AP Chemistry - Student Samples from the 2023 Exam Administration

2023

Chemistry

Sample Student Responses

and Scoring Commentary

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Inside:

Free-Response Question 1

 Scoring Guidelines

 Student Samples

 Scoring Commentary

AP® Chemistry 2023 Scoring Guidelines

Question 1: Long Answer

10 points

(a) (i)

For the correct answer:

Accept one of the following:

•

22 6262 5

122 33 43sspspsd

•

[Ar] 4 3sd

1 point

(ii)

For the correct answer, consistent with part (a)(i):

1 point

Total for part (a)

2 points

(b)

For the correct calculated value:

62.673 g 61.262 g 1.411 g Cl−=

1 point

(c)

For the correct calculated value, consistent with part (b):

1 mol Cl

1.411 g Cl 0.03980 mol Cl

35.45 g Cl

×=

1 point

(d)

For the correct answer, consistent with part (c):

0.03980 mol Cl 2 mol Cl

MnCl

0.0199 mol Mn 1 mol Mn

= ⇒

1 point

(e)

For the correct answer and a valid justification:

Less than. If some of the mass of aqueous

Mn Cl

is lost due to splattering, the final mass

of the dry beaker and

Mn Cl

will be decreased, which will decrease the calculated mass

and number of moles of chlorine in the dry solid.

1 point

AP® Chemistry 2023 Scoring Guidelines

(f) (i)

For the correct balanced equation:

2 2 23

2 23

2 MnO ( ) H O( ) 2 Mn O ( ) 2 OH ( )

Zn()2 OH() ZnO()HO()2

2 MnO ( ) Zn( ) Mn O ( ) ZnO( )

s l e s aq

s aq s l e

ss s s

−−

+ +→ +

+ →++

+→ +

1 point

(ii)

For the correct calculated value, consistent with part (f)(i):

0.15 V ( 1.28 V) = 1.43 V

cell

E = −−



1 point

(iii)

For the correct calculated value, consistent with part (f)(ii):

96,485 C

2 mol 1.43 J 1 kJ

276 kJ/mol

1 mol 1 C 1000 J

1 mol

rxn

G nFE

−

∆ °=− °=− × × × =−

1 point

(iv)

For the correct answer and a valid justification:

Accept one of the following:

• Disagree. The battery is enclosed, so no change in the total mass will occur.

• Disagree. All reactants and products are in the solid phase, so the mass of the

sealed battery will remain the same (no gases enter or exit the battery).

1 point

Total for part (f)

4 points

Total for question 1

10 points

Sample 1A 1 of 3

Sample 1A 2 of 3

Sample 1A 3 of 3

Sample 1B 1 of 3

Sample 1B 2 of 3

Sample 1B 3 of 3

Sample 1C 1 of 3

Sample 1C 2 of 3

Sample 1C 3 of 3

Chemistry 2023 Scoring Commentary

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Question 1

Note: Student samples are quoted verbatim and may contain spelling and grammatical errors.

Overview

Question 1 presented students with a variety of chemical situations involving manganese and its

compounds.

Part (a)(i) required students to provide an electron configuration for the transition element

manganese. The intent was for students to demonstrate understanding of the Aufbau principle using

an appropriate representation for electron configuration (Learning Objective SAP-1.A, Skill 3.B from

the AP Chemistry Course and Exam Description).

Part (a)(ii) required students to use the electron configuration determined in (a)(i) to identify which

subshell loses electrons first when manganese atoms form cations. The intent was for students to

demonstrate understanding of ion formation from the electronic structure of an atom (SAP-2.A, 4.A).

Parts (b), (c), (d), and (e) involved a chemical equation with unknown subscripts “x” and “y” for the

formation of a Mn

compound from the reaction between Mn(s) and HCl(aq). Students were given

a set of experimental data to analyze, which contains the mass of an empty beaker, the mass of the

empty beaker and Mn(s), and the mass of beaker and Mn

(s) heated to constant mass.

Part (b) required students to calculate the mass of Cl in the dry Mn

sample that remains in the

beaker. The intent of the question was for students to use the experimental data provided to find the

mass of Cl (SPQ-1.A, 5.F).

Part (c) required students to calculate the moles of Cl based on the mass of Cl determined in part (b)

(SPQ-1.A, 5.F).

Part (d) required students to determine the empirical formula of the Mn

sample using the moles

of Cl determined in part (c) and a given quantity of moles of Mn (SPQ-2.A, 3.B).

Part (e) required students to explain how the moles of Cl calculated in part (c) would be affected in

the event of an experimental error where a portion of the Mn

splattered out of the beaker during

the process of heating the product to dryness (SPQ-2.A, 6.G).

