FTCE Chemistry SAE
Preparation Course
Session 2
Lisa Baig
Instructor
Session Norms
• Respect
– No side bars
– Work on assigned materials only
– Keep phones on vibrate
– If a call must be taken, please leave the
room to do so
Course Outline
Session 1
Review Pre Test
Competencies 6, 7 and 8
Competencies 1 & 2
Competency 5
Session 2
Competency 3
Competency 4
Post Test
Required Materials
• Scientific Calculator
• 5 Steps to a 5: AP Chemistry
– Langley, Richard, & Moore, John. (2010).
5 steps to a 5: AP chemistry, 2010-2011
edition. New York, NY: McGraw Hill
Professional.
• Paper for notes
• State Study Guide
Chemistry Competencies
1. Knowledge of the nature of matter (11%)
2. Knowledge of energy and its interaction with
matter (14%)
3. Knowledge of bonding and molecular structure
(20%)
4. Knowledge of chemical reactions and
stoichiometry (24%)
5. Knowledge of atomic theory and structure (9%)
6. Knowledge of the nature of science (13%)
7. Knowledge of measurement (5%)
8. Knowledge of appropriate laboratory use and
procedure (4%)
Electronegativity
• Fluorine is the most electronegative
element.
• Pattern of increasing electronegativity
moves from bottom to top, and from
left to right across the periodic table.
Chemical Bond
Mutual electrical attraction between the
nuclei and valence electrons of different
atoms that bind the atoms together
Atoms would like to have 8 Valence electrons.
These bonds help the atoms to achieve
their full valence shells
Three Types
Ionic
Covalent
Metallic
Ionic Bond
• Force of attraction between oppositely
charged ions
• Occurs between Metal and Non-Metal
elements
• The Non-metal “steals” the valence
electron(s) from the Metal
• Forms a crystalline structure of these
positive and negative charges
• Typically solids at room temperature
Ionic Character
• Ionic Bonds are bonds with > 50%
ionic character
• Difference in Electronegativity of
involved atoms is >1.7
Covalent Bond
• Sharing of valence electron pairs by 2 atoms
• Occurs between 2 Non-metal elements
– Or the SAME non-metal element
• Can share one, two or three pairs of electrons
– Single Bond = 1 pair (1 sigma)
– Double Bond = 2 pairs (1 sigma, 1 pi)
– Triple Bond = 3 pairs (1 sigma, 2 pi)
• Sharing can also be “unequal”
– Called a POLAR covalent bond
• Typically liquids or gases at room
temperature
Character
• Ionic Character:
– Polar Covalent Bonds have between 5%
and 50% ionic Character
– Non-Polar Covalent Bonds have less than
5% ionic character
• Difference in Electronegativities
– Polar Covalent Bonds have between 0.3
and 1.7 as a difference in
electronegativities
– Non-Polar Covalent bonds have less than
0.3 difference in electronegativities
Break Time
Take a
10 minute
break
Ionic Compounds
• Ion names are used in combination
• Cation- same as the element
– Transitional Metals use Roman Numerals to
represent Charge
• Anion- replace the ending syllable of the
element name with –ide
• Polyatomic Ions- use the name of that iondo not try to rename.
Use “criss-cross” to determine charges
CuCl2
Copper (II) Chloride
CuO
Copper (II) Oxide
NaCl
Sodium Chloride
KI
Potassium Iodide
Mg3N2
Magnesium Nitride
PbO2
Lead (IV) Oxide
Lewis Structures
• A way to show the octet rule in
molecules
Practice
• Draw the lewis structures for
– Ammonia (NH3)
– Water (H2O)
– Phosphorus Trifluoride (PF3)
– Hydrogen Cyanide (HCN)
– Ozone (O3)
– Formaldehyde (CH2O)
VSEPR
•AB5
•Trigonal bipyramidal
•AB6
•Octahedral
VSEPR
• AB4
– Tetrahedral
– 109.50 Bond Angles
• AB3
– Trigonal Planar
– 1200 Bond Angles
• AB2
– Linear
– 1800 Bond Angles
VSEPR
• AB2E
– Bent or Angular
• AB2E2
– Bent or Angular
• AB3E
– Trigonal Pyramidal
Polarity?
• The potential for opposite charges at
different areas of a molecule
Shape and Polarity?
• What is the shape and polarity of
the following molecules?
