AP Chemistry Practice Test: Units 1 to 9
This is a free 15-question sampler in the style of the AP Chemistry exam. It touches all nine units, from atomic structure and bonding through gases, kinetics, thermochemistry, equilibrium, acids and bases, and thermodynamics with electrochemistry, mixing multiple choice, select all that apply and fill in the blank. Answer all 15 questions, then check your score and read the explanation for every question.
A compound is found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
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Prefer to read before you play? Every question in the sampler is listed below. Answers and explanations stay collapsed until you open them.
1. A compound is found to contain 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?
- A. CHO
- B. CH2O
- C. C2H4O
- D. C2H2O
Show answer and explanation
Answer: B. CH2O
Pretend you have 100 g of the compound so the percentages become grams, then convert each to moles: about 3.33 mol C, 6.7 mol H and 3.33 mol O. Dividing by the smallest value gives a ratio of 1 : 2 : 1, so the empirical formula is CH2O. The empirical formula is only the simplest ratio. To get the molecular formula you would also need the molar mass.
2. The first ionization energy of aluminum is lower than that of magnesium. Which explanation is best?
- A. Aluminum has a larger nuclear charge, so it holds its outer electron less tightly
- B. Magnesium has more protons than aluminum, so its outer electron is held more tightly
- C. Aluminum's outermost electron is in a 3p subshell, which is higher in energy and partly shielded by the 3s electrons
- D. Aluminum's outermost electron is in a half-filled subshell, which is unusually easy to remove
Show answer and explanation
Answer: C. Aluminum's outermost electron is in a 3p subshell, which is higher in energy and partly shielded by the 3s electrons
Across a period the nuclear charge rises, so the general trend is a rising ionization energy, but this pair is a known exception. Magnesium loses a 3s electron. Aluminum loses a 3p electron, which sits slightly farther out and is shielded a little by the filled 3s subshell, so it comes off with less energy. A larger nuclear charge on its own would raise the ionization energy, not lower it, and aluminum's 3p subshell holds a single electron, so it is not a half-filled subshell.
3. Complete the sentence.
The central xenon atom in XeF2 has _____ lone pairs of electrons, which gives the molecule a _____ shape.
Options for blank 1: 1, 2, 3, 4
Options for blank 2: bent, linear, trigonal planar, T-shaped
Show answer and explanation
Answer: 3, linear
Xenon brings 8 valence electrons and uses 2 of them to bond with the two fluorine atoms. The other 6 electrons form 3 lone pairs, so there are 5 electron domains and the electron geometry is trigonal bipyramidal. The three lone pairs sit in the equatorial plane, where they repel each other least, and that pushes the two fluorine atoms to opposite poles. The molecular shape is linear.
4. 1-Propanol (CH3CH2CH2OH) and methyl ethyl ether (CH3OCH2CH3) both have a molar mass of about 60 g/mol, yet 1-propanol boils at a much higher temperature. Why?
- A. Molecules of 1-propanol hydrogen bond to one another, and the ether cannot, so more energy is needed to separate them
- B. 1-Propanol has much stronger London dispersion forces because its molar mass is larger
- C. The covalent bonds inside 1-propanol are stronger, so they take more energy to break
- D. The ether has no intermolecular forces at all, so it vaporizes at almost any temperature
Show answer and explanation
Answer: A. Molecules of 1-propanol hydrogen bond to one another, and the ether cannot, so more energy is needed to separate them
Boiling separates molecules from each other and does not break the covalent bonds inside them. The two molar masses match, so dispersion forces are close to equal. The difference is that 1-propanol has an O-H bond, so one molecule's hydrogen can hydrogen bond to another molecule's oxygen. The ether has oxygen but no hydrogen attached to it, so it can accept hydrogen bonds from other molecules but cannot form them with its own kind. It still has dipole-dipole and dispersion forces.
