
1
P
3 orbitals - 6
2
D
5 orbitals - 10
3
F
7 orbitals - 14
o
Principle energy
# of sublevels
Types of sublevels
# of electrons
1
1S (1st energy
2
levels
n=1
level- s sublevel)
n=2
2
2S , 2P
2+6 =8
n=3
3
3S, 3P, 3D
2+6+10 =18
n=4
4
4S, 4P, 4D, 4F
2+5+10+14=
32
o 2n2 = the maximum number of electrons a energy level can
hold
o Magnetic quantum number (Ml) – l is a L
 Specific orientation of orbital’s
L

Ml
S 0
0
P 1
-1,0,1
D 2
-2,-1,0,1,2
F 3
-3,-2,-1,0,1,2,3
P. 315
o Q 58:
 Chart above

o Q:59
 A) N = 2, l = 1, Ml = -1 which works
 B) n = 3, l = 2, Ml = 0 which works
 C) n = 3, l = 3, Ml = 2 which doesn’t work
o Q:60
 1s – works
 2p – works
 4s – works
 2d – does not work
Ml magnetic quantum numbers – ranges from –l to l
 L angular momentum # ranges from 0 to (n-1)
Absorption / emission
 When it moves up it absorbes energy
 Come back down, emit energy
o 315: 64
 n = 3  n = 1 - emits
 n = 2  n = 4 - absorbs
Chapter 8.2-8.9
6/13/2009 10:13:00 PM
[2/23/09]
Electron configurations – way in which electrons are arranged in orbitals
around the nucleus
 Rules for writing electron configurations
o 1. Aufbau principle
 Electrons occupy orbital’s of lowest energy first
 1s,2s,2p,3s,3p,(*)4s,3d,4p,5s,4d,5p,6s,5d,4f,6p
 *doesn’t make sense anymore
 Picture – pg 325
o 2. Pauli exclusion principle
To occupy the same orbital, 2 electrons must have
opposite spins
 This is represented by a box which has an up
arrow and then a down arrow
o 3. Hund’s rule
 electrons occupy orbitals of the same energy in a way
that makes the number of electrons with the same spin
direction as large as possible
 ex.

[2/25/09]
Elements Orbital diagram
Electron configuration
H
1s1 (exponents represents
electrons)
He
1s2
Li
1s22s1
C
1s22s22p2
O
1s22s22p4
Na
1s22s22p63s1
1s22s22p63s23p64s23d2
Ti
[2/26/09]
Figuring out an electron configuration based periodic table
 SPD
o Right two columns are S block
o Middle columns (transition elements) are D block
o Columns to the left of middle (left six) are P block



o Inner transition elements are f Block
Period number (for S + P) = principle quantum number
o Ex. Ca  1s22s22p63s23p64s2
Period number (for D) = n – 1
o The block next to 4s would be 3d
o Ex. Fe = 1s22s22p63s23p64s23d6
 (its 3d not 4d since its in the d block)
Period number (for f) = n - 2



Other problems
o Ex. As  1s22s22p63s23p64s23d104p3
o Ex. Rb 1s22s22p63s23p64s23d104p65s1
If you don’t want to write the entire formula you can write a noble
gas and what comes after that
o Ex. [Kr]5s1
 Can only be done when specified, otherwise must write
entire thing
Exception to Aufbau principle
o All elements below Cr and below Cu (Mo, W, Unh, Cu, Ag, Au)

Since these two electrons are close to being half full or
filled the electron in the s before jumps to the d, so that
both will be half full or filled
 Ex. Cr  1s22s22p63s23p64s13d5
 Ex. Cu  1s22s22p63s23p64s13d10
 Or Cu  [Ar]4s3d10
 Close
Electrons
 Core
o Electrons in inner energy levels
 Valence
o Electrons in outermost principle energy level
 Example if the highest energy level is 4s1 then there is
one
o If it has four valance and it’s a metal it will lose electrons
 Other things if 1,2, or 3 it will lose, if 5,6,7 it will gain
 If 8 nothing will happen
[2/27/09]
Electron configuration for
 Mg2+  1s22s22p6
 O2-  1s22s22p6
o If are given any ion that figure out if it will lose or gain
electrons and then write the electron configuration
Periodic table
 Atomic radius


o One half of the distance between the nuclei of 2 atoms of
same element when atoms are joined
 Measured in picometers (pm) – x 10-12
If you go down a group the atomic size increases
o Since energy levels are added
If you go across a period the atomic size decreases
o Since a proton is added as well as a proton and there is a
greater charge between the proton and the electron
[3/3/09]
Periodic trends



Atomic size (did already)
Ionic size
o Size of cation always smaller then size of atom
o Size of anion always greater then size of atom
 Ex. Na will be larger then Na+
o Find chart to see trend (cant find picuture)
Ionization energy
o Energy required to remove an electron from an atom
o As you go down a group ionization energy decreases
o When you go across a period (left to right) ionization energy
increases
 (opposite trend as atomic size)
[3/5/09]
Notes on homework:
When transitions become ions and lsoe electrons they lose from the
outermost energy levels
 Exceptions under Cr and Cu
o ex. Mo3+  …4p64d3 (not – 5s24d1) or [Kr] 5s04d3
if two ions have the same number of electrons (isoelectronic), the one with
less protons has a larger radius
 Ex. Se2+ Kr Sr2+ Rb+ Br Increasing atomic radius: Sr2+ Rb+ Kr Br- Se2+
Trends continued
 Electro negativity
o Ability of an atom attract electrons when the atom is in a
compound
o As you go down a period electro negativity decreases
(decreases when size rises)
o When going across a group (left to right) it increases
 Metallic character
o When going down a period it increases
o When going across a group (left to right) it decreases
Group 8a
 nobles cases
 energy levels fixed
Group 1a – 7a
 representative elements
 group 7a called halogens
Transition elements
 Fill up d sub levels
Chapter 9
6/13/2009 10:13:00 PM
[3/6/09]
Types of chemical bonds
Type of atom
Type of bond
Characteristic of bond
Metal + non metal
Ionic
Transfer of electron from
metal to non metal
Non metal + non metal Covalent
Metal + metal
Share electrons
Metallic
Ionic compounds and Ionic bonds
 Valance electrons – electrons in the highest occupied energy level
o Octet rule

