IChO 1982 · Theory problem 2

Liebig combustion analysis and identification of an organic liquid C

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Figure 1

Quantitative analysis for carbon and hydrogen was originally carried out using a technique and apparatus (see figure) originally developed in 1831 by the famous chemist Justus Liebig. A carefully weighed sample of organic compound (C) is placed in a combustion tube (A) and vaporized by heating in a furnace (B). The vapours are swept by a stream of oxygen through a heated copper oxide packing (D) and through another furnace (E), which ensures the quantitative oxidation of carbon and hydrogen to carbon dioxide and water. The water vapour is absorbed in a weighed tube (F) containing magnesium perchlorate and the carbon dioxide in another weighed tube (G) containing asbestos impregnated with sodium hydroxide.

A pure liquid sample containing only carbon, hydrogen and oxygen is placed in a 0.57148 g platinum boat, which on reweighing weights 0.61227 g. The sample is ignited and the previously weighed absorption tubes are reweighed. The mass of the water absorption tube has increased from 6.47002 g to 6.50359 g, and the mass of the carbon dioxide tube has increased from 5.46311 g to 5.54466 g.

2.1 Calculate the mass composition of the compound.

2.2 Give the empirical formula of the compound.

To estimate the molar mass of the compound, 1.0045 g was gasified. The volume, measured at a temperature of 350 K and a pressure of 35.0 kPa, was 0.95 dm\(^3\).

2.3 Give the molar mass and the molecular formula of the compound.

2.4 Draw possible structures corresponding to the molecular formula excluding cyclic structures, stereo isomers, peroxides and unsaturated compounds. There are about 15 possibilities. Give 10 of them.

When the compound is heated with a sodium hydroxide solution, two products are formed. Fractional distillation of the reaction mixture yields one of the substances. The other substance is purified by distillation after acidification and appears to be an acid.

2.5 What structures are possible for compound C?

0.1005 g of the acid are dissolved in water and titrated with a sodium hydroxide solution with a concentration of 0.1000 mol dm\(^{-3}\). The indicator changes colour on addition of 16.75 cm\(^3\) of hydroxide solution.

2.6 What was the original substance C?

Official English text of the problem.

Answers

  • 2.1: Carbon 54.56%, hydrogen 9.21%, oxygen 36.23% by mass.
  • 2.2: Empirical formula: \(\boxed{\ce{C2H4O}}\).
  • 2.3: Molar mass: \(\boxed{88\ \mathrm{g\,mol^{-1}}}\); molecular formula: \(\boxed{\ce{C4H8O2}}\).
  • 2.4: Ten acceptable constitutional structures are listed below. Stereochemistry is unspecified because stereoisomers are excluded from the count.
Name Condensed structure SMILES
Butanoic acid \(\ce{CH3CH2CH2C(=O)OH}\) CCCC(=O)O
2-Methylpropanoic acid \(\ce{(CH3)2CHC(=O)OH}\) CC(C)C(=O)O
Methyl propanoate \(\ce{CH3CH2C(=O)OCH3}\) CCC(=O)OC
Ethyl ethanoate \(\ce{CH3C(=O)OCH2CH3}\) CC(=O)OCC
Propyl methanoate \(\ce{HC(=O)OCH2CH2CH3}\) CCCOC=O
Propan-2-yl methanoate \(\ce{HC(=O)OCH(CH3)2}\) CC(C)OC=O
2-Hydroxybutanal \(\ce{CH3CH2CH(OH)CHO}\) CCC(O)C=O
3-Hydroxybutanal \(\ce{CH3CH(OH)CH2CHO}\) CC(O)CC=O
1-Hydroxybutan-2-one \(\ce{HOCH2C(=O)CH2CH3}\) CCC(=O)CO
3-Hydroxybutan-2-one \(\ce{CH3C(=O)CH(OH)CH3}\) CC(=O)C(O)C
  • 2.5: The four ester candidates are methyl propanoate CCC(=O)OC, ethyl ethanoate CC(=O)OCC, propyl methanoate CCCOC=O, and propan-2-yl methanoate CC(C)OC=O.
  • 2.6: The original substance is ethyl ethanoate (ethyl acetate), \(\ce{CH3COOCH2CH3}\), SMILES CC(=O)OCC.

