AQA A-Level Chemistry Paper 2, June 2025: Question 8

13 marks · Hard difficulty · Practical Techniques & Data Analysis

Use experimental gas syringe data and the ideal gas equation to find the Mr of a volatile liquid, determine the uncertainty range, and calculate reacting gas volumes for hydrocarbon combustion.

Practise this question

Question

Question 8 consists of four parts related to gas volumes. Figure 9 diagrams a hypodermic syringe injecting a volatile liquid Y into a gas syringe located inside a heated oven. A nine-step method is given along with a table of results including initial air volume (5 cm³), hypodermic syringe masses before (3.873 g) and after injection (3.471 g), final gas volume (110 cm³), temperature (155 °C), and pressure (101 kPa). Part 08.1 (5 marks) asks for the Mr of Y using R = 8.31 J K⁻¹ mol⁻¹. Part 08.2 (3 marks) provides percentage uncertainties for volume (0.95%) and temperature (0.12%) with balance uncertainty of ±0.0005 g, requiring calculation of balance percentage uncertainty and the resulting range of Mr values. Part 08.3 (1 mark) asks why placing a seal over the needle improves the experiment. Part 08.4 (4 marks) presents combustion equations of methane and propane, asking for the volumes of methane and propane in a 40 cm³ mixture needing 170 cm³ of oxygen.
Question text

08 This question is about gas volumes.

Figure 9 shows some of the apparatus needed to determine an experimental value

for the Mr of an unknown volatile liquid Y.

Figure 9

A student uses this method to determine an experimental value for the Mr of Y.

1. Record the volume of air in a gas syringe.

2. Collect a sample of Y in a hypodermic syringe.

3. Record the mass of the hypodermic syringe and Y.

4. Push the needle of the hypodermic syringe through the rubber cap on the

gas syringe.

5. Inject a small amount of Y into the gas syringe.

6. Remove the hypodermic syringe from the gas syringe.

7. Record the mass of the hypodermic syringe and the remaining Y.

8. Record the volume of gas in the gas syringe.

9. Record the temperature in the oven and the atmospheric pressure.

Results

Initial volume of air in the gas syringe 5 cm3

Mass of hypodermic syringe and Y before injection 3.873 g

Mass of hypodermic syringe and Y after injection 3.471 g

Final volume of gas in the gas syringe 110 cm3

Temperature 155 °C

Pressure 27 101 kPa

08.1 Use the results to determine an experimental value for the Mr of Y.

The gas constant, R = 8.31 J K–1 mol–1

Show your working.

*26* [5 marks]

Mr

08.2 The percentage uncertainty in the volume of Y in the gas syringe is 0.95%

The percentage uncertainty in the temperature is 0.12%

*27The uncertainty in each balance reading is ± 0.0005 g*

Assume that the uncertainty in the value for pressure is negligible.

Calculate the percentage uncertainty in the use of the balance in this experiment.

Use your answer and the percentage uncertainties for volume and temperature to

calculate a range of values for the Mr of Y.

(If you were unable to calculate an experimental value for the Mr of Y in

Question 08.1, assume it is 141. This is not the correct answer.)

[3 marks]

Percentage uncertainty for balance

Mr range

08.3 A teacher suggests placing a seal over the needle of the hypodermic syringe after

collecting the sample of liquid Y, and then replacing the seal after injecting the liquid Y

into the gas syringe.

Give a reason why the teacher’s suggestion will improve the experiment.

[1 mark]

08.4 A mixture of methane and propane has a volume of 40 cm3

This mixture needs exactly 170 cm3 of oxygen for complete combustion.

The equations for these reactions are

CH4 + 2O2 ⟶ CO2 + 2H2O

C3H8 + 5O2 ⟶ 3CO2 + 4H2O

Calculate the volume

• of methane in the original mixture

• of propane in the original mixture.

All volumes are measured at the same temperature and pressure.

[4 marks]

Volume of methane cm3

Volume of propane cm3

Mark scheme

Show the mark scheme The mark scheme details marks for 08.1 to 08.4. For 08.1 (5 marks), M1 rearranges n = pV/RT or Mr = mRT/pV, M2 converts units (P = 101000 Pa, V = 105 x 10^-6 m^3, T = 428 K), M3 inserts numbers, M4 finds mass of Y = 0.402 g, M5 calculates Mr = 135 (or 134.8). For 08.2 (3 marks), M1 calculates % uncertainty for balance = (0.001 / 0.402) * 100 = 0.249%, M2 sums total apparatus uncertainty = 1.32%, M3 finds range 133.2 - 136.8 or 135 ± 1.8 (or 139.1 - 142.9 using dummy value). For 08.3 (1 mark), award for preventing evaporation/loss of Y or safety. For 08.4 (4 marks), sets up simultaneous equations based on reacting volume stoichiometry (vol O2 = 2x + 5y = 170 and x + y = 40) to deduce vol CH4 = 10 cm³ and vol C3H8 = 30 cm³.

