Chemistry Final Exam Review Sheet ANSWER KEY
Fill in the Conversions: molar mass = 1 mole = 6.02 x 1023 particles (atoms or molecules, etc)
The Mole
1. Calculate the molar mass for the following compounds: Use periodic table atomic masses (the decimal numbers)
and multiply by the number of atoms of each element. Do the atom inventory!
a) CuCl2
134.5 g/mol CuCl2
Cu = 1 x 63.5 = 63.5
Cl = 2 x 35.5 = 71
134.5 g/mol
b) K2SO4
174 g/mol K2SO4
K = 2 x 39 = 78
S = 1 x 32 = 32
O = 4 x 16 = 64
174 g/mol
2. How many moles are in 285 grams of CuCl2?
285 g CuCl2
1
x
1 mol CuCl2
= 285 =
134.5 g CuCl2
134.5
2.118959 = 2.12 mol CuCl2
3. How many moles are in 37.0 grams of K2SO4?
37.0 g K2SO4
1
x
1 mol CuCl2
174 g CuCl2
= 37
174
0.2126436 = 0.213 mol CuCl2
4. How many grams are in 3.00 moles of CuCl2?
3.00 mol CuCl2
1
x
134.5 g CuCl2
1 mol CuCl2
= 403.5 =
1
404 g CuCl2
ALWAYS DO A “NEED” AND “GIVEN” TO SOLVE THESE PROBLEMS!
Stoichiometry
In lab, copper (II) chloride reacts with aluminum metal in a single displacement reaction. The data collected during
the experiment is organized below.
Mass of copper (II) chloride which reacted 1210.5 g
Mass of aluminum metal (initial)
510. g
Mass of leftover aluminum wire (final)
132 g
Mass of aluminum metal which reacted
378 g
510 g – 132 g
Mass of copper actually produced
525 g
5. Write a balanced chemical equation for the single replacement reaction of copper (II) chloride with aluminum
metal to produce copper metal and aluminum chloride.
3 CuCl2 + 2 Al  3 Cu + 2 AlCl3
6. a) Calculate the number of moles of copper (II) chloride used.
Need: mol CuCl2
Given: 1210.5 g CuCl2 (from the data table)
1210.5 g CuCl2
1
x
1 mol CuCl2 =
134.5 g CuCl2
1210.5 = 9.0000 mol CuCl2
134.5
b) Calculate the number of moles of aluminum metal used.
Need: mol Al
Given: 378 g Al (from the data table)
378 g Al
1
x
1 mol Al
27.0 g Al
=
378
27
=
14.0 mol Al
7. Determine the limiting reactant.
1210.5 g CuCl2
1
378 g Al
1
x
x
1 mol CuCl2
134.5 g CuCl2
1 mol Al
27.0 g Al
x
x
1 rxn
3 mol CuCl2
1 rxn
2 mol Al
=
=
3.0000 reactions
with CuCl2
This is the limiting reactant
because it runs out after 3
reactions whereas Al runs out
later at 7 reactions.
7.00 reactions
with Al
8. What is the excess reactant? There are only two reactants: Al and CuCl2 so if the CuCl2 is the limiting
reactant, then the Al is the excess reactant.
9. How much excess reactant (in grams) is left over once the reaction is completed?
Need: g Al
Given: 1210.5 g CuCl2 (it’s the LR so it determines how much XS reactant is left over when the rxn stops)
1210.5 g CuCl2
1
x
1 mol CuCl2
134.5 g CuCl2
x
2 mol Al
x
3 mol CuCl2
27.0 g Al
1 mol Al
=
65367 = 162.00 g Al USED
403.5
Al to start with (in data table above problem #5) = 510 g Al
Al used when CuCl2 runs out =
162.00 g Al
348 g Al LEFT OVER (by subtracting)
10. Calculate the theoretical yield (in grams) of copper metal produced in this experiment.
Need: g Cu
Given: 1210.5 g CuCl2 (ALWAYS START WITH THE LIMITING REACTANT – IT DETERMINES THE AMOUNT
OF PRODUCTS MADE BECAUSE ONCE IT RUNS OUT, NO MORE PRODUCTS WILL BE MADE AND NO MORE OF
EXCESS REACTANT CAN BE USED).
