Download AP Chem Summer Assignment KEY

AP Chemistry
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ANSWER KEY
Summer Assignment
Other Ions
To memorize
Ion
Ion Name
O22-
peroxide ion
OH1-
hydroxide ion
HSO41-
bisulfate ion; hydrogen sulfate ion
NH4+
ammonium ion
O21-
superoxide ion
HCO31-
bicarbonate ion; hydrogen carbonate ion
HPO42-
hydrogen phosphate ion
dihydrogen phosphate ion
H2PO41-
Colors of Common Ions in Aqueous Solution
Most common ions are colorless in solution, however, some have distinctive colors. These
colors have appeared in questions on the AP exam.
Fe2+ and Fe3+
various colors
Cu2+
blue to green
Cr2+
blue
Cr3+
green or violet
Mn2+
faint pink
Ni2+
green
Co2+
pink
MnO41-
dark purple
CrO42-
yellow
Cr2O72-
orange
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AP Chemistry
Summer Assignment
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ANSWER KEY
Summer Assignment Part B:
Practice with Nomenclature, Balancing Equations, Oxidation Numbers,
Solubility Rules and Problem Solving
Nomenclature: Simple Inorganic Formulas and Nomenclature – Complete Exercise 1 in
the Appendix of this packet. Review the naming rules and commit the naming prefixes to memory!
Oxidation Numbers: Oxidation Numbers: Anions and Cations – Complete Exercise 2 in the
Appendix of this packet. Review/Memorize the Rules for Oxidation Numbers
More Nomenclature: Ternary Nomenclature: Acids and Salts - Complete Exercise 3 in
the Appendix of this packet.
Balancing Equations – Balancing Molecular Equations - Complete Exercise 4 in the
Appendix of this packet.
Reaction Types and Reaction Prediction - Predict the products, write the equation and then
balance, etc. – Complete Exercise 5 in the Appendix of this packet.
Solubility Rules – Using Solubility Rules Table – you must Memorize these! –Complete
Exercise 6 in the Appendix of this packet using the solubility rules listed in Part A of the
summer assignment.
Review Problem Set – Complete the problems on attached Review Problem Set. You will
need your own paper for parts of it. – show your work and clearly mark your final answers!!
Use correct significant figures for math answers and units where needed!
Arrive at Stadium with all of these items completed and you will be well on your way to a
terrific year in AP Chemistry!
2
AP Chemistry
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ANSWER KEY
Summer Assignment
Review Problems:
Nomenclature Review
Forming binary ionic compounds
A. In a binary ionic compound the total positive charges must equal the total negative
charges. The best way to write correct formula units for ionic compounds is to use the
“Criss Cross Method”.
B. Sample problem: What ionic compound would form when calcium ions combine with
bromide ions?
Steps to the Criss Cross Method:
1. Write the ions with their charges, cations are always first. Ca2+ Br12. Cross over the charges by using the absolute value of each ion’s charge as the
subscript for the other ion.
Ca1 Br2
3. Check to make sure the subscripts are in the lowest whole number ratio
possible. Then write the formula.
CaBr2
Naming binary ionic compounds
A. Combine the names of the cation and the anion.
B. Example: BaBr2 is named barium bromide.
C. First write the ions formed for the following elements. Then use the Criss Cross
method to determine the formula. Then name the compounds.
Naming binary ionic compounds that contain polyatomic ions
A. The polyatomic ions on your common ions list should be memorized.
B. The most common oxyanions – polyatomic anions that contain oxygen, end in –ate.
Oxyanions with one less oxygen end in –ite. For example:
NO3-1 is nitrate
SO42- is sulfate
NO2-1 is nitrite
SO32- is sulfite
C. Anions with one less oxygen than the –ite ion are given the prefix hypo-.
D. Anions with one more oxygen than the –ate ion are given the prefix per-.
ClO-1 is hypochlorite
ClO3-1 is chlorate
ClO2-1 is chlorite
ClO4-1 is perchlorate
E. Naming compounds with polyatomics is the same as naming other compounds, just
name the cation and then the anion. If there is a transition metal involved, be sure to
check the charges to identify which ion (+1, +2, +3, +4….) it may be so that you can put
the correct Roman numeral in the name. Name the following.
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AP Chemistry
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ANSWER KEY
Summer Assignment
Polyatomic Ions Ending in “ate”
BO3-3
CO3-2
NO3-1
O
F
SiO4-4
PO4-3
SO4-2
ClO3-1
AsO4-3
SeO4-2
BrO3-1
TeO4-2
IO3-1
Notes and Observations
The individual locations of the elements in the table correspond to their
relative locations on the periodic table.
The “legs” – green shaded areas all end in “O3”.
The “interior” – blue shaded areas all end in
“O4”.
