General chemistry is where DAT takers lose points they should have kept. Most remember the concepts, then burn ninety seconds dividing 0.0482 by 3.7 on a noteboard. The chemistry was never the problem. The execution was.
This post covers both halves: which topics in the official outline deserve most of your time, and how to work problems fast, by hand, with numbers that are friendlier than they look. Every sample problem below is original to ChairsideSource and has been checked twice. None are taken from actual DAT items.
Key takeaways
- General chemistry is 30 of the 100 items in the Survey of the Natural Sciences, which you get 90 minutes to finish. That averages out to about 54 seconds per question across the whole section.
- The on-screen calculator is only available in Quantitative Reasoning. For chemistry you have a periodic table, two noteboards, and your own arithmetic.
- Stoichiometry, gases, acids and bases, equilibrium, and thermodynamics carry the most reusable skills. Nail those first.
- DAT chemistry numbers are usually chosen to cancel cleanly. If your arithmetic is getting ugly, you probably set the problem up wrong.
- Three log shortcuts and scientific notation cover most "calculation" questions.
How general chemistry fits on the DAT
According to the ADA's current DAT Candidate Guide, the Survey of the Natural Sciences is the first scored section of the day and contains 100 items: 40 biology, 30 general chemistry, and 30 organic chemistry. You get 90 minutes for all of it. General chemistry receives its own scaled score on the 200 to 600 scale the ADA adopted on March 1, 2025, and it also feeds into your Survey of the Natural Sciences score and your Academic Average. If you want the full picture of how those scores are built and read, see What Is a Good DAT Score?
| Piece of the Natural Sciences section | Items | What that means for pacing |
|---|---|---|
| Biology | 40 | Mostly recall and reasoning, usually the fastest per item |
| General Chemistry | 30 | Mix of concept and calculation; calculations are the time sink |
| Organic Chemistry | 30 | Mechanisms, synthesis, structure; the ADA updated this outline in 2026 |
| Total | 100 items in 90 minutes | About 54 seconds per item on average |
The practical consequence: biology questions you know should take well under a minute, which banks time for chemistry calculations. For more on pacing, read how to take a full-length DAT practice test the right way.
No calculator in the sciences. The Candidate Guide describes the on-screen calculator as a tool for the Quantitative Reasoning section. Do all of your chemistry practice by hand from day one. Students who review with a phone calculator next to them are training for a test they will not take.
The official topic list, ranked by where your time pays off
The Candidate Guide lists thirteen general chemistry content areas. The ADA does not publish how many questions come from each one, so the priority column below is our editorial judgment, based on how many other topics depend on each skill and how often a single weakness costs you on multiple question types. It is not ADA data.
| Content area (from the Candidate Guide) | What to be able to do | Our priority |
|---|---|---|
| Stoichiometry and general concepts | Moles, molar mass, limiting reagent, percent composition, empirical formulas, density, balancing | Highest |
| Acids and bases | pH and pOH, strong vs. weak, Ka and Kb, conjugate pairs, buffers, titration curves | Highest |
| Chemical equilibria | Writing K, Q vs. K, Le Chatelier, Ksp and molar solubility | Highest |
| Gases | Ideal gas law, combined gas law, Dalton's law, kinetic molecular theory | High |
| Thermodynamics and thermochemistry | Signs of ΔH, ΔS, ΔG; Hess's law; heat capacity (q = mcΔT) | High |
| Solutions | Molarity, molality, dilution, colligative properties, solubility rules | High |
| Chemical kinetics | Rate laws from data, reaction order, activation energy, catalysts | High |
| Oxidation-reduction | Oxidation numbers, balancing, galvanic vs. electrolytic cells, E°cell | Medium to high |
| Atomic and molecular structure | Electron configurations, quantum numbers, Lewis structures, VSEPR shapes | Medium to high |
| Periodic properties | Radius, ionization energy, electronegativity, electron affinity trends | Medium |
| Liquids and solids | Intermolecular forces, phase diagrams, vapor pressure, boiling point comparisons | Medium |
| Laboratory | Glassware, significant figures, error analysis, safety | Medium |
| Nuclear reactions | Alpha, beta, gamma decay; balancing nuclear equations; half-life | Lower, but quick points |
Nuclear chemistry is ranked lower only because it is narrow. It is also an easy place to collect points, so do not skip it. Just do not start there.