1210.5 g CuCl2
1
x
1 mol CuCl2
134.5 g CuCl2
x
3 mol Cu
x
3 mol CuCl2
63.5 g Cu = 230600.25 = 571.5 g Cu (with sig figs)
1 mol Cu
403.5
this is THEORETICAL YIELD
11. What is the formula for percent yield? % YIELD = actual yield (from data table)
theoretical yield (from a problem I tell you to calculate)
12. Calculate the percent yield of copper metal produced in this experiment. Look in the data table to determine
how much copper metal was actually recovered.
% YIELD = actual yield (from data table)
= 525 g Cu
x 100 % = 91.9% Cu yield
theoretical yield (from question 11)
571.5 g Cu
Solutions
13. What is a solute? Substance doing the dissolving; present in the smaller amount
14. What is a solvent? Substance that does the dissolving; it’s the dissolving agnt; always present in the
larger amount
15. a) What is solvation? Solvent surrounds the solute before dissociation
b) List three ways to increase the rate of solvation.
- agitation (stirring)
-increase temperature
-increase surface area by pulverizing (crushing) a substance
--Molarity
16. What is the formula for finding molarity? M =
moles solute
Liter solution
17. Find the molarity of a 50.0 L solution that contains 5.00 moles of NaNO3 .
M = moles solute =
Liter solution
5.00 mol = 0.1 mol/L = 0.1 M
50.0L
18. Find the molarity of a solution with 165 g of SiF2 and a volume of 2000. mL.
M = moles solute =
Liter solution
2.496217852 mol =
2L
2000. mL x 1 L
= 2000
1
1000 mL
1000
165 g SiF2
1
x
1.2481089 mol/L = 1.25 M (with sig figs)
=2L
1 mol SiF2 = 165 = 2.496217852 mol SiF2
66.1 g SiF2
66.1
Si = 1 x 28.1 = 28.1
F = 2 x 19.0 = + 38
66.1 g/mol SiF2
19. How many grams of NaCl would you need to make 550 mL of 1.5 M solution?
M = moles solute =
Liter solution
1.5 mol/L =
x
0.55 L
CROSS MULTIPLY TO SOLVE
1.5 mol/L x 0.55 L = X = 0.825 mol NaCl
550 mL x 1 L
= 550 = 0.55 L
1
1000 mL 1000
0.825 mol NaCl
1
x
58.5 g NaCl = 48.2625 g NaCl = 48 g NaCl (with sig figs)
1 mol NaCl
1
Na = 1 x 23.0 = 23.0
Cl = 1 x 35.5 = + 35..5
58.5 g/mol NaCl
20. Use the following graph to answer the questions below:
Sometimes, you can’t just look at the graph to answer a
question; sometimes you have to do a proportion. The
proportions are set up in the following format:
How do you know when to use proportions with the graph?
a) problem asks you to find mass of water
b) problem has mass of water not same as graph
(100 g)
c) problem has solute mass that is higher than
highest number on the y-axis (150 g on the graph
to the left)
a) The x-axis is not labeled. What is on the x-axis? Temperature (in degrees Celsius)
b) How much NaNO3 dissolves in 100 g of water at 45C?
110 g NaNO3
c) How much KCl dissolves in 100 g of water at 60C?
~45 g KCl
d) What typically happens to the solubility of solids as the temperature increases?
Solubility increases
e) What typically happens to the solubility of gases as the temperature increases?
Solubility decreases
f) Which solutes on the graph are gases? Ce2(SO4)3 and Yb2(SO4)3 How do you know? Gases curve downward
because as the temperature goes up, solubility goes down making the downward-shaped curve and showing
the inversely proportional relationship between gas solubility and temperature.
g) If a saturated solution of potassium chloride dissolved in 100 g of water at 90C is cooled to 60C, how much
will precipitate out?
@90C the saturated point is 55 g KCl
@60C the saturated point is 45 g KCl
55 g - 45 g =10 g KCl that precipitates out (won’t dissolve)
h) How much sodium nitrate will dissolve in 450. g of water at 20.0C?
the graph is out of 100 g and this question refers to 450 g of water!