Question Answers Additional comments/Guidelines Mark

Method 1

PV M1 for rearrangement of PV = nRT

M1 n =

RT

M2 converting P to 101 × 103 AND V to 105 × 10–6 AND T to 428 K M2 three unit conversions

101 000 × 0.000105 -3 M3 insertion of their numbers into rearranged

M3 n = 8.31 × 428 (= 2.982 × 10 )

expression

M4 mass of Y = 0.402 g

M4

M5 Mr = = 135 (Allow 134.8)

M3

08.1 Method 2

(5 x AO2)

pV

M1 n = RT

mRT

M2 Mr = pV

M3 converting P to 101 × 103 AND V to 105 × 10–6 AND T to 428 K

M4 mass of Y = 0.402 g

0.402 × 8.31 × 428

M5 Mr = = 135 (Allow 134.8)

101 000 × 0.000105

– A-LEVEL CHEMISTRY – 7405/2 –

0.001 0.001

M1 % uncertainty for balance = M4 from 8.1 × 100 = 0.402 × 100 31

= 0.249% (Allow 0.25%)

M2 Total % apparatus uncertainty = 0.95 + 0.12 + M1 (= 1.32%) 3

08.2

M2 (3 x AO3)

M3 Mr Range M5 from Q8.1 ± (M5 × ) = (133.2 – 136.8 or 135 M3 ‘dummy’ answer gives range 139.1 – 142.9

100 or 141 ± 1.9

± 1.8)

To be marked with 8.1

Seal prevents evaporation / loss (of Y so the mass/volume

recorded will be more accurate)

08.3

OR (1 x AO3)

Seal will cover the sharp needle making the experiment safer

– A-LEVEL CHEMISTRY – 7405/2 –

M1 If vol CH = a cm3 then vol C H = (40 – a) cm3 Simultaneous equations where

43 8

vol CH = x cm3

M2 vol O2 = 2a + 5(40 – a) = 170 cm

vol C H = y cm3

M3 30 = 3 a so vol CH = 10 cm3

4 M1 x + y = 40 i.e. 2x + 2y = 80

M4 vol C H = 40 – M3 = 30 cm3 M2 2x + 5y = 170

32 M3 3y = 90 so y = 30

M4 x = 40 – y = 10

08.4 OR

(4 x AO2)

via M2 x + 2x + y + 5y = 210 i.e. 3x + 6y = 210

Alternative method

M1 If 100% CH4 then vol O2 = 80

M2 If 100% C3H8 then vol O2 = 200

M3 Since vol O2 = 170 ratio C3H8 : CH4 = 3:1

M4 vol CH = 10 cm3 and vol C H = 30 cm3

43 8

How to answer it

Gas Volumes, Ideal Gas Equation & Uncertainties

📋 What this question tests

This multi-skill question examines core Physical and Inorganic chemistry practical concepts:

  • Ideal Gas Equation (pV = nRT): Converting non-standard units (kPa, cm³, °C) to SI units and manipulating equations to find the molar mass (Mr) of a volatile liquid.
  • Practical Apparatus Corrections: Accounting for initial "dead volume" of air trapped in a syringe and volatile mass loss.
  • Uncertainty Analysis: Propagating two-reading balance uncertainties and summing percentage uncertainties to establish an experimental range.
  • Reacting Gas Stoichiometry: Using Avogadro's law and simultaneous equations to determine individual gas volumes in a hydrocarbon combustion mixture.

Question 08.1

Determining Experimental Mr from Gas Syringe Data (5 marks)

📐 Step-by-Step Calculation

Step 1: Calculate Mass of Liquid Y (m)
m = 3.873 g − 3.471 g = 0.402 g [M4]
Step 2: Correct Volume and Convert Units to SI
• Syringe already contained 5 cm³ air:
  V = 110 − 5 = 105 cm³
  V = 105 × 10⁻⁶ m³ (or 0.000105 m³)
• Pressure: P = 101 kPa = 101 × 10³ Pa = 101 000 Pa
• Temperature: T = 155 + 273 = 428 K
(All three conversions correct earns [M2] )
Step 3: Rearrange Ideal Gas Law for Moles (n)
pV = nRT  ⇒  n = pV / RT [M1]
n = (101 000 × 105 × 10⁻⁶) / (8.31 × 428)
n = 10.605 / 3556.68 = 2.982 × 10⁻³ mol [M3]
Step 4: Calculate Relative Molecular Mass (Mr)
Mr = m / n = 0.402 / 2.9817... × 10⁻³
Mr = 135 (or 134.8) [M5]

🧠 Exam Technique & Mark Scheme Breakdown

  • M1: Clear algebraic rearrangement: n = pV / RT (or direct substitution into Mr = mRT / pV).
  • M2: Unit conversions: All three must be spot on (Pa, m³, K).
  • M3: Correct insertion of converted numbers into n = pV / RT.
  • M4: Accurate subtraction of syringe masses to find mass injected (0.402 g).
  • M5: Final evaluation to give 135 or 134.8.

