Problem 2

Quantitative analysis for carbon and hydrogen was originally carried out with the technique developed in 1831 by Justus Liebig: a weighed sample of organic compound is vaporised and swept by oxygen through heated copper(II) oxide, which ensures quantitative oxidation of C to

\ce{CO2}

and H to

\ce{H2O}

; the water is absorbed in a weighed tube of magnesium perchlorate and the carbon dioxide in a weighed tube of sodium hydroxide on asbestos. A pure liquid sample containing only C, H and O is placed in a 0.57148 g platinum boat; the boat with sample weighs 0.61227 g. After ignition the water-absorption tube mass increased from 6.47002 g to 6.50359 g and the

\ce{CO2}

tube from 5.46311 g to 5.54466 g. 2.1 Calculate the mass percentage composition of the compound. 2.2 Give the empirical formula. 2.3 To estimate the molar mass, 1.0045 g of the compound was gasified; the volume, measured at 350 K and 35.0 kPa, was 0.95 dm

^3

. Give the molar mass and the molecular formula. 2.4 The compound is heated with sodium hydroxide solution; two products are formed. Fractional distillation gives one of them; the other, purified after acidification, is an acid. Which functional class can the compound belong to? 2.5 A 0.1005 g sample of the acid obtained is titrated with 0.1000 mol dm

^{-3}

NaOH; the indicator changes colour at 16.75 cm

^3

. Identify the original compound.

Step 5 of 5: Titration identifies the acid

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M(\text{acid}) = \dfrac{0.1005}{0.1000 \times 16.75\times 10^{-3}} = 60.0\ \text{g mol}^{-1} \;\Rightarrow\; \ce{CH3COOH}\;\Rightarrow\; \ce{CH3COOCH2CH3}

Analysis

A monoprotic acid obeys $n(\text{acid}) = c V$ : $n = 0.1000 \times 0.01675 = 1.675\times 10^{-3}$ mol, so $M = 0.1005/1.675\times 10^{-3} = 60$ g mol $^{-1}$ = acetic acid $\ce{CH3COOH}$ . The ester that saponifies to acetate among the $\ce{C4H8O2}$ candidates is ethyl acetate, $\ce{CH3COOCH2CH3}$ .