Part (f) of this question consisted of four parts that revolve around an alkaline battery containing

MnO

. Students were provided with a table containing three reduction half-reactions and the

accompanying standard reduction potentials. One half-reaction contains MnO

, and the other two

half-reactions contain Zn.

Part (f)(i) required students to use the half-reactions given in the table to write the balanced net ionic

equation representing the most thermodynamically favorable reaction (ENE-6.A, 5.E).

Part (f)(ii) required students to calculate the standard cell potential (E°

cell

) for the overall reaction

occurring in the battery (ENE-6.A, 5.F).

Chemistry 2023 Scoring Commentary

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Question 1 (continued)

Part (f)(iii) required students to calculate the change in Gibbs free energy (ΔG°

rxn

) for the reaction in

part (f)(i), in units of kJ/mol

rxn

, utilizing the mathematical relationship between ΔG°

rxn

and E°

cell

(ENE-

6.B, 5.F).

Part (f)(iv) required students to evaluate a claim (agree or disagree and then provide a justification)

that the total mass of the battery, a closed system, decreases during operation (ENE-6.A, 6.D).

Sample: 1A

Score: 10

This response earned 10 points. In part (a)(i) the point was earned for the correct electron

configuration. In part (a)(ii) the point was earned for correctly identifying the subshell from which the

electrons are lost first; the response is also consistent with the response to part (a)(i). In part (b) the

point was earned for correctly calculating the mass of Cl remaining in the dry product using the

experimental data provided; the answer is supported with work. In part (c) the point was earned for

correctly converting the mass of Cl from part (b) into moles using the molar mass; supporting work

is provided. In part (d) the point was earned for a correct empirical formula with supporting work

showing a mole ratio calculation. In part (e) the point was earned for correctly indicating that the

lower mass of the dry product results in fewer moles of Cl calculated. In part (f)(i) the point was

earned for correctly writing the net ionic equation for the reaction with the greatest thermodynamic

favorability; supporting work is provided. In part (f)(ii) the point was earned for correctly calculating

cell



for the overall reaction consistent with part (f)(i). In part (f)(iii) the point was earned for

correctly calculating ΔG° consistent with part (f)(ii). In part (f)(iv) the point was earned for

disagreeing and providing a correct justification.

Sample: 1B

Score: 5

This response earned 5 points. In part (a)(i) the point was earned for the correct electron configuration.

In part (a)(ii) the point was earned for correctly identifying the subshell from which the electrons are

lost first; the response is also consistent with the response to part (a)(i). In part (b) the point was

earned for correctly calculating the mass of Cl remaining in the dry product using the experimental

data provided; the answer is supported with work. In part (c) the point was earned for correctly

converting the mass of Cl from part (b) into moles using the molar mass; supporting work is provided.

In part (d) the point was earned for a correct empirical formula with supporting work. In part (e) the

point was not earned because the response incorrectly indicates that the number of moles of Cl

remains the same because the concentrations of Mn and Cl are the same. In part (f)(i) the point was not

earned because the net ionic equation provided is only the reverse of one half-reaction; it is not the net

ionic equation for the reaction with the greatest thermodynamic favorability. In part (f)(ii) the point was

not earned

because the

cell



calculated does not represent the reaction with the greatest

thermodynamic favorability and is inconsistent with part (f)(i). In part (f)(iii) the point was not earned

because the incorrect equation is given, and no answer or work is provided for ΔG°. In part (f)(iv) the

point was not earned because the justification is not correct.

Chemistry 2023 Scoring Commentary

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Question 1 (continued)

Sample: 1C

Score: 2

This response earned 2 points. In part (a)(i) the point was earned for the correct electron

configuration. In part (a)(ii) the point was earned for correctly identifying the subshell from which the

electrons are lost first. In part (b) the point was not earned because the calculation is incorrect. In

part (c) the point was not earned because there is no work to support the calculation of number of

moles of Cl, even though the answer given is consistent with part (b). The directions on the exam

say, “For calculations, clearly show the method used and the steps involved in arriving at your

answers. You must show your work to receive credit for your answer.” In part (d) the point was not

earned because there is no work to support the empirical formula. In part (e) the point was not

earned because the response incorrectly indicates that the number of moles of Cl will be equal to the

number calculated in part (c) because volume does not affect molarity. In part (f)(i) the point was not

earned because the net ionic equation provided is only a restatement of one half-reaction; it is not the

net ionic equation

for the reaction with the greatest thermodynamic favorability. In part (f)(ii) the

point was not earned because the

cell



calculated does not represent the reaction with the greatest

thermodynamic favorability and is inconsistent with part (f)(i). In part (f)(iii) the point was not earned

because although the

cell



is consistent with part (f)(ii), the calculation of ΔG° uses an incorrect n

value that is inconsistent with part (f)(i). In part (f)(iv) the point was not earned because the response

agrees that the mass decreases.