The charges of the ions become more positive as you go across a “period”. For
ions with the same root containing oxygen, the suffixes and prefixes are :
(Using chlorate as an example)
o
Ions starting with “per” will have one more oxygen. Ex. ClO4-1 = perchlorate
o
Ions ending in “ite” will have one less oxygen. Ex. ClO2-1 = chlorite
o
Ions starting with “hypo” and ending in “ite” will have two less oxygens.
Ex. ClO-1 = hypochlorite
Naming binary molecular compounds
A. With molecules, the prefix system is used.
Number
Prefix
Number
1
mono7
2
di8
3
tri9
4
tetra10
5
penta11
6
hexa12
Prefix
heptaoctanonadecaundecadodeca-
A. The less-electronegative element is always written first. It only gets a prefix if it has
more than one atom in the molecule.
B. The second element gets the prefix and the ending –ide.
C. The o or a at the end of the prefix is dropped when the word following the prefix begins
with another vowel, for example monoxide or pentoxide.
4
AP Chemistry
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Summer Assignment
ANSWER KEY
Exercise 1 - Nomenclature: Simple Inorganic Formulas and Nomenclature
I. In the first column, classify each of the following as molecular (M) or ionic (I). In the
second column, name each compound:
M
M
Name
or I
Name
or I
1) CaF2
I
Calcium fluoride
10) SrI2
I
Strontium Iodide
2) P4O10
M
Tetraphosphorus decoxide
11) CO
M
Carbon monoxide
3) K2S
I
Potassium Sulfide
12) Cs3P
I
Cesium phosphide
4) NaH
I
Sodium Hydride
13) ZnAt2
I
Zinc astitide
5) Al2Se3
I
Aluminum selenide
14) P2S3
M
Diphosphorus trisulfide
6) N2O
M
Dinitrogen monoxide
15) AgCl
I
Silver chloride
7) O2F
M
Dioxide monofluoride
16) Na3N
I
Sodium nitride
8) SBr6
M
Sulfur hexabromide
17) Mg3P2
I
Magnesium phosphide
9) Li2Te
I
Lithium telluride
18) XeF6
M
Xenon hexafluoride
II. In the first column, write the chemical formula (formula unit) for the compound formed
between the two given elements. In the second column, write the name for the compound:
Elements
Formula Unit
Name
1
magnesium and iodine
MgI2
Magnesium iodide
2
potassium and sulfur
K2S
Potassium sulfide
3
chlorine and aluminum
AlCl3
Aluminum chloride
4
zinc and bromine
ZnBr2
Zinc bromide
5
strontium and oxygen
SrO
Strontium oxide
6
calcium and nitrogen
Ca3N2
Calcium nitride
7
calcium and oxygen
CaO
Calcium oxide
8
copper(I) and oxygen
Cu2O
Copper(I) oxide
9
copper(II) and chlorine
CuCl2
Copper(II) Chloride
10 mercury(II) and oxygen
HgO
Mercury(II) oxide
11 nitrogen and aluminum
Al N
Aluminum nitride
5
AP Chemistry
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Summer Assignment
ANSWER KEY
Exercise 3 – More Nomenclature – Including Some Ternary Nomenclature: Acids and Salts
I. Name the following substances:
Formula
Name
Formula
Name
1) FeSO3
Iron(II) sulfite
16) Fe2O3
Iron(III) oxide
2) Cu(NO3)2
Copper(II) nitrate
17) (NH4)2SO3
Ammonium sulfite
3) Hg2Cl2
Mercury (I) chloride
18) Ca(MnO4)2
Calcium permanganate
4) AgBr
Silver bromide
19) PF5
Phosphorus pentafluoride
5) KClO3
Potassium chlorate
20) LiH
Lithium hydride
6) MgCO3
Magnesium carbonate
21) HIO3
Hydrogen iodate or iodic acid
7) BaO2
Barium oxide
22) NaBrO2
Sodium bromite
Potassium oxide
23) Ca3(PO4)2
Tin(IV) oxide
24) HIO4
8) K2O
9) SnO2
Calcium phosphate
Periodic acid
Acid name
10) Ni3(PO4)2
