PRACTICE Test for AP Chem – Summer Packet
Name: ___ANSWER KEY______
(NOTE – Answer Key needs to be checked – some answers “May” not be correct)
Complete the 20 answers below for Nomenclature (Naming) of (Acids, Ionic, Covalent compounds)
1)
2)
3)
4)
5)
6)
7)
8)
9)
10)
HCH3COO__Acetic Acid _______________
NO_______Nitrogen Monoxide________
Ca(IO)2___calcium hypoiodite_________
Sb2(C2O4)5_antimonic oxalate - antimony (V)
Fe(OH)3___ ferric hydroxide - iron (III)__
Mn(HSO4)4_manganese (IV) bisulfate ___
NH4CN____ammonium cyanide_______
SI6________silicon hexahydride_______
ZnS2O3____zinc thiosulfate __________
Sn(Cr2O7)2_stannic dichromate - tin (IV) __
lead (II) phosphate _____Pb3(PO4)____
diphosphorous pentoxide_P2O5_______
carbonic acid_________H2CO3_______
sodium hypobromite___NaBrO_______
bismuth perchlorate___Bi(ClO4)3______
cupric chloride_______CuCl_________
strontium bicarbonate__Sr(HCO3)2____
mercuric thiocyanate___Hg(SCN)2____
tetracarbon decahydride_C4H10_______
ferrous bromate_______Fe(BrO3)2____
11) Classify each of the substances as being soluble or insoluble in water (recommend that you state the rule).
a. CaS = “I” (sulfide is typically “I”)
b. HgCl2 = “I” (Cl = “S” except APH)
c. Rb3PO4 = “S” (1A = ALWAYS “S”)
d. BaF2 = “I” (F = “S” except CBS/APH)
e. NH4HCO3= “S” (NH4 = ALWAYS “S”)
f. calcium hydroxide = “S” (OH1- = “I” except CBS)
g. barium bromide = “S” (Br1- = “S” except APH)
h. aluminum sulfite = “I” (SO32- = typically “I”)
i. mercurous acetate =“S” (C2H2O31- = ALWAYS “S”
j. silver chlorate = “S” (ClO31- = “S” except Pb2+)
12) Complete the following table by identifying the two new compounds (products) which would be
produced if the two aqueous solutions were to be mixed and undergo a double replacement reaction. For
each reaction, CIRCLE the name of the product that would form an insoluble precipitate.
KOH
1A = “S”
(NH4)2CrO4
NH4 = “S”
AgCH3COO AgOH +
Acetate = “S” KCH3COO
AgCrO4 +
NH4CH3COO
OH- = I except CBS CrO4 typically “I”
Al2(SO4)3
Ag2SO4 +
Al(CH3COO)3
SO4 = “S” except
CBS/APH
Ca(OH)2 + KCl
CaCl2
CaCrO4 +
NR – No reaction NH4Cl
OH- = I except CBS CrO4 typically “I”
CaSO4 +
AlCl3
SO4 = “S” except
CBS/APH
Cl- = “S” except
APH
Pb(NO3)2
NO3 = “S”
AgNO3 +
Pb(CH3COO)2
NR – No reaction
PbCl2 +
Ca(NO3)2
Cl- = “S” except
APH
Cl1-, Br1-, I1- EXCEPT: (APH) = Ag1+, Pb2+, Hg2+
SO42- EXCEPT: (CBS/APH) = Ca2+, Ba2+, Sr2+ / Ag1+ Pb2+, Hg+,
OH1- EXCEPT: (CBS) = Ca2+, Ba2+, Sr2 (only “slightly” soluble)
Predict the product(s), write a balanced equation, and indicate the type of reaction: Synthesis
(Formation), Decomposition, Single Replacement, Double Replacement, or Combustion.
NOTE: None of the items in blue will be REQUIRED for this test. But, you should start thinking about them.
_SR_13. hydrobromic acid + magnesium metal  __magnesium bromide + hydrogen _____________
HBr
_2_ HBr (aq)
+
Mg
+ _1_ Mg(s)


Mg2+ Br1+ H2
_1_ MgBr2 (aq) + _1_ H2 (g)
(NOTE: production of hydrogen gas is evidence that a chemical reaction has occurred)
_S__14. potassium + oxygen  __potassium oxide ____________________________________________.
