Ap® CHEMISTRY2004 SCORING GUIDELINES (Form B)

Question 2

2. Answer the following questions related to hydrocarbons.

(a) Determine the empirical formula of a hydrocarbon that contains 85.7 percent carbon by mass.

7.14 mol C7.14

14.2 mol H7.14

1 point for correct formula

nc 85.7gC x ImolC = 7.14mo1C12.01 g C1 mol H

nH = 14.3 g H x 14.2 mol H1.008 g H

1 point for moles ofC and moles ofH

] mol C : 1.99 mol H 1 point for ratio of moles of C to moles ofH

The empirical formula is CH2

(b) The density of the hydrocarbon in part (a) is 2.0 g L-I at 50°C and 0.948 atm.

(i) Calculate the molar mass of the hydrocarbon.

PV = nRT = mass RTmolar massmass RT . RT-- x - = density x -

V P P

0.0821 L atm x 323 Kmolar mass = 2.0 g L-I X mol K

0.948 atm

molar mass = 1 point for correct substitution

1 point for the answermolar mass = 56 g mol"

(ii) Determine the molecular formula of the hydrocarbon.

1point for correct formula

empirical mass x n = molar mass

empirical mass for CH2 = 14 g mol-I

14 g mor ' x n = 56 g mol"!

n = 4The molecular formula is C4Hg.

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4

Ap® CHEMISTRY2004 SCORING GUIDELINES (Form B)

Question 2 (cont'd.)

(c) Two flasks are connected by a stopcock as shown below. The 5.0 L flask contains CH4 at a pressure of3.0 atm, and the 1.0 L flask contains C2H6 at a pressure of 0.55 atm. Calculate the total pressure of thesystem after the stopcock is opened. Assume that the temperature remains constant.

5.0L3.0 atm

1.0 L0.55 atm

~~ (3.0 atm) (5.0 L) - 25 CH- . atm 4VI 6.0 L

P f C H - ~V; - (0.55 atm) (1.0 L) - 0092 C HI 0 2 6 - -- - -. atm 2 6VI 6.0 L

Pr = PICH4 + PIC2H6 = 2.5 atm + 0.092 atm = 2.6 atm

1 point for final pressure of CH4 or C2H6

1 point for the total pressure

(d) Octane, CgH1g(l), has a density of 0.703 g mL-1at 20°C. A 255 mL sample of CgH1g(l) measured at20°C reacts completely with excess oxygen as represented by the equation below.

Calculate the total number of moles of gaseous products formed.

_ 0.703 g CgHlg 1 mol CgH1Snproducts-255mLCgHISX ImLCgH1S Xl14gCsH1S x

34 mol products= 26.7 mol products

2 mol CSHIS

1 point for substitution of any of theseconversion factors

1 point for the correct answer

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5

Ap® CHEMISTRY2005 SCORING GUIDELINES

Question 7

Use principles of atomic structure, bonding, and/or intermolecular forces to respond to each of the following.Your responses must include specific information about all substances referred to in each question. .

(a) At a pressure of 1 atm, the boiling point of NH3(l) is 240 K, whereas the boiling point of NF3(l)is 144 K.

(i) Identify the intermolecular forces(s) in each substance.

NH3 has dispersion forces and hydrogen-bonding forces.NF3 has dispersion forces and dipole-dipole forces.(Credit earned for hydrogen-bonding and dipole-dipole forces)

One point is earned for the correctintermolecular attractive forces for both

NH3 and NF3.

(ii) Account for the difference in the boiling points of the substances.

The higher boiling point for NH3 is due to the greaterstrength ofthe hydrogen-bonding intermolecular attractiveforces among NH3 molecules compared to that of thedipole-dipole attractive forces among NF3 molecules.

One point is earned for correctly identifyingNH3 as having stronger intermolecular forces

than NF3.

(b) The melting point of KCI(s) is 776°C, whereas the melting point of NaCJ(s) is 801°C.

(i) Identify the type of bonding in each substance.

One point is earned for naming ionic bonds as the bonds inKCI and NaCl.Both KCI and NaCI have ionic bonds.

(ii) Account for the difference in the melting points of the substances.

The difference in the melting points is due to the different strengths ofionic bonding in the substances. The charges on the cationsand anions are the same in both compounds, therefore the relative size ofthe ions is the determining factor. Since Na+ has a smaller ionic radiusthan K+, the lattice energy of NaCI is higher than that of KCl. Thusmore energy is required to overcome the ionic forces in solid NaCl thanin solid KCl, and NaCI has the higher melting point.

One point is earned for a correctexplanation of the cause of thedifference in melting points of

KCl and NaCl.

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16

Ap® CHEMISTRY2005 SCORING GUIDELINES

Question 7 (continued)

(c) As shown in the table below, the first ionization energies of Si ,P, and CI show a trend.