o Every atom wants to have 8 valence electrons (this results in
them either gaining or giving up electrons)
 Na  Na+ + electron
 1s22s22p63s1  1s22s22p6
Lewis structure = electron dot structure
o A diagram that shows the valance electrons as dots and
bonds as dashes
 Go above, below, left and to the right (two on each
side)
The first two electrons go together on top
The next three go on different sides
 Pauli exclusion rule
 After that you add a second to each side (that has one)
o When an atom loses an electron, dots are not show (but +
and –‘s are)
o When it gains electrons the dots are shown (as well as + or –
‘s)
Properties of compounds



o Crystalline solids at room temperature
o Ions arranged in repeating 3D pattern
o Large attractive force (very strong bonds)
 High melting point and boiling point
o Conduct electric current when melted or dissolved in water
[3/11/09]
Metallic bonding


Metals: made of closely packed cations. Valance electrons of metal
atoms can be modeled as sea of electrons. Electrons are mobile.
o Since they are mobile they are:
 Good conductors
 Ductile
 Malleable
Alloys: mixture of two or more elements.
o Properties are similar to those of the elements they are made
up of
 Ex. brass = alloy of copper + zinc

Ex. Sterling silver = silver (92.5%) + copper (7.5%)
[3/12/09]
Covalent bonding (form molecular bonds)
 Lower boiling and melting points
 Weaker than ionic
 Molecular vs. ionic
o Molecular
 Collection of molecules
o Ionic

 Array of cations + anions
Want to share electrons to fill outermost levels
o H2  the two H’s share the two electrons
 Lewis structure H = H
 The dash (-) is the shared, bonded pair
 The dots are unshared lone pairs
o F2  Share two electrons (can be seen by using dot
structures)
 Lewis structure F – F (each F has six dots around it)
o H2O  H-O-H (O has a lone pair above and below)
Rules for drawing Lewis structure
 1. Count number of valance electrons in the entire compound
 2. Use dashes between central atom and surrounding atoms
(dashes represent a pair)
o Central atom – atom that there is only one of
 3. Draw lone pairs on surrounding atoms, then on central
o keep in mind, H only needs 2, rest needs 8

4. Check that every atom has 8 (except H) and that we used the
correct number of valance electrons
[3/13/09]
 *put brackets around ions and write the charge outside on the
upper right hand corner
 *when a structures could have more then one structure, all
structures must be drawn (put all possibilities in brackets and “”
between the brackets)
o Resonance – actual bonding is a hybrid, or mixture, of the
extremes represented by the resonance structure

Exceptions to octet rule
o 1. Odd number of valance electrons
 cant have an odd number greater then 8
[3/16/09]
Exceptions to octet rule (continued…)
o 2. Expanded octet
 When more then 4 atoms are with the central atom
 Ex. PCl5
 The central atom may have more then 8 electrons
through bonds, or bonds and lone pairs
Chapter 10
6/13/2009 10:13:00 PM
[3/16/09]
VSEPR theory
 Valence shell electron pair repulsion
 Explains 3D shape of molecules
 Repulsion between electron pairs causes molecular shapes to adjust
so that valance electrons pairs stay as far apart as possible
o When electrons have lone pairs the angle is smaller (w/o –
109.5 w/ - 107)
 Pictures 414-415
[3/17/09]
Number of
Number of
Electron pair
Example Molecular
Bond angle
electrons pairs
(contains lone
pairs)
lone pairs
(on central
atom)
geometry
2
0
Linear
BeCl2
Cl – Be - Cl
180°
3
0
Trigonal planer
BF3
Trigonal planar
120°
3
1
Trigonal planer
SO2
Bent
<180°
4
0
Tetrahedral
CCl4
Tetra
109.5°
4
1
Tetrahedral
NH3
Trigonal
< 109.5°
geometry
(actual shape)
pyramidal
4
2
Tetrahedral
H2O
Bent
< 109.5°
4
3
Tetrahedral
HF
Linear
180°
5
0
Trigonal
bipyromidal
PCl5
Trigonal
bipyromidal
90° and
120°
5
1
Trigonal
bipyromidal
SF4
Distorted
tetrahydron
<90°,
<120°
5
2
Trigonal
bypyromidal
ICl3
T-shaped
<90°,
<120°
5
3
Trigonal
bypyromidal
I3-
Linear
180°
6
0
Octahedral
SF6
Octahedral
90°
6
1
Octahedral
BrF5
Square
pyramidal
<90°
6
2
Octahedral
XeF4
Square planar
<90°
[3/23/09]
Polar bonds + molecules
 non polar covalent bond
o when atoms in a bond pull equally, the bonding electrons are
shared equally
 ex. all diatomic molecules are non polar
 polar covalent bonds
o between atoms where electrons are shared unequally
o *the more EN (electric negative) atom attracts electrons more
strongly and gains a slightly negative charge, the less EN
atom gains a slightly positive charge
 ex. Hδ+-Clδ one way is to put a lower case delta - δ on top of
each atom, the one on top of the H will receive a
δ+ and the Cl a δ–
 or you can make a positive sign on the H and an
arrow continue to the Cl
Difference in EN
Type of bond (all
covalent)
Example
0-0.4
Non polar covalent
H2 (0)
0.4-1.0
Moderately polar
H-Cl (0.9)
1.0-2.0
Very polar
H-F (1.9)
≥ 2.00
Ionic
Na-Cl (2.1)