K + O2 
K1+ O2_4_K(s) + _1_ O2(g)  _2_ K2O (s)
NOTE: This is an REDOX RXN 
(NOTE: group 1A metals are HIGHLY reactive (1 Ve-) and will react with oxygen in the air)
_DR_15. ferric chloride + plumbous bisulfate  __iron (III) bisulfate + lead (II) chloride___________
Fe 3+ Cl1- + Pb2+ HSO41-  Fe 3+ HSO41- + Pb2+ Cl1_2_FeCl3 (aq) + _3_ Pb(HSO4)2 (aq)  _2_ Fe(HSO4)3 (aq) + _3_PbCl2 (s)
(NOTE: production of a solid precipitate is evidence of a RXN. (PbCl2 Cl- = “S” except with APH)
_DR_16. ammonium hypoiodite + cupric chlorate  __ ammonium chlorate + copper (II) hypoiodite__
NH4 1+ IO 1- + Cu 2+ ClO31-  NH4 1+ ClO31- + Cu 2+
IO 1_2_NH4 IO (aq) + _1_Cu(ClO3)2 (aq)  _2_NH4ClO3 (aq) + _1_Cu(IO)2 (s)
(NOTE: an “assumption” is being made that the polyatomic ion hypoiodite has similar solubility to
other polyatomic ions. Where most polyatomic ions are typically insoluble.)
_C__17. LGP (liquid gas propane C3H8) is used to power the engines of most Korean taxi cabs.
C3H8
+ O2
 CO2 + H2O
+ Heat
_1_ C3H8 (l) + _5_O2 (g  _3_ CO2 (g) + _4_ H2O (g)
(Exothermic RXN)
(Note: H2O is a gas)
(NOTE: burning anything is a chemical RXN, production of heat is “usually” evidence of a RXN)
_SR_18. Hydrogen gas can be created by dropping a piece of zinc metal into hydrochloric acid.
Zn +
HCl

Zn2+ Cl1- +
_1_ Zn (s) + _2_ HCl(aq)  _1_ ZnCl2
H2
(aq)
NOTE: This is an REDOX RXN 
+ _1_ H2 (g)
(NOTE: production of hydrogen gas is evidence that a chemical reaction has occurred)
_SR_19. The CIA has experimented with a suicide pill created by mixing gaseous hydrogen cyanide (HCN)
with a solution of potassium hydroxide. After evaporating off the water, what solid is remaining?
H1+ CN1- + K1+ OH1-  H1+ OH1- + K1+ CN1_1_ HCN (g) + _1_ KOH (aq) 
_1_ H2O (l) + _1_ KCN (aq)
(NOTE: potassium cyanide is aqueous, but if the water is evaporated out of the solution, the ionic
salt potassium cyanide is remaining. There will not be any “visible” RXN, no solid precipitate, no
bubbling from a gas. Because KOH is already aqueous, the production of water will not be visible).
20. Explain the difference between an empirical formula and a molecular formula.
An empirical formula is the “simplest” formula for a compound. It shows the
different types of atoms (elements) in the simplest (lowest) whole number ratio.
The molecular formula has the same whole number ratio of atoms, but the
molecular formula indicates the ACTUAL number of each atom which is
necessary for balancing RXNs and calculating the atomic mass of the molecule.
CH2 = Empirical (lowest ratio), C8H18 = Molecular formula (actual numbers)
21. Complete the chart below for both Empirical Formulas and Molecular Formulas:
Empirical
Formula
C5H11O2N
C9H17O
HO
Al2(Cr2O7)3
CH2O
Molecular
Formual
C10H22O4N2
C45H85O5
Common substance
Al2(Cr2O7)3
Think Biology
H2O2
Hydrogen Peroxide
C6H12O6
Glucose
22. An organic compound with a molecular mass of 140g/mol is 68.54% carbon, 8.63% hydrogen, and
22.83% oxygen. What is the empirical formula and molecular formula of the compound?
Mass Atom
68.54g C x
8.63 g H x
22.83 g O x
X 1/atomic mass = Moles Atom / (lowest)
(whole number?)