Element First Ionization Energy(kJ mol'")

Si 786P '1,012

CI 1,251

(i) For each of the three elements, identify the quantum level (e.g., n = 1, n = 2, etc.) of the valence electronsin the atom.

One point is earned for the principal quantumlevel for alI three elements.

The valence electron is located in the n = 3 levelfor all three atoms.

(ii) Explain the reasons for the trend in first ionization energies.

One point is earned for explainingthat greater ionization energy isdue to increased nuclear charge.

Note: Explanations of the trendon the basis of effective nuclearcharge are acceptable.

Because the valence electrons in all three elements are shielded by thesame number of inner core electrons and the nuclear charge increasesgoing from Si to P to CI, the valence electrons feel an increasingattraction to the nucleus going from Si to P to Cl. Valence electronshaving a greater attraction to the nucleus, as in CI, wilI be moredifficult to remove, so CI has the highest ionization energy. P has thesecond highest ionization energy, and Si has the lowest ionizationenergy.

(d) A certain element has two stable isotopes. The mass of one of the isotopes is 62.93 amu and the mass of theother isotope is 64.93 amu.

(i) Identify the element. Justify your answer.

One point is earned for the element and theexplanation.

Copper. The relative average atomic mass is betweenthe two isotopic masses given.

(ii) Which isotope is more abundant? Justify your answer.

One point is earned for the correct choice andexplanation.

The isotope with mass 62.93 amu must be more abundantbecause its mass is closer to 63.55 amu (the relativeweighted average atomic mass for copper) than is the massof the other isotope.

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17

Ap® CHEMISTRY2006 SCORING GUIDELINES (Form B)

(a) Atomic size decreases from Na to CI in the periodic table.

Across the periodic table from Nato CI, the number of electronsin the s- and p- orbitals of the valence shell increases, as does thenumber of protons in the nucleus. The added electrons onlypartially shield the added protons, resulting in an increasedeffective nuclear charge. This results in a greater attraction forthe electrons, drawing them closer to the nucleus, making theatom smaller.

Question 7

7. Account for each of the following observations in terms of atomic theory and/or quantum theory.

One point is earned for indicatingthe increase in nuclear charge.

One point is earned for attributingthe size decrease to the greaterattraction of the nucleus for the

electrons caused by the increase innuclear charge.

(b) Boron commonly forms molecules of the type BX3. These molecules have a trigonal planar structure.

Boron has three valence electrons, each of which canform a single covalent bond with X. The three singlecovalent bonds of the boron atom orient to minimizeelectron-pair interaction, resulting in bond angles of1200 and a trigonal planar structure.

One point is earned for describing thevalence electrons and the bonds.

One point is earned for acorrect VSEPR argument.

(c) The first ionization energy of K is less than that of Na.

Both Na and K have an s' valence-shell electronconfiguration (Na: [Ne] 3s1 ; K: [Ar] 4s1). The K atomvalence electron has a higher n quantum number, placing itfarther from the nucleus than the Na atom valence electron.The greater distance results in less attraction to the nucleus.Because its valence electron is less attracted to its nucleus,the K atom has the lower ionization energy.

One point is earned for the sizeexplanation.

One point is earned for describingthe attraction to the nucleus.

(d) Each element displays a unique gas-phase emission spectrum.

Each element has a unique set of quantized energy states forits electrons (because of its unique nuclear charge and uniqueelectron configuration). As the electrons of an element absorbquanta of energy, they change to higher energy states(are excited) - during de-excitation, energy is released as EMradiation as the electrons cascade to lower energy states.Each photon of the EM radiation is associated with a specificwavelength (A. = hclE), a flux of which produces the lines ofthe emission spectrum.

One point is earned for describingthe quantized energy states and

emission phenomenon.

One point is earned for describing theeffect of the uniqueness of the nucleus

and/or electron configuration.

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13

Ap® CHEMISTRY2006 SCORING GUIDELINES (Form 8)

Question 8

8. Use chemical and physical principles to account for each of the following.

(a) An aluminum container filled with an aqueous solution of CuS04 eventually developed a leak.Include a chemical equation with your answer.

Al(s) + Cu2+(aq) ~ Al3+(aq) + Cu(s)

Cu2+ has a higher reduction potential than does Al3+, whichresults in the oxidation and eventual disappearance of the Almetal (depending upon the amount of Cu2+).

One point is earned for the correctequation (phases not required).

One point is earned for theexplanation of relative reactivity.

(b) The inside of a metal container was cleaned with steam and immediately sealed. Later, the containerimploded.

