Download 4 - College of Arts and Sciences

6/27/2014
Chem 1311H Final Exam Review
• 50 multiple choice questions ( 2 pts each) = 100 pts
– 30 questions from chapters 1 – 6
– 20 questions from chapters 7 – 9
Your Grade Is Up To You Not Me
I just report what you tell me to
Final Average = Exam (1+2+3+4) + Lab
5
• 15 nomenclature questions ( 1 pt each) = 15 pts
• Total points on Final Exam = 100 + 15 + SI
• Final Exam Grade will automatically replace any
exam grade (1, 2, or 3) that is lower
Final Average
Grade
90  up = A
80  89 = B
70  79 = C
60  69 = D
0  59 = F
Key to final exam will be posted in glass case
outside room 100 after exam is over
WHAT TO STUDY ?
At the end of each chapter
1. Strategies in Chemistry
2. Summary & Key Terms
Also
1. Old Exams
2. Web problems & Exams
30 questions from “old” material
Chap 1 - Measurements {laws, properties, & dimensional analysis
Chap 2 - Atoms, Molecules and Ions {84Po210
Chap 3 - Stoichiometry {% yield, Limiting reagent etc
Chap 4 - Solution Stoichiometry {Molarity; Solubility, Oxidation
Chap 5 – Thermochemistry {Specific heat, Calorimetery, Hess’ Law
Chap 6 - Electron configuration, Quantum numbers, Hund’s Rule ...
20 questions from “new” material
Chap 7 - Periodic Properties of the Elements size of atoms & ions
Chap 8 - Chemical Bonding Lewis Structures / Resonance Structures
Chap 9 - Molecular Geometry { VSEPR & V.B. Models
Chapter 1
1.
2.
3.
4.
5.
6.
Intensive Properties & Extensive Properties
Atoms, Elements & Compounds
Law of Mass Conservation
Explained
Law of definite proportions
by
Law of multiple proportions
ATOMS
Fundamental particles
Name
Charge
Mass (amu)
Proton
+1
1
Neutron
0
1
Electron
-1
0
Notation For Atoms
Mass Number
Symbol
A
Protons + Neutrons
 X
Z
Atomic Number
Number of Protons
Protons = ?
Neutrons = ?
For
31 P 3-
Electrons = ?
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Dimensional–Analysis method for
SOLVING PROBLEMS makes
use of conversion factors
Use UNITS To Solve Problems !!!!!!
Density relates mass to volume.
What is the mass (in kg) of the air in a room that is
12.0 x 17.0 x 8.50 ft if the density of air= 0.936 g/L
• How do you find mass ?
• From the density !
Density 
SIGNIFICANT FIGURES: The Number
Of Digits That Are KNOWN With Certainty
Chapter 2
• Atomic Theory of Matter by Dalton
• Atomic Structure by Rutherford
weight
volume
• How do you convert ft3 to Liters ?
Answer 3.45x103 Kg
Types of Radiation
alpha() beta() gamma()
– atomic numbers, mass numbers, and isotopes
– notation
• The Periodic Table
– metals, nonmetals, and metaloids
• Names and Formulas (Bonus question)
More EXPERIMENTS
The Molecule & the Mole
• THOMSON’S Experiments
Discovered ELECTRONS
Mole: A counting unit for objects such
as atoms , molecules , ions, …
• MILLIKAN’S Experiments
The MASS Of ELECTRON
A mole represents 6.02x1023 objects
RUTHERFORD’S Experiment
STRUCTURE of ATOMS:
PROTONS and NEUTRONS
Molecule: A chemical combination of two
or more atoms. For example C6H12O6
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Periodic Table of the
ELEMENTS
PERIODIC TABLE
Divided Into
PERIODS …….ROWS
GROUPS………COLUMNS
and
I. METALS
II. METALLOIDS
III. NONMETALS
Chapter 3 Interpretation of Chemical Reactions
•
•
•
•
•
•
Chemical Equations
Formula Weights
Avogadro’s number and the Mole
Empirical Formulas
Limiting Reactants
Stoichiometry
CHEMICAL REACTIONS
( YOU Need To KNOW )
•
•
• MOLECULAR - Uses The Full
Formula Of Reactants and Products
• IONIC – ONLY Ions Are Shown
• NET IONIC – SPECTATOR Ions Are
Removed From Ionic Equation
• SPECTATOR Ions – Ions that do not take
part in the chemical reaction
WHY DOES A REACTION OCCUR ?