Nickel (II) phosphate
25) Fe(IO2)3
Iron(III) iodite
11) Pb(OH)2
Lead(II) hydroxide
26) HBr(aq)
Hydrobromic acid
Acid Name
12) CuCH3COO
Copper(I) acetate
27) C6H5COOH
Benzoic acid
13) N2O4
Dinitrogen tetroxide
28) Hg2(IO)2
Mercury(I) hypoiodite
14) Rb3P
Rubidium phosphide
29) H3PO3
Trihydrogen phosphite or
Phosphorous acid
15) S8
Sulfur (octatomic)
30) NH4BrO3
Ammonium bromate
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AP Chemistry
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ANSWER KEY
II. Write formulas for the following substances:
Name
Formula
Name
Formula
1) vanadium (V) oxide
V2O5
16) francium dichromate
Fr2Cr2O7
2) dihydrogen monoxide
H2O
17) calcium carbide
Ca2C
3) ammonium oxalate
(NH4)2C2O4
18) mercury (I) nitrate
HgNO3
4) polonium (VI)
thiocyanate
Po (SCN)6
19) cerium (IV) benzoate
Ce (C6H5COO)4
5) tetraphosphorus
decoxide
P4O10
20) potassium hydrogen
Phosphate
KH2PO4
Or K2HPO4
6) zinc hydroxide
Zn(OH)2
21) carbonic acid
H2CO3
7) potassium cyanide
KCN
22) calcium hypochlorite
Ca(ClO)2
8) cesium thiosulfate
Cs2S2O3
23) hydrotelluric acid
H2Te
9) oxygen molecule
O2
24) copper (II) nitrite
Cu(NO2)2
10) mercury (II) acetate
Hg(CH3COO)2
25) nitrous acid
HNO2
11) silver chromate
Ag2CrO4
26) hypoiodous acid
HIO
12) tin (II) carbonate
SnCO3
27) phosphoric acid
H3PO4
13) sodium hydrogen
carbonate
NaHCO3
28) Hydrocyanic acid
HCN
14) manganese (VII) oxide Mn2O7
29) tin(IV) chromate
Sn (CrO4)2
15) copper(II) hydrogen
phosphate
30) Acetic acid
Cu(H2PO4)2
CH3COOH or
HC2H3O2
7
AP Chemistry
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Summer Assignment
ANSWER KEY
III. Practice with acids! Remember:
-IC from –ATE
-OUS from –ITE
HYDRO-, -IC from -IDE
Complete the Following Table:
Name of Acid
Formula of Acid
Name of Anion
hydrochloric
HCl
chloride
sulfuric acid
H2SO4
sulfate
HI
Hydroiodic acid
Sulfurous acid
chlorous acid
Nitric acid
Acetic acid
hydrobromic acid
Hydrosulfuric acid
Phosphoric acid
sulfite
H2SO3
HClO2
chlorite
nitrate
HNO3
HC2H3O2 or CH3COOH
HBr
acetate
bromide
sulfide
H2S
HNO2
Nitrous acid
chromic acid
iodide
H2CrO4
H3PO4
nitrite
chromate
phosphate
8
AP Chemistry
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Summer Assignment
ANSWER KEY
Exercise 4 – Balancing Equations
I. Balance the following equations by adding coefficients as needed. Some equations
may already be balanced.
1)
2 C6H6 + 15 O2  6 H2O + _12 CO2
2)
4 NaI + __ Pb(SO4)2  __ PbI4 + 2 Na2SO4
3)
4 NH3 + 5 O2 4 NO + 6 H2O
4)
2 HNO3 + __ Mg(OH)2  2 H2O + __ Mg(NO3)2
5)
__ H3PO4 + 3 NaBr  3 HBr + __ Na3PO4
6)
2 CaO + __ MnI4  __ MnO2 + 2 CaI2
7)
__ C2H2 + 2 H2  __ C2H6
8)
2 VF5 + 10HI  __ V2I10 + 10 HF
9)
__ OsO4 + 2 PtCl4  2 PtO2 + __ OsCl8
10)
2 Hg2I2 + __ O2  2 Hg2O + 2 I2
Exercise 5 – Reaction Prediction Practice - Chemistry Review
I.
Predict the products, write the equation and then balance.
COMBUSTION
1. C4H9OH + oxygen 6 O2 
4 CO2 +
5 H2O
2. 2 Mg + O2 2MgO
3. 2C7H14 + oxygen 21 O2

14 CO2 +
14 H2O
SYNTHESIS
1. sodium 4 Na + oxygen O2 2 Na2O
2. calcium 3 Ca + nitrogen N2 
Ca3N2
3. potassium 2 K + bromine Br2
 2KBr
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AP Chemistry
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ANSWER KEY
DECOMPOSITION- for now take all decomposition all the way to the elements.
