Chemistry Labs

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 COX2\ce{CO2} and H to HX2O\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 COX2\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 dm3^3. 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^{-3} NaOH; the indicator changes colour at 16.75 cm3^3. Identify the original compound.
Step 1 of 5: Mass composition
← PreviousNext →
Intuition

Every gram of COX2\ce{CO2} collected hides 12/44 of carbon; every gram of HX2O\ce{H2O} hides 2/18 of hydrogen.

m(C)=0.08155×12.044.0=0.02224 g;m(H)=0.03357×2.0218.02=3.76×10−3 g;%C=54.56, %H=9.21, %O=36.23m(\mathrm C) = 0.08155 \times \tfrac{12.0}{44.0} = 0.02224\ \text{g};\quad m(\mathrm H) = 0.03357 \times \tfrac{2.02}{18.02} = 3.76\times 10^{-3}\ \text{g};\quad \%\mathrm C = 54.56,\ \%\mathrm H = 9.21,\ \%\mathrm O = 36.23
Analysis

Sample mass =0.61227−0.57148=0.04079= 0.61227 - 0.57148 = 0.04079 g. From the tube gains m(COX2)=0.08155m(\ce{CO2}) = 0.08155 g and m(HX2O)=0.03357m(\ce{H2O}) = 0.03357 g: carbon is 0.08155×12/44=0.022240.08155 \times 12/44 = 0.02224 g (54.6 %) and hydrogen 0.03357×2.02/18.02=3.76×10−30.03357 \times 2.02/18.02 = 3.76\times 10^{-3} g (9.2 %); oxygen is obtained by difference, 0.04079−0.02600=0.014790.04079 - 0.02600 = 0.01479 g (36.2 %).

Common pitfall. Never read oxygen off the absorption tubes — the trapped COX2\ce{CO2} and HX2O\ce{H2O} contain oxygen pulled from the combustion gas stream; O must come from the mass difference.