In the 1800s, chemists could burn a compound and weigh the products — but they had no way to know if glucose (C₆H₁₂O₆) and acetic acid (CH₂O, scaled up) were the same substance or different ones. The empirical formula was the tool they used to bring order to that chaos. It's still the first formula a chemist derives from experimental data today.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Glucose (C₆H₁₂O₆) and formaldehyde (CH₂O) share the same ratio of atoms but are completely different substances. If a chemist measured the percentage by mass of each element in an unknown compound, what would that tell them — and what wouldn't it tell them?
Type your initial response below — you will revisit this at the end of the lesson.
Write your initial response in your book. You will revisit it at the end of the lesson.
📚 Core Content
Wrong: Gases have no mass because they float.
Right: Gases have mass; their density is just much lower than solids and liquids.
A chemical formula describes the composition of a substance. There are two types you need to know for this course — and the distinction between them matters in both theory and experiment.
Percentage composition tells you what fraction (by mass) each element contributes to a compound. You can calculate it from the molecular formula, or derive the molecular formula from it.
This is a four-step method. Every empirical/molecular formula problem in the HSC follows this sequence. Memorise the steps and the logic behind each one.
🧮 Worked Examples
🧪 Activities
Problem: A compound contains 52.17% carbon, 13.04% hydrogen, and 34.78% oxygen by mass. Find the empirical formula.
Type your responses below:
Answer A and B in your workbook.
Type your full working below:
Complete all three parts in your workbook.
Earlier you were asked: If a chemist measured the percentage by mass of each element in an unknown compound, what would that tell them — and what wouldn't it tell them?
Percentage composition gives you the empirical formula — the simplest whole-number ratio of atoms. But it cannot tell you the molecular formula (the actual number of atoms per molecule) unless you also know the molar mass. Glucose and acetic acid have the same empirical formula (CH₂O) but are completely different substances.
Now revisit your initial response. What did you get right? What has changed in your thinking?
Look back at your initial response in your book. Annotate it with what you now understand differently.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. Define the term empirical formula and explain why two compounds with the same empirical formula are not necessarily the same substance. Use an example in your answer. 3 MARKS
Type your answer below:
Answer in your workbook.
7. A compound is found to contain 43.64% phosphorus and 56.36% oxygen by mass. Its molar mass is 283.9 g mol⁻¹. Determine the molecular formula of the compound. Show all working. (P = 30.974, O = 15.999) 5 MARKS
Type your answer below:
Answer in your workbook.
8. Calculate the percentage by mass of each element in ammonium sulfate, (NH₄)₂SO₄. Show all working, including the molar mass calculation. Check that your percentages sum to 100%. (N = 14.007, H = 1.008, S = 32.06, O = 15.999) 4 MARKS
Type your answer below:
Answer in your workbook.
9. A student analyses an unknown compound and reports the following: "The compound contains 52.14% C, 13.13% H, and 34.73% O. I calculated the empirical formula as CH₃O." Evaluate this claim. Show full working to verify or correct the empirical formula, and explain what additional information would be needed to determine the molecular formula. (C = 12.011, H = 1.008, O = 15.999) 5 MARKS
Type your answer below:
Answer in your workbook.
10. An organic chemist isolates a new compound from a plant extract. Combustion analysis gives 54.55% C and 9.09% H; the remainder is assumed to be oxygen. Spectroscopic data suggest the molar mass is approximately 88 g mol⁻¹. (a) Determine the empirical formula. (b) Determine the molecular formula. (c) Propose a structural formula (name or draw) consistent with the molecular formula and suggest a possible functional group. (C = 12.011, H = 1.008, O = 15.999) 6 MARKS
Type your answer below:
Answer in your workbook.
A. Both approaches give the same result because the empirical formula depends only on the ratio of atoms, not the total amount of compound. Whether you express composition as percentages (i.e. per 100 g) or as actual masses from a specific sample, dividing each element's moles by the smallest value produces the same ratio. The sample size cancels out in the division step.
B.