90 g NaNO3
100 g H2O
=
x g NaNO3
450 g H2O
(90 g NaNO3)(450 g H2O) = (x)(100 g H2O)
100 g H2O
100 g H2O
Do a proportion b/c the water from
CROSS MULTIPLY TO FIND X!
405 g NaNO3
=
x
i) Which solute is the least dependent on temperature? NaCl
j) Which solute is the most dependent on temperature? Ba(OH)2
Acids/Bases
21. How did Arrhenius classify acids and bases? Acids = have H+
Bases = have OH-
22. a) How does the Bronsted-Lowry theory treat acids? Hydrogen ion (proton) donors
b) How does the Bronsted-Lowry theory treat bases? Hydrogen ion (proton) acceptors
23. On the pH scale, acids have a value of 0-6.9 and bases have a value of 7.1-14. Acids taste sour while bases
taste bitter.
24. Complete the chart by indicating the color change for each type of solution in the specified indicator.
red litmus
blue litmus
type of solution
phenolphthalein
pH paper
paper
paper
no change; stays
acidic
stays red
turns red
red to orange
clear
basic
turns blue
stays blue
turns pink
blue to purple
neutral
stays red
stays blue
no change; stays
clear
green
25. Fill in the mathematical formulas below based on the information given.
pH = -log[H+]
pH + pOH = 14
pOH = -log[OH-]
26. Calculate the pH and pOH of the following solutions, and indicate whether the solution will be acidic, basic, or
neutral.
pH = - log [H+]
pOH = - log [OH-]
pH + pOH = 14
a) [H+] = 3.6  10-9M
pH = 8.4
pOH = 5.6
Acid/Base/Neutral? Base (see pH > 7)
b) [H+] = 2.7  10-6M
pH =5.6
pOH = 8.4
Acid/Base/Neutral? Acid (see pH < 7)
c) [OH-] = 8.8  10-3M
pH =11.9
pOH = 2.1
= 12 w/SF
Acid/Base/Neutral? Base (see pH > 7)
d) [OH-] = 4.0  10-3M
pH =11.6
pOH = 2.4
= 12 w/SF
Acid/Base/Neutral? Base (see pH > 7)
27. a) What type of reaction takes place when an acid reacts with a base? neutralization
b) What are the TWO products in this type of reaction? Water and ionic salt
28. Write, balance, and predict the products from the following reactants:
a) hydrobromic acid reacts with cesium hydroxide
1 HBr + 1 CsOH  1 H2O + 1 CsBr
b) chlorous acid reacts with potassium hydroxide
HClO2 + KOH  1 H2O + KClO2
29. a) A 25.0 mL solution of sulfuric acid is neutralized by 18.0 mL of 1.0 M sodium hydroxide using
phenolphthalein as an indicator. What is the concentration of the sulfuric acid solution?
STEP 1: WRITE AND BALANCE NEUTRALIZATION EQUATION
1 H2SO4 + 2 NaOH → 2 H2O + 1 Na2SO4
STEP 2: ORGANIZE KNOWN MOLARITY DATA FOR ACID AND BASE FROM THE GIVEN PROBLEM
H2SO4
M =
mol =
L = 0.025 L
NaOH
M = 1.0 mol/L
mol = ?
L = 0.018 L
25.0 mL x 1 L
= 25 = 0.025 L
1
1000 mL
1000
18.0 mL x
1 L
= 18 = 0.018 L
1
1000 mL
1000
STEP 3: SOLVE FOR UKNOWN OF NaOH
M = mol/L
1.0 mol/L =
x
0.018 L
x = 0.018 mol NaOH
STEP 4: PERFORM STOICHIOMETRY
-
Need: mol H2SO4
Given: 0.018 mol NaOH
0.018 mol NaOH x 1 mol H2SO4 = 0.018 = 0.009 mol H2SO4
1
2 mol NaOH
2
STEP 5: SOLVE FOR UKNOWN OF H2SO4
M = mol/L
M = 0.009 mol H2SO4
0.025 L
M = 0.36 mol/L
M = 0.36 M
b) What do you call the procedure above used to neutralize an acid with a base? titration
c) What is the purpose of this method? Explain. Helps to determine the concentration (molarity) of an
unknown solution from the known information about molarity of the other solution
Calorimetry
30. Explain the Law of Conservation of Energy.
Energy can neither be created nor destroyed only converted from one form to another (like kinetic
energy to potential and vice versa).