❌ Common Traps

  • Forgetting Initial Air: Using 110 cm³ instead of (110 − 5) = 105 cm³. The syringe wasn't empty at the start!
  • Unit Slip-ups: Leaving temperature in °C or pressure in kPa will completely derail the calculation.
  • Premature Rounding: Rounding n to 0.003 mol gives Mr = 134, losing precision marks. Keep calculator values in memory.
Total: 5 marks (AO2)

Question 08.2

Percentage Uncertainties & Mr Range (3 marks)

📐 Step-by-Step Calculation

Step 1: Balance Percentage Uncertainty
Mass is calculated by difference (two readings: before and after injection).
Absolute uncertainty = 2 × (±0.0005 g) = ±0.001 g
% uncertainty = (0.001 / 0.402) × 100 = 0.249% (or 0.25%) [M1]
Step 2: Total Percentage Uncertainty
Sum all individual percentage uncertainties:
Total % = %volume + %temp + %balance
Total % = 0.95% + 0.12% + 0.249% = 1.32% (or 1.319%) [M2]
Step 3: Determine Experimental Range for Mr
Uncertainty value = 135 × (1.32 / 100) = ±1.78 (or ±1.8)
Range = 135 ± 1.8
Lower bound = 135 − 1.78 = 133.2
Upper bound = 135 + 1.78 = 136.8
133.2 to 136.8 (or 135 ± 1.8 ) [M3]

💡 Key Knowledge

Two Readings = Double the Uncertainty: When a quantity is found by subtraction (e.g. initial & final mass, initial & final burette readings, temperature rise), the total reading uncertainty is twice the instrument tolerance.

Combined Percentage Uncertainty: For quantities that are multiplied or divided (like pV = nRT and Mr = m/n), individual percentage uncertainties are added together.

❌ Common Errors

  • Using 0.0005 g instead of 0.001 g in the numerator, forgetting that mass was obtained by weighing by difference.
  • Calculating the total uncertainty but not applying it to the Mr value from 08.1 to produce an actual numerical range.
  • Note: If you could not calculate Mr in 08.1, the prompt provided a "dummy" value of 141, which gives 141 ± 1.9 (range 139.1 to 142.9).
Total: 3 marks (AO3)

Question 08.3

Practical Improvement: Needle Seal (1 mark)

✅ Accepted Answers (Either 1)

  • Accuracy explanation: The seal prevents evaporation / loss of volatile liquid Y, so the measured mass (or gas volume) will be more accurate.
  • Safety explanation: The seal covers the sharp needle, making the experiment safer / preventing needle-stick injury.

🧠 Examiner Insight

The stem notes that liquid Y is volatile (it easily vaporises at room temperature). Top students immediately link volatile liquids to evaporation during handling and transfer before injection.

Total: 1 mark (AO3)

Question 08.4

Reacting Gas Volumes & Combustion Stoichiometry (4 marks)

📐 Step-by-Step Calculation

By Avogadro's Law, equal volumes of gases at the same temperature and pressure contain equal numbers of moles. Therefore, gas volume ratios directly equal stoichiometric mole ratios!

Step 1: Set Up Variables
Let volume of methane (CH₄) = x cm³
Let volume of propane (C₃H₈) = y cm³
Equation 1: x + y = 40  ⇒  y = 40 − x [M1]
Step 2: Relate to Oxygen Volume
From the equations:
• 1 mol CH₄ reacts with 2 mol O₂ ⇒ requires 2x cm³ O₂
• 1 mol C₃H₈ reacts with 5 mol O₂ ⇒ requires 5y cm³ O₂
Equation 2: 2x + 5y = 170 [M2]
Step 3: Solve Simultaneously for x (CH₄)
Substitute (40 − x) for y:
2x + 5(40 − x) = 170
2x + 200 − 5x = 170
30 = 3x
x = 10 cm³ (Volume of methane) [M3]
Step 4: Solve for y (C₃H₈)
y = 40 − 10 = 30 cm³ (Volume of propane) [M4]

✅ Final Answers

• Volume of methane = 10 cm³

• Volume of propane = 30 cm³

Check: (10 × 2) + (30 × 5) = 20 + 150 = 170 cm³ O₂. Perfectly consistent!

💡 Alternative Logic (Extremes Method)

  • If the 40 cm³ were 100% CH₄, it would need 40 × 2 = 80 cm³ O₂.
  • If the 40 cm³ were 100% C₃H₈, it would need 40 × 5 = 200 cm³ O₂.
  • Actual O₂ needed = 170 cm³.
  • The difference from 80 is 90; difference from 200 is 30. Ratio of propane to methane is 90 : 30 = 3 : 1.
  • 40 cm³ divided in a 3 : 1 ratio gives 30 cm³ C₃H₈ and 10 cm³ CH₄.
Total: 4 marks (AO2)

Topics

Physical Chemistry · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.