1. Strontium carbonate
2 SrCO3  2 Sr
+ 2 C + 3 O2
2. Mercury (II) oxide
2HgO 2Hg + O2
3. aluminum chlorate
2Al(ClO3)3  2Al + 3Cl2 + 9 O2
DOUBLE REPLACEMENT (all starting chemicals are aqueous solutions)
1. Iron (III) sulfate + calcium hydroxide 
Fe2 (SO4)3
+ 6LiOH2 Fe(OH)3 (s)
+
3Li2SO4
2. Sodium hydroxide + sulfuric acid 
2NaOH
+ H2SO4  Na2SO4 + 2H2O (l)
3. sodium sulfide + manganese (VI) acetate 
3Na2S + Mn(CH3COO)6  6NaCH3COO + MnS3
4. chromium(III) bromide + sodium sulfite 
2 CrBr3 + 3 Na2SO3
 Cr2(SO3)3
+ 6 NaBr
5. barium hydroxide + chlorous acid 
Ba(OH)2 + HClO2

BaClO2 + 2 H2O
SINGLE REPLACEMENT
Use the activity series in your book (or online) to complete and balance these equations. If no reaction
occurs, write NR.
1. Nickel + steam 
NR
2. Chlorine 3Cl2 + aluminum iodide 2AlI3  2AlCl3
+ 3 I2
3. Potassium 2 K + water 2 H2O  2 KOH + H2
4. Lead Pb + copper (II) chloride CuCl2  PbCl2 + Cu
5. Zinc Zn + hydrochloric acid 2 HCl  ZnCl2
+ H
10
AP Chemistry
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ANSWER KEY
Summer Assignment
Reaction Prediction Practice – Net Ionic Equations
For each of the following reactions(both double and single replacement, use your solubility rules and the
examples in this booklet and your notes to write the:
Molecular equation
Complete ionic equation
Net ionic equation
DOUBLE REPLACEMENT (all starting chemicals are aqueous solutions)
1. Iron (III) sulfate + calcium hydroxide 
Molecular: Fe2 (SO4)3
+ 6LiOH
2 Fe(OH)3 (s)
Complete ionic: 2Fe+3 + 3SO4-2 + 6 Li+ + 6OH-1
Net ionic:
Fe+3 + + 3OH-1
+
3Li2SO4
2Fe(OH)3 (s)
+
6Li+ + 3 SO4-2
Fe(OH)3 (s)
2. Sodium hydroxide + sulfuric acid 
Molecular:
2NaOH + H2SO4  Na2SO4 + 2H2O (l)
Complete ionic: 2Na+1 + OH-1
Net Ionic:
OH-1
+ H+1 
+ 2H+1 + SO4-2
 2Na+1 + SO4-2 + 2H2O (l)
H2O (l)
3. Calcium nitrate + lithium phosphate 
Molecular: 3Ca(NO3)2 + 2Li3PO4
Complete ionic:
 Ca3(PO4)2 (s) +
3Ca+2 + 6 NO3-1 + 6Li+1 + 2PO4-3
6 LiNO3
 Ca3(PO4)2 (s) + 6 NO3-1 + 6Li+1
Net ionic:
3Ca+2 + 2PO4-3  Ca3(PO4)2 (s)
SINGLE REPLACEMENT
1. Chlorine gas + aluminum iodide (aqueous) 
Molecular: 3Cl2 (g) +
Complete ionic:
3Cl2 (g) +
2AlI3  2AlCl3
2Al+3 +
+ 3 I2 (s)
6 I-1  2Al+3 + 6Cl-1 + 3 I2 (s)
Net ionic: 3Cl2 (g) + 6 I-1  6Cl-1
+ 3 I2 (s)
2. Potassium metal + water 
Molecular: 2 K (s) + 2 H2O  2 KOH (aq) + H2 (g)
Complete ionic: K (s) +
H2O
 K+1 + OH-1 (aq) + H2 (g)
Ionic: K (s) + H2O  K+1 + OH-1 (aq) + H2 (g)
3. Zinc metal + hydrochloric acid (aqueous) 
Molecular: Zn (s) + 2HCl
 ZnCl2 (aq) + H2 (g)
Complete ionic: Zn (s) + 2H+1 + 2Cl-1  Zn+2 + 2Cl-1 (aq) + H2 (g)
Net ionic: Zn (s) + 2H+1 +  Zn+2 + H2 (g)
11
AP Chemistry
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Summer Assignment
ANSWER KEY
Exercise 6 – Solubility Rules: Using Solubility Rules that you received in first year chem do the
following.
For the compounds in the table, write the formula for each compound in the first
column and then use the solubility rules to determine if each compound is soluble or
insoluble in water. In the second column write an (S) for those that are soluble and
an (I) for those that are insoluble in water.
Name
(aq) or (s)
Formula
silver nitrate
AgNO3
cobalt (II) sulfate
CoSO4
aq
zinc hydroxide
Zn(OH)2
s
iron (III) iodide
FeI3
aq
nickel (II) chloride
NiCl2
aq
lead (II) iodide
PbI2
s
sodium carbonate
Na2CO3
aq
barium sulfate
BaSO4
s
lead (II) sulfide
PbS
s
silver phosphate
Ag3PO4
s
lithium phosphate
Li3PO4
aq
nickel (II) carbonate
NiCO3
s
copper (II) hydroxide
Cu(OH)2
s
tin (IV) sulfate
Sn(SO4)2
s
lead (II) nitrate
Pb(NO3)2
aq
aq
12
AP Chemistry
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Summer Assignment
ANSWER KEY
Review Problem Set: Complete the problems below – show your work and clearly mark your
final answers!! Report your answers with correct significant figures and don’t forget units!!
