FINAL EXAM REVIEW
Fall Semester
FINAL FORMAT
 35
multiple choice questions
 2 free response
Electrochemistry
 Chemical equilibrium

FREE RESPONSE 2000
FREE RESPONSE 2000
FREE RESPONSE 2000
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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

ELECTROLYTIC VERSUS VOLTAIC
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
o
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
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.
FINAL EXAM REVIEW
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
AuCl41.314gAu
I=?
1.314g Au 1 mol Au
1 mol AuCl4-1
196.97g Au
1 mol Au
= 1930.9736 = 1931C
3 mole e96485F
1 mol AuCl4- 1 mol e-
1931C = 3.2183 = 3.22amps
600.s
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
CHEMICAL EQUILIBRIUM
Doing an ICE problem
1. Determine the molarities of the given
substances.
2. Fill out the ice chart with what you know.
3. 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.
4. Then solve for x (and then determine
concentrations) or determine K.
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 ↔
I
4.0
C
-2x
E 4.0-2x
2x = 1.0M SO2
2SO2 + O2
0
0
+2x
+x
2x
x
(from 3.0mol/3.0L), so x = 0.50M
ICE
K = [O2][SO2]2 = (0.50)(1.0)2 = 0.056
[SO3]2
(3.0)2
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?
FINAL EXAM REVIEW
10.
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) = 0.0600mol
NaOH
mol = MV = (0.2500L)(0.500M) = 0.125mol
0.0600mol Fe(NO3)3 1 mol Fe(OH)3 = 0.0600mol Fe(OH)3
1 mol Fe(NO3)3
0.125mol NaOH 1 mol Fe(OH)3 = 0.0417mol Fe(OH)3
LR
1 mol NaOH
0.0417mol Fe(OH)3 106.87g Fe(OH)3 = 4.56g Fe(OH)3
1 mole Fe(OH)3
DETERMINING ION
CONCENTRATION

1.
2.
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  PbI4 + 4KNO3
Determine if there is a change in moles of ions
from reactant to product.
4K+  4K+
no change
DETERMINING ION
CONCENTRATION
Convert molarity to moles
15mL
1L
1.0mol KI 1 mol K+ = 0.015mol K+
1000mL
1L
1 mol KI
4. Convert back to molarity by putting moles over whole
volume.
0.015mol K+ = 0.43M K+
(0.020L + 0.015L)
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.
3.
DETERMINING ION
CONCENTRATION
Determine limiting reactant
LR
15mL
1L
1.0mol KI 1 mol PbI4 = 0.0038mol PbI4
1000mL
1L
4 mol KI
20mL
1L
6.0mol Pb(NO3)4
1 mol PbI4
1000mL
1L
1 mol Pb(NO3)4
= 0.12mol PbI4
6. Determine moles of the ions in the LR: 0.038mol Pb+4
7. Determine how much it should make
20mL
1L
6.0mol Pb(NO3)4
1 mol Pb+4
1000mL
1L
1 mol Pb(NO3)4
= 0.12mol Pb+4
5.
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+4 - 0.038mol Pb+4 = 0.12mol Pb+4
8.
0.12mol Pb+4 = 3.42857 = 3.4M Pb+4
0.035L
AVERAGE ATOMIC MASS

1.
2.
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.
AVERAGE ATOMIC MASS
The atomic masses of iridium-191 is an 191.0 amu
and iridium-193 is 193.0 amu. The percentage
abundance for each is 37.58% ( iridium-191) and
62.42% ( iridium-193). Calculate the average
atomic mass.
191Ir
193Ir
191.0 amu x 0.3758 = 71.78 amu
193.0 amu x 0.6262 = 120.9 amu
192.68 = 192.7 amu
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
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.
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
1 mol Al
26.98g Al
= 1.96071 = 1.96 mol Al
1 mol O
16.00g O
1.96mol Al = 1
1.96 mol Al
= 2.94375 = 2.94 mol O
47.1g O
2.94 mol O = 1.5
1.96 mol Al
Al2O3 aluminum oxide
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 12.01 = 36.03
H 4 x 1.01 = 4.04
O 3 x 16.00 = 48.00
88.07
180 / 88.07 = 2
C3H4O3 x 2 
C6H8O6
formula
emp. C3H4O3
molec.
?
mass
88.07g
180
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.
REDOX REACTIONS
STEPS TO FOLLOW:
1. 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.
+1 -2 +5 -2
-1
0
+1 -2
HS-1 + IO3-1  I-1 + S + H2O
2.
Write the separate half-reactions.
-2
0
+5
HS-1  S
IO3-1
-1
 I-1
REDOX REACTIONS
3.
Balance all the elements except O and H
(already balanced in this one).
HS-1
4.

S
IO3-1
 I-1
If the oxygen is unbalanced, add enough water
(H2O) to the side deficient in oxygen.
HS-1

S
IO3-1
 I-1 + 3H2O
REDOX REACTIONS
5.
Add sufficient hydrogen ions (H+) to the side
deficient in hydrogen to balance the hydrogen.
HS-1

S + H+1
6H+1 + IO3-1
 I-1 + 3H2O
REDOX REACTIONS
Write the electrons in each half reaction.
-2
0
HS-1  S + H+1 + 2e+5
-1
6e- + 6H+1 + IO3-1  I-1 + 3H2O
7.
Determine the least common multiple and multiply
each to get it so that the number of electrons gained
equals the number of electrons lost.
6.
(x3)
3HS-1

3S + H+1 + 6e-
(x1) 6e- + 6H+1 + IO3-1
 I-1 + 3H2O
REDOX REACTIONS
8.
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-1 + 6e- + 6H+1 + IO3-1  3S + 3H+1 + 6e- + I-1
+ 3H2O
becomes 3H+1
3HS-1 + 3H+1 + IO3-1  3S + I-1 + 3H2O
REDOX REACTIONS

If acidic, you:
Add water to balance the oxygens
 Add H+ to balance the hydrogens


If Basic, you:
Add water to balance the oxygens
 Add H+ to balance the hydrogens
 Add OH- to get rid of all the H+ by making water.
Add the same number of OH- to both sides.
