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) HCH3COO__Acetic Acid ______lead (II) phosphate _____Pb3(PO4)____
2) NO______Nitrogen Monoxide______diphosphorous pentoxide_P2O5______
3) Ca(IO)2___calcium hypoiodite______carbonic acid______H2CO3______
4) Sb2(C2O4)5_antimonic oxalate - antimony (V) sodium hypobromite___NaBrO______
5) Fe(OH)3___ ferric hydroxide - iron (III)__ bismuth perchlorate___Bi(ClO4)3______
6) Mn(HSO4)4_manganese (IV) bisulfate ___ cupric chloride______CuCl______
7) NH4CN____ammonium cyanide______strontium bicarbonate__Sr(HCO3)2____
8) SI6______silicon hexahydride______mercuric thiocyanate___Hg(SCN)2____
9) ZnS2O3____zinc thiosulfate ______tetracarbon decahydride_C4H10______
10) Sn(Cr2O7)2_stannic dichromate - tin (IV) __ 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”) / f. calcium hydroxide = “S” (OH1- = “I” except CBS)b. HgCl2 = “I” (Cl = “S” except APH) / g. barium bromide = “S” (Br1- = “S” except APH)
c. Rb3PO4 = “S” (1A = ALWAYS “S”) / h. aluminum sulfite = “I” (SO32- = typically “I”)
d. BaF2 = “I” (F = “S” except CBS/APH) / i. mercurous acetate =“S” (C2H2O31- = ALWAYS “S”
e. NH4HCO3= “S” (NH4 = 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.
KOH1A = “S” / (NH4)2CrO4
NH4 = “S” / Al2(SO4)3 / Pb(NO3)2
NO3 = “S”
AgCH3COO
Acetate = “S” / AgOH + KCH3COO
OH- = I except CBS / AgCrO4 +
NH4CH3COO
CrO4 typically “I” / Ag2SO4 +
Al(CH3COO)3
SO4 = “S” except CBS/APH / AgNO3 +
Pb(CH3COO)2
NR – No reaction
CaCl2 / Ca(OH)2 + KCl
NR – No reaction
OH- = I except CBS / CaCrO4 +
NH4Cl
CrO4 typically “I” / CaSO4 +
AlCl3
SO4 = “S” except CBS/APH
Cl- = “S” except APH / 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 à __magnesium bromide + hydrogen ______
HBr + Mg à Mg2+ Br1- + H2
_2_ HBr (aq) + _1_ Mg(s) à _1_ MgBr2 (aq) + _1_ H2 (g)
(NOTE: production of hydrogen gas is evidence that a chemical reaction has occurred)
_S__14. potassium + oxygen à __potassium oxide ______.
K + O2 à K1+ O2-
_4_K(s) + _1_ O2(g) à _2_ K2O (s) NOTE: This is an REDOX RXN J
(NOTE: group 1A metals are HIGHLY reactive (1 Ve-) and will react with oxygen in the air)
_DR_15. ferric chloride + plumbous bisulfate à __iron (III) bisulfate + lead (II) chloride______
Fe 3+ Cl1- + Pb2+ HSO41- à Fe 3+ HSO41- + Pb2+ Cl1-
_2_FeCl3 (aq) + _3_ Pb(HSO4)2 (aq) à _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 à __ ammonium chlorate + copper (II) hypoiodite__
NH4 1+ IO 1- + Cu 2+ ClO31- à NH4 1+ ClO31- + Cu 2+ IO 1-
_2_NH4 IO (aq) + _1_Cu(ClO3)2 (aq) à _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 à CO2 + H2O + Heat (Exothermic RXN)
_1_ C3H8 (l) + _5_O2 (g à _3_ CO2 (g) + _4_ H2O (g) (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 à Zn2+ Cl1- + H2 NOTE: This is an REDOX RXN J
_1_ Zn (s) + _2_ HCl(aq) à _1_ ZnCl2 (aq) + _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- à H1+ OH1- + K1+ CN1-
_1_ HCN (g) + _1_ KOH (aq) à _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 / CH2OMolecular Formual / C10H22O4N2 / C45H85O5 / Common substance
H2O2 / Al2(Cr2O7)3 / Think Biology
C6H12O6
Hydrogen Peroxide 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 X 1/atomic mass = Moles Atom / (lowest) (whole number?) Empirical Formula
68.54g C x 1 mol C/12.01g C = 5.707 mol C / 1.427 mol = 4 C x1 C4H6O
8.63 g H x 1 mol H/1.008 g H = 8.561 mol H / 1.427 mol = 6 H x1
22.83 g O x 1 mol O/16.00 g O = 1.427 mol O / 1.427 mol = 1 O x1
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 à 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 à Molecular Formula
294.20 g/mol K2Cr2O7 K2Cr2O7
24. Gaseous hydrogen cyanide (HCN) can be produced as follows: 2 CH4+ 2 NH3+ 3 O2à 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 à H2O + Ca2+ PO43-
_3_ Ca(OH)2 (s) + _2_ H3PO4 (aq) à _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
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) à _1_Mn(NO3)4 (aq) + _2_PbCl2 (s) Cl1- = “S” except APH
1 Mn4+(aq) + 4 Cl1-(aq) + 2 Pb2+(aq) + 4 NO31-(aq) à 1 Mn4+(aq) + 4 NO31-(aq) + 2_PbCl2 (s)
2 Cl1-(aq) + 1 Pb2+(aq) à + 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 Cl1-
27. 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 Cl-
AlCl3= 3 mol Cl- per 1 mol AlCl3 0.39 M AlCl3 * 3mol Cl- /1 mol AlCl3 = 1.17 M Cl-
FeCl2 = 2 mol Cl- per 1 mol FeCl2 0.62 M FeCl2 * 1mol Cl- /1 mol FeCl2 = 1.24 M Cl-
The 0.39 M AlCl3 will have the lowest chloride ion concentration at 1.2 M Cl-
NOTE: 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 à _4_CO2 + _6_H2O _2_ C2H6 + _7_O2 à _4_CO2 + _6_H2O
1.50 g 1.50g _____g 0.403 g + 1.50g à 1.18g + 0.72 g
Law of Conservation of Mass
MM: C2H6 = 2(12.01g) + 6(1.01g) = 30.08 g/mol C2H6 Not required, just for review
MM: O2 = 2(16.00g) = 32.00 g/mol O2
MM: CO2 = 1(12.01g) + 2(16.00g) = 44.01 g/mol CO2
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