Calculations, Sig Figs, and Conversions
1. Perform the following calculations with correct significant figures:
A. 4.6584 x 48.34 = 53.9
4.18
B. (5.02 – 4.68 + 38.760 + 14.0) / 3.1416 = 16.9
Density:
2. Calculate the mass of a sample of copper that occupies 5.3 x 10 -2 cm3 if the density of copper is
8.94 g/cm3.
m=Dxv
mass = 8.94 g/cm3 x (5.3 x 10-2 cm3) = 0.47 cm3
3. A 9.46 g sample of a solid is placed in a 25.00 ml flask. The remaining volume in the flask is
filled with benzene in which the solid is insoluble. The solid and the benzene together weigh
26.83 g. The density of the benzene is 0.879 g/ml. What is the density of the solid?
Complicated puzzle like problem!
Organize (make a variable-relationship chart) what you are given and
the relationships between variables.
Find as many variables as you can using the relationships and known numbers, until you get to the answer.
Solid:
Vol solid = ?
Mass solid = 9.46 g
Density solid = ?
Benzene:
Vol B =?
Mass B = ?
Density B =0.879 g/mL
Relationship
Total together:
Vol s + Vol B =
25.00 ml
mass solid + mass B = 26.83 g
D = m/v
not meaningful
Find mass of Benzene first: total mass – mass of solid = 26.83 g - 9.46 = 17.37 g Benzene
Find volume of benzene: Using mass and density of benzene: v = m/D = 17.37 g / 0.879g/mL = 19.76 mL B
Find volume of solid: Vol s + Vol B = 25.00 ml becomes- 25.00mL – 19.76 mL B = 5.24 mL solid
Find density of solid: D = m/v
9.46 g / 5.24 mL = 1.81g/mL
Answer
13
AP Chemistry
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Summer Assignment
Electromagnetic Spectrum
c = 
E=h
c = speed of light = 3.00x 108 m/sec
h = planck’s constant = 6.626 x 10-34 J sec
4. What is the wavelength (of light with a frequency ( of 3.2 x 1014 Hz (Hz = 1/sec).
Choose correct equation, rearrange mathematically, put in known numbers, calculate, include sig figs and units.
c =  becomes c/ v = 3.00x 108 m/sec / 3.2 x 1014 1/sec = 9.375 x 10-7 m
5. How much energy (in KJ) is associated with a radio wave of wavelength ( 1.2 X 102m?
Choose correct equation, rearrange mathematically, put in known numbers, calculate, include sig figs and units.
E = h and c =  go together mathematically, related through to become E = h c/
E = 6.626 x 10-34 J sec x 3.00x 108 m/sec = 1.6565 x 10-27 J (notice how the units work out!)
1.2 X 102m
Atomic Theory, Electron Configuration & Periodicity
6. Copy and fill in the following table:
Element/ion
56Fe
24Na+
57
Co
# of protons
26
30
11
13
30
+2
31S251Cr3+
# of neutrons
27
16
24
17
27
# of electrons
26
10
25
18
21
Write the electron configurations for Ca2+=1s2, 2s2, 2p6, 3s2, 3p6
Br-1 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p6
7. For Se write:
A. the complete electron configuration 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d10, 4p4
B. the noble gas electron configuration [Ar] 4s2, 3d10, 4p4
C. the orbital diagram from the noble gas electron configuration
D. the dot diagram
8. For the following elements, draw the Lewis dot structures.
A. Pb:
B. N
C. F
D. Ca
E. He
Given these elements:
S, Se, I, Ca and Be
Place these elements in order of:
A. increasing atomic radius Be, S, Se, Ca, I
B. decreasing ionization energy Be, S, Se, Ca, I
Average Atomic Mass
10. Find the mass of an element, if, out of a sample of 100:
5 % have a mass of 176,
0.05 x 176 = 8.8 Identify this element by symbol and name?
19 % have a mass of 177,
0.19 x 177 = 33.6
27 % have a mass of 178,
0.27 x 178 = 48.1
Hafnium - Hf
14 % have a mass of 179 and
0.14 x 179 = 25.1
35 % have a mass of 180?
0.35 x 180 = 63.0
Total = average atomic mass = 178.6 g/mol
9.
