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 3 of 5: Molar mass and molecular formula

← Previous Next →

M = \dfrac{mRT}{pV} = \dfrac{1.0045 \times 8.314 \times 350}{35.0\times 10^{3} \times 0.95\times 10^{-3}} = 87.8\ \text{g mol}^{-1} \approx 88\;\Rightarrow\; \ce{C4H8O2}

Analysis

The ideal-gas law on the vapourised sample gives $n = pV/RT = 0.01143$ mol, hence $M = 1.0045/0.01143 = 87.8$ g mol $^{-1}$ — exactly twice the empirical formula mass, so the molecular formula is $\ce{C4H8O2}$ .