1 mol C/12.01g C = 5.707 mol C / 1.427 mol = 4 C
x1
1 mol H/1.008 g H = 8.561 mol H / 1.427 mol = 6 H
x1
1 mol O/16.00 g O = 1.427 mol O / 1.427 mol = 1 O
x1
Empirical Formula
C4H6O
Molecular Mass of C4H6O = 4(12.01g) + 6(1.008g) + 1(16.00g) = 70.01 g/mol C4H6O
140 g/mol organic compound = 2x empirical 
Molecular Formula
70.01 g/mol C4H6O
C8H12O2
23. A sample of a substance is determined to be composed of 0.89 grams of potassium, 1.18 grams of
chromium, and 1.27 grams of oxygen. Calculate the empirical formula of this substance. If the molecular
mass is 294.18 g/mol, determine the molecular formula of this substance.
Mass Atom X 1/atomic mass = Moles Atom / (lowest)
(whole number?) Empirical Formula
0.89g K x 1 mol K/39.10 g K = 0.02276 mol K / 0.02269 mol = 1 K x2
K2Cr2O7
1.18g Cr x 1 mol Cr/52.00 g Cr = 0.02269 mol Cr / 0.02269 mol = 1 Cr x2
1.27g O x 1 mol O/16.00 g O = 0.07938 mol O / 0.02269 mol = 3.5 O x2
Molecular Mass of K2Cr2O7 = 2(39.10g) + 2(52.00g) + 7(16.00g) = 294.20 g/mol K2Cr2O7
294.18 g/mol substance = 1x empirical 
Molecular Formula
294.20 g/mol K2Cr2O7
K2Cr2O7
24. Gaseous hydrogen cyanide (HCN) can be produced as follows: 2 CH4 + 2 NH3 + 3 O2  2 HCN + 6 H2O
_____g
12.0 mol
Based on this equation, how many grams of ammonia gas would be required to produce 12.0 moles of HCN?
MM: NH3 = 1(14.01g) + 3(1.01g) = 17.04 g/mol NH3
MM: HCN = the molar mass of HCN is not required as the question is asking for moles not grams.
2 mol NH3
17.04 g NH3
12.0 mol HCN x ------------------ x ----------------- = 204.48 g NH3
2 mol HCN
1 mol NH3
25. During a classroom experiment, a reaction of 4.75 g of calcium hydroxide crystals mixed with an excess of
0.50 M phosphoric acid produced 5.31 g of calcium phosphate. Calculate the theoretical yield and the
percent yield from this experiment. Ca2+ OH1- + H3PO4 
H2O + Ca2+ PO43_3_ Ca(OH)2 (s) + _2_ H3PO4 (aq)  _6_ H2O (l) + _1_Ca3(PO4)2 (s)
4.75 g
excess
______ g
MM: Ca(OH)2 = 1(40.08g) + 2(16.00g) + 2(1.01g) = 74.10 g/mol Ca(OH)2
MM: Ca3(PO4)2 = 3(40.08g) + 2(30.97g) + 8(16.00g) = 310.18 g/mol Ca3(PO4)2
1 mol Ca(OH)2
1 mol Ca3(PO4)2
310.18 Ca3(PO4)2
4.75 g Ca(OH)2 x --------------------- x ---------------------- x ---------------------- = 6.628 g Ca3(PO4)2
74.10 g Ca(OH)2
3 mol Ca(OH)2
1 mol Ca3(PO4)2
Theoretical Yield
Actual yield
5.31 g Ca3(PO4)2
Percent Yield = ----------------------- x 100% = ------------------------ x 100 % = 80.1 % Ca3(PO4)2
Theoretical Yield
6.63 g Ca3(PO4)2
NOTE: Because 0.50 M phosphoric acid is excess, we do not need to use it for any of the calculations
26. Write a balanced net ionic equation for the following reaction. Then identify the spectator ions present.
_1_MnCl4 (aq) + _2_Pb(NO3)2 (aq)  _1_Mn(NO3)4 (aq) + _2_PbCl2 (s)
Cl1- = “S” except APH
1 Mn4+(aq) + 4 Cl1-(aq) + 2 Pb2+(aq) + 4 NO31-(aq)  1 Mn4+(aq) + 4 NO31-(aq) + 2_PbCl2 (s)
2 Cl1-(aq) + 1 Pb2+(aq)  + 1_PbCl2 (s)
Spectator Ions = 1 Mn4+(aq) + 4 NO31-(aq)
NOTE: for the balanced net ionic equation, I reduced the coefficient to the lowest terms 2 Cl1- vs 4 Cl127. Which will have the lowest concentration of chloride ions: 1.31 M KCl, 0.39 M AlCl3, or 0.62 M FeCl2?