31. Name and define two types of energy:
a) Kinetic energy (energy of motion)
b) potential energy (stored energy)
32. a) What is specific heat? Amount of energy (J) needed to raise the temp of 1 g of a substance by 1C.
This is why the unit for specific heat is J/gC.
b) How is specific heat like a fingerprint? Every substance has a unique specific heat. Every state of
matter for the same substance also has different specific heats. Example: liquid water’s specific
heat is 4.184 J/gC but solid water (ice) has a specific heat of 2.03 J/gC.
33. Given the following calorimetry data, perform the needed calculations. Show all work!
Mass of Unknown Metal
51.0 g
Mass of Water in Calorimeter
60.0 g
Specific Heat of Water (use J/gC)
4.184 J/gC
Initial Temperature of Water in Calorimeter
18.0C
Initial Temperature of Unknown Metal
100. C
Final Temperature of Metal and Water
20.0C
Fill in blank!
a) Calculate the change in temperature of the water.
T = Tfinal - Tinitial
20.0C – 18.0C = 2.0C
b) Calculate the change in temperature of the metal.
T = Tfinal – Tinitial
20.0C – 100.C = 80.0C
c) Calculate the heat of the water.
Q = mcT
Q = (60.0 g)(4.184 J/gC)(2.0C)
Q = 502.08 J
Q = 502 J (with 3 SF)
d) Use the heat of the water to find the specific heat of the metal.
Q = mcT
502 J = (51.0 g)(x)(80.0C)
502 J = (4080 gC)(x)
x = 0.1230392 J/gC
x = 0.123 J/gC
This is the experimental value of metal’s specific heat b/c it’s calculated from data in a lab!
e) Identify the unknown metal using the data table below. LEAD because 0.123 J/gC is closest to 0.129 J/gC
Substance
Aluminum
Iron
Copper
Tin
Lead
Specific Heat
0.902 J/gC
0.451 J/gC
0.385 J/gC
0.222 J/gC
0.129 J/gC
accepted value b/c it is what you
SHOULD get
f) Determine the percent error of the experimentally determined specific heat.
% error = accepted – experimental x 100%
accepted
0.129 J/gC - 0.123 J/gC
0.129 J/gC
x 100%
= 0.006 J/gC = 4.651162 = 4.65% Error
0.129 J/gC
Absolute value menas no negative answers – make it
positive – always!
34. If the temperature of 24.4 g of ethanol increases from 25.0C to 78.8C, how much heat has been absorbed
by the ethanol? Assume the specific heat of ethanol is 2.44 J/g∙C).
Q = mcT
Q = (24.4 g)(2.44 J/gC)(53.8C)
Q = 3203.0368 J
Q = 3.20 x 103 J
Endo/Exo? Endothermic because the temperature increased meaning heat was ABSORBED/TAKEN IN.
35. a) Suppose you put 125 g of water into a calorimeter and find that its initial temperature is 25.6C. You then
heat a 50.0 g sample of an unknown metal to a temperature of 115.0C and put the metal sample into the
water. Heat flows from the hot metal to the cooler water until both have attained a final temperature of
29.3C. Assuming no heat is lost to the surroundings, calculate the specific heat of the metal.
Metal
m = 50.0 g
c = ??
T Tf – Ti =
29.3C – 115.0C
Water
m = 125 g
c = 4.184 J/gC
T Tf – Ti =
29.3C – 25.6C
Qmetal = Qwater
mcT = mcT
(50.0 g)(x)(85.7C) = (125 g)(4.184 J/gC)(3.7C)
(4285 gC)(x) = 1935.1 J
4285 gC
4285 gC
X = 0.451598 J/C (need 3 SF) = 0.452 J/gC
b) Use the data table below to help you identify the unknown substance.