Polar molecule – one end of the molecule is slightly negative and
the other is slightly positive
o sometimes called a dipole
Symmetry
o polar bonds and not symmetric shapes produce a polar
molecule
 ex. H2O is polar even though its symmetric because it is
bent
 needs to be polar along x and y axis
o polar bonds that are symmetric produce non polar molecules
 ex. O=C=O (with a plus on top of the C with arrows
going both ways)

since its symmetric and linear the bond polarity
cancels each other out
Chapter 11
6/13/2009 10:13:00 PM
[3/25/09]
Classifying reactions (not part of chap 11)
 Combination/synthesis
o 2MG(s) + O2(g)  2MGO
o when two or more substances react to produce one new
substance
 Decomposition
o 2MGO(s)  2MG(s) + O2
 Single replacement/double replacement
o One element replaces another in a compound


o Zn(s) + Cu(NO3)  Cu (s) + Zn(NO3)
Double replacement
o Exchange of cat ions between 2 compounds
o Na2S + Cd(NO3)2  CdS + N NaNO3
Combustion
o Burning
o An element or a compoundreacts with O2
 Usually products are CO2 H2O
o 2C8H18 + 25 O2  19 CO2 + 18 H2O
Chapter 11
Solid, liquid, gas
 Solid
o Most dense, except for ice
o Definite shape, definite volume.
 Liquid
o Medium density
o Indefinite shape, definite volume
 Gas
o Low density.
o Indefinite volume and shape
Attraction and forces
 Attractions between molecules.
o Intermolecular forces
o Weaker then intra molecular forces
o Defines the state of the matter
 Van der Waals (London dispersion forces)




o Weakest form
o Caused by the motion of the electrons.
Occurs between non-polar ad polar molecules
Dipole-Dipole forces
o Between 2 Polar molecules
o Slightly negative region of one molecule is attracted to the
positive region of another.
o Higher molar mass has higher boiling point, if they have the
same type of molecular bond
Hydrogen bond
o Attractive forces in which a hydrogen covalently bonded to a
highly electro-negative atom (N, O, F)
o Strongest form – highest boiling point
Surface tension
o Resistance of liquid to an increase in surface area.
o Energy required to overcome intermolecular forces in liquid
 The stronger the intermolecular force the higher the
surface tension.
Viscosity
o High viscosity= high resistance to flowing.
 Capillary Action
o Spontaneous rising of a liquid in narrow tube.
 Cohesive
 Intermolecular forces among molecules of liquid
 Adhesive
 Intermolecular forces between liquid molecules
and its container.
[3/26/09]
Capillary action
 cohesive – intermolecular forces among molecules of liquid
 adhesive – forces between liquid molecular and container
Heating curve for water



B – 0° C
D – 100° C
E – gas is really vapor
[3/27/09]
ΔHfus – molar heat of fusion

Heat absorbed by one mole of a solid as it melts to a liquid at its
melting point
 Water: 6.01 KJ / mol (to melt one mole to water- B)
o How many grams of ice at 0 C will melt if 2.25 KJ of heat is
added?
 2.25 KJ / 6.01 (KJ/mol) = .374 mol H2O x 18 g / mol =
6.74 g ice
ΔHsolid = - ΔHfus
 Go from liquid to solid
ΔHvap = molar heat of vaporization
 Amount of heat to vaporize 1 mol of liquid to vapor at boiling point
 Water: 40.7 KJ / mol (to vaporize one mol of water – D)
o This plateau is much greater, since more energy is required
 Since its easier to break bonds of a solid then of a liquid
ΔHvap = - ΔHcond
 Go from vapor to liquid
Heat: q

Measured in joules or calories
o 4.184 J = 1 cal
Heat capacity
 Amount of heat required to increase the temperature of an object
by 1 degree Celsius
Specific heat (c)
 Amount of heat required to increase temperature by 1 g of
substance by 1 degree Celsius
J/g° C
Cal /g° C
Water
4.184
1
Ice
2.1
.5
Iron
0.46
.11
q = mcΔT
 q – heat absorbed/ released
o if positive absorbed
o if negative released
 m – mass (g)
 c – specific heat
 ΔT – change in temperature (always final – initial )
o Tf - Ti
 Ex. 27 g H2O from 10 C to 90 C (answer in J)
o (27 g)*(4.184 j/g°C)*(80°C) = 9037.44 J
 changing temperature not phase
[3/31/09]
Sublimation = solid  gas
 Heat = molΔH
o Change phase not temperature
49.5 g liquid water boiled at 100°C. then heat vapor to 120° C. how much
heat is required? Heat of vaporization for water is 2.1 j/g° C
 49.5 g /18 g/mol = 2.75 mol
 2.75 mol x 40.7 kj/mol = 11.93 KJ
 q = 49.5 (2.1 j/g C)(20 C) = 2079 J = 2.079 KJ
 answer: 14.009 KJ
33.3 g ice at 0 C heated until steam at 150 C. Calculate heat required.
 Info