KCl = 1 mol Cl- per 1 mol KCl
1.31 M KCl * 1mol Cl- /1 mol KCl = 1.31 M ClAlCl3= 3 mol Cl- per 1 mol AlCl3
0.39 M AlCl3 * 3mol Cl- /1 mol AlCl3 = 1.17 M ClFeCl2 = 2 mol Cl- per 1 mol FeCl2
0.62 M FeCl2 * 1mol Cl- /1 mol FeCl2 = 1.24 M ClThe 0.39 M AlCl3 will have the lowest chloride ion concentration at 1.2 M ClNOTE: My final answer was very clear and easy to understand AND I provided mathematical
evidence to support my answer. My answer is also in units of concentration M.
28. Provide mathematical evidence to determine which of the following two solutions will result in a
higher sodium ion concentration. 2.57 g of NaCl added to 25.0 mL of water, or 18.12 g of Na2CO3 added
to 150.0 mL of water. Molarity = moles of solute/Liter of solution
MM: NaCl = 1(22.99g) + 1(35.45g) = 58.44 g/mol NaCl
MM: Na2CO3 = 2(22.99g) + 1(12.01g) + 3(16.00g) = 105.99 g/mol Na2CO3
2.57 g NaCl x
1 mol NaCl / 58.44 g NaCl
= 0.04397 mol NaCl
M = mol/L = 0.04397 mol NaCl/0.0250 L = 1.759 M NaCl
(1 mol Na1+ / 1mol NaCl) = 1.759 M Na1+
18.12 g Na2CO3 x 1 mol Na2CO3 / 105.99 g Na2CO3 = 0.17096 mol Na2CO3
M = mol/L = 0.17096 mol Na2CO3/0.1500 L = 1.1397 M Na2CO3 (2 mol Na1+ / 1mol Na2CO3) = 2.2794
M Na1+
The 18.12 g of Na2CO3 added to 150.0 mL of water produced the higher 2.279 M Na1+ sodium ion
concentration compared to only 1.76 M Na1+ for the other solution.
29. 1.50 g of ethane gas (C2H6) is mixed with 1.50 g of oxygen. When exposed to a flame, the two gasses
explode to releasing heat. What is the maximum mass of carbon dioxide gas that could theoretically be
created? Be sure that your chemical equation is properly balanced before you start your calculations.
_2_ C2H6 + _7_O2  _4_CO2 + _6_H2O
1.50 g
1.50g
_____g
MM: C2H6 = 2(12.01g) + 6(1.01g) = 30.08 g/mol C2H6
MM: O2 = 2(16.00g) = 32.00 g/mol O2
MM: CO2 = 1(12.01g) + 2(16.00g) = 44.01 g/mol CO2
_2_ C2H6 + _7_O2  _4_CO2 + _6_H2O
0.403 g + 1.50g  1.18g + 0.72 g
Law of Conservation of Mass
Not required, just for review
1 mol C2H6
4 mol CO2
44.01 g CO2
1.50 g C2H6 x ------------------- x ----------------- x ---------------------- = 4.389 g CO2 Excess
30.08 g C2H6
2 mol C2H6
1 mol CO2
Reactant
1 mol O2
4 mol CO2
44.01 g CO2
1.50 g O2 x ------------------ x ----------------- x -------------------- = 1.179 g CO2 Limiting
32.00 g O2
7 mol O2
1 mol CO2
Reactant
The maximum amount of CO2 possible (based on limiting reactant) = 1.179 g CO2
Not required, but good practice: What is the amount of excess reactant?
1 mol O2
2 mol C2H6
30.08 g C2H6
1.50 g O2 x ------------------ x ----------------- x -------------------- = 0.4029 g C2H6 Needed
32.00 g O2
7 mol O2
1 mol C2H6
Reactant
Excess =
Have Need
= 1.50g C2H6 - 0.403g C2H6 = 1.097g C2H6 in excess