Over the last two years of your high school career you’re typically faced with an onslaught of standardized tests. While many of these, such as the SAT, ACT, or SAT II tests are required as part of your college application package, the AP exams are somewhat unique in that they can actually affect your college coursework by placing you into higher level courses. Perhaps not surprisingly then, the AP exams are among the hardest of the standardized tests you’ll take, and require a much greater depth of knowledge in order to succeed. Today we’ll take a look at the format of the AP Chemistry exam and go over some strategies to help guide you through it.
The AP Chemistry exam is a 3 hour and 5 minute exam, in two parts. Part I is a 90 minute, 75 question multiple choice section. The second part consists of six short answer or open-ended questions, and lasts for 1 hour and 35 minutes. You may use a calculator only for certain questions in part II, and a periodic table is provided for both parts. For part II only, you will also be provided with a table of reduction potentials as well as several commonly used equations and constants.
Part I: Multiple Choice
The multiple choice questions in the first part of the test can be divided into three main categories: calculations, concepts, and facts. Let’s look at examples of each:
EX1. Which of the following is the correct name for the compound with the molecular formula Mg3As2?
(A) Magnesium arsenic
(B) Trimagnesium diarsenic
(C) Magnesium arsenite
(D) Magnesium arsenate
(E) Magnesium arsenide
This is a typical factual question. You either know the answer off the top of your head, or you can get to the answer quickly based on your knowledge of chemistry in general. In this case, we’re dealing with a binary salt of magnesium and arsenic; these are named by writing the name of the cation (magnesium) followed by the anion whose ending is replaced by –ide (arsenic becomes arsenide). Thus the correct answer is E, magnesium arsenide.
EX2. Which of the following changes to a reaction system at equilibrium would affect the value of the equilibrium constant (Keq) for the reaction?
(A) Adding more reactants to the system
(B) Adding a catalyst to the reaction system
(C) Increasing the pressure of the system
(D) Increasing the temperature of the system
(E) Removing a product of the reaction from the system
This is a conceptual problem. There are no calculations involved, but the question requires you to analyze all of the reaction choices carefully to determine the correct answer. These are perhaps the trickiest questions in this section, partially because they often test the minute details of your understanding. In this case, for example, it’s easy to get misled—at first glance it may seem that several of the reaction choices are correct, since many affect the position of the reaction equilibrium (that is, the relative amounts of reactants and products). However, the question asks specifically about the equilibrium constant, which does not depend on the amounts of substances present, but is temperature-dependent (but not pressure-dependent). The correct answer, therefore, is D.
In general, if several answer choices for a questions seem correct, it helps to re-read the question, paying particular attention to the details in the wording. You may find a key word was overlooked, making solving the questions impossible.
EX3. A steady electric current is passed through molten MgCl2 for 1.00 hour, yielding 243 g of Mg metal. If an identical current is passed through molten AlCl3, the mass of Al metal produced is closest to:
(A) 27.0 g
(B) 120. g
(C) 180. g
(D) 243. g
(E) 270. g
This is a simple calculation question – you’ll get a conceptual set-up followed by a straightforward numerical calculation. As we’ll see, the numbers given are often easy to work with (remember, no calculator on this section!). From your periodic table, you’ll find that the atomic mass of Mg is 24.30, so over the 1 hour experiment 10 moles of Mg was produced. But, however, note that this does not mean that 10 moles of Al will be produced! Rather, this means that 2 * 10 = 20 moles of electrons were supplied to the metal. Working with this number, the number of moles of Al produced will thus be 20/3. The atomic mass of Al is 26.98, or roughly 27.0. So the correct answer is 20*27/3 = 180. g, choice C.
Part II: Part II usually contains two types of questions—“dry-lab” questions, where you’ll go through the steps involved in a lab activity, and reaction questions, where you’ll have to write a chemical equation, and answer a short question about the reaction.
Let’s look at some examples.
EX4. The following question deals with the titration of 10.00 mL of 0.0732 M HBr with a solution of NaOH of unknown concentration.
a. Describe the steps you would take to prepare and fill a buret for the titration.
b. Calculate the pH of the starting HBr solution.
c. A volume of 8.39 mL of the unknown NaOH solution was needed to reach the endpoint of the titration. Calculate the molarity of the NaOH solution used for the titration.
d. Assuming that the actual concentration of the NaOH solution was 0.0900 M, calculate the percent error in your result.
e. Calculate the volume of 12.0 M NaOH needed to prepare 400.0 mL of a 0.0900 M NaOH solution.
This is a fairly standard dry-lab question—it basically takes you through all the steps you would take to determine the concentration of an unknown base (or acid) by titration. If you’ve done this experiment in lab, you know exactly what to do here! So, step by step:
a. Rinse the buret with distilled water, then with some of the unknown base solution to avoid diluting it in the buret. Then fill the buret with the unknown base and drain it through the stopcock until all air bubbles have been released. Record the initial volume of solution in the buret.
b. Since HBr is a strong acid, it is fully ionized in water, and so [H+] = [HBr]initial. Thus, [H+] = 0.0732 M. Finally, pH = -log[H+], so pH = - log (0.0732) = 1.13.
c. At the end point, moles (HBr) = moles (NaOH).
Moles (HBr) = (Volume HBr) * [HBr]
Moles (HBr) = (10.00 * 10-3 L) * (0.0732 M) = 7.32 * 10-4 mol = Moles (NaOH)
[NaOH] = Moles (NaOH) / Volume (NaOH) = (7.32 * 10-4 mol)/(0.00839 L)
[NaOH] = 0.0872 M.
d. % error = (|measured – actual|/actual) * 100%
% error = (|0.0872 – 0.0900)/0.0900) * 100%
% error = 3.11%
e. For mixing two solutions, C1V1 + C2V2 = CfVf
In this case, C2 = 0, since we are diluting the 12.0 M NaOH solution with water
So C1V1 = CfVf
(12.0 M)V1 = (0.0900 M)(400.0 mL)
V1 = (0.0900 M)(400.0 mL)/(12.0 M) = 3.00 mL.
In general, these questions are focused on concepts more than simply getting the right numerical answer. Thus, if you’re running out of time, it’s often best to work out all of the parts of a dry-lab question, even if you don’t have numbers to plug in from previous parts. Similarly, even if you know you got the wrong answer for part c, for example, work out part d anyway—the grading policy is that as long as the question is worked out correctly, using a wrong value from a previous part will not count against you. In the same vein, math errors and significant figures will incur a penalty, but it is relatively small and can only be deducted once per question.
EX5a. Write a balanced equation for the reaction that occurs when copper powder is added to a solution of silver nitrate (ignore any species that are unchanged in the reaction).
EX5b. Which species is reduced in this reaction?
a. Cu + 2 Ag+ à Cu2+ + 2 Ag
b. Ag+ is reduced.
As these answers suggest, the questions can be answered quickly and without extensive explanation. The key is to balance all your equations and make sure you leave out anything that isn’t necessary (nitrate ion in this case).