14
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Summer Assignment
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ANSWER KEY
Mole Calculations
11. Convert 3.48 x 1020 molecules of SO2 to moles. What is the mass of this quantity?
3.48 x 1020 molecules of SO2
1 𝑚𝑜𝑙𝑒 𝑆𝑂2
64.06 𝑔 SO2
6.02 𝑥 10 23 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 1 𝑚𝑜𝑙𝑒 SO2
=
3.70 x 10-2 g SO2
12. Calculate the following for quantities for 4.68g of Ca3(PO4)2:
6.02 𝑥 10 23 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑢𝑛𝑖𝑡𝑠
a. formula units 4.68g of Ca3(PO4)2 310.181 𝑚𝑜𝑙𝑒
= 9.08 x1021 f.u. Ca3(PO4)2
𝑔 𝐶𝑎3(𝑃𝑂4)2
1 𝑚𝑜𝑙𝑒 𝐶𝑎3(𝑃𝑂4)2
b. Ca2+ ions 3 Ca+2 ions x 9.08 x1021 f.u. Ca3(PO4)2 = 2.72 x 1022 ions Ca+2
c. PO43- ions 2 PO43- ions x 9.08 x1021 f.u. Ca3(PO4)2 = 1.82 x1022 ions PO43d. O atoms 8 O atoms x 9.08 x1021 f.u. Ca3(PO4)2 = 7.26 x 1022 O atoms
Empirical & Molecular Formula
13. The koala bear dines exclusively on eucalyptus leaves. The chief constituent in eucalyptus oil is a
substance called eucalyptol, which contains 77.87 % C, 11.76 % H and the remainder O. If the
molecular weight of eucalyptol is 154 amu, what is the empirical and molecular formula of this
compound?
Find % O. Change % to grams. Change grams to moles. Divide by
smallest # of moles. Get to whole numbers for empirical formula. Compare mass of
empirical formula to molecular to determine the whole number multiple between
empirical and molecular. Find molecular formula.
Find % O: 77.87 % C + 11.76 % H + ? %O = 100%
so, % O = 100 – 77.87 – 11.76 =
77.87 % C  77.87 g C x 1 mole C/12.01 g = 6.484 mol C ÷ 0.6481 = 10 C
11.76 % H  11.76 g H x 1 moleH/ 1.01 g = 11.64 mol H ÷ 0.6481 = 18.0 H
10.37 %O  10.37 g O x 1 mol O/ 16.00g = 0.6481 mol O ÷ 0.6481 = 1 O
Empirical formula = C10H18O empirical mass = 154.28 g/mol
Empircal formula = Molecular formula because the molar masses are the same.
Bonding & Lewis Dot Structures
14. Draw the Lewis structures for the following and identify its VSEPR Shape (molecular
geometry) and polarity (polar or nonpolar):
A. CH4, methane tetrahedral, nonpolar B. H2O Bent, polar C. SO2 bent, polar
D. Ozone, O3 bent, nonpolar
B. phosphate ion tetrahedral, nonpolar
Stoichiometry (mass, solution, and gas)
15. Suppose a solution containing 4.50g of sodium phosphate is mixed with a solution containing 3.75g of
barium nitrate. How many grams of barium phosphate can be produced?
Need balanced equation, Double replacement rxn.
2Na3PO4 + 3Ba(NO3)2  6NaNO3 + Ba3(PO4)2
Recognize it as a limiting reactant/ theoretical yield problem, because it has the quantities of
two reactants.
Change mass of each reactant to the mass of product (Barium phosphate) asked for.
4.50 g Na3PO4 x 1mol Na3PO4 1 mol Ba3(PO4)2 601.93 g Ba3(PO4)2 = 8.26 g Ba3(PO4)2
163.94 g Na3PO4 2 moleNa3PO4
1 mole Ba3(PO4)2
3.75 g Ba(NO3)2
x 1mol Ba(NO3)2
261.35 g Ba(NO3)2
1 mol Ba3(PO4)2
3 mole Ba(NO3)2
601.93 g Ba3(PO4)2
1 mole Ba3(PO4)2
= 2.88 g Ba3(PO4)2
Choose the lowest number from the two answers: 2.88 g Ba3(PO4)2
15
AP Chemistry
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Summer Assignment
ANSWER KEY
16. Over the years, the thermite reaction (mixing of solid iron (III) oxide with aluminum metal) has been used for
welding railroad rails, in incendiary bombs, and to ignite solid-fuel rocket motors.
Write a balanced equation representing the reaction. Fe2O3 (s) + 2Al(s)  Al2O3 (s) + 2Fe (s)
What masses of iron (III) oxide and aluminum must be used to produce 15.0 g of iron?
This is the reverse of a limiting reactant problem.
Start with the mass of product and find masses of each reactant.