o Specifc heat for liquid = 4.184 j / g C
o Heat of fusion for water = 6.01 KJ/mol
o Specific heat of water vapor = 2.1 j /g C
o Heat of vaporization for water = 40.7 kj/mol
Steps
o S  L - 0° C
 1.85 * 6.01 kj/mo = 11.1185 KJ
o L - 0°  100° C
 33.g (4.184 j/gC)(100°C) = 13932 j  13.9 KJ
o L  V - 100° C
 1.85 mol * 40.7 kj/mol = 75.3 KJ
o V – 100°  150° C
 33.3 g (2.1 J/gC)(50°) = 3496.5 J  3.5 KJ
o answer: 103.8
Vaporization
 liquid  vapor
Evaporation
 When vaporization occurs at the surface of a liquid that’s not boiling
[4/6/09]
X axis – temperature (Celsius)
Y axis – pressure
Phases
 on the right is vapor
 In the middle its water
 on the left is ice
Triple point
 Where all three lines meet, pressure and temperature where all
three phases exist in equilibrium with each other
Along the lines there is equilibrium between the phases they separate
Critical point
 Liquid and gas become super critical fluid
Questions – 514
 Q 84:
o A. Normal bp: 184.4 C
o B. Normal mp: 113.6 C
C. Solid
Chapter 12
6/13/2009 10:13:00 PM
Chapter 12 (solutions) sections 4,5, and 7
[4/2009]
Solubility
 Maximum quantity of a substance that will dissolve in a certain
amount of water at a specified temperature
Graphs
 X = temperature
 Y = grams dissolved in 100g H2O
 The curve will show a saturated solution
o Unsaturated solution
Contains less dissolved solute then the amount of a that
solvent can normally hold at that temperature
o Supersaturated solution
 Solution that contains more solute than could be
dissolved
Concentration can be expressed in different ways
 Molarity = mol solute / L solution
o Symbol (M)
 Molality = mol solute / Kg solvent


o Symbol (m)
 500 grams H2O produce .06 molal KI solution. How
many grams KI dissolve
 x / .5 = .06  .03 mol *166 g / mol = 4.98 g KI
o since solvent find g solution and subtract g solute to get g
solvent (and change into kg)
Mole fraction (X) = mol solute / totally mol solution
o C2H6O2 – 1.25 mol
 XC2H6O2 = 1.25/5.25 = .238
o H2O – 4 mol
 4/5.25 = .762
 Mass percent = (mass of solute / mass of solution) * 100
Colligative properties
 Properties that depend only on the amount of solute and not on
their identity
 1. Vapor pressure lowering
o a solution has lower vapor pressure then its pure solvent

2. Boiling point elevation
o difference in temperature between boiling point of solution
and boiling point of pure solvent
o ΔTb = Kb * M
 Kb – molal boiling point elevation constant
 Water: Kb = .512 °C/M
o Q: 1.5 m NaCl solution in H2O. what is boiling point?
 ΔTb = (.512 °c/m)(1.5 m) = .77 °C  100.77 °C (pure
water boiling point is 100 so you add .77
[4/21/09]

3. Freezing point depression
o ΔTf – Kf * m
 Kf: molal freezing point depression constant
 H2O – Kf = 1.86 °C/m
o Q. 100 g C2H6O2 in .5 kg H2O. calculate freezing point
 100g / 62 g/ mol = 1.612 mol  1.612 mol/ .5 g =
3.224 m  3.224 * 1.86 °C / m = 6°C  0C – 6°C = 6 °C
[4/22/09]
Q: suppose that there are two water solutions, one of glucose (180 g/mol),
the other of sucrose (342 g/mol) each contains 50 grams of solute in 1,000
grams of water. Which has the higher boiling point? The lower freezing
point?
 Glucose:
o 50 g/180 g/mol = .27 mol
o ΔTb = .512 (.27/1) = .14 + 100 = 100.14
o ΔTf =1.86(.27) = .52  -.52 C
 Sucrose:
o 50 g / 342 g/mol = .146 mol
o ΔTb = .512(.146/1) = .07 + 100 = 100.07
o ΔTf = 186(.146) = .27  -.27 C
Q: Calculate the freezing point depression of a benzene solution containing
400 grams of solvent and 200 grams of solute (C3H6O). Kf for benzene is
5.12 C/m
 200g /58 g/mol =3.44 mol  (3.44 mol /.4 kg)5.12 C/m = 44.14 C
o if given freezing point, subtract this number from that
Heterogeneous mixtures:
 Suspension
o Mixture that particles begin to settle out of solution upon
standing
o greater then 1000 nm diameter
 In a solution they are about 1 nm
 Colloid
o Particles range in size from 1 nm to 1000 nm
chapter 15
6/13/2009 10:13:00 PM
[4/24/09]
Acids and Bases
 Acids
o Sour
o Dissolve many metals
o Litmus paper turns red
o HCl, H2SO4, HNO3
 Base
o Bitter
o Slippery feel
o
o
o
o
o
Litmus paper turns blue
NaOH
KOH
NH3
NaHCO3
Definitions
Acid
Base
Arrhenius
Produces H+ in solutions
Produces OH- in solution
Bronsted Lowry
H+ donor
H+ acceptor
Lewis
Accept electrons pair
Donate electron pair

formula
o NH4+ + OH-  NH3 + H2O
 NH4+ - acid (H+ donor)
 OH- - base (H+ acceptor)
 NH3 - conjugate base (loses an H+)
 H2O - conjugate acid (gains an H+)
o Some substances can be both acids and base, these are
called amphoteric

concentration
o [H+] = 1 x 10-7 M
o [OH-] = 1 x 10-7 M
 They are neutral
o Kw = [H+] x [OH-] = 1 x 10-14 M2
 Ion product constant for H2O
o [H+] > [OH-]  acidic



o [H+] < [OH-]  base
 ex. [H+] = 1 x 10-5 M
 then the [OH-] = 1 x 10-9 M
 divide - 10-15 / 10-5
 its acidic
pH scale
o pH = -log[H+] (H+ = H3O+)
o range from 1 to 14
 1 to 7 – acid
 7 – neutral
 7 to 14 – basic
o ex. OH- = 4 x 10-11 M
 14-10.4 = 3.6
pOH
o pOH = -log[OH-]
o 1 to 7 – basic
o 7 – neutral
o 7 to 14 –acidic
pH + pOH = 14 the hardest
[4/27/09] be
Questions:
 1. Calculate pH if [OH-] = 4.3 x 10-11 M
o pOH = -log (4.3 x 10-11)  14-ans = 3.65 ph
 2. [OH-] = 1 x 10-3 M, what Is [H+]? Basic, acidic, natural?
o 1 x 10-11 M
Log scale
 Every pH change of 1 is a concentration change of 10x
 Given pH, find [H+]