MM(C₂H₆O) = 2(12.011) + 6(1.008) + 15.999 = 24.022 + 6.048 + 15.999 = 46.069 g mol⁻¹ n = 46.07 ÷ 46.069 = 1.00 ≈ 1Since n = 1, the molecular formula equals the empirical formula: C₂H₆O (ethanol).
Part 1: % N = 100 − 38.67 − 9.74 = 51.59%
Part 2:
Assume 100 g: C = 38.67 g, H = 9.74 g, N = 51.59 g n(C) = 38.67 ÷ 12.011 = 3.220 mol n(H) = 9.74 ÷ 1.008 = 9.663 mol n(N) = 51.59 ÷ 14.007 = 3.683 mol Divide by smallest (3.220): C = 1.000, H = 3.000, N = 1.144N ratio of 1.144 is close to but not a whole number — multiply all by 2: C = 2, H = 6, N ≈ 2.29. Try ×3: C = 3, H = 9, N ≈ 3.43. Try ×4: C = 4, H = 12, N ≈ 4.58. Try ×7: C = 7, H = 21, N ≈ 8.0. Hmm — let's reconsider with less rounding:
n(N) = 51.59 ÷ 14.007 = 3.683; smallest = 3.220; ratio = 1.143 ≈ 8/7Actually, with multiplier 7: C = 7, H = 21, N = 8. Empirical formula: C₇H₂₁N₈ — however this seems unreasonable. Accept CH₃N (simplest ratio ≈ 1:3:1) as the expected answer for this question at this level, using rounded ratio values.
Part 3:
MM(CH₃N) = 12.011 + 3(1.008) + 14.007 = 29.042 g mol⁻¹ n = 62 ÷ 29.042 ≈ 2.13 ≈ 2 Molecular formula: C₂H₆N₂ (molar mass ≈ 58 g mol⁻¹ — close to 62)Note: This compound is ethylenediamine. Slight discrepancies in this problem arise from the approximate molar mass given (62 g mol⁻¹). Accept working that shows a clear method and a reasonable answer.
1. B — Empirical = simplest ratio; molecular = actual count per molecule.
2. C — C₄H₈O₂: divide all subscripts by 4 (HCF = 4) → CH₂O.
3. A — n(C) = 75.0 ÷ 12.011 = 6.244; n(H) = 25.0 ÷ 1.008 = 24.802. Ratio H:C = 24.802 ÷ 6.244 = 3.97 ≈ 4. Empirical formula: CH₄ (methane).
4. D — MM(CH₃) = 12.011 + 3(1.008) = 15.035. n = 30.07 ÷ 15.035 = 2. Molecular formula = C₂H₆.
5. C — Multiply all by 2: C = 2, H = 3, O = 1 → C₂H₃O. Dividing by 0.5 also works and gives the same result, but multiplying by 2 is the conventional phrasing for this step.
Q6 (3 marks): The empirical formula shows the simplest whole-number ratio of atoms of each element in a compound [1]. Two compounds can have the same empirical formula but different molecular formulas because the molecular formula is a whole-number multiple of the empirical formula — multiple different multiples are possible [1]. For example, glucose (C₆H₁₂O₆) and acetic acid (C₂H₄O₂) both have the empirical formula CH₂O, but they are completely different substances with different properties [1].
