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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 Predict 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]
a.

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]
b.

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]
c.

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

[2]
d.

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]
e.

Markscheme

«8.802 g44.01 g mol-1=» 0.2000 «mol of C/CO2»

AND «3.604 g18.02 g mol-1=» 0.2000 «mol of H2O»/0.4000 «mol of H»

OR

 «8.802 g44.01 g mol-1×12.01 g mol-1» 2.402 «g of C»

OR

«3.604 g18.02 g mol-1×2×1.01 g mol-1=» 0.404 «g of H» ✔

 

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


«2.402 g12.01 g mol-1=0.2000 mol C; 0.404 g1.01 g mol-1=0.400 mol H; 1.600 g16.00 g mol-1=0.1000 mol O»

C2H4O ✔

 

Award [3] for correct final answer.

a.

«88.12 g mol-144.06 g mol-1=2» C4H8O2


C2S2 if CS used.

b.


Award [1 max] for correctly identifying all 3 compounds without valid reasons given.

Accept specific values of wavenumbers within each range.

c.

d.

CH3O


Accept any structure i.e. “CH2OH+”.

e.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.

Syllabus sections

Core » Topic 11: Measurement and data processing » 11.3 Spectroscopic identification of organic compounds
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