1. FINAL EXAM REVIEW
Fall Semester

2. FINAL FORMAT 35 multiple choice questions 2 free response
Electrochemistry
Chemical equilibrium

3. FREE RESPONSE 2000

4. FREE RESPONSE 2000

5. FREE RESPONSE 2000

6. FREE RESPONSE 2000

7. FREE RESPONSE 2000

8. FREE RESPONSE 2000

9. Topics you wanted to cover
Compound Formulas - in packet
Periodicity - in packet
Electrochemistry – difference between galvanic and electrolytic cell calculations and drawings; calculating plating
Chemical equilibrium: ICE problems, Partial pressure, Kp
Solution Stoichiometry
Calculating average atomic mass
Basic stoichiometry – percent composition, empirical and molecular formulas.
Redox reactions – half reactions, esp. acidic vs basic

10. ELECTROLYTIC VERSUS VOLTAIC

11. Stoichiometry of Electrolytic Processes
Step 1 – convert current and time to quantity of charge in coulombs
(amps)(time) = total charge transferred in coulombs
(Coulomb/sec)(sec) = coulombs
Step 2 – convert quantity of charge in coulombs to moles of electrons
coulombs ÷ (96,485 coulomb/mol e-) = mol e-
Step 3 – Convert moles of electrons to moles of substance
(mol e-) (mole substance/mol e-) = mol substance
Step 4 – Convert moles of substance to grams of substance
(mol substance)(formula mass of substance) = mass of substance

12. CALCULATING TIME, MASS, OR CURRENT
Basic Structure:
Coulombs = time (s) x current
Take Coulombs and determine moles of electrons using Faradays’ constant.
Determine the ratio of moles of electrons to the solid (based on the charge).
Convert moles of solid to mass of solid plated.
If they give you the mass, work backwards to time or current.

13. FINAL EXAM REVIEW = 1930.9736 = 1931C 1931C = 3.2183 = 3.22amps 600.s
24. A constant current was passed through a solution of AuCl4-1 ions between gold electrodes. After a period of 10.0min, the cathode increased in mass by 1.314g. What was the current?
10.0min AuCl gAu I = ?
1.314g Au 1 mol Au mol AuCl mole e F
196.97g Au mol Au 1 mol AuCl mol e-
= = 1931C 1931C = = 3.22amps
600.s

14. CHEMICAL EQUILIBRIUM Equilibrium Expressions involving Pressure
KP = K(RT)∆n
Reverse the reaction: K = 1/K = K′
If the original is multiplied by some factor, n, then K = Kn

15. CHEMICAL EQUILIBRIUM Doing an ICE problem
Determine the molarities of the given substances.
Fill out the ice chart with what you know.
If solving for K, they have to give you a way to solve for x. If solving for x, they have to give you K.
Then solve for x (and then determine concentrations) or determine K.

16. ICE
At a particular temperature, 12.0mol SO3 is placed in a 3.0L rigid container and the SO3 dissociates by the reaction 2SO3(g) ↔2SO2(g) + O2(g) At equilibrium, 3.0mol of SO2 is present. Calculate K for this rxn.
?MSO3 = 12.0mol/3.0L = 4.0mol SO3
2SO3 ↔ 2SO2 + O2 I
C -2x x x
E x 2x x
2x = 1.0M SO2 (from 3.0mol/3.0L), so x = 0.50M

17. ICE
K = [O2][SO2]2 = (0.50)(1.0)2 = [SO3]2 (3.0)2

18. SOLUTION STOICHIOMETRY
Write the balanced equation and the net ionic equation. Figure out what is the precipitate.
Convert molarity to moles: mol = MV
Convert each mole to moles of the precipitate.
Detemrine LIMITING REACTANT – smallest number.
Find the mass produced from the limiting moles
GOING FURTHER: Need to find the ions?

19. 3NaOH(aq) + Fe(NO3)3(aq)  Fe(OH)3(s) + 3NaNO3(aq)
FINAL EXAM REVIEW
What mass of solid iron(III) hydroxide can be produced when 150.0mL of a 0.400M Fe(NO3)3 is added to 250.0mL of a 0.500M NaOH?
3NaOH(aq) + Fe(NO3)3(aq)  Fe(OH)3(s) + 3NaNO3(aq)
Fe(NO3)3 mol = MV = (0.1500L)(0.400M) = mol
NaOH mol = MV = (0.2500L)(0.500M) = 0.125mol
0.0600mol Fe(NO3)3 1 mol Fe(OH)3 = mol Fe(OH)3
1 mol Fe(NO3)3
0.125mol NaOH 1 mol Fe(OH)3 = mol Fe(OH)3
1 mol NaOH
0.0417mol Fe(OH) g Fe(OH)3 = 4.56g Fe(OH)3
1 mole Fe(OH)3 LR

20. DETERMINING ION CONCENTRATION
15mL of a 1.0M potassium iodide solution reacts with 20.0mL of 6.0M lead(IV) nitrate. Calculate the concentration of potassium and lead(IV) ions in the supernatant (the clear liquid above the precipitate).
Write the balanced equation.
4KI + Pb(NO3)4  PbI KNO3
Determine if there is a change in moles of ions from reactant to product.
4K+  4K+ no change

21. DETERMINING ION CONCENTRATION
Convert molarity to moles
15mL L mol KI mol K+ = mol K+
1000mL L mol KI
Convert back to molarity by putting moles over whole volume.
0.015mol K = 0.43M K+
(0.020L L)
Back at #2, if there is a change in the number of ions
Lead goes from 1 ion (reactant) to 0 ions (product)
So, you must do stoichiometry to determine what is limiting.

