# Mass of beaker + metal M + I

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1 To determine the molar mass of an unknown metal, M, a student reacts iodine with an excess of the metal to form the water-soluble compound MI, as represented by the equation above. The reaction proceeds until all of the I is consumed. The MI (aq) solution is quantitatively collected and heated to remove the water, and the product is dried and weighed to constant mass. The experimental steps are represented below, followed by a data table. Data for Unknown Metal Lab Mass of beaker Mass of beaker + metal M Mass of beaker + metal M + I Mass of MI, first weighing Mass of MI, second weighing g g g g g

2 Question (a) Given that the metal M is in excess, calculate the number of moles of I that reacted = g I 1 mol I g I g I 2 = mol I 1 point is earned for the number of moles. (b) Calculate the molar mass of the unknown metal M. Number of moles of I = number of moles of M g MI g I = g M Molar mass of M = g M mol M = 65.4 g/mol 1 point is earned for the number of grams of M. 1 point is earned for the molar mass. The student hypothesizes that the compound formed in the synthesis reaction is ionic. (c) Propose an experimental test the student could perform that could be used to support the hypothesis. Explain how the results of the test would support the hypothesis if the substance was ionic. The student could dissolve the compound in water or melt the compound and see if the solution/melt conducts electricity. If the solution/melt conducts electricity, mobile ions capable of carrying charge must be present, thus the compound is likely to be ionic. OR The student could heat the compound until it melts or boils. If the melting/boiling point is very high, then the compound is likely to be ionic. 1 point is earned for an appropriate test. 1 point is earned for explaining how the results would support the hypothesis. The student hypothesizes that Br will react with metal M more vigorously than I liquid at room temperature. did because Br is a

3 Question (d) Explain why I is a solid at room temperature whereas Br is a liquid. Your explanation should clearly reference the types and relative strengths of the intermolecular forces present in each substance. Both Br and I molecules are nonpolar molecules, therefore the only possible intermolecular forces are London dispersion forces. The London dispersion forces are stronger in I because it is larger in size with more electrons and/or a more polarizable electron cloud. The stronger London dispersion forces in I result in a higher melting point,which makes I 2 a solid at room temperature. 1 point is earned for identifying the forces in each substance as London dispersion forces. 1 point is earned for explaining why the forces are stronger in I than in Br. While cleaning up after the experiment, the student wishes to dispose of the unused solid I in a responsible manner. The student decides to convert the solid I to (aq) anion. The student has access to three solutions, H O (aq), Na S O (aq), and Na S O (aq), and the standard reduction table shown below. Half-reaction E (V) S O (aq) + 2 e 2 S O (aq) 0.08 I (s) + 2 e 2 I (aq) 0.54 O (g) + 2 H (aq) + 2 e H O (aq) 0.68 (e) Which solution should the student add to I (s) to reduce it to I (aq)? Circle your answer below. Justify your answer and include a calculation of E for the overall reaction. H O (aq) Na S O (aq) Na S O (aq) [Na S O (aq) should be circled.] The reaction between S O (aq) and I (s) will be thermodynamically favorable because or the reaction is positive ( = = V), from which it follows that is negative because = nf. 1 point is earned for the correct choice. 1 point is earned for a correct justification. (f) Write the balanced net-ionic equation for the reaction between I and the solution you selected in part (e). I + 2 S O 2 I + S O 1 point is earned for the correct equation.

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10 AP CHEMISTRY 2016 SCORING COMMENTARY Question 3 Overview Question 3 evaluated students ability to analyze data from a common laboratory experiment. Students were given pictures of the experimental steps to synthesize and collect the ionic solid, MI 2. In part (a) students calculated the moles of I 2 that reacted with an excess of metal M. In part (b) the value from part (a) was used to calculate the molar mass of the metal M. In part (c) students proposed an experimental test that could be used to determine if MI 2 is an ionic solid, and then explained how the results of the test supported this claim. Students explained why I 2 is solid at room temperature, but Br 2 is a liquid in part (d). In part (e) students utilized a standard reduction potential chart to choose which species, H 2 O 2, Na 2 S 2 O 3 or Na 2 S 4 O 6 could reduce the leftover I 2, and then to justify their claim. Students wrote the net-ionic equation for the reaction between I 2 and the solution they chose in part (e). Sample: 3A Score: 10 The point was earned in part (a) for correctly calculating the number of moles of I 2 that reacted with metal M. Both points were earned in part (b) for determining the mass and molar mass of metal M. Both points were earned in part (c): the student proposes to test the conductivity of an aqueous MI 2 solution and states that if the solution conducts electricity this confirms that MI 2 is ionic. In part (d) the first point was earned for indicating that London dispersion forces are present in both I 2 and Br 2. The second point in part (d) was earned for explaining that I 2 is a larger molecule with more electrons and, therefore, stronger dispersion forces. In part (e) 2 points were earned for correctly choosing Na 2 S 2 O 3 solution to reduce I 2 to I. The student justifies the choice with the reaction cell voltage and by stating that a positive E indicates a spontaneous reaction. In part (f) 1 point was earned for the balanced net-ionic equation between I 2 and Na 2 S 2 O 3. Sample: 3B Score: 8 The student earned 1 point in part (a) for the correct number of moles of I 2. In part (b) 2 points were earned for the correct molar mass of the metal M. In part (c) the student earned both points for the proposal to test the conductivity of an aqueous solution of MI 2 and the explanation of how the test supports the hypothesis. In part (d) only 1 of the points was earned. The student does not indicate what type of intermolecular forces are present in I 2 and Br 2 ; however, the student explains that the forces in I 2 are stronger because I 2 has a bigger electron cloud that is more polarizable. Only 1 point was earned in part (e). The student selects Na 2 S 2 O 3 to reduce I 2 to I ; however, the second point was not earned because there is no calculation with the justification. In part (f) 1 point was earned for the correct net-ionic equation The College Board.

11 AP CHEMISTRY 2016 SCORING COMMENTARY Question 3 (continued) Sample: 3C Score: 6 In part (a) 1 point was earned because the student calculates the number of moles of I 2 correctly. Part (b) earned 0 points; the mass of the metal, M, is incorrect and the student does not use the correct number of moles of M to determine the molar mass. Both points were earned in part (c): 1 point was earned for suggesting that the boiling point of the MI 2 could be used to determine if MI 2 is ionic and another point was earned for stating that ionic compounds have high boiling points. Both points were earned in part (d) for the identification of the correct intermolecular forces in I 2 and Br 2 and for explaining why the forces are stronger in I 2. In part (e) the student choses the incorrect solution to reduce I 2 and earned 0 points. In part (f) 1 point was earned for a correct net-ionic equation that is consistent with the student s choice in part (e) The College Board.

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