𝟏 𝒎𝒐𝒍𝑭𝒆 𝟏𝒎𝒐𝒍 𝑭𝒆𝟐𝑶𝟑
𝟏𝟓𝟗.𝟕𝟎 𝒈 𝑭𝒆𝟐𝑶𝟑
𝟐𝒎𝒐𝒍 𝑭𝒆
𝟏𝒎𝒐𝒍 𝑭𝒆𝟐𝑶𝟑
𝟏 𝒎𝒐𝒍𝑭𝒆 𝟐 𝒎𝒐𝒍 𝑨𝒍 𝟐𝟔.𝟗𝟖 𝒈 𝑨𝒍
𝟓𝟓.𝟖𝟓𝒈𝑭𝒆 𝟐𝒎𝒐𝒍 𝑭𝒆 𝟏𝒎𝒐𝒍 𝑨𝒍
15g Fe 𝟓𝟓.𝟖𝟓𝒈𝑭𝒆
15g Fe
=
43 g Fe2O3
=
7.3 g Al
What is the maximum mass of aluminum oxide that could be produced? Mass to mass calc.
𝟏 𝒎𝒐𝒍𝑭𝒆 𝟏 𝒎𝒐𝒍 𝑨𝒍𝟐𝑶𝟑
𝟏𝟎𝟏.𝟗𝟔 𝒈 𝑨𝒍𝟐𝑶𝟑
15g Fe
= 14 g Al2O3
𝟓𝟓.𝟖𝟓𝒈𝑭𝒆
𝟐𝒎𝒐𝒍 𝑭𝒆
𝟏𝒎𝒐𝒍 𝑨𝒍𝟐𝑶𝟑
17. Given: 50 ml of 2.0 M nitric acid reacts with 15 g of aluminum hydroxide
Write a balanced equation for the reaction.
3HNO3(aq) + Al(OH)3(s)  Al(NO3)3 (aq) + 3H2O (l)
Write the balanced net ionic equation.
3 H+1 + Al(OH)3  3 H2O + Al+3
Which of the reactants would be the limiting reagent?
Use molarity formula to find moles of nitric acid. Molarity = M = mol/L, so mol = M x L
50 ml = 0.050 L; mol HNO3 = 2.0 M x 0.050 L = 0.10 mol HNO3
Then change mol HNO3 to mass of water (look at next question). Be careful, you are starting with moles!
0.10 mol HNO3 3 mol H2O
18.02 g H2O =
3 mole HNO3
1 mol H2O
1.8 g H2O
That’s one reactant. Now use the other reactant info to find mass of H 2O. You are starting with g.
15 g Al(OH)3 1 mole Al(OH)3
3 mole H2O
18.02 g H2O =
10. g H2O
78.01 g Al(OH)3
1 mole Al(OH)3 1 mole H2O
The lower of the two numbers indicates the theoretical yield. The limiting reactant produced
that lower number. So, the limiting reactant is HNO3
How many grams of water would be produced? Because we looked ahead and calculated the
theoretical yield of water… The answer is- 1.8 g H2O produced.
18.
3AgNO3 (aq) + Na3PO4 (aq)  Ag3PO4 (s) + 3 NaNO3 (aq) Balance the equation!
How many grams of silver phosphate are produced when 210.4 g of silver nitrate and 210.4 g sodium
phosphate are reacted? This is a classic limiting reactant and theoretical yield problem. Start with
the masses of the reactants and find the mass of the product asked for.
210.4 g AgNO3 1 mole AgNO3 1 mole Ag3PO4 418.58 g Ag3PO4
169.88 g AgNO3 3 mole AgNO3 1 mole Ag3PO4
=
172.8 g Ag3PO4
210.4 g Na3PO4 1 mole Na3PO4 1 mole Ag3PO4 418.58 g Ag3PO4 =
163.94g Na3PO4 1 mole Na3PO4 1 mole Ag3PO4
Choose the lower of the two answers. 172.8 g Ag3PO4 are produced.
537.2 g Ag3PO4
16
AP Chemistry
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ANSWER KEY
19. How much, in grams, of the excess reactant in Problem 18 is left over?
Notice that the reactant that makes the lower amount of product is the limiting.
So, the other one is the excess reactant, Ag3PO4. We need to find out how much of this excess reactant will
react with the limiting reactant. Then subtract the reacted amount from the starting amount of excess.
Work with the starting amount of the limiting reactant, AgNO3, this time changing it to the excess reactant.
210.4 g AgNO3 1 mole AgNO3 1 mole Na3PO4 163.94g Na3PO4 = 67.7 g Na3PO4 would be used up
169.88 g AgNO3 3 mole AgNO3
1 mole Na3PO4
To find excess left over: 210.4 g Na3PO4 - 67.7 g Na3PO4 = 142.7 g Na3PO4 left over
20. S8(s) + 8 Na2SO3(aq) + 40 H2O(l)  8 [Na2S2O3. 5H2O](s) Balance the equation!
a. What is the theoretical yield of sodium thiosulfate pentahydrate when 3.25 g of sulfur is boiled with
13.1g of sodium sulfite? Sodium thiosulfate pentahydrate is very soluble in water.
