Edexcel A-Level Chemistry AS Paper 2, November 2020: Question 1

10 marks · Medium difficulty · Calculations

Calculate the empirical formula and molecular formula of fluorine-containing organic refrigerants using mass spectrometry data and percentage composition, and write equations for free-radical substitution reactions.

Practise this question

Question

A multi-part chemistry exam question about organic compounds containing fluorine and chlorine. Part (a) asks for the molecular formula of dichlorodifluoromethane. Part (b)(i) provides percentage compositions by mass for another refrigerant and asks for the empirical formula. Part (b)(ii) includes a mass spectrum graph plotting relative intensity against m/z from 0 to 120, showing a prominent molecular ion peak, and asks to show that empirical and molecular formulae are the same. Part (b)(iii) asks for the species responsible for the peak at m/z = 69. Part (c)(i) asks for an equation for the formation of trifluoromethane from difluoromethane and fluorine. Part (c)(ii) asks for the initiation and two propagation steps for this free-radical substitution reaction.
Question text

1 This question is about organic compounds containing fluorine and chlorine.

(a) The use of chlorofluorocarbons as refrigerants has ceased due to concerns about

their effects on the ozone layer. One such compound is dichlorodifluoromethane.

Give the molecular formula of dichlorodifluoromethane.

(1)

(b) (i) A different refrigerant contains 34.0% chlorine and 54.5% fluorine by mass,

with the remainder carbon.

Calculate the empirical formula of this compound.

(3)

(ii) Use the mass spectrum to show that the empirical and the molecular formulae

of this compound are the same.

(1)

relative

intensity

0 20 40 60 80 100 120

m/z

(iii) Suggest the species responsible for the peak at m/z = 69.

(1)

*P62307A0228*

(c) Compounds containing carbon and fluorine but no chlorine can be used as

refrigerants as they are not harmful to the ozone layer.

These can be made by the reaction of fluorine with alkanes or fluoroalkanes.

A refrigerant currently in use contains the compound trifluoromethane, CHF3.

(i) Write the equation for the formation of trifluoromethane by the reaction of

difluoromethane with fluorine. State symbols are not required.

(1)

(ii) The mechanism for this reaction is similar to that of the reaction between

chlorine and methane.

Give the equations for the following steps in the mechanism for the reaction

between fluorine and difluoromethane. Curly arrows are not required.

(3)

Initiation step *P62307A0328*

First propagation step

Second propagation step

(Total for Question 1 = 10 marks)

Mark scheme

Show the mark scheme The mark scheme provides answers for all parts of Question 1. Part (a) accepts CCl2F2 or CF2Cl2. Part (b)(i) outlines the 3-mark calculation for percentage of carbon, division by atomic mass, and empirical formula. Part (b)(ii) requires reference to the molecular ion peak at 104/106. Part (b)(iii) gives CF3+ as the ion for m/z = 69. Part (c)(i) gives the equation CH2F2 + F2 -> CHF3 + HF. Part (c)(ii) gives the initiation step F2 -> 2F., first propagation step F. + CH2F2 -> .CHF2 + HF, and second propagation step .CHF2 + F2 -> CHF3 + F.

How to answer it

Organic Compounds Containing Fluorine and Chlorine

What this question tests

This question assesses core AS Level organic and physical chemistry principles, including translating names to molecular formulas, calculating empirical formulas from mass percentages, interpreting mass spectra (identifying molecular ion peaks and fragmentation), writing balanced stoichiometric equations, and recalling free-radical substitution mechanisms (initiation and propagation steps).

Part (a) — Molecular Formula from Name

✅ Correct Answer

CCl₂F₂ (or CF₂Cl₂ )

💡 Key Knowledge

Translating IUPAC-style chemical nomenclature into standard molecular formulas. Prefixes like "dichloro" and "difluor" tell you the exact count of each halogen atom bonded to the carbon chain.

❌ Common Errors

Using incorrect elemental symbols, such as writing Fl instead of the correct symbol for fluorine ( F ).

Mark: (1) available

Part (b)(i) — Calculating Empirical Formula

📐 Step-by-Step Calculation

  1. Find % of Carbon: 100 - (34.0 + 54.5) = 11.5%
  2. Divide by Atomic Mass:
    Cl: 34.0 / 35.5 = 0.95775
    F: 54.5 / 19.0 = 2.8684
    C: 11.5 / 12.0 = 0.95833
  3. Find Simplest Ratio: Divide all by the smallest value (0.95775):
    Cl = 1
    F = 2.8684 / 0.95775 = 2.9949 ≈ 3
    C = 0.95833 / 0.95775 = 1.0006 ≈ 1
  4. Empirical Formula: CF₃Cl (or CClF₃ )

🧠 Exam Technique

Always state the missing percentage clearly first. Never round your intermediate decimal values too early—keep at least 3 or 4 significant figures during working out to avoid rounding errors in the final ratio.

Marks: (3) available — 1 for % carbon, 1 for division by Ar, 1 for correct ratio/formula.

Part (b)(ii) — Mass Spectrum and Molecular Formula

✅ Correct Answer

The molecular ion peak is at m/z = 104 / 106, which matches the relative molecular mass calculated from the empirical formula ( CF₃Cl ).

💡 Key Knowledge

The molecular ion peak (furthest peak to the right, ignoring minor isotope peaks) corresponds to the relative molecular mass (Mr) of the intact molecule. Dual peaks at 104 and 106 are characteristic due to chlorine isotopes (³⁵Cl and ³⁷Cl).

Mark: (1) available

Part (b)(iii) — Identifying Fragment Peaks

✅ Correct Answer

CF₃⁺ (must include the positive charge).

❌ Common Errors

Omitting the positive charge sign ( + ). In mass spectrometry, fragments are ions, so missing charge notation loses the mark.

Mark: (1) available

Part (c)(i) — Stoichiometric Equation

✅ Correct Answer

CH₂F₂ + F₂ → CHF₃ + HF

🧠 Exam Technique

Check your balancing carefully. State symbols are not required here unless explicitly requested, saving you time. Make sure substitution products align with the question context (substitution of H by F producing HF).

Mark: (1) available

Part (c)(ii) — Free-Radical Substitution Mechanism

✅ Correct Answers (Steps)

  • Initiation: F₂ → 2F•
  • First propagation: F• + CH₂F₂ → •CHF₂ + HF
  • Second propagation: •CHF₂ + F₂ → CHF₃ + F•

❌ Common Errors & Examiner Guidance

Watch out for radical dots ( • )—missing a radical dot will be penalized (penalised once across the response). Ensure propagation steps alternate between radicals and molecules correctly, and do not mix up chlorine into a fluorine-based mechanism!

Marks: (3) available — 1 mark per correct step.

Topics

Physical Chemistry · Organic Chemistry · Topic 5: Formulae, Equations and Amounts of Substance · Topic 6: Organic Chemistry I · Topic 7: Modern Analytical Techniques I

Question and mark scheme from the Edexcel A-Level Chemistry examination, AS Paper 2, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.