5. A fixed amount of an ideal gas in a rigid container is heated from 300 K to 600 K. Which statement is correct?
- A. The average kinetic energy and the typical molecular speed both double
- B. The average kinetic energy is unchanged because the volume is unchanged
- C. The typical speed doubles and the average kinetic energy quadruples
- D. The average kinetic energy doubles and the typical speed increases by a factor of about 1.4
Show answer and explanation
Answer: D. The average kinetic energy doubles and the typical speed increases by a factor of about 1.4
Average kinetic energy is directly proportional to the Kelvin temperature, so doubling T doubles it. Kinetic energy depends on the square of the speed, so the typical (root-mean-square) speed grows by the square root of 2, about 1.4. The pressure doubles too, since the amount of gas and the volume are fixed. Using Celsius instead of Kelvin is the usual mistake on this kind of question.
6. A student uses a spectrophotometer to measure the absorbance of a colored solution. According to the Beer-Lambert law, which changes would increase the measured absorbance?
Select all that apply.
- A. Using a solution with a higher concentration of the absorbing species
- B. Using a cuvette with a shorter path length
- C. Using a cuvette with a longer path length
- D. Setting the instrument to a wavelength the species absorbs more strongly
- E. Diluting the solution with water
Show answer and explanation
Answer: A. Using a solution with a higher concentration of the absorbing species, C. Using a cuvette with a longer path length, D. Setting the instrument to a wavelength the species absorbs more strongly
The Beer-Lambert law is A = abc, where a is the molar absorptivity: absorbance rises with the absorptivity at the chosen wavelength, the path length and the concentration. That is why you measure near the wavelength of maximum absorbance, and why a straight-line calibration of absorbance against concentration works. Diluting the solution or shortening the path length lowers the absorbance.
7. A reaction vessel contains 0.60 mol of Al(s) and 0.60 mol of Cl2(g), which react as 2 Al + 3 Cl2 -> 2 AlCl3. Which reactant is limiting, and how many moles of AlCl3 can form?
- A. Al is limiting; 0.60 mol of AlCl3
- B. Cl2 is limiting; 0.40 mol of AlCl3
- C. Cl2 is limiting; 0.60 mol of AlCl3
- D. Al is limiting; 0.40 mol of AlCl3
Show answer and explanation
Answer: B. Cl2 is limiting; 0.40 mol of AlCl3
Use the mole ratio from the balanced equation. The 0.60 mol of Al could make 0.60 mol of AlCl3, but the 0.60 mol of Cl2 can make only 0.60 x (2/3) = 0.40 mol. The reactant that gives the smaller amount runs out first, so Cl2 is limiting and 0.40 mol of AlCl3 forms. Check it the other way as well: 0.60 mol of Al would need 0.90 mol of Cl2, and there is only 0.60 mol.
8. Complete the sentence.
In a reaction between A and B, doubling [A] while holding [B] constant doubles the initial rate, and doubling [B] while holding [A] constant quadruples the initial rate. The reaction is _____ order in A and _____ order in B.
Options for blank 1: zero, first, second, third
Options for blank 2: zero, first, second, third
Show answer and explanation
Answer: first, second
Compare how the rate changes when one concentration changes and the other stays fixed. A factor of 2 in [A] gave a factor of 2 in rate, so the order in A is 1 (2^1 = 2). A factor of 2 in [B] gave a factor of 4, so the order in B is 2 (2^2 = 4). The rate law is rate = k[A][B]^2, and the overall order is 3.
9. Which statements about a catalyst are correct?
Select all that apply.
- A. It increases the value of the equilibrium constant
- B. It provides an alternative pathway with a lower activation energy
- C. It leaves the enthalpy change of the overall reaction unchanged
- D. It is permanently used up, so it has to be replaced as the reaction proceeds
- E. It can be consumed in one elementary step and regenerated in a later step
Show answer and explanation
Answer: B. It provides an alternative pathway with a lower activation energy, C. It leaves the enthalpy change of the overall reaction unchanged, E. It can be consumed in one elementary step and regenerated in a later step
A catalyst speeds up a reaction by opening a pathway with a lower activation energy, often through a different mechanism, and it takes part in that mechanism: it is used in one step and made again in another. It does not appear in the net equation. Because it changes only the route, not the starting and ending energies, the enthalpy change is the same and the equilibrium constant is the same. It reaches equilibrium faster and does not shift where equilibrium lies.