Solution

2.1 — Mass composition

The mass of the sample is

\[ m_{\text{sample}}=0.61227-0.57148 =0.04079\ \mathrm{g}. \]

The absorber mass increases give the amounts of water O and carbon dioxide O=C=O:

\[ m(\ce{H2O})=6.50359-6.47002 =0.03357\ \mathrm{g}, \]

\[ m(\ce{CO2})=5.54466-5.46311 =0.08155\ \mathrm{g}. \]

Using atomic molar masses of \(12.01\), \(1.008\), and \(16.00\ \mathrm{g\,mol^{-1}}\) for carbon, hydrogen, and oxygen, respectively,

\[ m(\ce{C}) =m(\ce{CO2})\frac{12.01}{44.01} =0.0222544\ \mathrm{g}, \]

\[ m(\ce{H}) =m(\ce{H2O})\frac{2(1.008)}{18.016} =0.00375650\ \mathrm{g}. \]

Because the compound contains only carbon, hydrogen, and oxygen, the oxygen mass follows by difference:

\[ m(\ce{O}) =0.04079-0.0222544-0.00375650 =0.0147791\ \mathrm{g}. \]

Therefore,

\[ \boxed{ \begin{aligned} w(\ce{C})&=\frac{0.0222544}{0.04079}\times100\%=54.56\%,\\ w(\ce{H})&=\frac{0.00375650}{0.04079}\times100\%=9.21\%,\\ w(\ce{O})&=\frac{0.0147791}{0.04079}\times100\%=36.23\%. \end{aligned}} \]

2.2 — Empirical formula

The amounts of atoms in the sample are

\[ \begin{aligned} n(\ce{C})&=\frac{0.0222544}{12.01} =1.85299\times10^{-3}\ \mathrm{mol},\\ n(\ce{H})&=\frac{0.00375650}{1.008} =3.72669\times10^{-3}\ \mathrm{mol},\\ n(\ce{O})&=\frac{0.0147791}{16.00} =9.23695\times10^{-4}\ \mathrm{mol}. \end{aligned} \]

Dividing by the smallest amount gives

\[ n(\ce{C}):n(\ce{H}):n(\ce{O}) =2.006:4.035:1 \approx2:4:1. \]

Thus the empirical formula is

\[ \boxed{\ce{C2H4O}}. \]

2.3 — Molar mass and molecular formula

Treating the vapour as an ideal gas,

\[ pV=nRT,\qquad M=\frac{mRT}{pV}. \]

With \(R=8.314\ \mathrm{kPa\,dm^3\,mol^{-1}\,K^{-1}}\),

\[ M= \frac{ (1.0045\ \mathrm{g}) (8.314\ \mathrm{kPa\,dm^3\,mol^{-1}\,K^{-1}}) (350\ \mathrm{K}) }{ (35.0\ \mathrm{kPa})(0.95\ \mathrm{dm^3}) } =87.91\ \mathrm{g\,mol^{-1}}. \]

The volume has two significant figures, so the reported result is

\[ \boxed{M\approx88\ \mathrm{g\,mol^{-1}}}. \]

The empirical formula mass is

\[ M_{\mathrm{emp}} =2(12.01)+4(1.008)+16.00 =44.052\ \mathrm{g\,mol^{-1}}. \]

Hence

\[ \frac{M}{M_{\mathrm{emp}}} =\frac{87.91}{44.052} \approx2, \]

and the molecular formula is

\[ \boxed{\ce{C4H8O2}}. \]

Its complete combustion with oxygen O=O is balanced as

\[ \ce{C4H8O2 + 5O2 -> 4CO2 + 4H2O}. \]

2.4 — Ten possible structures

The degree of unsaturation is

\[ \mathrm{DBE}=\frac{2(4)+2-8}{2}=1. \]

With rings and carbon–carbon multiple bonds excluded, this requires one carbonyl group. The second oxygen can occur in a carboxylic acid, ester, hydroxyl, or ether group.

