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Date	November 2021	Marks available	1	Reference code	21N.2.hl.TZ0.1
Level	HL	Paper	2	Time zone	TZ0
Command term	Determine	Question number	1	Adapted from	N/A

Question

A 4.406 g sample of a compound containing only C, H and O was burnt in excess oxygen. 8.802 g of CO₂ and 3.604 g of H₂O were produced.

The following spectrums show the Infrared spectra of propan-1-ol, propanal and propanoic acid.

NIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. All rights reserved. Available at: https://webbook.nist.gov/cgi/cbook.cgi?ID=C71238&Units=SI&Type=IRSPEC&Index=3#IR-SPEC [Accessed 6 May 2020]. Source adapted.

NIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. Available at: https://webbook.nist.gov/cgi/cbook.cgi?ID=C79094&Units=SI&Mask=80#IR-Spec [Accessed 6 May 2020]. Source adapted.

NIST Mass Spectrometry Data Center Collection © 2021 copyright by the U.S. Secretary of Commerce on behalf of the United States of America. Available at: https://webbook.nist.gov/cgi/cbook.cgi?Name=propanal&Units=SI&cIR=on&cTZ=on#IRSpec [Accessed 6 May 2020]. Source adapted.

Determine the empirical formula of the compound using section 6 of the data booklet.

[3]

Determine the molecular formula of this compound if its molar mass is 88.12 g mol⁻¹. If you did not obtain an answer in (a) use CS, but this is not the correct answer.

[1]

Identify each compound from the spectra given, use absorptions from the range of 1700 cm⁻¹ to 3500 cm⁻¹. Explain the reason for your choice, referring to section 26 of the data booklet.

[3]

Predict the number of ¹H NMR signals, and splitting pattern of the –CH₃ seen for propanone (CH₃COCH₃) and propanal (CH₃CH₂CHO).

[2]

Predict the fragment that is responsible for a m/z of 31 in the mass spectrum of propan‑1‑ol. Use section 28 of the data booklet.

[1]

Markscheme

« $\frac{8.802 g}{44.01 g {mol}^{- 1}} =$ » 0.2000 «mol of C/CO₂»

AND « $\frac{3.604 g}{18.02 g {mol}^{- 1}} =$ » 0.2000 «mol of H₂O»/0.4000 «mol of H»

« $\frac{8.802 g}{44.01 g {mol}^{- 1}} \times 12.01 g {mol}^{- 1}$ » 2.402 «g of C»

« $\frac{3.604 g}{18.02 g {mol}^{- 1}} \times 2 \times 1.01 g {mol}^{- 1} =$ » 0.404 «g of H» ✔

«4.406 g − 2.806 g» = 1.600 «g of O» ✔

« $\frac{2.402 g}{12.01 g {mol}^{- 1}} = 0.2000 mol C; \frac{0.404 g}{1.01 g {mol}^{- 1}} = 0.400 mol H; \frac{1.600 g}{16.00 g {mol}^{- 1}} = 0.1000 mol O$ »