Stoichiometry: the skill everything else sits on
Gases, solutions, equilibrium, and thermochemistry all start by converting something into moles. Spend your first review days here even if it feels basic.
Worked problem 1: limiting reagent
Question (original, hypothetical). Methane burns according to CH₄ + 2 O₂ → CO₂ + 2 H₂O. If 8.0 g of CH₄ reacts with 16.0 g of O₂, what mass of water can form? (C = 12, H = 1, O = 16)
(A) 4.5 g (B) 9.0 g (C) 12 g (D) 18 g (E) 36 g
Solution. Convert both reactants to moles. CH₄ is 16 g/mol, so 8.0 g is 0.50 mol. O₂ is 32 g/mol, so 16.0 g is 0.50 mol. The equation needs 2 mol O₂ for every 1 mol CH₄, so 0.50 mol CH₄ would need 1.0 mol O₂. You only have 0.50 mol, which makes O₂ the limiting reagent. Each mole of O₂ produces one mole of water (2 : 2), so you get 0.50 mol H₂O, or 0.50 × 18 = 9.0 g. Answer: B.
Choice D is what you get if you assume CH₄ is limiting. Wrong answers are built from exactly these predictable mistakes, so identify the limiting reagent before multiplying anything.
Empirical formulas in thirty seconds
Given mass percentages, assume 100 g. Example (hypothetical): 40.0% C, 6.7% H, 53.3% O gives about 3.33 mol C, 6.7 mol H, and 3.33 mol O, a 1 : 2 : 1 ratio, so the empirical formula is CH₂O. Divide any given molar mass by 30 g/mol to get the molecular formula multiplier.
Gases: one equation, several shortcuts
You need PV = nRT, the combined gas law, and Dalton's law. Also know what kinetic molecular theory assumes (negligible particle volume, no attractions) and therefore when real gases deviate: high pressure and low temperature.
Worked problem 2: combined gas law
Question (original, hypothetical). A gas occupies 6.0 L at 27 °C and 1.0 atm. What is its volume at 127 °C and 2.0 atm?
(A) 1.5 L (B) 3.0 L (C) 4.0 L (D) 9.0 L (E) 12 L
Solution. Convert to kelvin first: 27 °C is 300 K and 127 °C is 400 K. Then reason with ratios instead of plugging into a formula. Doubling the pressure halves the volume: 6.0 L becomes 3.0 L. Raising temperature from 300 K to 400 K multiplies volume by 4/3: 3.0 L becomes 4.0 L. Answer: C.
Ratio thinking beats plugging in. Ask what each change does on its own (up or down, by what factor) and multiply the factors. It is faster and lets you sanity-check direction first. Forgetting kelvin is the most common gas-law error.
Solutions and colligative properties
Colligative properties (boiling point elevation, freezing point depression, vapor pressure lowering, osmotic pressure) depend on the number of dissolved particles. The van't Hoff factor, i, is particles per formula unit: about 1 for glucose, 2 for NaCl, 3 for CaCl₂. Ranking questions only need i × concentration.
Worked problem 3: ranking freezing points
Question (original, hypothetical). Which 0.10 m aqueous solution has the lowest freezing point? (A) glucose (B) NaCl (C) KNO₃ (D) CaCl₂ (E) sucrose
Solution. Lowest freezing point means the most particles. Glucose and sucrose give i = 1, NaCl and KNO₃ give i = 2, CaCl₂ gives i = 3. Answer: D. With Kf for water of 1.86 °C/m, ΔTf = 3 × 1.86 × 0.10, about 0.56 °C, so it freezes near -0.56 °C.
Molarity is moles per liter of solution; molality is moles per kilogram of solvent, and colligative formulas use molality. For dilution, M₁V₁ = M₂V₂: 25 mL of 2.0 M diluted to 250 mL is 0.20 M.
Acids, bases, and buffers
Know the strong acids (HCl, HBr, HI, HNO₃, H₂SO₄, HClO₄) and strong bases (Group 1 and heavier Group 2 hydroxides). Know that pH + pOH = 14 at 25 °C, Ka × Kb = Kw for a conjugate pair, and a stronger acid has a weaker conjugate base.
The three log shortcuts you need
- If [H⁺] = 1 × 10⁻ⁿ, then pH = n. A concentration of 1 × 10⁻⁴ M gives pH 4.