o Method 1
 pH = -log x  -ph = log x  x = log-1 (-pH)
o Method 2
 pH = -log x  -ph = log x  10^(-pH) = x
ex. pH = 4.8
o –10^-4.8 = 1.6 x 10-5 M
[5/4/09]
Acid base indicator
 Its acid form and base form have different color in solution
Strength of acid + bases
 Strong acids – completely ionize in solution
o Ex. HCl, H2SO4, HNO3
 HCl + H2O  H3O+ + Cl- - 100 % ionized
 Weak acid – partially ionized in solution
o Ex. Acetic acid
CH3COOH + H2O  H3O+ + CH3COO- (partial ionization –
can tell by double arrow)
Strong Base




o Ex. Mg(OH)2 , Ca(OH)2
Weak base
o Ex. NH3
Concentration / strength
o Concentration
 How much acid is dissolved in solution
 Concentrated/ dilute
o Strength
 Extend of ionization

Strong weak
Neutralization reaction
 Acid +Base  Salt + Water
o Mix solution of strong acid and strong base and we get a
neutral solution results
o Salt = compound consisting of anion from aid and cation from
base
 Ex. HCl + NaOH  H2O + NaCl
 Ex. H2SO4 + 2KOH  2H2O + K2SO4

Q: How many moles of potassium hydroxide are needed to
completely neutralize 1.56 mol of phosphoric acid?
o H3PO4 + 3KOH  3H2O + K3PO4
 1.56 mol x (3 KOH / 1 H3PO4)
Titration
 The process of adding a known amount of solution of known
concentration to determine the concentration of another solution


Standard solution (know concentration of) we kept titrating until
neutralization occurred, which is called the end point
Ex. a 25 ml solution of H2SO4 is neutralized by adding 18 ml of 1 M
NaOH what is concentration H2SO4?
o H2SO4 + 2NaOH  2H2O + Na2SO4

1M =



x mol / .018 L
.018 mol NaOH
.009 mol H2SO4
.009 mol / .025 L = .36 M
[5/5/09]
Buffer



A solution where the pH remains relatively the same
Made of weak acids and its salt or a weak base and its salt
Ex. CH3COOH + NaCH3COO = acid buffer
o Acetic acid + salt
o Ex. add acid: if H+ were added to this then the salt would split
up into its ions and it would combine with CH3COOo Ex. Add Base: OH- + CH3COOH  H2O + CH3COO-
6/13/2009 10:13:00 PM
[5/5/09]
Thermo chemistry
 Potential energy
o Energy stored in chemical bonds
 Kinetic energy
o Energy of motion
 Heat transfer or work
o 1. Heat – symbol “q”
 amount of energy transferred from one object to
another, higher temp. to lower temp.



o W=




o ΔE =


endothermic – reaction that absorbs heat
 q>0
exothermic - reaction that releases heat
 q<0
can be found by MCΔT
-PΔV
ΔV > 0  W < 0
 Work being done by system
ΔV < 0  W > 0
 Work done on system
Unit is jules
V – final – initial
q+w
Internal energy in system
Unit is jules
+
-
Q
Endothermic
Exothermic
W
Work done on
Work done by
system
system
Energy in
Energy out
ΔE
o 101.3 J = 1 L*atm
[5/6/09]
Q: System compresses 5L to 3.5 L pressure 740 mmHg realeses 20.6 KJ
heat find ΔE
 -(740/760)*-1.5L = 1.46


Heat of

1.56 * 101.3 J = 148 J
148 j - 20.6 kj = -20.47 KJ
reaction
Exothermic
o CaO (s) + H2O  Ca(OH)2 (s)
 ΔH = -65.2 KJ (of heat must be released)
o Ca(OH)2 (s) + 65.2 KJ
 Endothermic
o Ca(OH)2 (s)  CaO (s) + H2O
 ΔH = 65.2 KJ
CaO (s) + H2O – 65.2 Kj

NaHCO3 (s)  Na2CO3 (s) + H2O (g) + CO2 (g)
ΔH = 129 KJ
Calculate heat required to decompose 2.24 mol NAHCO3?
 2.24 mol * (129 KJ / 2 mol) = 144.5 KJ
 Q: 3H2 +N2  2NH3, ΔH = -4.5 KJ
o If 16 g NH3, find heat released?
o (16/17) = .941 Mol  .941 mol (-4.5 KJ/ 2) = -2.118 KJ
Heat of combustion
o
Q: 2
o
o

Heat of reaction for complete burning of one mole of substance
o Ex. Methan CH4
 CH4 + 2O2  CO2 + 2H2O , ΔH = -802.3 KJ
 267 KJ heat
o mass CH4?
 = 276 KJ (1 mol / 802.3 Kj) = .333 mol
 .33 mol X 16 g / mol
5.32 g
Enthalpy = ΔH
[5/11/09]
Hess’ law
 C(s, diamond)  C(s, graphite)
o A. C (s, graphite) + O 2  CO 2 ΔH = -393.5

o
Want
o
o
B. C (s, diamond) + O2  CO 2 ΔH = -395.4
to use reactions a and b to get first reaction
Reaction B stays the same
Reaction A is flipped