The high temperature of the steam causes the air/watermixture in the container to be at an elevated temperature.When the container is sealed and the temperature decreases,the pressure of the residual gases decreases below theexternal pressure, causing the implosion. The decrease inpressure occurs because pressure is proportional totemperature and/or vapor pressure of water decrease withtemperature, which means that condensation occurs uponcooling with a resultant pressure drop.

One point is earned for explaining theimplosion in terms of internal pressure

decrease.

One point is earned for the explanationof the change of pressure (either cause

is accepted).

(c) Skin feels cooler after rubbing alcohol has been applied to it.

Rubbing alcohol evaporates rapidly. Evaporation isendothermic so heat energy is absorbed from the skin in theprocess, which causes the cooling sensation.

One point is earned for reference tothe volatility of the alcohol.

One point is earned for discussing theendothermic nature of the process.

(d) The redness and itching of the skin caused by ant bites (injections of methanoic acid, HC02H) canbe relieved by applying a paste made from water and baking soda (solid sodium hydrogen carbonate).Include a chemical equation with your answer.

HC02H + NaHC03 ~ NaHC02 + H20 + CO2

Methanoic acid is neutralized by the HC03 - ion; with theneutralization of the acid; the redness and itching of theant bites subside.

One point is earned for the equation.

One point is earned for the explanation.

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14

Ap® CHEMISTRY2007 SCORING GUIDELINES

Question 6

Answer the following questions, which pertain to binary compounds.

(a) In the box provided below, draw a complete Lewis electron-dot diagram for the IF3 molecule .

...F'. .

'F" •• ·F··. : I" ... .--...

One point is earned for a correct Lewis diagram (can be done with dots or lines).

(b) On the basis of the Lewis electron-dot diagram that you drew in part (a), predict the molecular geometry ofthe IF3 molecule.

T-shapedOne point is earned for the molecular geometry consistent with the Lewis diagram

in part (a).

(c) In the S02 molecule, both of the bonds between sulfur and oxygen have the same length. Explain thisobservation, supporting your explanation by drawing in the box below a Lewis electron-dot diagram (ordiagrams) for the S02 molecule.

.s'0'" "0". . ..- .. - s.

"0" "'0'., ... ..

One point is earned for a correct diagram (can be done with dots or lines).One point is earned for some indication or discussion of resonance (but the point is not earned for a

description of resonance as a dynamic process).

OR

- - - One point is earned for a correct diagramO=S=O- - (can be done with dots or lines).

The bonds are the same length because they are both One point is earned for stating that both bondsdouble bonds. are double bonds.

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Ap® CHEMISTRY2007 SCORING GUIDELINES

Question 6 (continued)

(d) On the basis of your Lewis electron-dot diagram(s) in part (c), identify the hybridization of the sulfuratom in the S02 molecule. .

One point is earned for hybridization consistent with part (c).

The reaction between S02(g) and 02(g) to form S03(g) is represented below.

The reaction is exothermic. The reaction is slow at 2YC; however, a catalyst will cause the reaction toproceed faster.

(e) Using the axes provided below, draw the complete potential-energy diagram for both the catalyzed anduncatalyzed reactions. Clearly label the curve that represents the catalyzed reaction.

Catalyzed

Reaction Progress

One point is earned for an uncatalyzed reaction curve that must show that Ea > 0 and m < O.

One point is earned for a catalyzed reaction curve that must show Ea < uncatalyzed Ea 'must be clearly labeled, and must begin and end at the same energies as the uncatalyzed curve.

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Ap® CHEMISTRY2007 SCORING GUIDELINES (Form B)

Question 6

First Ionization Second Ionization Third IonizationEnergy Energy Energy

(kJ mol-l) (kJ mol") (kJ rnol")Element 1 1,251 2,300 3,820Element 2 496 4,560 6,910Element 3 738 . 1,450 7,730Element 4 1,000 2,250 3,360

The table above shows the first three ionization energies for atoms of four elements from the third period of theperiodic table. The elements are numbered randomly. Use the information in the table to answer the followingquestions.

(a) Which element is most metallic in character? Explain your reasoning.

Element 2. It has the lowest first-ionization energy. Metallicelements lose electron(s) when they become ions, and element2 requires the least amount of energy to remove an electron.

(b) Identify element 3. Explain your reasoning.

Magnesium. Element 3 has low first and second ionization energiesrelative to the third ionization energy, indicating that the element hastwo valence electrons, which is true for magnesium.(The third ionization of element 3 is dramatically higher, indicatingthe removal of an electron from a noble gas core.)

(c) Write the complete electron configuration for an atom of element 3.

One point is earned forthe identification.

One point is earned forthe justification.

One point is earned forthe identification.

One point is earned forthe justification.

One point is earned for the correct electron configuration.

(d) What is the expected oxidation state for the most common ion of element 2?

1+ One point is earned for the correct oxidation state.