22. DETERMINING ION CONCENTRATION
Determine limiting reactant
15mL L mol KI mol PbI4 = mol PbI4
1000mL L mol KI
20mL L mol Pb(NO3) mol PbI mL L mol Pb(NO3)4
= 0.12mol PbI4
Determine moles of the ions in the LR: 0.038mol Pb+4
Determine how much it should make
20mL L mol Pb(NO3) mol Pb+4
1000mL L mol Pb(NO3)4
= 0.12mol Pb+4 LR

23. DETERMINING ION CONCENTRATION
Take was you started with and subtract what you used to make the solid to get the amt of ion left over.
0.12mol Pb mol Pb+4 = 0.12mol Pb+4
0.12mol Pb+4 = = 3.4M Pb+4
0.035L

24. AVERAGE ATOMIC MASS
The average atomic mass on the periodic table comes from the weighted averages of the isotopes. To determine the mass, you need the following:
The masses of the isotopes
Their percent abundance
Multiple those two together, add all the numbers to get the average atomic mass.

25. AVERAGE ATOMIC MASS
The atomic masses of iridium-191 is an amu and iridium-193 is amu. The percentage abundance for each is 37.58% ( iridium-191) and % ( iridium-193). Calculate the average atomic mass.
191Ir amu x = amu
193Ir amu x = amu = amu

26. STOICHIOMETRY Percent composition: g part x 100 = answer g whole
Empirical Formula:
If they give you a percentage, change it to mass out of 100.
Convert to moles
Do mole ratios, dividing all by the smallest number
Determine formula from whole numbers

27. STOICHIOMETRY Molecular Formula:
Once you have the empirical formula, determine its mass from the periodic table.
Divide that mass into the given molecular mass to get the ratio. That number is what all elements in the empirical formula get multiplied by.

28. EMPIRICAL FORMULA
While trace impurities of iron and chromium in natural corundum form the gemstones ruby and sapphire, they are basically a binary compound of aluminum and oxygen, with 52.9% Al and 47.1% O. Find the empirical formula and give the chemical name for corundum.
52.9g Al mol Al = = mol Al
26.98g Al
47.1g O mol O = = mol O
16.00g O
1.96mol Al = mol O = 1.5
1.96 mol Al mol Al
Al2O3 aluminum oxide

29. Molecular formula
The simplest formula for vitamin C is C3H4O3. Experimental data indicates that the molecular mass of vitamin C is 180 amu.
C 3 x =
H 4 x = formula mass
O 3 x = emp. C3H4O g
molec ?
180 / = 2
C3H4O3 x 2  C6H8O6

30. REDOX REACTIONS
The fundamental principle in balancing redox equations is that the number of electrons lost in an oxidation process (increase in oxidation number) must equal the number of electrons gained in the reduction process (decrease in oxidation number).
Write the oxidation numbers and find out what two elements changes charge.

31. REDOX REACTIONS STEPS TO FOLLOW:
Write out the equation. Then change it to a net ionic equation if it is not one already, omit the spectator ions, and assign oxidation numbers to all atoms in the equation. Write them above the element.  
HS IO3-1  I S + H2O
Write the separate half-reactions.
HS  S IO  I-1

32. REDOX REACTIONS
Balance all the elements except O and H (already balanced in this one).
HS  S IO  I-1
If the oxygen is unbalanced, add enough water (H2O) to the side deficient in oxygen.
HS  S IO  I H2O

33. REDOX REACTIONS
Add sufficient hydrogen ions (H+) to the side deficient in hydrogen to balance the hydrogen.
HS  S + H+1
6H IO  I H2O

34. REDOX REACTIONS Write the electrons in each half reaction. -2 0
HS  S + H e-
6e H IO  I H2O
Determine the least common multiple and multiply each to get it so that the number of electrons gained equals the number of electrons lost.
(x3) 3HS  3S + H e-
(x1) 6e H IO  I H2O

35. REDOX REACTIONS
Add the two half reactions together and return the spectator atoms. Delete anything that exactly occurs on both sides. (Notice how the water showed back up!)
3HS e H IO3-1  3S + 3H e- + I-1
+ 3H2O 
becomes 3H+1
3HS H IO3-1  3S + I H2O

36. REDOX REACTIONS If acidic, you: If Basic, you:
Add water to balance the oxygens
Add H+ to balance the hydrogens
If Basic, you:
Add OH- to get rid of all the H+ by making water. Add the same number of OH- to both sides.