10. Given N2(g) + O2(g) -> 2 NO(g), delta H = +180 kJ, and 2 NO(g) + O2(g) -> 2 NO2(g), delta H = -114 kJ, what is delta H for N2(g) + 2 O2(g) -> 2 NO2(g)?
- A. -294 kJ
- B. -66 kJ
- C. +66 kJ
- D. +294 kJ
Show answer and explanation
Answer: C. +66 kJ
Add the two steps. The 2 NO on the product side of the first equation cancels the 2 NO on the reactant side of the second, and one O2 from each side combines to leave N2 + 2 O2 -> 2 NO2. By Hess's law the enthalpy changes add: +180 + (-114) = +66 kJ. A positive value means the overall reaction is endothermic.
11. For A(g) + B(g) <=> C(g), K = 5.0 at a certain temperature. A mixture has [A] = 0.20 M, [B] = 0.20 M and [C] = 0.40 M. What happens next?
- A. Q = 2.0, so the reaction shifts toward the products
- B. Q = 10, so the reaction shifts toward the products
- C. Q = 5.0, so the mixture is already at equilibrium
- D. Q = 10, so the reaction shifts toward the reactants
Show answer and explanation
Answer: D. Q = 10, so the reaction shifts toward the reactants
The reaction quotient has the same form as K but uses the current concentrations: Q = [C] / ([A][B]) = 0.40 / (0.20 x 0.20) = 10. Since Q is larger than K = 5.0, the mixture has too much product to be at equilibrium, and the net reaction runs in reverse until Q falls to K. If Q were smaller than K, the shift would go toward the products instead.
12. Complete the sentence.
A buffer contains equal concentrations of acetic acid and acetate ion. The pKa of acetic acid is 4.74, so the pH of the buffer is _____. When a small amount of strong base is added, the pH _____.
Options for blank 1: 3.74, 4.74, 5.74, 7.00
Options for blank 2: rises slightly, rises sharply, falls slightly, does not change at all
Show answer and explanation
Answer: 4.74, rises slightly
The Henderson-Hasselbalch equation is pH = pKa + log([A-]/[HA]). With equal concentrations the ratio is 1 and log(1) = 0, so the pH equals the pKa, 4.74. Added hydroxide reacts with acetic acid and turns a little of it into acetate. That nudges the ratio up, so the pH rises slightly, but both buffer components are still present in large amounts, which is why the change is small and not sharp.
13. A weak monoprotic acid is titrated with a strong base. At the equivalence point, the solution is:
- A. basic, because the conjugate base of the weak acid reacts with water
- B. neutral, because the moles of acid and base are equal
- C. acidic, because some weak acid is always left over
- D. at a pH equal to the pKa of the acid
Show answer and explanation
Answer: A. basic, because the conjugate base of the weak acid reacts with water
At the equivalence point all of the weak acid has been converted to its conjugate base, and that base reacts with water to make a little hydroxide, so the pH is above 7. Equal moles of acid and base does not mean a pH of 7 unless both are strong. The pH equals the pKa at the half-equivalence point, where half of the acid has been converted.
14. Which processes would you expect to have a positive entropy change for the system?
Select all that apply.
- A. I2(s) -> I2(g)
- B. H2O(g) -> H2O(l)
- C. 2 NO2(g) -> N2O4(g)
- D. CaCO3(s) -> CaO(s) + CO2(g)
- E. H2O(s) -> H2O(l) at the melting point
Show answer and explanation
Answer: A. I2(s) -> I2(g), D. CaCO3(s) -> CaO(s) + CO2(g), E. H2O(s) -> H2O(l) at the melting point
Entropy rises when matter becomes more dispersed or has more ways to arrange itself. Sublimation, melting and a solid breaking down into a solid plus a gas all spread the particles out, so each is positive. Condensing a gas and joining two gas molecules into one both reduce the number of free particles, so those are negative.