Limiting reactant-theoretical yield problem.
3.25 g S8 1 mole S8 8 mole Na2S2O3. 5H2O 248.20g Na2S2O3. 5H2O = 25.2 g Na2S2O3. 5H2O
256.48g S8 1 mole S8
1 mole Na2S2O3. 5H2O
13.1g Na2SO3 1 mole Na2SO3 8 mole Na2S2O3. 5H2O 248.20g Na2S2O3. 5H2O = 25.8 g Na2S2O3. 5H2O
126.04 g Na2SO3 8 mole Na2SO3
1 mole Na2S2O3. 5H2O
Choose the lower of the two numbers: 25.2 g Na2S2O3. 5H2O (notice the two numbers are close to each other)
b.
What is the percent yield of the synthesis if a student doing this experiment is able to isolate (collect)
only 5.26 g of the product? % yield = actual yield (= measured mass of product)
x 100
Theoretical yield (= calculated from limiting reactant)
% yield = [5.26 g Na2S2O3. 5H2O / 25.2 g Na2S2O3. 5H2O ] x 100 = 20.8 %
21. What is the molarity of a solution in which 10.0 g of AgNO3 is dissolved in 550.0 mL of solution?
Molarity = mole of solute
Change grams to moles. Change mL to Liters. Use formula.
Liter of solution
10.0 g AgNO3 1 mole AgNO3 = 0.589 mole AgNO3
550.0 mL = 0.550 L
169.88 g AgNO3
Molarity = M = 0.589 mole AgNO3 / 0.550 L = 0.107 M AgNO3
22. How many grams of KNO3 should be used to prepare 2.50 L of a 0.500 M solution?
Molarity = mole of solute
So, solve for moles. Then change moles to grams.
Liter of solution
Mole = Molarity x Liters = 0.500 M x 2.50 L = 1.25 mole KNO 101.11 g KNO3 = 126.4 g KNO3
1 mole KNO3
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ANSWER KEY
23. To what volume should 5.0g of KCl be diluted in order to prepare a 0.25 M solution?
Change grams to moles. Then use the molarity equation, M = mol/ L to find volume in liters.
5.0 g KCl 1 mole KCl =
74.55 g
0.067 mole KCl.
L = mol / M = 0.067 mole / 0.25 M = 0.27 L
KMT, States of Matter, & Gas Laws
24. A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has
a pressure of 345 torr at a temperature of –150C and a volume of 3.48 L. If conditions are
changed so that the temperature is 360C and the pressure is 268 torr, what will be the volume of
the sample?
Notice it is not burning in this case and there is
no chemical reaction.
This is a gas law problem. All gas laws come from PV = nRT. R = universal gas constant.
Temperatures must be in Kelvin. The units on the variables must match the units on R.
With so much information, it is a good technique to list the variables. Then decide which formula to use.
P1 = 345 torr
V1 = 3.48 L
T1 = -15 oC = 258 K
P2 = 268 torr
V2 = ? what we are solving for
T2 = 36 oC = 309 K
R = 0.821 L atm/mol K
or R = 62.364 L torr/mol K
If you only had R in atm (pressure unit) you would need to change the torr to atm [1 atm = 760 torr].
Since you could look up R in torr. Use that R.
Hopefully you notice that this is a Combined Gas Law: P1V1 = P2V2
T1
T2
And then you notice you don’t need R!
also:
P1V1 T2 = P2V2 T1
Solve for V2 =
P1V1 T2 = 345 torr 3.48 L 309 K = 5.37 L
P2 T1
268 torr 258 K
Thermochemistry
25. The specific heat capacity of graphite is 0.71 J/oC-g. Calculate the energy (in calories) required to
raise the temperature of 1.8 kg of graphite by 100.0 oC.
The equation to use is Q = s x m x T; where Q = energy in calories or joules;
s = specific heat capacity in J/oC g OR cal/ oC g; m = mass in grams( change kg to g) ;
T = change in temperature = [ T final – T initial]. As always, pay attention to the units.
The trick is that “s” is given in joules and the answer is asked for in calories. You need the conversion between
these. 1 calorie = 4.184 Joules.
Use the formula first to find Q, then convert to Joules.
o
o
Q = 0.71 J/ C-g x 1800 g x 100.0 C = 127,800 J 1 calorie = 30545 cal = 31,000 cal [2 sf]
4.184 J
18
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ANSWER KEY
19