Chemistry • Year 11 • Module 2 • Lesson 3
Empirical and Molecular Formulas
Build HSC Band 5–6 extended-response technique on evaluating experimental formula claims, designing combustion analysis investigations, and reasoning from combustion data.
1. Evaluate a student’s empirical formula claim (Band 5–6)
8 marks Band 5–6
Scenario. A Year 11 student analyses an unknown compound by combustion analysis and reports: “The compound contains 52.14% C, 13.13% H, and 34.73% O. I applied the four-step method and got the empirical formula CH2O. I concluded that the compound must be glucose because glucose has the formula C6H12O6, which simplifies to CH2O.”
The student was given additional information: the compound’s molar mass is approximately 46 g mol−1. (C = 12.011, H = 1.008, O = 15.999)
Q1. Evaluate the student’s claim. In your response you must:
- Apply the four-step method to verify or correct the student’s empirical formula, showing full working.
- Use the given molar mass to determine the multiplier n and the molecular formula of the compound.
- Assess whether the student’s conclusion (that the compound is glucose) is correct; justify your answer with reference to the molecular formula and molar mass.
- Explain the conceptual error in the student’s reasoning — what does sharing an empirical formula actually tell us about two compounds?
- State one additional piece of information (beyond molar mass) that would help confirm the identity of the unknown compound.
2. Experimental design — determining the empirical formula of a candle wax (Band 5–6)
7 marks Band 5–6
Research question. A chemistry teacher wants students to determine the empirical formula of a paraffin wax candle, which is believed to contain only carbon and hydrogen. The combustion products (CO2 and H2O) can be collected and weighed using standard Year 11 laboratory equipment.
Constraints: You have access to a balance (0.01 g precision), a burning candle, anhydrous CaCl2 (to absorb H2O), NaOH solution (to absorb CO2), and glass tubing. The investigation must be completed in a single lesson period.
Q2. Design the investigation and present it in the format below.
- State a hypothesis in the form: “If paraffin wax has the formula CxHy, then combustion of a known mass will produce a predictable ratio of CO2 mass to H2O mass.” Identify the independent and dependent variables.
- Describe the procedure in at least four numbered steps, including how the CO2 and H2O masses will be determined from the absorber mass changes.
- Explain how you would calculate n(C) and n(H) from the collected masses of CO2 (M = 44.009 g mol−1) and H2O (M = 18.015 g mol−1).
- State two sources of error that could affect accuracy and one improvement to address each.
- Predict the empirical formula if n(C) : n(H) = 1 : 2.25, and explain how you would handle the non-integer ratio.
Q1 — Sample Band 6 response (8 marks), annotated
Step 1 — Assume 100 g: C = 52.14 g, H = 13.13 g, O = 34.73 g.
Step 2 — Convert to moles: n(C) = 52.14 ÷ 12.011 = 4.341; n(H) = 13.13 ÷ 1.008 = 13.025; n(O) = 34.73 ÷ 15.999 = 2.171. [1 mark — correct mole calculations]
Step 3 — Divide by smallest (2.171): C = 4.341 ÷ 2.171 = 1.999 ≈ 2; H = 13.025 ÷ 2.171 = 5.999 ≈ 6; O = 2.171 ÷ 2.171 = 1.000. Empirical formula = C2H6O. The student’s claim of CH2O is incorrect. [1 mark — correct EF with evidence of error identification]
Step 4 — Molecular formula: MM(C2H6O) = 2(12.011) + 6(1.008) + 15.999 = 24.022 + 6.048 + 15.999 = 46.069 g mol−1. n = 46 ÷ 46.069 ≈ 0.999 ≈ 1. Molecular formula = C2H6O (consistent with ethanol). [1 mark — correct MM and n; 1 mark — molecular formula]
Assessment of glucose claim: The student’s conclusion is incorrect. Glucose has the molecular formula C6H12O6 with a molar mass of 180.16 g mol−1. The unknown compound has a molar mass of only ~46 g mol−1 and a molecular formula of C2H6O — it cannot be glucose. [1 mark — correct rejection of claim with MM evidence]