1. Precipitation Reaction
NEUTRALIZATION (Acid + Base)
[MUST KNOW ARRHENIUS Acids/Bases]
2. Acid-Base
Reaction
PRECIPITATION REACTIONS
[MUST KNOW SOLUBILITY RULES]
•
CHEMICAL EQUATIONS
OXIDATION REDUCTION (Redox)
3. Oxidation-Reduction
Reaction
An INSOLUBLE Solid
is Formed
Slightly IONIZED
Substance Formed
Change in Oxidation
State
[MUST KNOW OXIDATION “RULES”]
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Stoichiometry & Balancing Equations
The fat stored in the hump of a camel is a source of both
energy and water. Calculate the grams produced by metabolism
of 5.599 kg of the fat (C57H110O6) stored in the hump
1st Write the reaction
C57H110O6 + ? O2  CO2 + H2O
2nd Balance reaction
C57H110O6 + ? O2  57 CO2 + 55 H2O
3rd Interpret Reaction
Answer 345.6 grams
Use Molecular Weight
To Convert
To Convert
Grams to Moles
or
Moles to Grams
___________________________________
Use AVOGARDRO’S Number
For NUMBER of Atoms
Or NUMBER of Molecules
Calculate moles of fat (C57H110O6)
•
•
•
•
•
Molecular weight of C57H110O6 = 890 g / mol
5.599 kg of (C57H110O6) x 1/890 mol/g = 5599/890
55 x 5599/890 mol water formed
(55x5599/890) x 18g/mol water formed
Answer 345.6 grams of water formed
Grams, moles, molecules & atoms
A sample of acetominophen (C8H9O2N) has
6.02 x 1023 atoms of Hydrogen. What is the
mass in grams of the sample?
How many atoms of H in one mole of C8H9O2N ?
9 x (6.02 x 1023) atoms of H
Therefore have 1/9 of a mole of acetominophen
What is the molecular weight of acetominophen ?
151 grams / mole
Empirical / Molecular Formulas
Chapter 4
Determine the empirical formula of a
compound that has (by mass)
21.7% C ; 9.6% O ; and 68.7% F.
• answer C3OF6
Determine the molecular formula of the
compound if its molar mass = 166
• Answer: same as empirical formula
1. Reactions in Aqueous Solutions
& Solution Stoichiometry
2. Molarity is a CONVERSION FACTOR
3. Solubility Rules
4. Oxidation Rules
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Solution Stoichiometry
MOLARITY is a
A solution of 285 mL of 1.35 M KOH is mixed
with a solution of 100 mL of 0.500 M NiSO4
Is there a limiting reagent ? If so, what is it?
How many grams of the precipitate form ?
CONVERSION FACTOR
Write and Balance Reaction
KOH(aq) + NiSO4 (aq)  ?????????
moles of reactants and products = ?
BETWEEN
NUMBER OF MOLES OF SOLUTE
AND
THE VOLUME OF SOLUTION
MUST KNOW OXIDATION “RULES”
What is the oxidation number ?
What is oxidized ?
What is causing oxidation?
What is reduced ?
What is causing reduction ?
Chapter 5 Thermochemistry
HEAT LOST = HEAT GAIN
Something is gaining Heat
While Something else looses Heat.
Energy Changes Involved
1. From One Phase To Another
Phase Changes {Gas/Liquid/Solid
-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
1. q
2. H
Two (2) Ways to express heat
heat liberated or required
Enthalpy (∆H Change in Enthalpy)
2 WITHIN One Phase
Specific Heat
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HEATS
OF
Calorimetry
REACTIONS
1. DIRECT METHOD
EXPERIMENTAL Calorimeter
Heat LOST = Heat GAINED
2. INDIRECT METHOD -> HESS’S
Law
If the calorimeter contained 2.77 kg of water, what is
the heat capacity of the calorimeter in kJ/oC ?
Heat Liberated
By Chemical
Reaction
=
Heat absorbed by Water
+
Heat absorbed by calorimeter
(15.57) x (2.03) kJ = (4.184)(2.77)(2.3) + calorimeter
Heat absorbed by calorimeter in kJ/oC = ?????
The heat of combustion of glucose is 15.57
kJ/g. A 2.03 g sample of glucose is burned
in a bomb calorimeter. The temperature of
the calorimeter and the water increased
from 20.3 to 22.6 oC.