The ten structures in the Answers table comprise two carboxylic acids, four esters, two hydroxyaldehydes, and two hydroxyketones. Each has exactly four carbon atoms, eight hydrogen atoms, and two oxygen atoms, with no ring or peroxide bond.

For the requested drawings, the condensed formulas specify the connectivity: every bond is single except the indicated carbonyl double bond. The terminal aldehyde group \(\ce{-CHO}\) contains a carbonyl carbon bonded to hydrogen. In the hydroxyaldehydes, numbering begins at this aldehyde carbon; in the hydroxyketones, the carbonyl is at carbon 2. No separate stereoisomers are counted.

2.5 — Structures consistent with alkaline hydrolysis

Heating an ester with sodium hydroxide [Na+].[OH-] produces an alcohol and a sodium carboxylate. The neutral alcohol can be distilled from the reaction mixture. The carboxylate remains in solution until acidification releases the corresponding carboxylic acid.

For an ester with four carbon atoms, the distribution between the acid and alcohol portions can be \(3+1\), \(2+2\), or \(1+3\). The three-carbon alcohol portion has two possible connectivities. This gives four ester candidates:

Possible compound C Alcohol obtained Acid obtained after acidification
Methyl propanoate CCC(=O)OC Methanol CO Propanoic acid CCC(=O)O
Ethyl ethanoate CC(=O)OCC Ethanol CCO Ethanoic acid CC(=O)O
Propyl methanoate CCCOC=O Propan-1-ol CCCO Methanoic acid O=CO
Propan-2-yl methanoate CC(C)OC=O Propan-2-ol CC(O)C Methanoic acid O=CO

The balanced hydrolysis equations, written ionically, are

\[ \ce{CH3CH2COOCH3 + OH- -> CH3CH2COO- + CH3OH}, \]

\[ \ce{CH3COOCH2CH3 + OH- -> CH3COO- + CH3CH2OH}, \]

\[ \ce{HCOOCH2CH2CH3 + OH- -> HCOO- + CH3CH2CH2OH}, \]

\[ \ce{HCOOCH(CH3)2 + OH- -> HCOO- + (CH3)2CHOH}. \]

Here the ions are hydroxide [OH-], propanoate CCC(=O)[O-], ethanoate CC(=O)[O-], and methanoate [O-]C=O.

2.6 — Identification of the original substance

All three candidate acids are monocarboxylic and consume one mole of hydroxide per mole of acid. Thus

\[ \begin{aligned} n_{\mathrm{acid}} &=n_{\mathrm{NaOH}}\\ &=(0.1000\ \mathrm{mol\,dm^{-3}}) (16.75\times10^{-3}\ \mathrm{dm^3})\\ &=1.675\times10^{-3}\ \mathrm{mol}. \end{aligned} \]

The acid’s experimental molar mass is therefore

\[ M_{\mathrm{acid}} =\frac{0.1005\ \mathrm{g}} {1.675\times10^{-3}\ \mathrm{mol}} =\boxed{60.00\ \mathrm{g\,mol^{-1}}}. \]

This identifies ethanoic acid (acetic acid) CC(=O)O, whose formula is \(\ce{CH3COOH}\) and whose formula molar mass is approximately \(60.05\ \mathrm{g\,mol^{-1}}\).

Its titration reaction is

\[ \ce{CH3COOH + OH- -> CH3COO- + H2O}. \]

The ethanoate portion contains two carbon atoms, leaving two of the original four carbon atoms for the alcohol portion. That alcohol is therefore ethanol CCO, and the original ester is

\[ \boxed{\text{ethyl ethanoate (ethyl acetate), }\ce{CH3COOCH2CH3}} \]

with SMILES CC(=O)OCC.

Its alkaline hydrolysis is

\[ \ce{CH3COOCH2CH3 + NaOH -> CH3COONa + CH3CH2OH}, \]

where sodium ethanoate has SMILES [Na+].CC(=O)[O-]. Ethanol is the first distilled product.

Acidification protonates ethanoate; using hydronium [OH3+] to represent the acidifying solution,

\[ \ce{CH3COO- + H3O+ -> CH3COOH + H2O}. \]

The resulting ethanoic acid is the second isolated product.

The answers were checked against the official answers published by the organisers.