 The sign is flipped so its now 393.5 KJ
o The two reactions can now be added
 (393.5) + (-395.4) = ΔH= -1.9 KJ
When adding two or more reactions, you can also add heats of
reactions
o Flip a reactions  ΔH: change sign
o Multiply a reaction  multiply everything by same coefficient
o Sometimes reactions will have to be multiplied and fliped
Q: C + ½O2  CO, find ΔH
o C + O2  CO2, ΔH = -393.5 KJ
o CO + ½O2  CO2, ΔH = -283 KJ
 Flip second reaction: CO2  CO + ½O2 , ΔH = 283 KJ
 Add both reactions together: C + O2 + CO2  CO +
½O2 + CO2
 Cross out: C + ½O2  CO
 ΔH = -110.5 KJ
Standard heats of formation
 Symbol- ΔH°f
 Change in enthalpy that accompanies the formation of one mole of
a compound from its elements
 ΔH°f = 0  for free elements and diatomic
[5/13/09]
ΔH° = Σ(np ΔH°f,product) – Σ (nr ΔH°f,reaction)
 standard heat of reaction = Σ( # mol product x standard heat of
formation) – Σ(# mol reaction x standard heat of formation)
 Ex. 2 CO + O2  2CO2, ΔH =
o ΔH = 2(ΔH°f CO2) – (2ΔH°f CO + 1ΔH°f O2)
o = 2(-393.51) – (2 (-110.53) + 0)
o = -566 KJ
 Ex. 2NO + O2  2NO2
o 2(22.18) – (2(90.25) - 0) = -113 KJ
Gibbs free energy
 energy available to do work
 symbol ΔG
Spontaneous reaction
 Occurs naturally
 Releases free energy
 ΔG < 0
Non spontaneous reaction
 Doesn’t occur naturally
 Requires energy input
 ΔG > 0
Entropy
 Measure of a disorder of a system
 Law of disorder
o Natural tendency for system is to move in the system of more
disorder (randomness)
 Symbol ΔS
o Increases in S  favors spontaneous reaction
 For a given substance the entropy for the Ssolid < Sliquid < Sgas
 S increases when divide substances
 In reaction when # mol product > # mol reactant  S increases
 When t increases , s increases
[5/14/09]
Gibbs Helmholtz Equation

ΔSrxn
ΔG = ΔH – TΔS
o Temperature is in Kelvin
 ΔG < 0
o Spontaneous
 ΔG > 0
o Non spontaneous
= Σ(npS°products) – Σ(nrS°reactnats)
6/13/2009 10:13:00 PM
Rates and Equilibrium
 Rate = measure of speed of any change in some time interval
o Usually- Δ[ ] / time
 Activation
o Minimum energy that colliding particles must obtain in order
to react
o
Ea is activation energy
 This is exergonic
o Large activation energy  slower process

o

endergonic

o the top of activation energy is the activated complex
 which is a transition state
What factors affect the rate of a reaction
o Temperature increases
 Rate increases
o Particle size is smaller (surface area)
 rate increases
o Increases in concentration
 increase in[ ] and increase in rate
o Catalyst



Speeds up reaction, without being used
Lower activation energy
(catalyst supposed written on time of arrows)

Rate law
o Expression for rate of reaction in terms of [ ] of reactions
o Rate constant (K)
[5/15/09]
equilibrium
 conversion of reactants to products and conversion of products to

reactants occur simultaneously
double arrow symbolizes a reversible reaction
o 
 one reaction is the forward reaction, the other is the reverse
reaction
 rate of forward reaction is same as rate of reverse reaction
o called dynamic equilibrium
Le Châtelier’s principle
 A system at equilibrium responds to a stress by moving in the
direction that relieves the stress
o Had graph
 Stresses: temperature, pressure, concentration, volume
o If any of these are changed, the reaction will shift in another
direction to balance the stress
[5/18/09] cont.
 Concentration
o Ex. H2CO3  CO2 + H2O
At equilibrium. If you add H2CO3 forward reaction is
favored, shifts to right
 At equilibrium, add CO2. Shifts to left
 At equilibrium remove H2O, shift to right
o Changing the concentration any reactant or product disturbs
the equilibrium and the system adjusts to minimize the effect
Temperature
o Increases the temperature causes equilibrium position to shift
in direction that absorbs heat
 If decreases, then shifts to direction that releases heat



o Ex. 2SO2 + O2  2SO3 (ΔH= -x  exothermic)
 If exothermic then heat is product, if endothermic then
heat is a reactant
 So reaction shifts left
Pressure
o If increases in pressure, system wants to decrease it, so it
shifts in direction of fewer moles of gas
o Ex. N2(g) + 3H2(g)  2NH3(g)
 Only look at moles of gas



If pressure increases, shifts right
If pressure decreases, shift left
Volume
o If volume rises, pressure decreases
o If volume decreases, pressure increases
o Ex. 2KClO3(s)  2KCl(s)+3O2(g)
 if volume decreases, shifts left

Q. PCl5(g)  PCl3(g) + Cl2(g) ΔH= 92.5 KJ
o Add Cl2  left
o Increases pressure  left
o Decrease temperature  left
o Increases volume  right

Q. CaCO3(s)  CaO(s) + CO2(g)
o Add CaCO3  no effect
Equilibrium Constant (Keq)

Ratio of product concentrations, to reactant concentrations, with
each concentration raised to a power equal to the number of moles
of the substance in balanced equation

Ex. aA +bB  cC + dD
o Keq = [C]c[D]d / [A]a[B]b
Keq > 1  products favored over reactants
Keq < 1  reactants


N2O4(g)  2NO2(g)
 A liter at equilibrium at 10 C, .0045 mol N2O4, .03 mol NO2
 Calculate Keq

o .032 M2 / .0045 M = .02 M
 unit will be different every time
at equilbirum: 2 L flask
o .2 mol PCl5
o .3 mol H2O
o .6 mol HCl
o .3 mol POCl3
 PCl5(g)+H2O(g)  2HCl(g)+POCl3(g)
o Find Keq

