Skip to main content
A-LevelChemistryAtoms, molecules and stoichiometry9701/21 · May/June 2017 · Q110 Marks

Combustion data can be used to calculate the empirical formula, molecular formula and relative molecular mass of many organic compounds.

  1. (a)Define the term relative molecular mass.[2]
  2. (b)(i)T is an alcohol, CₓHᵧO. A gaseous sample of T occupied a volume of 20 cm³ at 120°C and 100 kPa. The sample was completely burned in 200 cm³ of oxygen (an excess). The final volume, measured under the same conditions as the gaseous sample, was 250 cm³. Under these conditions, all water present is vaporised. Removal of the water vapour from the gaseous mixture decreased the volume to 170 cm³. Treating the remaining gaseous mixture with concentrated alkali, to absorb carbon dioxide, decreased the volume to 110 cm³. The equation for the complete combustion of T can be represented as shown. CₓHᵧO + zO₂ → xCO₂ + (y/2)H₂O Use the data given to calculate the value of x.[1]
  3. (b)(ii)Use the data given to calculate the value of y.[1]
  4. (b)(iii)If you were unable to calculate values for x and y then use x = 4 and y = 10 for the remaining parts of this question. These are not the correct values. Complete the equation for the complete combustion of the alcohol, T. ................. + .........O₂ → ................CO₂ + ................H₂O[1]
  5. (b)(iv)Give the skeletal formulae for two possible structures of T. Name each alcohol.[2]
  6. (b)(v)Use the general gas equation to calculate the mass of T present in the original 20 cm³ gaseous sample, which was measured at 120°C and 100kPa. Give your answer to three significant figures. Show your working.[3]

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

  1. [1] The mass of a molecule
  2. OR the (weighted) average / (weighted) mean mass of the molecules
  3. [1] Relative / compared to 1/12 (the mass) of an atom of carbon-12
  4. OR on a scale in which a carbon-12 atom / isotope has a mass of (exactly) 12 (units)
  5. [1] 3
  6. [1] 8
  7. [1] C₃H₈O + 4½O₂ → 3CO₂ + 4H₂O
  8. [1] OH AND propan-2-ol / 2-propanol (skeletal formula shown in mark scheme)
  9. [1] OH AND propan-1-ol / 1-propanol (skeletal formula shown in mark scheme)
  10. Alternative answers (any two):
  11. OH AND butan-1-ol / 1-butanol (skeletal formula shown in mark scheme)
  12. OH AND butan-2-ol / 2-butanol (skeletal formula shown in mark scheme)
  13. OH AND (2-)methylpropan-1-ol / (2–)methyl-1-propanol (skeletal formula shown in mark scheme)
  14. OH AND (2-)methylpropan-2-ol / (2–)methyl-2-propanol (skeletal formula shown in mark scheme)
  15. [1] correct conversions of data to SI/consistent units p = 100 000 ; V = 20 × 10⁻⁶ ; T = 393
  16. [1] calculation of n ( = pV/RT) from M1 values n = 100×10³×20×10⁻⁶ / 8.31×393
  17. [1] calculation of mass m ( = n × Mr) AND answer correct to 3sf m = 6.12 × 10⁻⁴ × 60 = 0.0367 (g)
  18. Alternative answer for using C₄H₁₀O: m = 6.12 × 10⁻⁴ × 74 = 0.0453 (g)

Write your answer and get it marked

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.

Sign up free →

About this question

Topic
Atoms, molecules and stoichiometry, A-Level Chemistry (syllabus 9701). Structured question, 10 marks.
Source
A-Level Chemistry, May/June 2017 Paper 2 Variant 1, question 1.

Questions and mark schemes are from past examination papers and remain the copyright of the exam board. How Oracle Prep builds its content

Related Chemistry questions

© 2026 Oracle Prep