– Water
– Ammonia
– Carbon Tetrachloride
– Carbon Dioxide
– Hydrogen Chloride
Hybrids
• Atoms “don’t like” to have empty orbitals
• Hybridization
– Mixing of 2 or more atomic orbitals of similar
energies to produce new hybrid orbitals of
equal energies
• It works like this
– Methane: CH4 Normally: 1s22s22p2
– Through hybridization- it forms an “sp” orbital,
with 4 electrons total
– New arrangement: 1s22(sp3) 4
Hybrid Orbitals
Atomic
Orbitals
Type of
Hybrid
Number Molecular
of Orbitals Geometry
s p
sp
2
s p p
sp2
3
s p p p
sp3
4
1800
Linear
1200
Trigonal
Planar
109.50
Tetrahedral
What type of hybrid?
• Beryllium fluoride
– BeF2
– sp
• Ammonia
– NH3
– sp2
• Methane
– CH4
– sp3
Break Time
Take a
10 minute
break
Spectroscopy
• Devices that measure the interaction
between matter and energy
• Absorption
– Measures the wavelengths of
electromagnetic waves absorbed by a
substance
• X-Ray spectroscopy
– Used to determine elemental
composition and types of bonding
Spectroscopy
• UV
– Used to quantify DNA and Protein
concentration
• Infrared
– Used to determine bond type
• Bonds resonate when exposed to the
radiation
• Nuclear Magnetic Imaging (NMR)
– Used to determine bond structure
Simple Organics
• Alkanes (end in –ane)
– Containing only single bonds
– CnH2n+2
• Alkenes (end in –ene)
– Containing at least one double bond
– CnH2n
• Alkynes (end in –yne)
– Containing at lease one triple bond
– CnH2n-2
Simple Organics
Name
Number of Carbons
Meth-
1
Eth-
2
Prop-
3
But-
4
Pent-
5
Hex-
6
Hept-
7
Oct-
8
Non-
9
Dec-
10
Functional Groups
Compound
Type
Alcohol
Suffix or
Prefix
Image
-OH
-ol
Haloalkane
-X
Halo-
Amine
-CN-
-amine
Aldehyde
-COH
-al
Ketone
-
-COC-
-one
Carboxylic Acid
-COOH
-oic acid
Ester
-COOC-
-oate
Naming and Formulas
• Numbers are used in the name to
designate locations of the following
– Types of bonds
– Branches
– Attached functional groups
• For Example
–
–
–
–
–
2,2,4- trimethylpentane
1-pentyne
2,3,4- trimethylnonane
2-methyl 3-hexene
2- propanol
Macromolecules
• Carbohydrates
– Chains of carbon, hydrogen and oxygen.
– Isomers
• Lipids
– Fatty acids- Chains of Carbon and Hydrogen
• Proteins
– Chains of Amino acids
– Differ in their R group
• Nucleic Acids
– Chains of Nucleic Acids
Organic Compound Naming
• Numbers are used in the name to
designate locations of the following
– Types of bonds
– Branches
– Attached functional groups
• For Example
–
–
–
–
–
2,2,4- trimethylpentane
1-pentyne
2,3,4- trimethylnonane
2-methyl 3-hexene
2- propanol
Lunch Time
We start
Again
In
ONE HOUR
Determining Empirical Formulas
• Say you have 65.0g of compound
containing Na and Cl.
• Determine the Empirical Formula if
the compound is 39.3% Na and
60.7%Cl
Higher Level Practice
• 1st Step: Convert your percentages to
mass of each element present
• Na: (.393)(65.0g)= 25.545g Na
• Cl: (.607)(65.0g) = 39.455g Cl
Higher Level Practice
• 2nd Step: Determine number of moles
of each element in the sample
25.545g Na 1 mole
= 1.11 mol Na
22.989 g/mol
39.455g Cl
1 mole = 1.11 mol Cl
35.453 g/mol
Higher Level Practice
• 3rd Step: Use these moles to
determine the smallest whole number
ratio of elements to each other. That
is your empirical formula!