STANDARD ENTHALPIES
OF FORMATION
THE STANDARD ENTHALPY OF
FORMATION (Hf ) OF ANY
ELEMENT IN ITS MOST STABLE
FORM IS ZERO (BY DEFINATION)
Using Hess’s Law Calculate the standard
enthalpy of formation of acetylene (C2H2)
from the following combustion reactions
The combustion of
C(graphite)
Hydrogen
Acetylene
H
-393.5 kJ
-285.5 kJ
-1299.4 kJ
Calculate  H for
2 C(graphite) + H2(gas)  C2H2(gas)
C (graphite) + O2 (g)  CO2 (g)
 H = -393.5 kJ
H2 (g) + ½ O2 (g)  H2O
 H = -285.5 kJ
C2H2 (g) + 5/2 O2 (g) 2 CO2 (g) + H2O  H = -1299.4
2 CO2 (g) + H2O  C2H2 (gas) + 5/2 O2 (g)  H = +1299.4
answer  H = + 226.6 kJ
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Aufbau Principle:
Chapter 6
Quantum numbers
Orbital diagrams
Electron configuration
Electromagnetic radiation
Energy, wavelength, frequency
Orbitals only
hold two
electrons,
and each
should have
different
spin.
Electron Configuration
How many e- ?
(a)1s22s22p63s2
12
therefore Mg
(b) [Ne] 2s22p1
13
therefore Al
24
therefore Cr
52
therefore Te
(d) [Kr]
5s2
4d10
5p4
Core
[He]
[Ne]
[Ar]
[Kr]
[Xe]
[Rn]
1s
2s
3s
4s
5s
6s
7s
2p
3p
4p
5p
6p
7p
3d
4d 4f
5d 5f
6d
ORBITAL DIAGRAMS
Identify Element
(c) [Ar] 4s1 3d5
Increasing Energy
1.
2.
3.
4.
5.
Lower energy orbitals fill first.
1S
H ( )
He ( )
Li ( )
Be ( )
B ( )
C ( )
N ( )
2S
( )
( )
( )
( )
( )
( )
( )
2P
3S
( ) ( ) ( )
( )
( ) ( ) ( )
( )
( ) ( ) ( )
( )
( ) ( ) ( )
( )
( )( ) ( )
( )
( )( )( ) NOTE!
( )( )() NOTE!
Hund’s Rule
QUANTUM NUMBERS (Four)
How many UNPAIRED electrons in
n
l
n = 1, 2, 3, 4, 5,
(a) Mg
{1s2
2s2
2p6
2s2
3s2
3p1
(b) Al
{ [Ne]
(c) Cr
{ [Ar] 4s1 3d5
0
s
6, or 7
l = n – 1, n – 2, ….
1
m=
6
m
-l
S= +½
to
+l
or – ½
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Identify the GROUP of elements that corresponds
to each of the
following generalized electron configurations
[noble gas] ns2 np5
Halogens (Group 17)
[noble gas] ns2 (n-1)d2
Group 4
RELATIONSHIP BETWEEN
C  and 
C =  x 
UNITS Solve Problems !!
meters
1
 meters 
sec
sec
Electromagnetic Radiation
The amount of energy depends on the
frequency ( )
E = h
Where h is a constant
Periodic Trends
• Arises from the Periodic patterns in
ELECTRON CONFIGURATIONS.
• Elements in the same GROUP have the
same number of VALENCE electrons
Chapter 7
7.2 Effective Nuclear Charge
7.3 Size of Atoms & Ions
7.4 Ionization
7.5 Electron Affinity
In an isoelectronic series
Ionic size decreases with
an increasing nuclear charge
{See Text page 269}
• VALENCE electrons = electrons in the
outermost orbital
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IONIZATION ENERGY
Energy required to “get rid” of an electron
Na  Na+ + e−
ELECTRON AFFINITY
Energy change accompanying addition of
an electron
Cl + e−  Cl−
ELECTRONEGATIVITY
The ability of one atom in a molecule to attract
electrons to itself “a tug of war” H + -- F  -
Why are H through F REACTIVE ?
While Ne is NOT REACTIVE (ie Stable)
ELECTRON CONFIGURATION PREDICTS STABILITY
------------------------------------------------------------------------------------------------------------------
Element
1S2
Li (3e)
1 S2 2 S1
Li1+(2e)
[He]
Mg (12e)
Mg2+(10e)
1S2 2S2 2P6 3S2
[Ne]
Ionic
Polar Covalent
Covalent
Two Extremes
---------------------------------------------------------------------------------------------------------------
A difference in electronegativity
forms a polar covalent bond
H+  F the molecule, HF has a di pole
DISCRIPTION OF ELECTRONS
1. How Many Electrons ?
2. Electron Configuration
Electron configurations
He (2e)
Chapter 8 Basic Concepts of Chemical Bonding
Stable
3. Orbital Diagram
REACTIVE
4. Quantum Numbers
Stable
REACTIVE
5. LEWIS SYMBOLS
Stable
Drawing Lewis Structures
1. What TYPE Of Bonds Will be Formed
Ionic ? Covalent? or Polar Covalent?
