FTCE Chemistry SAE
Preparation Course
Session 4
Lisa Baig
Instructor
Course Outline
Session 1
Review Pre Test
Competencies 1 & 2
Session 2
Competency 5
Session 3
Competency 3
Session 4
Competency 4
Session 5
Competencies 6, 7 and 8
Post Test
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
Homework Review
Any questions from last night?
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.
6.
7.
8.
Knowledge of atomic theory and structure (9%)
Knowledge of the nature of science (13%)
Knowledge of measurement (5%)
Knowledge of appropriate laboratory use and
procedure (4%)
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
Homework
• Diagnostic Exam in your AP Chem
Prep book- Page 17-26
• Only answer the questions for these
Chapters & Questions
– Chapter 6 #6-7, 11
– Chapter 7 #14, 16
– Chapter 8 #20
– Chapter 13 #59
– Chapter 14 #63
– Chapter 15 #66