1.11 mol Na : 1.11 mol Cl
1 mol Na : 1 mol Cl
Empirical Formula = NaCl
Balancing Equations
•
__ C3H8 + __ O2  __ CO2 + __ H2O
•
__ Ca2Si + __ Cl2  __ CaCl2 + __ SiCl4
•
__ C7H5N3O6  __ N2 + __ CO + __ H2O + __ C
•
__ C2H2 + __ O2  __ CO2 + __ H2O
•
__ Fe(OH)2 + __ H2O2  __ Fe(OH)3
•
__ FeS2 + __ Cl2  __ FeCl3 + __ S2Cl2
• __ Al + __ Hg(CH3COO)2  __ Al(CH3COO)3 + __ Hg
•
__ Fe2O3 + __ H2  __ Fe + __ H2O
• __ NH3 + __ O2  __ NO + __ H2O
Types of Chemical Reactions
• Synthesis
– A+B  AB
• Decomposition
– AB  A + B
• Combustion
– Burn in the presence of O2, to form dioxide
gas, and other products **(CO2 + H2O)
• Single Displacement
– ACTIVITY SERIES
– AB + C  AC + B
• Double Displacement
– AB + CD  AD + CB
Predict the Product
CaO + H2O 
H2SO3 + O2 
CaCO3 
KClO3 
C6H10 + O2 
C6H12O6 + O2 
Al + CuCl2 
Ca + KCl 
Na2SO4 + CaCl2 
KCl + NaOH 
Ca(OH)2
H2SO4
CaO + CO2
KCl + O2
CO2 + H2O
CO2 + H2O
AlCl3 + Cu
No Reaction
NaCl + CaSO4
KOH + NaCl
Identifying Redox Reactions
Redox
Redox
Not
Redox
Redox
Not
Redox
2 KNO3(s)  2 KNO2(s) + O2(g)
+1 -1
+1 -1
0
H2(g) + CuO(s)  Cu(s) + H2O(l)
0
-2 +2
0 2(+1) -2
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
+1 -1
+1 -1
+1 -1 2(+1) -2
H2(g) + Cl2(g)  2HCl(g)
0
0
+1 -1
SO3(g) + H2O(l)  H2SO4(aq)
+6 3(-2) 2(+1) -2
2(+1) -2
Balancing Redox Reactions
• The following unbalanced equation
represents a redox reaction that takes
place in a basic solution containing KOH.
Balance the redox reaction.
Br2(l) + KOH(aq)  KBr(aq) + KBrO3(aq)
Br2(l) + KOH(aq)  KBr(aq) + KBrO3(aq)
Ionic Reaction: Br2  Br- + BrO30
-1 +5 3(-2)Reduction ½ Rxn:
Br2  BrBr2 + 2e-  2Br5(Br2 + 2e-  2Br-)
Oxidation ½ Rxn:
Br2  BrO312OH- + Br2  2BrO3- + 6H2O + 10eCombined Rxn:
5Br2 + 12OH- + Br2 + 10e-  10Br- + 2BrO3- + 6H2O + 10e6Br2 + 12KOH  10KBr + 2KBrO3 + 6H2O
3Br2 + 6KOH  5KBr + KBrO3 + 3H2O
Standard Reduction Potentials
in Voltaic Cells
Write the overall cell reaction and
calculate the cell potential for a voltaic
cell consisting of the following half-cells:
an Iron electrode in an Iron (III) Nitrate
solution, and a Silver electrode in a
Silver(I) Nitrate solution.
• Fe3+(aq)+3e-Fe(s) E0=-0.04V
• Ag+(aq)+e-Ag(s)
E0=+0.80V
• E0cell= E0cathode- E0anode
• E0cell= (+0.80 V)- (-0.04 V)= +0.84 V
• E0cell= positive = spontaneous
Acid/Base Properties
• Strong Acids and Bases
– Will ionize completely in a solvent
• Weak Acids and Bases
– Will ionize partially in a solvent
• Buffer Systems
– Solution containing a weak acid, and a salt
of the weak acid
• Acetic Acid and Sodium Acetate
• Carbonic Acid and Bicarbonate
Break Time
Take a
10 minute
break
Mass-Mass Stoichiometry
3 Cu + 8 HNO3  3 Cu(NO3)2 + 4 H2O +
2NO
• Copper Nitrate is used in creation of
some light sensitive papers
• Specialty photographic film
• Your company needs 150 grams of
Copper nitrate to fill an order. How
many grams of Nitric Acid are needed
to undergo reaction?
• Step 3: Compute
150g Cu(NO3)2 1 mole
187.554g
8 mol HNO3
63.012 g
3 mol Cu(NO3)2 1 mole
134 g HNO3
=
Gas Stoichiometry
Xenon gas reacts with fluorine gas
according to the shown reaction. If a
researcher needs 3.14L of XeF6 for an
experiment, what volumes of Xenon
and Fluorine should be reacted?
