Problem-Solving Playbook

Most quantitative chemistry problems in this course yield to the same five-step frame. The frame is not a trick — it is what experienced problem-solvers do automatically, written down so you can do it on purpose until it becomes automatic for you too.

Use this alongside Extra Practice (problems to run the frame on) and Common Misconceptions (the wrong models that make the frame produce wrong answers).

The five-step frame

1. Read it twice and write down what you have

Read once for the story, once for the numbers. Then, before anything else, write every given quantity with its units and a label, and write what the question is actually asking for (a mass? a concentration? a formula? a temperature change?). Underline the target. Half of all wrong answers are answers to a question that was not asked, or setups missing a number that was in the problem.

2. Translate the chemistry into a relationship

Decide which chemical idea connects what you have to what you want:

• a balanced equation (for anything about how much reacts or forms);
• a defined ratio — molar mass, density, Avogadro's number, a conversion factor, molarity as mol/L;
• a formula — q = mcΔT, PV = nRT, M₁V₁ = M₂V₂, percent composition;
• a conservation statement — mass, atoms, charge, energy.

If the problem is about a reaction, balance the equation now, before you touch the numbers. If it is a mixture of ideas (a gas produced by a reaction, then measured), you will need more than one relationship, chained.

3. Plan the path (units first)

Sketch the route from what you have to what you want, as a chain of conversions, and check that the units cancel to the target unit before you compute anything:

grams A  →  mol A  →  mol B  →  grams B
        (÷ M_A)   (× ratio)   (× M_B)

If the units do not cancel to what the answer should be, the plan is wrong — fix it here, on paper, where it costs nothing. This single habit catches more errors than any other.

4. Compute — carefully, once

Carry extra significant figures through the intermediate steps and round only the final answer, to the number of significant figures the least-precise given allows. Keep the units attached at every line. Do not round 35.453 to 35 in step one and wonder why the answer is off.

5. Check that the answer is reasonable

Ask three questions:

• Units: are they the units the question wanted?
• Size: is the number plausible? A few grams of product from a few grams of reactant is reasonable; 4,000 g is not. A pH of 23 is impossible. A negative mass is impossible. A percent yield over 100% is an error.
• Direction: if you diluted a solution, is the new concentration lower? If you added heat, did the temperature go up? A sign or a direction that contradicts the chemistry means a step is inverted.

If step 5 fails, the fastest fix is usually to recheck step 3 (the plan) and the setup of step 4, not to redo the arithmetic.

The frame on each problem type

Stoichiometry (mass to mass)

What mass of CO₂ forms when 10.0 g of CH₄ burns completely?

1. Have: 10.0 g CH₄ (three sig figs). Want: grams CO₂.
2. Relationship: the balanced equation. CH₄ + 2O₂ → CO₂ + 2H₂O.
3. Path: g CH₄ → mol CH₄ (÷ 16.04) → mol CO₂ (× 1/1) → g CO₂ (× 44.01). Units cancel to grams.
4. 10.0 / 16.04 = 0.6234 mol → 0.6234 mol CO₂ → × 44.01 = 27.4 g.
5. Units: grams ✓. Size: ~27 g of CO₂ from 10 g of fuel — plausible (CO₂ is heavier). Direction: n/a. 27.4 g.

Limiting reactant

5.40 g Al and 12.0 g Cl₂ react. What mass of AlCl₃ forms?

1. Have: 5.40 g Al, 12.0 g Cl₂. Want: grams AlCl₃.
2. Relationship: 2Al + 3Cl₂ → 2AlCl₃.
3. Path: convert each reactant to moles, divide each by its coefficient, the smallest quotient is limiting; then do stoichiometry from that one.
4. Al: 5.40/26.98 = 0.2001 mol, ÷ 2 = 0.1001. Cl₂: 12.0/70.90 = 0.1693 mol, ÷ 3 = 0.0564. Cl₂ is limiting. AlCl₃: 0.1693 × 2/3 = 0.1129 mol × 133.33 = 15.0 g.
5. Units: grams ✓. Size: less than the total reactant mass ✓. 15.0 g. (A common wrong turn: assuming Al is limiting because there are fewer grams — the coefficient-adjusted comparison in step 3 is the point.)