Keq = (.3 M)2(.15 M) / (.1 M)(.15 M) = .9 M
[5/19/09]
Q. 1 mol H2 and 1 mol I2. In a 1 L flask. At eqm, 1.56 mol HI


H2+I2  2HI
To solve this we need to make an ice chart (make one when given
initial concentrations)
H2
2HI
Initial
1
I2
1
Change
-1.56/2
-1.56/2
+ 1.56
Equilibrium
.22
.22
1.56

0
 Keq = 1.56 m2 / (.22m)(.22m) = 50.3
Q. at eqm [SO3] = .25 M, calculate K
 Given: .6 mol SO2, .06 mol O2, in 2 L container


2SO2(g) + O2(g)  2SO3(g)
Keq = [SO3]2 / [SO2]2[O2]
2SO2
O2

2SO3
Initial
.3
Change
.3-.25 = .05 (.3-.25)/2 = + .25
.125
Equilibrium
m]2
.05
m]2
.3
0
.175
.25
m-1
 Keq = [.25
/ [.05
[.175 m] = 143
Q. Initial - 3 mol H2, 3 mol I2, 1 L container. Keq = 45.9
 Find all at eqm
H2
2SHI
Initial
3
I2
3
Change
-x /2
-x /2
+x
Equilibrium
3 – (x/2)
3 – (x/2)
X

0

45.9 = x2 / (3-(x/2))2  radicalize entire thing  √45.9 (3-(x/2)) =
x  (6.77*3) – 6.77x/2 = x  20 - 3.85x = x  20 = 4.86x  x =
4.11
 4.11 is concentration of HI
o So 3-(4.11/2)/ = .95 M
 Which is concentration of H2 and I2
[5/22/09]
N2O4(g)  2NO2(g)
 Kc = 0.513
 Initially:
o [N2O4] = .05 M
 Find: eqm concentrations
o Keq =


N2O4
2NO2
Initial
.05
0
Change
-X
2X
Equilibrium
.05-X
2X
(2x)2
/ (.05–x) = .513
4x2+.513x-.026 = 0
 x = .0395 or a negative number (but cant be
negative)
Plug in x in equilibrium .01 M, and .08 M
At a certain temperature Keq = .05 for the reaction: N2(g) + O2(g)  2NO(g)
 What is the concetration of NO at equilibrium when placed in a flask
with .8 M N2 and .2 M O2 initially?
N2
O2
2NO
Initial
.8
.2
0
Change
-x
-x
2x
Equilibrium
.8-x
.2-x
2x
6/13/2009 10:13:00 PM
[5/26/09]
Redox








Oxidation
o Loss of electrons
Reduction
o Gain of elections
(OIL RIG - Oxidation involved Loss, Reduction involves gain)
Mg + S  Mg2+ + S2o Sulfur is reduced  oxidizing agent
o Magnesium gains  reducing agent
Ex. Silver nitrate reacts with copper to form copper (II) nitrate and
silver. What is oxidized? What is reduced?
o Copper – reduced
o AgNO3 – oxidized
Oxidation number
o A positive or negative number assigned to an element to
indicate its degree of oxidation or reduction
 Chang of S  S2 Oxidation number  -2
Rules for assigning oxidation number
o 1) Oxidation number of monatomic ion is equal to its charge
 Ex. Br-  -1
o 2) Oxidation number of hydrogen in a compounds is +1,
except when in metal hydrides, such as NaH, where it is -1
o 3) Oxidation number of oxygen in a compound is -2, except in
peroxides, H2O2, where it is -1
o 4) Oxidation number of uncombined element is zero (ex.
diatomic molecules…)
o 5) For neutral compound, sum of oxidation numbers = 0
o 6) For polyatomic ions, sum of oxidation numbers = charge of
polyatomic ions
ex.
o SO2
 S: +4
 O: -2
o CO32+
 C: +4

O: -2
o Na2SO4
 (do one on outside first, then figure out one in middle)
 Na: +1
 S: +6
 O -2
o (NH4)2S
 N: -3
 H: +1
 S: -2
o KMnO4
 K: +1
 Mn: +7
 O: -2
[5/27/09]
Use oxidation numbers to identify oxidizing agent + reducing agent
 2AgNO3 + Cu  Cu(NO3)2 + 2Ag
o AgNO3 – oxidizing agnet
o Cu – reducing agent

Cl2 + 2HBr  2HCl + Br2
o Cl2 – Oxidizing agent
o HBr – reducing agent
Balancing redox reactions in acidic solution
 1) Use oxidation numbers to write out half reactions
o Ex. MnO4- + Fe2+  Fe3+Mn2+
 MnO4-  Mn2+
 Fe2+  Fe3+
 2) Balance each half reaction separately
o a) Balance all elements except H and O
o b) Balance O by adding H2O
 MnO4-  Mn2+ + 4H2O
o c) Balance H by adding H+
 8H+ + MnO4-  Mn2+ + 4H2O
o d) Balance charge by adding e 5e- + 8H+ + MnO4- Mn2+ + 4H2O
 Fe2+  Fe3+ + 1e-



3) Multiply half reactions so that they have equal number of
electrons
o 5e- + 8H+ + MnO4- + 5Fe2+  Mn2+ + 4H2O + 5Fe3+ + 5e4) Add, simply and check
o The 5e- cancel each other  8H+ + MnO4- + 5Fe2+  Mn2+ +
4H2O + 5Fe3+
 Check that charges balance
Ex. HNO3 + H2S  S + NO
o HNO3  NO
  HNO3  NO + 2H2O
 3H+ + HNO3  NO + 2H2O + 3e must multiply this by 2
o H2S  S
  H2S  S + 2H+ + 2e must multiply this by 3
+
 6H + 2HNO3 + 3H2S  +6H+ + 6e- +2NO + 4H2O +
6e
[6/1/09]
Electrochemistry