Assume all volumes are measured
under the same temperatures and
pressures.
Xe (g) + 3 F2 (g)  XeF6 (g)
Gas Stoichiometry
• Xenon
3.14L XeF6 1mole 1Xe
22.4L =
22.4L 1XeF6 1 mole
3.14L Xe
• Fluorine
3.14L XeF6 1 mole 3 F2
22.4L =
22.4L 1 XeF6 1 mole
9.42L F2
Solution Stoichiometry
• How many milliliters of 18.0M Sulfuric
Acid are required to react with 250mL of
2.50M Aluminum Hydroxide?
• H2SO4 + Al(OH)3  H2O + Al2(SO4)3
• 3 H2SO4 + 2 Al(OH)3  6 H2O + Al2(SO4)3
250mL Al(OH)3
1L 2.5 mol 3 H2SO4
1L
1000mL
1000mL 1 L
2 Al(OH)3 18.0 mol 1L
52.1 mL H2SO4
Titrations
• In a titration, 27.4mL of 0.0154M Ba(OH)2 is added
to a 20.0mL sample of HCl solution with unknown
concentration until the equivalence point is
reached. What is the molarity of the acid solution?
0.0154M Ba(OH)2 x 27.4mL Ba(OH)2 x 2 mol HCl x 1
=
1
1 mol Ba(OH)2 20.0mL
4.22 x 10-2 M HCl
Limiting Reactant
• The reaction of Ozone with Nitrogen
Monoxide to form Oxygen and
Nitrogen Dioxide in the atmosphere is
responsible for the Ozone hole over
Antarctica.
• If 0.960g of Ozone reacts with 0.900g
of Nitrogen Monoxide, how many
grams of Nitrogen Dioxide are
produced?
Limiting Reactant
0.960g O3
1 mole
48g O3
1 NO2 44.0g NO2
1 O3
1 mole
0.880g NO2
0.900gNO 1 mole
30g O3
1 NO2 44.0g NO2
1 O3
1 mole
1.32g NO2
Break Time
Take a
10 minute
break
Chemical Equilibrium
• Chemical Equilibrium
– Point in a reversible chemical reaction
when the rate of the forward reaction
equals the rate of the reverse reaction.
– The concentrations of its products and
reactants remain unchanged
• Le Chatelier’s Principle
– If a system at equilibrium is stressed, the
equilibrium is shifted in the direction
that relieves the stress
How to Affect Equilibrium
• Change in Pressure
– Only affects reactions with gases
– Increased pressure increases concentration
– Decreased pressure decreases concentration
• Change in Concentration
– Of reactants or products.
• Increase one- it moves to the other
• Decrease one- it moves towards the one you lowered
• Change in Temperature
– Exothermic
• Increase temperature will direct in reverse
• Decrease temperature will direct forward
– Endothermic
• Increase temperature will direct forward
• Decrease temperature will direct in reverse
Equilibrium Constant
nA + mB ↔ xC + yD
K= [C]x[D]y
[A]n[B]m
Factors affecting Reaction Rates
Rate Laws
A chemical reaction is expressed by the
balanced chemical equation
A + 2B  C
Three reaction rate experiments yield the
following data.
Experiment
Number
Initial
[A]
Initial
[B]
Initial Rate of
Formation of C
1
0.20 M
0.20 M
2.0 x 10-4 M/min
2
0.20 M
0.40 M
8.0 x 10-4 M/min
3
0.40 M
0.40 M
1.6 x 10-3 M/min
What is the Rate Law for the Reaction?
What is the Order of the reaction with respect
to B?
Rate Law for the Reaction
A + 2B  C
R = k[A][B]2
Order of the Reaction with respect to B
B is of a 2nd order reaction
A is of a 1st order reaction
Calculating pH and pOH
pH + pOH = 14 pH = -log[H+] pOH = -log[OH-]
• What is the pH of a 2.5x10-6M HNO3
solution?
• pH = -log [2.5x10-6]
• pH = 5.6
Break Time
Take a 10 minute Break
When we return…
POST TEST
Post-Test
• You will have one and a half hours to
complete the post-test
• This test will include examples from
all the competencies.
• Scores will be posted on the Quia
Website tomorrow as a class file.
• Also to be posted- a reference key of
the correct answers AND which
competency and skill were covered
for each question.
Good Luck!
When finished, turn in test to
instructor, and you may leave.