2. The LEAST Electronegative atom is in the center
3. COUNT Number of Valence electrons Adding or
Subtracting e- For IONS
4. Draw a SINGLE Bond between central atom and
Each of surrounding atoms
5. Complete the OCTETS Of ALL Atoms
DRAW THE LEWIS STRUCTURE FOR
Cyanide Ion CN How many valence electrons? C N
4 + 5 + 1 = 10
First draw a bond between the C and N
C–N
Next, eight (8) electrons around each atom
|C – N|
Count electrons. Correct number? Now what?
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Cyanide Ion CN -
Cyanide Ion CN -
How many valence electrons? 10
How many valence electrons? 10
Reduce electrons by putting in a double bond
|C = N|
Reduce electrons by putting in a triple bond
|C  N| ?
Count electrons. Correct number?
Count electrons. 12
Correct number? No!
Now what?
C N
Hypo chlorite ion
ClO -
Lewis Structures
&
Formal Charge
| O – Cl |
Number of Valence e-
Formal charge is a charge assigned to each
Number of Nonbonding e-
atom in a Lewis structure that helps to
½ Number of Bonding
distinguish among competing structures.
e-
6
7
-6
-6
-1
-1
-1
0
Formal Charge
Where does the H go on HClO?
| O – Cl – H
Valence
e-
Nonbonding
e½ Bonding e-
or
H – O – Cl |
Resonance In Nitrate Ion
In Nitrate Ion [NO3-] the extreme possibilities
have one double and two single bonds
O
Cl
O
Cl
6
7
6
7
-6
-4
-4
-6
O
-1
-2
-2
-1
-1
-1
0
0
The resonance structure has three identical
bonds of intermediate character.
O
O
N
O
N
O
O
N
O
O
O
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Chap 9 Molecular Geometry
Two (2) Theories
1.Valence Shell Electron Pair
Repulsion
(VSEPR) THEORY
2.
The
Molecular Shapes
There are five fundamental geometries for molecules
1. 2 Bonds Linear…vs Bent
2. 3 Bonds
Trigonal Planar vs pyramid
3. 4 Bonds
Tetrahedral
4. 5 Bonds Trigonal bipyramidal
Valence Bond
5. 6 Bonds Octahedral
(VB) THEORY
VSEPR Theory
Electron pairs on the central atom are called
ELECTRON DOMAINS
Single, double or triple bonds all count as
One ELECTRON DOMAIN
The atom A (which violates
the octect rule) in this
molecule, has four (4)
electron domains
THE GEOMETRY OF A MOLECULE
IS
DETERMINED
BY THE
REPULSIVE INTERACTION OF
ELECTRONS
PAIRS IN THE
VALENCE SHELL OF ITS
CENTRAL ATOM
Hybridized MOLECULAR
VALENCE BOND THEORY
ORBITALS
VALENCE SHELL ORBITALS
HYBRIDIZE
SP
Linear 180 o angles
THE
SP2
Flat triangle 120 o angles
SP3
Tetrahedral 109 o angles
ORIENTATION OF ALL
HYBRID VALENCE
SHELL
ORBITALS DETERMINES THE
GEOMETRY OF THE MOLECULE
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SP3 HYBRIDIZATION
AB4 Molecules Such as CCl4 are Tetrahedral
Carbon TetraChloride
AB3E (Molecules With UnPaired Electrons On
the Central Atom) Such as NH3 are Not Planar
AB2E2 (Molecules With UnPaired Electrons
On the Central Atom) Such as H2O are Bent
Draw Lewis Structure of C3F6O
Determine Geometry of C3F6O
by VSEPR Theory & VB Theory
C3 F6
O
Number of valence e- 3(4) + 6(7) + 1(6) = 60
F
O
F
F
C
C
C
F
F
F
Does each carbon have 4 bonds
Does each fluorine have 1 bond
Does oxygen have 2 bonds
SP2
F
F
C
F
O
C
F
C
F
F
SP3
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15 Nomenclature Bonus Questions
• All multiple choice
• Name the following
• Formula for the following
13