Identifying organic compounds (Core Practical 15)

Formula from combustion

Burn a known mass of the compound and measure the masses of CO2 and H2O.

• mass of C = mass of CO2 x 12.0 / 44.0
• mass of H = mass of H2O x 2.0 / 18.0
• mass of O = total mass - C - H

Convert to moles and find the empirical formula, then use Mr (from the molecular ion in the mass spectrum) to get the molecular formula.

Functional group tests

Using spectra together

• Mass spectrum: Mr from M+; fragments; accurate mass gives the formula.
• IR: functional groups (broad O-H, C=O near 1700 cm^-1, N-H).
• 13C NMR: number and type of carbon environments.
• 1H NMR: proton environments, ratios and neighbours.

Check that your final structure fits every piece of evidence.

Core Practical 15

Unknowns are tested systematically. Inorganic: flame tests, NaOH and NH3(aq) (especially for transition metal ions), acidified silver nitrate, barium chloride, and heating. Organic: the tests above, focusing on carbonyls and carboxylic acids, supported by spectra. Record observations precisely (colours, gases, precipitates) and use the results to eliminate possibilities.

Worked example

0.440 g of a compound containing C, H and O burns to give 0.880 g CO2 and 0.360 g H2O. Its Mr is 88. Find the molecular formula.

1. C: 0.880 x 12.0 / 44.0 = 0.240 g; H: 0.360 x 2.0 / 18.0 = 0.040 g; O: 0.440 - 0.240 - 0.040 = 0.160 g
2. Moles: C 0.020, H 0.040, O 0.010; ratio 2 : 4 : 1
3. Empirical formula C2H4O (44.0); 88 / 44 = 2

Answer: C4H8O2

Check yourself

1. A compound has empirical formula C2H4O and its molecular ion is at m/z 88. How many carbon atoms does one molecule contain?
   Show answer

   4. C2H4O has mass 44.0; 88 / 44.0 = 2, so C4H8O2 has 4 carbons.

2. Complete combustion of an organic compound gives 1.32 g of CO2. What mass of carbon did the sample contain? (C = 12.0, O = 16.0)
   Show answer

   0.360 g. 1.32 x 12.0 / 44.0 = 0.360 g.

3. A compound C4H8O2 fizzes with sodium hydrogencarbonate and its IR spectrum shows a very broad absorption from 2500 to 3300 cm^-1 and a peak at 1710 cm^-1. Its 1H NMR spectrum has three peaks with areas 6 : 1 : 1. What is it?
   • 2-methylpropanoic acid
   • Butanoic acid
   • Ethyl ethanoate
   • Methyl propanoate
   Show answer

   2-methylpropanoic acid. A carboxylic acid with two equivalent CH3 groups (6H, doublet), one CH (1H) and one COOH (1H): (CH3)2CHCOOH.

4. An unknown C4H8O gives an orange precipitate with 2,4-DNPH and a silver mirror with Tollens' reagent. Which compound could it be?
   • Butanone
   • Butan-1-ol
   • But-2-en-1-ol
   • Butanal
   Show answer

   Butanal. Carbonyl (DNPH) and aldehyde (Tollens').

5. An unknown C4H8O gives an orange precipitate with 2,4-DNPH and a pale yellow precipitate with iodine in alkali, but no change with Fehling's solution. What is it?
   • Butanal
   • Butanone
   • 2-methylpropanal
   • Butan-2-ol
   Show answer

   Butanone. Carbonyl, not an aldehyde, with a CH3CO- group.

6. A compound has M+ at m/z 58, a strong IR absorption at 1715 cm^-1 and a single peak in its 1H NMR spectrum. What is it?
   • Propanal
   • Butane
   • Propan-2-ol
   • Propanone
   Show answer

   Propanone. C3H6O with C=O and six equivalent H: CH3COCH3.

7. In Core Practical 15, an unknown solution gives a green precipitate with NaOH(aq) that dissolves in excess to a dark green solution. Which ion is present?
   • Fe2+
   • Cr3+
   • Ni2+
   • Cu2+
   Show answer

   Cr3+. Chromium(III) hydroxide is amphoteric.

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Sources

• Pearson Edexcel IAL Chemistry (2018) specification
• Pearson Edexcel IAL Chemistry (2018) scheme of work
• Pearson IAL Chemistry teacher practical guide

Last updated 7 November 2026. Spotted a mistake? Tell us.