- 41-page printable PDF
- 181 terms with examples
- 52 practice questions, answers explained
- 2-page cram sheet and study plans
CLEP Chemistry Study Guide
Everything the CLEP Chemistry exam tests, in one PDF you can print. Built from the College Board's official outline, heaviest topics first.
- 181 key terms, each with an example
- 52 practice questions with every answer explained
- A 2-page cram sheet for the night before
- 2-, 4- and 8-week study plans built from the exam weights
Last updated 2026 · exam details checked against the College Board's published exam information in 2026
What you get for $5
One chapter for each official content area, in the order the exam weighs them.
- 📘41pages, printable
- 🧭9chapters, heaviest first
- 🔑181terms with examples
- ✍️52practice questions, answers explained
- ⚠️27exam traps to avoid
- 📄2page cram sheet
Try 10 free Chemistry questions
In the style of the real exam, from all 9 areas. About 10 minutes, with every answer explained and a score by area.
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Question 1 · Structure of Matter
A radioactive isotope has a half-life of 12.0 days. How much of a 64.0 mg sample remains after 36.0 days?
Show the answer
Answer: B. 8.00 mg. 36.0 days is 3 half-lives (36.0 ÷ 12.0), so the sample is halved three times: 64.0 → 32.0 → 16.0 → 8.00 mg.
The trap: 21.3 mg comes from dividing 64.0 by 3, which treats decay as a steady loss. Each half-life halves whatever is left.
36.0 days is 3 half-lives (36.0 ÷ 12.0), so the sample is halved three times: 64.0 → 32.0 → 16.0 → 8.00 mg.
The trap: 21.3 mg comes from dividing 64.0 by 3, which treats decay as a steady loss. Each half-life halves whatever is left.
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Question 2 · States of Matter
A rigid 8.21 L tank holds a gas at 2.00 atm and 400 K. How many moles of gas does it contain? (R = 0.08206 L·atm/(mol·K))
Show the answer
Answer: C. 0.500 mol. n = PV/RT = (2.00 × 8.21) / (0.08206 × 400) = 16.42 / 32.82 = 0.500 mol.
The trap: 0.125 mol comes from dividing by the pressure instead of multiplying by it; PV = nRT puts P in the numerator.
n = PV/RT = (2.00 × 8.21) / (0.08206 × 400) = 16.42 / 32.82 = 0.500 mol.
The trap: 0.125 mol comes from dividing by the pressure instead of multiplying by it; PV = nRT puts P in the numerator.
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Question 3 · Descriptive Chemistry
What is the name of the compound Fe₂(SO₄)₃?
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Answer: D. Iron(III) sulfate. Three sulfate ions (2− each) total 6−, shared by two iron ions, so each iron is 3+. The name is iron(III) sulfate.
The trap: Iron(II) sulfate mistakes the subscript 2 on iron for its charge; the charge comes from balancing three sulfate ions.
Three sulfate ions (2− each) total 6−, shared by two iron ions, so each iron is 3+. The name is iron(III) sulfate.
The trap: Iron(II) sulfate mistakes the subscript 2 on iron for its charge; the charge comes from balancing three sulfate ions.
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Question 4 · Reaction Types
In the equilibrium HCO₃⁻ + H₂O ⇌ H₂CO₃ + OH⁻, which species is the conjugate acid of HCO₃⁻?
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Answer: A. H₂CO₃. HCO₃⁻ acts as a base here by accepting a proton from water. The conjugate acid is the base plus one proton, which is H₂CO₃.
The trap: CO₃²⁻ is the conjugate base of HCO₃⁻, one proton fewer, not its conjugate acid.
HCO₃⁻ acts as a base here by accepting a proton from water. The conjugate acid is the base plus one proton, which is H₂CO₃.
The trap: CO₃²⁻ is the conjugate base of HCO₃⁻, one proton fewer, not its conjugate acid.
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Question 5 · Equations and Stoichiometry
N₂ + 3H₂ → 2NH₃. A mixture of 28.0 g N₂ and 8.00 g H₂ reacts until one reactant runs out. What is the maximum mass of NH₃? (Molar masses: N₂ 28.0, H₂ 2.02, NH₃ 17.0 g/mol.)
Show the answer
Answer: E. 34.0 g. 28.0 g N₂ is 1.00 mol and would need 3.00 mol H₂. The 8.00 g of H₂ is 3.96 mol, so N₂ is limiting. 1.00 mol N₂ gives 2.00 mol NH₃ × 17.0 g/mol = 34.0 g.
The trap: 44.9 g is based on hydrogen as if it were limiting. It would need more nitrogen than the mixture contains.
28.0 g N₂ is 1.00 mol and would need 3.00 mol H₂. The 8.00 g of H₂ is 3.96 mol, so N₂ is limiting. 1.00 mol N₂ gives 2.00 mol NH₃ × 17.0 g/mol = 34.0 g.
The trap: 44.9 g is based on hydrogen as if it were limiting. It would need more nitrogen than the mixture contains.
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Question 6 · Experimental Chemistry
A 0.5106 g sample of potassium hydrogen phthalate (KHP, KHC₈H₄O₄, 204.22 g/mol) is titrated with NaOH, and 25.00 mL of base reaches the endpoint. What is the molarity of the NaOH?