15. A galvanic cell has a positive standard cell potential. What does that tell you about the cell reaction?
- A. delta G is positive and K is greater than 1
- B. delta G is positive and K is less than 1
- C. delta G is negative and K is greater than 1
- D. delta G is negative and K is less than 1
Show answer and explanation
Answer: C. delta G is negative and K is greater than 1
The standard free energy change is delta G = -nFE, so a positive E gives a negative delta G. The same quantity is related to the equilibrium constant by delta G = -RT ln K, so a negative delta G means ln K is positive and K is greater than 1. All three signs point the same way: the reaction is thermodynamically favorable as written, and a spontaneous reaction is what a galvanic cell uses to do work.
What the AP Chemistry exam looks like and how it is weighted
The AP Chemistry exam is scored 1 to 5 and has two sections of equal weight. Section I is 60 multiple-choice questions in 90 minutes, some of them standalone and some grouped around a table, graph or experiment. Section II is seven free-response questions in 105 minutes: three long questions worth 10 points each and four short questions worth 4 points each. Calculators are permitted, and a reference sheet is provided, so the exam is not testing whether you memorized a formula. It is testing whether you know which relationship to use and why. The multiple-choice part is now taken in the Bluebook app and the free-response answers are handwritten, so check the College Board page for the current rules before test day.
The course is organized into nine units, and the College Board publishes how much of the multiple-choice section each one gets. Atomic structure, compound structure, chemical reactions, kinetics, thermochemistry, equilibrium and thermodynamics with electrochemistry each carry roughly 7 to 9 percent. Two units are heavier: properties of substances and mixtures, which covers intermolecular forces, gases, solutions and spectroscopy, at roughly 18 to 22 percent, and acids and bases at roughly 11 to 15 percent. The course description has been reorganized in recent years, so unit names and topic lists in an older review book may not match. Use the current Course and Exam Description as your reference.
A 5 is the top score, but what a score is worth is up to each college. Some give credit for a 3, many ask for a 4 or a 5, and some give none for AP Chemistry. Look up the policy at the schools you care about before you decide how much to push.
How to use this sampler
The 15 questions are spread across all nine units, with the most questions on the unit the exam weights most. They are meant to show you which unit to open next rather than to predict a score. Each one asks for the reason, not only the answer: why aluminum gives up an electron more easily than magnesium, why two molecules with the same mass boil at different temperatures, what a catalyst changes and what it cannot touch. That is the kind of reasoning the free-response section rewards.
Take it in one sitting with a calculator and a periodic table, but without notes, then read the explanation for every question, including the ones you got right. Treat a miss as a topic. If you missed the buffer question, the answer is not a pH to memorize, it is the habit of asking whether the acid and its conjugate base are present in equal amounts. Write your units on every number when you check your work, since a lost unit is the cheapest way to lose a point on the free-response section.
The questions are original. They are written to the published course outline and do not use any released exam question. This sampler covers the multiple-choice style of reasoning. It does not practice the long, multi-part free-response questions, so pair it with past free-response questions from the College Board for that side of the exam.
How StudyPDF builds AP Chemistry practice tests from your class notes
Fifteen questions can show you which units are shaky. They cannot get you ready on their own, and a generic question bank rarely matches the labs and problem sets your teacher actually assigned.
StudyPDF works from your material instead. Upload your textbook chapters, your teacher's slides, a lab write-up or your review packet, press New material, and StudyPDF builds practice tests from those exact pages. Every question is grounded in your material and every explanation cites where the answer comes from, so you can go straight back to the page when a step does not make sense.
You do not need to upload anything to start. Name a topic instead, for example buffers and titration curves or reaction mechanisms, and Bo sets it up with its key concepts so you can build quizzes and flashcards on it. Your results track which concepts you keep missing, so the next practice test can target them. Starting is free.
Written by the StudyPDF team. Last updated 2026-08-19.