OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2022: Question 17

6 marks · Hard difficulty · Calculations

Determine the molar mass and molecular formula of an unknown organic compound using ideal gas equation data, and draw a possible chemical structure for it.

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Question

An exam question asking to determine the molar mass, molecular formula, and structure of Compound X. A table gives its molecular formula as C_x H_y F_6 O, functional groups as C–F and C–O–C, and chirality as 1 chiral carbon. Further data states that at a pressure of 1.07 × 10^5 Pa at 30 °C, 1.327 g of Compound X is a gas with a volume of 186 cm^3. Lines are provided for molar mass and molecular formula answers, followed by a large box for the structure of Compound X.
Question text

17 This question is about an analysis of an unknown organic Compound X.

Some properties of Compound X are shown in the table.

Molecular formula Functional groups Chirality

C–F

CxHyF6O 1 chiral carbon

C–O–C

At a pressure of 1.07 × 105 Pa at 30 °C, 1.327 g of Compound X is a gas with a volume of

186 cm3.

Determine the molar mass of Compound X and its molecular formula.

Draw a possible structure for a molecule of Compound X.

molar mass … g mol–1

molecular formula …

Structure of Compound X

[6]

Mark scheme

Show the mark scheme The mark scheme details the allocation of 6 marks for the question. It shows the use of the ideal gas equation pV = nRT with unit conversions to find moles, molar mass calculation leading to 168 g mol^-1, the molecular formula C3H2F6O, and two alternative acceptable structures containing F6, an ether linkage, and a chiral carbon center.

AO

Question Answer Marks Guidance

element

17 FIRST CHECK ANSWER LINES 6 AO1.2 ALLOW ECF throughout

If M=168(.0) Award 4 marks for calculation providing ×1

unit conversions are correct ALLOW calculator value of 167.968115 (using

AO2.4 8.314) for M

--------------------------------------------------------------------- ×3 ALLOW calculator value of 167.8873033 (using

Use of ideal gas equation 8.31) for M

pV AO2.5

pV= nRT OR n = RT

×2

pV

SI Unit conversions AND substitution into n = RT :

• R = 8.314 OR 8.31

• V = 186 × 10–6

• T in K: 303 K

1.07 × 105 × 186 × 10–6

e.g. Calculator value of n:

8.314 × 303 –3

from 8.314 = 7.900308915 × 10

from 8.31 = 7.904111711 × 10–3

Calculation of n

n = 7.90 × 10–3 (mol)

Calculation of M

1.327

M = –3 = 168(.0)

7.90 × 10

Molecular formula ALLOW ECF that matches M but the formula

C3H2F6O MUST contain F O

------------------------------------------------------------

Use of 24 dm3:

186.0 –3

e.g. n = = 7.75 × 10 No mark

24000

(calculation much simpler)

1.327

M = –3 = 171(.2) ECF

7.75 × 10

C3H5F6O ECF

AO

element

Structure ALLOW ECF for a feasibile chemical structure

that matches M AND contains F6O AND has a

chiral carbon

DO NOT ALLOW

OR

F F

F

F

F O

no chiral carbon

F

Total 6

How to answer it

Overall Difficulty Judgement: Medium

Analysis of Unknown Organic Compound X

What this question tests

This 6-mark synoptic question combines physical chemistry calculations with organic structural determination. It tests your ability to rearrange and apply the ideal gas equation (pV = nRT) using correct SI units, deduce molar mass and molecular formulae, and translate analytical data (functional groups, chirality, and formula constraints) into a valid 3D or displayed structural isomer.

Part 1: Determining Molar Mass and Molecular Formula

Calculations and Analytical Deduction (4 Marks)

📐 Step-by-Step Calculation

  1. State and rearrange the ideal gas equation:
    pV = nRT → n = pV / (RT)
  2. Convert into strict SI units:
    Pressure ( p ) = 1.07 × 10⁵ Pa
    Volume ( V ) = 186 cm³ = 186 × 10⁻⁶ m³
    Temperature ( T ) = 30 °C + 273.15 = 303 K
    Gas constant ( R ) = 8.314 J mol⁻¹ K⁻¹
  3. Calculate moles ( n ):
    n = (1.07 × 10⁵ × 186 × 10⁻⁶) / (8.314 × 303)
    n = 7.90 × 10⁻³ mol (or 7.903 × 10⁻³ mol)
  4. Calculate Molar Mass ( M ):
    M = mass / n = 1.327 g / (7.90 × 10⁻³ mol)
    M = 168 g mol⁻¹

✅ Final Answers & Mark Scheme

Molar mass: 168 g mol⁻¹ (allow 167.9 – 168.9)

Molecular formula: C₃H₂F₆O

Mark Breakdown:
• 1 mark for correct pV = nRT rearrangement
• 1 mark for correct unit conversions (cm³ to m³, °C to K)
• 1 mark for calculating n
• 1 mark for calculating M (168 g mol⁻¹)
• 1 mark for molecular formula deduction

❌ Common Calculation Traps

  • The 24.0 dm³ Trap: Using n = V / 24000 assumes the gas is at room temperature and pressure (RTP). The question specifies 30 °C and 1.07 × 10⁵ Pa —you must use the ideal gas equation.
  • Unit Conversion Failures: Forgetting to convert cm³ to m³ (multiplying by 10⁻⁶) or failing to convert Celsius to Kelvin (+273) are the most frequent places students lose accuracy marks.

Part 2: Drawing the Structure of Compound X

Structural Isomerism, Functional Groups, and Chirality (2 Marks)

💡 Key Knowledge & Constraints

  • Given Formula: C₃H₂F₆O
  • Functional Groups: Must contain a C–F bond (specifically six fluorine atoms distributed across carbons) and a C–O–C linkage (an ether group).
  • Chirality Rule: Must possess exactly 1 chiral carbon (a carbon atom bonded to four completely different groups/atoms).

🧠 Exam Technique & Examiner Guidance

When drawing fluorinated ethers with a chiral center:

  • Make sure the central oxygen bridges two carbon chains (e.g., CH(F)₂–O–CH(F)CF₃ or its constitutional isomer equivalent).
  • Verify the chiral center: look for a carbon bonded to –H , –F , –CF₃ (or –CHF₂ ), and the –O– group.
  • Double-check that all 6 fluorine atoms, 2 hydrogens, 3 carbons, and 1 oxygen from your formula are fully accounted for in the drawn structure.

✅ Accepted Structures (Mark Scheme)

Any valid displayed formula matching C₃H₂F₆O containing an ether ( C–O–C ), six fluorine atoms ( F₆ ), and exactly one chiral center is fully rewarded.

Example description: CHF₂–O–CH(F)CF₃ (where the middle carbon –CH(F)– is the chiral center bonded to H, F, CF₃, and O–CHF₂ ).

Topics

Module 2: Foundations in chemistry · Module 6: Organic chemistry and analysis · 6.2 Nitrogen compounds, polymers and synthesis · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.