Chemistry • Year 11 • Module 2 • Lesson 4

Gases & Molar Volume

Build HSC Band 5–6 technique on multi-step gas calculations, evaluating student proposals, and designing a procedure to identify an unknown gas by molar mass.

Master · Extended Response

1. Data + evaluation, two students identify an unknown gas (Band 5–6)

9 marks   Higher-order

Scenario. A chemistry laboratory at a NSW school produces an unknown gas during a reaction. Two students each collect a sample and make the following measurements at SATP (25 °C, 100 kPa).

Student AStudent B
Volume of gas collected4.96 L9.92 L
Mass of gas collected8.80 g17.60 g
ConditionsSATPSATP

Illustrative data. Molar masses of common gases (g mol−1): CO2 = 44.01, SO2 = 64.06, N2 = 28.01, O2 = 32.00, CH4 = 16.04, C3H8 = 44.10.

Q1. Using both students’ data, determine the molar mass of the unknown gas and identify it. In your response you must:

  • Calculate the number of moles in each student’s sample using n = V ÷ Vm, showing full working for both.
  • Calculate the molar mass from each student’s data using MM = m ÷ n, and compare the two results.
  • Identify the unknown gas and justify your identification by reference to the molar mass table.
  • Explain why using two independent samples strengthens confidence in the identification.
  • State one assumption you made and identify one source of error that could cause the calculated molar mass to differ from the true value.
Plan: n(A) = 4.96 ÷ 24.8; MM(A) = 8.80 ÷ n. Repeat for B. Compare to table. Assumption: gas behaves ideally. Error: incomplete drying of gas (moisture adds to mass).

2. Experimental design, identify an unknown gas using molar volume (Band 5–6)

8 marks   Higher-order

Research question. A laboratory technician supplies you with a sealed gas-collection tube of a pure gas that the technician has already confirmed is safe to handle. You are told only that it is either carbon dioxide (CO2, MM = 44.01 g mol−1) or propane (C3H8, MM = 44.10 g mol−1). Because the molar masses are almost identical, a molar mass calculation alone may not distinguish between them reliably.

Equipment available: digital balance (±0.001 g), gas syringe (0–100 mL), barometer, thermometer, limewater (Ca(OH)2(aq)), distilled water, gas-washing bottles.

Safety boundary. Never release, smell or ignite an unknown or compressed gas yourself, and never test an unlabelled storeroom cylinder. Design your investigation around a teacher- or technician-supplied sample that has already been confirmed safe. The limewater test is the safe chemical discriminator for these two gases. Any flammability evidence must come from a teacher demonstration behind a safety screen or from supplied secondary data, not from a student igniting the gas.

Q2. Design an investigation to identify the gas. In your response you must:

  • State a hypothesis that includes the independent and dependent variables.
  • Describe at least two separate tests you would perform (include a molar-volume molar-mass calculation AND at least one chemical identification test), with enough procedural detail for another student to follow.
  • Predict the results for each test if the gas is CO2, and contrast with the predicted results if the gas is C3H8.
  • Explain what result would falsify your hypothesis.
  • State two limitations and one way to improve reliability.
Tests: (1) Collect 50 mL at SATP, weigh, calculate MM. (2) Bubble gas through limewater, CO2 turns it milky (Ca(OH)2 + CO2 → CaCO3↓); C3H8 does not. (3) Flammability is provided as secondary data or a teacher demonstration only (C3H8 is flammable; CO2 is not), never as a student ignition test.
Answers, Do not peek before attempting

Q1, Sample Band 6 response (9 marks), annotated

Moles, Student A: Vm = 24.8 L mol−1 (SATP, 25 °C, 100 kPa). n = V ÷ Vm = 4.96 ÷ 24.8 = 0.200 mol [1]. MM = m ÷ n = 8.80 ÷ 0.200 = 44.0 g mol−1 [1].

Moles, Student B: n = 9.92 ÷ 24.8 = 0.400 mol [1]. MM = 17.60 ÷ 0.400 = 44.0 g mol−1 [1].