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Ap® CHEMISTRY2007 SCORING GUIDELINES (Form B)

Question 6 (continued)

(e) What is the chemical symbol for element 2?

Na One point is earned for the correct symbol.

(f) A neutral atom of which of the four elements has the smallest radius?

Element 1One point is earned for the correct

identification of the element.

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Ap® CHEMISTRY2009 SCORING GUIDELINES

Question 2 (continued)

After the experiment was completed, the instructor informed the student that the unknown gas was carbon dioxide(44.0 g mol:"). .

(e) Calculate the percent error in the value of the molar mass calculated in part (d).

144.0 g mol" - 40.1 gmol-IIpercent error = x 100 = 8.9%

44.0 g mol'"

One point is earned for thecorrect setup and answer.

(f) For each of the following two possible occurrences, indicate whether it by itself could have beenresponsible for the error in the student's experimental result. You need not include any calculations withyour answer. For each of the possible occurrences, justify your answer.

Occurrence 1: The flask was incompletely flushed with COig), resulting in some dry air remainingin the flask.

This occurrence could have been responsible.

The dry air left in the flask is less dense (or has a lowermolar mass) than CO2 gas at the given T and P. Thiswould result in a lower mass of gas in the flask and a lowerresult for the molar mass of the unknown gas.

One point is earned for the correctreasoning and conclusion.

Occurrence 2: The temperature of the air was 23.0DC, but the temperature of the CO2(g) was lowerthan the reported 23.0DC.

This occurrence could not have been responsible.

The density of CO2 is greater at the lower temperature. Alarger mass of CO2 would be in the flask than if the CO2

had been at 23.0DC, resulting in a higher calculated molarmass for the unknown gas.

One point is earned for the correctreasoning and conclusion.

(g) Describe the steps of a laboratory method that the student could use to verify that the volume of the rigidflask is 843 rnL at 23.0DC. You need not include any calculations with your answer.

Valid methods include the following:

L Find the mass of the empty flask. Fill the flask with a liquid ofknown density (e.g., water at 23DC), and measure the mass oftheliquid-filled flask. Subtract to find the mass of the liquid. Using theknown density and mass, calculate the volume.

2. Measure 843 rnL ofa liquid (e.g., water) in a 1,000 rnL graduatedcylinder and transfer the liquid quantitatively into the flask to see ifthe water fills the flask completely.

One point is earnedfor a valid method.

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Ap® CHEMISTRY2009 SCORING GUIDELINES

Question 2 (continued)

Note: Significant figures were checked in tills problem: parts (a) and (d) were scored with ±1 significant figureneeded, and parts (b) and (c) were scored with the correct number of significant figures needed for the subtraction.

Ap® CHEMISTRY2009 SCORING GUIDELINES

Question 4 (15 points)

(a) A sample of solid iron(lll) oxide is reduced completely with solid carbon.

(i) Balanced equation:

2 Fe203 + 3 C ~ 4 Fe + 3 CO2

ORFe203 + 3 C ~ 2 Fe + 3 CO

One point is earned for both correct reactants.

Two points are earned for the correct products(1 point each).

One point is earned for correctly balancing(mass and charge) the equation.

(ii) What is the oxidation number of carbon before the reaction, and what is the oxidation number ofcarbon after the reaction is complete?

The oxidation number of C before the reaction is 0,and the oxidation number of C after the reaction is +4.

One point is earned for both oxidationnumbers consistent with part (i).

(b) Equal volumes of equirnolar solutions of ammonia and hydrochloric acid are combined. I

(i) Balanced equation:

NH3+W~NH/OR

NH3 + H30+ ~ NH4 + + H20

,Two points are earned for the correct reactants.

One point is earned for the correct product(s).

One point is earned for correctly balancing (massand charge) the equation.

(ii) Indicate whether the resulting solution is acidic, basic, or neutral. Explain.

The resulting solution is acidic because of thehydrolysis of the NH4 + ion, which reacts with water

to form NH3 and W.

OR

The mixing of a strong acid and a weak base resultsin an acidic solution.

One point is earned for a correctanswer consistent with part (i).

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Ap® CHEMISTRY2009 SCORING GUIDELINES

Question 4 (continued)

(c) Solid mercti.ry(ll) oxide decomposes as it is heated in an open test tube in a fume hood.

(i) Balanced equation:

2 HgO ~ 2 Hg + 02

One point is earned for the correct reactant.

Two points are earned for the correct products(1 point each).

One point is earned for correctly balancing(mass and charge) the equation.

(ii) After the reaction is complete, is the mass of the material in the test tube greater than, less than, orequal to the mass of the original sample? Explain.

The mass of the contents of the test tube will decreaseowing to the loss of 02 gas to the atmosphere.

One point is earned for a correctanswer consistent with part (i).

,/I