- If [H⁺] = a × 10⁻ⁿ with a between 1 and 10, the pH is between n - 1 and n. Specifically it is n minus log a. With log 2 ≈ 0.30 and log 3 ≈ 0.48, a concentration of 3 × 10⁻⁴ M gives a pH of about 4 - 0.48 = 3.52. Often you only need to know it falls between 3 and 4 to pick the answer.
- Each tenfold change in [H⁺] moves pH by exactly 1. This is also the key to buffer problems.
Worked problem 4: buffer pH
Question (original, hypothetical). A buffer contains 0.20 M acetic acid (pKa = 4.74) and 0.020 M sodium acetate. What is its pH?
(A) 2.74 (B) 3.74 (C) 4.74 (D) 5.74 (E) 6.74
Solution. Use Henderson-Hasselbalch: pH = pKa + log([A⁻]/[HA]). The ratio of base to acid is 0.020 / 0.20 = 0.10, and log 0.10 = -1. So pH = 4.74 - 1 = 3.74. Answer: B. Sanity check: more acid than base means pH below pKa, which eliminates C, D, and E.
For a weak acid alone, [H⁺] ≈ √(Ka × C). Example (hypothetical): 0.010 M acid with Ka = 1.0 × 10⁻⁶ gives [H⁺] ≈ 1.0 × 10⁻⁴, so pH 4. For titration curves, know that pH equals pKa at the half-equivalence point and that a weak acid titrated with strong base has an equivalence point above 7.
Equilibrium and solubility
Write K as products over reactants raised to their coefficients, leaving out pure solids and liquids. If Q is smaller than K, the reaction proceeds forward. Only temperature changes the value of K; adding reactant or changing pressure shifts position but leaves K alone.
Worked problem 5: molar solubility from Ksp
Question (original, hypothetical). A sparingly soluble salt MX₂ has Ksp = 4.0 × 10⁻¹². What is its molar solubility in pure water?
(A) 1.0 × 10⁻⁶ M (B) 2.0 × 10⁻⁶ M (C) 1.0 × 10⁻⁴ M (D) 2.0 × 10⁻⁴ M (E) 1.6 × 10⁻³ M
Solution. MX₂ dissolves into one M²⁺ and two X⁻. If s mol/L dissolves, [M²⁺] = s and [X⁻] = 2s. So Ksp = s × (2s)² = 4s³. Set 4s³ = 4.0 × 10⁻¹², so s³ = 1.0 × 10⁻¹² and s = 1.0 × 10⁻⁴ M. Answer: C. Choice A is what you get if you treat the salt as 1 : 1 and take a square root.
Also know the common ion effect: a shared ion lowers solubility.
Thermodynamics and kinetics
Thermodynamics tells you whether a reaction is favorable (ΔG, ΔH, ΔS, K). Kinetics tells you how fast it goes (rate laws, activation energy). A catalyst lowers activation energy and speeds both directions but changes neither ΔG nor K.
Worked problem 6: when does a reaction become spontaneous?
Question (original, hypothetical). A reaction has ΔH = +40 kJ/mol and ΔS = +100 J/(mol·K). Above what temperature is it spontaneous, assuming ΔH and ΔS stay constant?
(A) 0.4 K (B) 40 K (C) 250 K (D) 400 K (E) 4,000 K
Solution. ΔG = ΔH - TΔS crosses zero at T = ΔH / ΔS. Convert first: 40 kJ is 40,000 J, so T = 40,000 / 100 = 400 K. Answer: D. Choice A is the unit trap.
| ΔH | ΔS | Spontaneous when |
|---|---|---|
| Negative | Positive | Always |
| Positive | Negative | Never |
| Negative | Negative | At low temperature |
| Positive | Positive | At high temperature |
Worked problem 7: rate law from experimental data
Question (original, hypothetical). For A + B → products, three trials give the following initial rates. What is the rate law and the value of k?
| Trial | [A] (M) | [B] (M) | Initial rate (M/s) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 8.0 × 10⁻³ |
| 3 | 0.10 | 0.20 | 2.0 × 10⁻³ |
Solution. Compare trials 1 and 2: [A] doubles, [B] is constant, and the rate quadruples, so the reaction is second order in A (2² = 4). Compare trials 1 and 3: [B] doubles and the rate does not change, so it is zero order in B. The rate law is rate = k[A]². Using trial 1: 2.0 × 10⁻³ = k × (0.10)² = k × 0.010, so k = 0.20 M⁻¹s⁻¹. For an overall order of n, k has units of M^(1 - n) per second.