Show the answer
Answer: B. 0.1000 M. Moles KHP = 0.5106 g / 204.22 g/mol = 0.002500 mol. KHP has one acidic hydrogen, so it reacts 1:1 with NaOH. M = 0.002500 mol / 0.02500 L = 0.1000 M.
The trap: 0.2000 M assumes each KHP gives two H⁺, but only the one carboxylic acid hydrogen in KHC₈H₄O₄ is acidic.
Moles KHP = 0.5106 g / 204.22 g/mol = 0.002500 mol. KHP has one acidic hydrogen, so it reacts 1:1 with NaOH. M = 0.002500 mol / 0.02500 L = 0.1000 M.
The trap: 0.2000 M assumes each KHP gives two H⁺, but only the one carboxylic acid hydrogen in KHC₈H₄O₄ is acidic.
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Question 7 · Equilibrium
For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the forward reaction is exothermic. Which change shifts the equilibrium toward more SO₃?
Show the answer
Answer: D. Decreasing the container volume. The reactants are 3 mol of gas and the product is 2 mol. A smaller volume raises the pressure, and the system shifts toward fewer gas moles, which means more SO₃.
The trap: Raising the temperature favors the endothermic direction. Here that is the reverse reaction, so SO₃ would decrease.
The reactants are 3 mol of gas and the product is 2 mol. A smaller volume raises the pressure, and the system shifts toward fewer gas moles, which means more SO₃.
The trap: Raising the temperature favors the endothermic direction. Here that is the reverse reaction, so SO₃ would decrease.
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Question 8 · Thermodynamics
N₂(g) + O₂(g) → 2 NO(g) has ΔH = +180.5 kJ. 2 NO(g) + O₂(g) → 2 NO₂(g) has ΔH = −114.1 kJ. What is ΔH for N₂(g) + 2 O₂(g) → 2 NO₂(g)?
Show the answer
Answer: E. +66.4 kJ. Adding the two steps cancels 2 NO and gives the overall equation. ΔH = +180.5 + (−114.1) = +66.4 kJ.
The trap: +294.6 kJ comes from adding the magnitudes and dropping the negative sign of the second step.
Adding the two steps cancels 2 NO and gives the overall equation. ΔH = +180.5 + (−114.1) = +66.4 kJ.
The trap: +294.6 kJ comes from adding the magnitudes and dropping the negative sign of the second step.
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Question 9 · Kinetics
Initial rates were measured for A + B → products. Run 1: [A] = 0.10 M, [B] = 0.10 M, rate = 2.0 × 10⁻³ M/s. Run 2: [A] = 0.20 M, [B] = 0.10 M, rate = 8.0 × 10⁻³ M/s. Run 3: [A] = 0.20 M, [B] = 0.30 M, rate = 2.4 × 10⁻² M/s. What is the rate law?
Show the answer
Answer: C. rate = k[A]²[B]. From run 1 to 2, [A] doubles and the rate quadruples, so second order in A. From run 2 to 3, [B] triples and the rate triples, so first order in B.
The trap: rate = k[A][B] would fit only if doubling [A] doubled the rate; the data show it quadrupled.
From run 1 to 2, [A] doubles and the rate quadruples, so second order in A. From run 2 to 3, [B] triples and the rate triples, so first order in B.
The trap: rate = k[A][B] would fit only if doubling [A] doubled the rate; the data show it quadrupled.
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Question 10 · Structure of Matter
Element X has two stable isotopes: one of mass 10.0 u with 20.0% abundance and one of mass 11.0 u with 80.0% abundance. What is the average atomic mass of X?
Show the answer
Answer: A. 10.8 u. (10.0 × 0.200) + (11.0 × 0.800) = 2.00 + 8.80 = 10.8 u. The heavier isotope is more abundant, so the average lies closer to 11.0.
The trap: 10.5 u is the simple average of 10.0 and 11.0. It ignores the abundances, which must weight each mass.
(10.0 × 0.200) + (11.0 × 0.800) = 2.00 + 8.80 = 10.8 u. The heavier isotope is more abundant, so the average lies closer to 11.0.
The trap: 10.5 u is the simple average of 10.0 and 11.0. It ignores the abundances, which must weight each mass.
The CLEP Chemistry exam at a glance
The CLEP Chemistry exam is a 90-minute, computer-based test from the College Board with about 75 multiple-choice questions. A score of 50 on the 20–80 scale earns an ACE credit recommendation of 6 semester hours, though each college sets its own policy. The exam covers 9 content areas; the heaviest is Structure of Matter at 20%.
- Questions
- About 75multiple choice
- Time
- 90 minabout 72 seconds a question
- Passing score
- 50on a 20–80 scale
- ACE credit
- 6semester hours
- Exam fee
- $97paid to College Board
- Calculator
- Yeson-screen calculator provided
Source: official CLEP Chemistry exam page.
What the exam covers
9 areas, heaviest first. The guide's chapters follow this order.
- Structure of Matter20%
- States of Matter19%
- Descriptive Chemistry14%
- Reaction Types12%
- Equations and Stoichiometry10%
- Experimental Chemistry9%
- Equilibrium7%
- Thermodynamics5%
- Kinetics4%
How it works
- 1Buy once, $5Secure checkout. No subscription needed.
- 2Download the PDFFrom your TransferCredit.org account, right away and any time later.
- 3Study the heaviest areas firstFollow the 2-, 4- or 8-week plan, then read the cram sheet the night before.
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