Substance
Aluminum
Iron
Copper
iron
Specific Heat
0.902 J/gC
0.451 J/gC
0.385 J/gC
Entropy
36. a) What is the Law of Disorder? States that the natural order of the universe is to proceed in the order
of disorder (increased entropy)
b) What is another, more technical term for disorder or randomness? entropy
37. Predict whether entropy is increasing or decreasing.
a. water is heated from 55◦C to 95◦C
increasing (+S)
b. solid CO2 sublimes
increasing (+S)
c. solid silver is melted
increasing (+S)
d. CO2 (g)  CO2 (aq)
decreasing (-S)
e. C2H5OH (l)  C2H5OH (aq)
increasing (+S)
f. 2 H2 (g) + O2 (g)  2 H2O (g)
decreasing (-S)
Reaction Rates
38. Draw and label all the parts (reactants, products, Ea, activated complex, enthalpy) of an energy diagram for an
exothermic reaction.
39. Draw and label all the parts (reactants, products, Ea, activated complex, enthalpy) of an energy diagram for an
endothermic reaction.
Use the graphs below to answer the questions that follow.
Diagram A
Diagram B
40. Which graph represents an endothermic reaction?
B
41. Which graph represents an exothermic reaction?
A
42. What is the energy of the reactants in Diagram A?
30 J
(assume J b/c SI unit for heat since no unit given on graph)
43. What is the energy of the products in Diagram A?
10 J
44. What is the energy of the activated complex in Diagram A?
40 J
45. What is the activation energy, Ea, of Diagram B? 300 J – 75 J = 225 J
46. What is the change in energy, ΔE, of Diagram B?
b/c reactants already have some of the energy
225 J – 75 J = 150 J
47. Which reaction would benefit most from the addition of a catalyst? Diagram B because it has a higher
activation energy; catalysts reduce Ea so it would benefit a reaction with a high Ea barrier.
48. List four ways to speed up a chemical reaction.
a) increase temperature
c) increase concentration (molarity)
b) add a catalyst
d) increase surface by pulverizing (crushing) substance
49. a) You have two Alka-Seltzer tablets. You also have a glass of water at 20C and a glass of water at 70C.
You want to dissolve the Alka-Seltzer quickly to relieve your upset stomach, so which glass do you choose and
why? Explain using evidence from the Collision Theory to justify your answer.
Water @ 70C – higher temperature increases kinetic energy (motion) of the particles making frequent
collisions and more collisions with proper orientation and with more kinetic energy it is easier to
overcome the required activation energy.
b) Will the tablet dissolve faster as one whole piece or if you break it into smaller pieces? Why? Explain using
evidence from the Collision Theory to justify your answer.
If you break the tablet into smaller pieces it will dissolve faster because there is more surface
exposed during crushing. Even though the pieces are smaller, surface area has increase because
more surface is exposed. This will allow more frequent collisions among particles and the more
collisions that take place increases the odds that more collisions will occur with proper orientation.
50. Given the following experimental data, use the method if initial rates to determine the rate law for the
reaction aA + bB → products AND calculate the specific rate constant.
Experimental Data
Initial [A]
Initial [B]
Initial Rate
Trial
(M)
(M)
(mol/(L∙s))
1
0.100 M 2
0.100 M
2.00 × 10-3 mol/L∙s 1
2
0.200 M
0.100 M 2
2.00 × 10-3 mol/L∙s
2
3
0.200 M
0.200 M
4.00 × 10-3 mol/L∙s
2A = 1
A = 0
2B = 2
B = 1
rate law: R = k[A]0[B]1
R = k[A]0[B]1
2.00 x 10-3 mol/L∙s = k [0.100 mol/L]0[0.100 mol/L]1
2.00 x 10-3 mol/L∙s = k [0.100 mol2/L2]
rate law constant: k = 0.02 L/mol∙s
Equilibrium
51. How would each of the following changes affect the equilibrium position of the system used to produce
methanol from carbon monoxide and hydrogen? Would the reaction shift left, right, or exhibit no change?
CO(g) + 2H2(g)  CH3OH(g) + heat
a) adding CO to the system
RIGHT
b) cooling the system
RIGHT
c) adding a catalyst to the system
NO CHANGE
d) removing CH3OH from the system
RIGHT
e) decreasing the volume of the system
RIGHT