Solution dilution

Dilute 25.0 mL of 6.00 M HCl to 500.0 mL. New concentration?

1. Have: M₁ = 6.00 M, V₁ = 25.0 mL, V₂ = 500.0 mL. Want: M₂.
2. Relationship: dilution conserves moles of solute → M₁V₁ = M₂V₂.
3. Path: solve for M₂ = M₁V₁ / V₂. (Volumes in the same unit cancel.)
4. M₂ = (6.00)(25.0) / 500.0 = 0.300 M.
5. Units: molarity ✓. Direction: you added water, so the concentration should be lower — 0.300 < 6.00 ✓. 0.300 M.

Calorimetry (specific heat)

A 45.0 g metal at 100.0 °C is dropped into 100.0 g of water at 23.0 °C; the final temperature is 26.8 °C. Specific heat of the metal?

1. Have: metal 45.0 g, 100.0 → 26.8 °C. Water 100.0 g, 23.0 → 26.8 °C, c = 4.184 J/g·°C. Want: c_metal.
2. Relationship: energy conservation → heat lost by metal = heat gained by water, −q_metal = q_water, with q = mcΔT.
3. Path: compute q_water, set q_metal equal in magnitude, solve for c_metal = |q| / (m ΔT)_metal.
4. q_water = (100.0)(4.184)(3.8) = 1590 J. c_metal = 1590 / (45.0 × 73.2) = 0.483 J/g·°C.
5. Units: J/g·°C ✓. Size: less than water's 4.184 — true of nearly every metal ✓. 0.483 J/g·°C.

Empirical formula from percent composition

40.00% C, 6.71% H, 53.29% O; molar mass ≈ 180. Molecular formula?

1. Have: mass percents, molar mass ≈ 180. Want: the molecular formula.
2. Relationship: the atom ratio is the mole ratio; molecular = whole-number multiple of empirical.
3. Path: assume 100 g → grams of each element → moles of each (÷ molar mass) → divide all by the smallest → clear to whole numbers → get empirical mass → 180 / empirical mass = the multiplier.
4. C 40.00/12.01 = 3.331, H 6.71/1.008 = 6.657, O 53.29/16.00 = 3.331. Divide by 3.331: 1 : 2 : 1 → empirical CH₂O (30.03 g/mol). 180 / 30.03 ≈ 6. Molecular formula C₆H₁₂O₆.
5. Size check: the multiplier came out to a clean whole number (≈ 6), which is what should happen — a multiplier of 5.7 means an arithmetic slip. C₆H₁₂O₆.

If you are stuck

Work down this list:

• You cannot start. Do step 1 properly — physically write every given with units and box the target. Often the path becomes obvious once it is all on paper.
• You do not know which formula or relationship. Ask what kind of quantity the answer is (mass, moles, concentration, energy, temperature change) and what kind you were given. The relationship is whatever connects those two. Reaction word ("reacts", "forms", "produced", "burns") → you need a balanced equation.
• The units will not cancel. You have an inverted conversion factor (multiplied where you should divide) or a missing step. Rebuild the path in step 3, writing each factor as a fraction with units.
• You get a number but it seems wrong. Run step 5. A pH over 14, a yield over 100%, a negative mass, a diluted solution that got more concentrated — each points to a specific inverted step.
• You are stuck on the algebra, not the chemistry. Solve the formula for the unknown symbolically first (c = q / (mΔT)), then substitute numbers. The Study Toolkit has the rearrangement templates.
• It still will not come. That is a good, specific question for office hours, a tutor, or Supplemental Instruction: "I've set it up like this and I'm stuck here — can you watch me try one?" Bring your step-1 and step-3 work; it shows exactly where the gap is.