Comparison: Both students obtain MM = 44.0 g mol−1, consistent across a doubled sample, which increases confidence [1].

Identification: From the table, CO2 = 44.01 g mol−1 and C3H8 = 44.10 g mol−1. The calculated value (44.0 g mol−1) is consistent with CO2; however, the molar masses of CO2 and C3H8 are so close that this calculation alone cannot definitively distinguish them [1].

Why two samples strengthen confidence: Two independent samples of different volumes give the same molar mass, showing the result is reproducible and not the result of a random measurement error in a single trial [1].

Assumption: The gas behaves as an ideal gas at SATP; real gases deviate slightly from ideal behaviour, especially at high pressure or for polar molecules [1].

Source of error: If water vapour (from the reaction) is not removed before weighing, the measured mass will be higher than the true gas mass, making the calculated molar mass too large [1].

Marking criteria (9 marks): 1 = n(A) correct with working; 1 = MM(A) correct; 1 = n(B) correct; 1 = MM(B) correct; 1 = comparison of two results noting agreement; 1 = identification with reference to table; 1 = explanation of why two samples increase confidence; 1 = one valid assumption; 1 = one specific source of error.

Q2, Sample Band 6 response (8 marks), annotated

Hypothesis: If the gas is CO2, it will have a calculated molar mass of approximately 44.01 g mol−1, will turn limewater milky, and (from supplied flammability data) will be non-flammable. IV: gas identity (CO2 vs C3H8). DV: (i) calculated molar mass; (ii) effect on limewater; (iii) flammability classification from secondary data [1].

Test 1, Molar mass: (i) Record temperature and pressure to confirm SATP conditions. (ii) Collect exactly 50.0 mL (0.0500 L) of gas from the cylinder into a pre-weighed gas syringe. Seal the syringe. Weigh the syringe + gas and subtract the empty mass to find mass of gas. (iii) Calculate: n = 0.0500 ÷ 24.8 = 2.02 × 10−3 mol. MM = m ÷ n [1]. Predicted result: both CO2 and C3H8 give approximately 44 g mol−1, so this test alone is inconclusive for distinguishing them [1].

Test 2, Limewater: Bubble a stream of the gas through 20 mL of limewater (Ca(OH)2(aq)) in a test tube for 30 s. Record whether the solution turns milky. CO2(g) + Ca(OH)2(aq) → CaCO3(s) + H2O(l) [1]. If CO2: limewater turns milky (white precipitate of CaCO3). If C3H8: limewater stays clear [1].

Test 3, Flammability (teacher demonstration / secondary data only): Students must not ignite the gas. The teacher may demonstrate flammability behind a safety screen, or supply a reference data card. Secondary data: C3H8 is flammable (lower flammability limit ~2.1% in air); CO2 is non-flammable and will extinguish a flame. This evidence confirms the chemical test but is never collected by a student handling an unknown gas [1].

Falsification: If the limewater stays clear AND the supplied flammability data show the gas burns, the hypothesis (gas is CO2) would be falsified; the data would indicate C3H8 [1].

Limitations: (1) The molar-mass calculation alone cannot distinguish CO2 from C3H8 because their molar masses differ by only 0.09 g mol−1, within experimental uncertainty. (2) The limewater test cannot quantify the amount of CO2; trace amounts of CO2 contamination in C3H8 could give a false positive. Improvement: repeat each test three times (triplicate trials) and use a calibrated gas analyser (e.g. IR spectroscopy or a CO2 sensor) for a definitive chemical identification [1].

Marking criteria (8 marks): 1 = hypothesis with IV and DV; 1 = molar-mass test described with procedure; 1 = acknowledgement that MM test is inconclusive for these two gases; 1 = limewater test described with chemical equation; 1 = predicted contrasting results for CO2 vs C3H8; 1 = what would falsify the hypothesis; 1 = two limitations identified; 1 = one specific improvement.

HSCScience • Chemistry • Year 11 • Module 2 • Lesson 4 • Worksheet 3 (Master)