Conceptual error: The student made the error of assuming that two compounds with the same empirical formula are the same substance. Sharing an empirical formula only means the compounds have the same simplest whole-number ratio of atoms; the actual number of atoms per molecule (the molecular formula) can be any whole-number multiple of the empirical formula. Many different compounds can have the same empirical formula (e.g. CH2O represents formaldehyde, acetic acid, glucose, fructose and others). [1 mark — correct conceptual explanation; 1 mark — example or further elaboration]
Additional information: Any one of the following: spectroscopic data (IR, NMR or mass spectrometry) to confirm functional groups or molecular structure; chemical tests (e.g. Tollens’ reagent to test for an aldehyde vs. a primary alcohol); boiling point measurement (ethanol bp 78 °C; glucose is a solid at room temperature); solubility behaviour; or combustion data for a larger sample to confirm molar mass more precisely. [1 mark — any one valid and specific additional test]
Marking criteria summary (8 marks): 1 = correct four-step working with mole values; 1 = correctly identifies C2H6O as the empirical formula (not CH2O); 1 = correct MM(C2H6O) and n = 1; 1 = states molecular formula C2H6O; 1 = correctly rejects glucose claim citing molar mass difference; 1 = identifies conceptual error (same EF ≠ same compound); 1 = elaborates on the conceptual error with an example or further reasoning; 1 = states one valid additional test to confirm identity.
Q2 — Sample Band 6 response (7 marks), annotated
Hypothesis: If paraffin wax contains only C and H, then combustion of a known mass will produce CO2 and H2O in a ratio reflecting the empirical formula CnH2n+2. The masses of CO2 (absorbed by NaOH) and H2O (absorbed by CaCl2) will allow calculation of n(C) and n(H). IV: mass of wax burned (or combustion time). DV: masses of CO2 and H2O collected. [1 mark — testable hypothesis with IV and DV]
Procedure: (1) Weigh the candle before burning (m1). Connect the combustion tube to a CaCl2 U-tube (to absorb H2O), then to a NaOH absorber (to absorb CO2). Weigh both absorbers before burning. (2) Burn the candle for a set time (e.g. 5 minutes) while drawing combustion gases through the absorbers using a pump. Ensure all gas passes through both absorbers in the correct order (CaCl2 before NaOH). (3) Weigh both absorbers after burning. Calculate: m(H2O absorbed) = mass gain of CaCl2 absorber; m(CO2 absorbed) = mass gain of NaOH absorber. (4) Weigh the candle again (m2); mass of wax burned = m1 − m2. Record all data. [1 mark — four clear steps with absorber order and mass difference method]
Calculation of n(C) and n(H): n(C) = m(CO2) ÷ MM(CO2) = m(CO2) ÷ 44.009 (since each CO2 contains one C atom). n(H) = 2 × [m(H2O) ÷ MM(H2O)] = 2 × [m(H2O) ÷ 18.015] (since each H2O contains two H atoms). Divide both n values by the smallest to get the whole-number ratio. [1 mark — correct calculation method for both elements]
Sources of error and improvements: Error 1: Incomplete combustion (if O2 supply is insufficient) produces CO instead of CO2, so the CO2 mass underestimates carbon content. Improvement: use a pure O2 supply or ensure excess air flow. Error 2: H2O absorbed by the CaCl2 tube may also partially absorb CO2 if the order of absorbers is wrong. Improvement: always place CaCl2 (H2O absorber) before NaOH (CO2 absorber) in the gas flow. [1 mark per error-improvement pair → 2 marks]
Non-integer ratio C : H = 1 : 2.25: Multiply all values by 4 to obtain whole numbers: C = 4, H = 9. Empirical formula = C4H9. This corresponds to a branched-chain alkyl group (e.g. butyl). Paraffin waxes typically have long-chain empirical formulas approximating CH2 to CH2.2 depending on chain length. [1 mark — correct multiplication by 4 and formula C4H9]
Marking criteria summary (7 marks): 1 = hypothesis with IV and DV; 1 = four clear steps with correct absorber order; 1 = correct n(C) and n(H) calculation method; 1 = Error 1 with valid improvement; 1 = Error 2 with valid improvement; 1 = correct multiplication by 4 and empirical formula C4H9 (or equivalent); 1 = precise chemical terminology and reasoning throughout.