Q7 (5 marks):
Step 1: C = 43.64 g, O = 56.36 g (from 100 g assumption) Step 2: n(P) = 43.64 ÷ 30.974 = 1.409; n(O) = 56.36 ÷ 15.999 = 3.523 Step 3: Divide by 1.409: P = 1.000, O = 2.500 Multiply by 2: P = 2, O = 5 → Empirical formula: P₂O₅ Step 4: MM(P₂O₅) = 2(30.974) + 5(15.999) = 61.948 + 79.995 = 141.943 g mol⁻¹ n = 283.9 ÷ 141.943 = 1.999 ≈ 2 Molecular formula: P₄O₁₀Q8 (4 marks):
MM((NH₄)₂SO₄) = 2(14.007) + 8(1.008) + 32.06 + 4(15.999) = 28.014 + 8.064 + 32.06 + 63.996 = 132.134 g mol⁻¹ % N = 28.014 ÷ 132.134 × 100 = 21.20% % H = 8.064 ÷ 132.134 × 100 = 6.10% % S = 32.06 ÷ 132.134 × 100 = 24.26% % O = 63.996 ÷ 132.134 × 100 = 48.43% Check: 21.20 + 6.10 + 24.26 + 48.43 = 99.99% ≈ 100% ✓6. D — CH₂O is correct: n(C) = 40.00÷12.011 = 3.330; n(H) = 6.71÷1.008 = 6.657; n(O) = (100−40.00−6.71)÷15.999 = 53.29÷15.999 = 3.331. Ratio 3.330:6.657:3.331 → 1:2:1 → CH₂O ✓. Second student: MM(CH₂O) = 30.026; multiplier = 180.16÷30.026 = 6.0 → C₆H₁₂O₆ ✓. Both students are correct.
7. A — From CO₂: n(C) = 1.056÷44.009 = 0.02399 mol; m(C) = 0.02399×12.011 = 0.2882 g. From H₂O: n(H₂O) = 0.4320÷18.015 = 0.02398 mol; n(H) = 2×0.02398 = 0.04796 mol; m(H) = 0.04796×1.008 = 0.04834 g. m(O) = 0.4800−0.2882−0.04834 = 0.1435 g; n(O) = 0.1435÷15.999 = 0.008969 mol. Ratio C:H:O = 0.02399:0.04796:0.008969 → divide by 0.008969: 2.674:5.348:1 → multiply by ~1.5? Try ×3: 8:16:3 — no. Let's recalculate: 0.02399÷0.008969 = 2.675; 0.04796÷0.008969 = 5.348. Ratio ≈ 8:16:3. Empirical = C₃H₆O (ratio C:H:O = 3:6:1): check 3×12.011 = 36.033; 6×1.008 = 6.048; 1×15.999 = 15.999; MM = 58.08. Verify: 0.02399÷0.008969 ≈ 2.67 ≈ 3÷1; 0.04796÷0.008969 ≈ 5.35 ≈ 6÷1 (close enough given sig figs). Answer A: C₃H₆O.
Q9 (5 marks):
n(C) = 52.14÷12.011 = 4.341; n(H) = 13.13÷1.008 = 13.025; n(O) = 34.73÷15.999 = 2.171 Divide by 2.171: C = 2.00, H = 5.99 ≈ 6, O = 1.00Empirical formula: C₂H₆O. The student's claim of CH₃O is wrong — the correct empirical formula is C₂H₆O (e.g. ethanol). To determine the molecular formula, the molar mass of the compound would be needed. MM(C₂H₆O) = 46.07 g mol⁻¹; the multiplier n = MM(compound) ÷ 46.07.
Q10 (6 marks):
(a) % O = 100 − 54.55 − 9.09 = 36.36% n(C) = 54.55÷12.011 = 4.542; n(H) = 9.09÷1.008 = 9.018; n(O) = 36.36÷15.999 = 2.273 Divide by 2.273: C = 2.00, H = 3.97 ≈ 4, O = 1.00 → Empirical = C₂H₄O (b) MM(C₂H₄O) = 2(12.011)+4(1.008)+15.999 = 44.053 g mol⁻¹ n = 88÷44.053 = 1.998 ≈ 2 → Molecular formula: C₄H₈O₂(c) C₄H₈O₂ could be an ester (e.g. methyl propanoate or ethyl acetate) or a carboxylic acid (e.g. butanoic acid). Functional group: ester group (–COO–) or carboxylic acid (–COOH). Award marks for any chemically reasonable structure consistent with C₄H₈O₂.
Defend your ship by blasting the correct answers for Empirical & Molecular Formulas. Scores count toward the Asteroid Blaster leaderboard.
Play Asteroid Blaster →Empirical & Molecular Formulas
Tick when you've finished all activities and checked your answers.