Electrochemical conversion between electric and chemical energy
o Voltaic cells (galvanic cells) is where this process takes place
 Convert chemical energy into electric energy
o Anode is where oxidation takes places
 Ex. Zn
o Cathode were reduction takes place
 Ex. Cu
Way to remember
o An Ox Red Cat
o

Each are called half cells
o Together they produce energy
 Connected by a salt bridge, which is the redox reaction
Electrolytic cells
 Supply energy to bring about a chemical change


electrons are pushed from anode to cathode
o redox reanction is non spontaneous
[6/2/09]
Organic chemistry
 Compounds containing carbon

hydrocarbons
o Contain hydrogen and carbon
o 3 classes
 saturated
 Alkanes
 All carbon – carbon bonds are single bonds
o Simplest form is one c with four H
called Methane
 all alkanes end with –ane
 formula for all is CnH2n+2

methane, ethane, propane, butane,
pentane, hexane, heptane, octane, nonane,
decane
 unsaturated
 aromatic
o structural isomers
 have same molecular formula but different structures
[6/3/09]

continued hydrocarbons

saturated
o Branched chain alkanes
 Substituent group – atom or group of atoms that take
place of hydrogen atom in hydrocarbon molecules
 Called alkyl group = hydrocarbon substituent
 To name group
 Get rid of -ane and add –yl
o Ex. ethyl
 Naming branched chain alkanes




1) find the longest chain of carbons
 not linearly
2) number the carbons, starting at end closest to
substituent groups
3) name substituent groups and add numbers
 form: number - group
o ex. 2 – methyl
4) use prefixes to indicate appearance of more
then one of the same group



ex. if you have two methyls it becomes 
dimethyl
o sprefixes:
 di, tri, tetra, penta
5) list substituent groups in alphabetical order
(not including prefixes)
 ex. 4 – ethyl 2,3 – dimethyl
6) put this together with rest of group (no spaces,
hyphens between number and letters, commas
between numbers)

ex. 4-ethyl-2,3-dimethylheptane
[6/4/09]
 unsaturated hydrocarbon
o 1) Alkanes
 have a C = C (double bond)
 end in –ene,
 formula is CnH2n

form: Number double bond same way we number
substituent groups
 If double or triple bond, number so that the double
bond or triple bond would have the lowest number
 Put the number where double bond is where you would
place the substituent group before the name of
hydrocarbon
o 2) Alkynes
 have a carbon carbon triple bond
 formula is CnH2n-2


end in –yne
Put the number where triple bond is where you would
place the substituent group before the name of
hydrocarbon
[6/5/09]
 Hydrocarbon classes
o Aromatic hydrocarbons
 Derived from benzene (C6H6)
 Ring shape
Drawn as a hexagon with a circle in the middle
Each corner has a carob with a hydrogen
coming off of it
 Naming
 Start with substituent group closest to lowest
numbers
 If benzene in the substitution group then it has the
name Phenyl
Replacing Hydrogens
o Functional group - Can be a nonmetal atom or a small group


of atoms
 1) Halogen
 anything in group 7
 R- X (rest of chain + this)
 naming with chloro, fluoro, bromo, ioh
o ex. 1,3-dichlorobutane
 2) alcohol
 replacing H with –OH : a hydroxyl group




R- OH ( rest of chain + OH)
3) Carboxylic acid
 C = O, C – OH, C - R (same C)
 Carboxyl group
 -COOH
4) Ester
 R – (C double bonded with O and single bodn with
another O which is bonder with) – R
5) Amine
 R – NH2

Amino group
[6/8/08]
 continued
o functional group
 6) Amide
 R – (C double bonded with O and bonded with) –
N - (bonded a H and a R)
 7) Keytone
 O = C (which is bonded to two diffrent R)


8) Aldehyde
 O = C (bonded with one H and one R)
9) Ether
 R-O–R
6/13/2009 10:13:00 PM
Final









Significant figures
Changing units
Accuracy vs precision
models
Atomic mass
moles
Avogados number
Covalent and ionic bonds
Molecular and empirical formula



Percent composition
Naming acids
Stoichiometry, limiting reactant
o Theoretical
o Excess reactant
Molarity
Concentration
Solubility rules
o Precipitations reactions












Complete molecular, complete ionic, net ionic
Gas laws
o Know direct or indirect
Kinetic molecular theory
What is an ideal gas
o Real gases act like them under high temperatures and low
pressure
o 1 mol = 22.4 L at STP (273, 22.4, 1 )
law of effusion
partial pressure
waves
o frequencies, ray length, type of rays – don’t need to know
order
as go further from nucleus
o ladder – spaces, higher energies
o electron must jump – quantum of energy
o absorb and release
orbitals
o electron configurations







o orbital energy diagrams
o afbau principal
o exclusion
o hunds rule
o exceptions
trends
groups
o elements
bonds
lewis structures
o resonance structures
VSEPR theory
Polar vs non polar
o Drawing polar bonds
 If they are on same axis they cancel out and are not
polar
Forces
o Van der wells, dipole, hydrogen










Single and Double replacement
Heating Curve
Phase diagram
Solubility curve
Molarity and Molality
Boiling point elevation
Colligative properties
Conjugate acids and bases
Strength of acids and bases
Concentration of Acids and Bases


Titration
ThermoChemistry
o Probably can lay off buffers (that’s what she said(not direct
quote))
o Still haveto know how to do standard heat of formatio
questions (she will give it)
o Entropy
o Gibbs German guy equation
o Raes if reation




o Equilibrium
o Pics of such
o Chattlears principle
o equation for euilibrium and ice charts(only for gasses)
Redox
Electrochemistry
Organic Chemistry
last semester
o midterm – 30
o
o
o
o
test -20
quiz – 20
lab – 15
hw -15