Redox and electrochemistry
Start with oxidation numbers. Example (hypothetical): in K₂Cr₂O₇, 2(+1) + 2x + 7(-2) = 0, so chromium is +6. Oxidation (loss of electrons) happens at the anode and reduction at the cathode, in both galvanic and electrolytic cells.
Worked problem 8: standard cell potential
Question (original, hypothetical). Given the standard reduction potentials Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V and Zn²⁺ + 2e⁻ → Zn, E° = -0.76 V, what is E°cell for a galvanic cell built from these half-cells, and which metal is the anode?
Solution. The more positive reduction potential runs as the reduction, so copper is reduced at the cathode and zinc is oxidized at the anode. E°cell = 0.34 - (-0.76) = +1.10 V. Never multiply E° values by coefficients; potential is intensive.
A spontaneous reaction has negative ΔG, positive E°cell, and K greater than 1. Expect to be given one and asked about another.
Atomic structure, periodic trends, and bonding
These are fast recall questions that bank time. Know electron configurations through the first transition series (including the chromium and copper exceptions), the four quantum numbers, and VSEPR shapes up to six electron domains.
| Trend | Across a period (left to right) | Down a group |
|---|---|---|
| Atomic radius | Decreases | Increases |
| First ionization energy | Generally increases | Decreases |
| Electronegativity | Increases | Decreases |
| Metallic character | Decreases | Increases |
Watch the ionization energy dips from group 2 to 13 and from group 15 to 16. Cations are smaller than their neutral atoms and anions are larger. The on-screen periodic table gives you masses, not trends.
Nuclear chemistry and lab: the quick points
Alpha decay lowers mass number by 4 and atomic number by 2. Beta-minus raises atomic number by 1. Positron emission and electron capture lower it by 1. Gamma changes neither.
Worked problem 9 (original, hypothetical). A radioactive isotope has a half-life of 5 days. How much of a 64 mg sample remains after 20 days? Twenty days is four half-lives, so the sample halves four times: 64, 32, 16, 8, 4. Answer: 4 mg.
Lab questions test judgment: precise glassware (volumetric pipette or buret, not a beaker), significant figures, accuracy vs. precision, and safety such as adding acid to water.
Common mistakes that cost gen chem points
The repeat offenders. Forgetting to convert Celsius to kelvin. Using grams instead of moles in a ratio. Mixing kilojoules and joules in ΔG = ΔH - TΔS. Treating a 1 : 2 salt as 1 : 1 in a Ksp problem. Multiplying E° values by stoichiometric coefficients. Assuming a catalyst changes K. Every one of these is a predictable wrong-answer choice on a well-built multiple-choice question.
Two habits fix most of them: write units on every number on your noteboard, and predict the direction of the answer before calculating. If the result disagrees with the prediction, you just caught an error.
General chemistry readiness checklist
- I can find a limiting reagent and theoretical yield without a calculator.
- I can get an empirical formula from mass percentages in under a minute.
- I solve gas-law changes with ratios and always convert to kelvin.
- I can rank colligative effects using i × concentration.
- I know the six strong acids and can estimate pH from any [H⁺] using log 2 and log 3.
- I can use Henderson-Hasselbalch and know what happens at the half-equivalence point.
- I can set up Ksp for 1 : 1, 1 : 2, and 2 : 1 salts.
- I can find the crossover temperature from ΔH and ΔS with correct units.
- I can derive a rate law and the units of k from a table of trials.
- I can assign oxidation numbers and compute E°cell from two half-reactions.
- I know periodic trends and their main exceptions.
- I can balance alpha, beta, and positron decay equations and count half-lives.
How to fit gen chem into your study plan
Start with our DAT study schedule for the overall phases. Within it, a sensible gen chem order is stoichiometry and solutions, then gases, then acids, bases, and equilibrium as one block (shared math), then thermodynamics and kinetics, then redox, and finally the recall topics.
Pair each block with timed sets, because only timed practice shows whether your arithmetic is fast enough. Our free DAT practice questions let you drill general chemistry without signing up. Sort every miss into one of three buckets: did not know the concept, set it up wrong, or arithmetic slip. Most students find the third bucket is bigger than expected.
Where to go next
For organic chemistry, see the reactions worth memorizing first, and note that the ADA reorganized the organic chemistry outline in 2026. For biology, see DAT biology high-yield topics. For math that carries into QR, see the QR formulas students forget. Near test day, read DAT Test Day: What to Expect.
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