Principle: A known mass of an organic compound is strongly heated with dry cupric oxide (CuO), when carbon and hydrogen are quantitatively oxidized to CO2 and H2O respectively. The masses of CO2 and H2O thus formed are determined. From this, the percentages of carbon and hydrogen can be calculated.
Procedure: Pure and dry oxygen is passed through the entire assembly of the apparatus (Fig) till the CO2 and moisture is completely removed.
A boat containing weighed organic substances is introduced inside from one end of the combustion tube by opening it for a while. The tube is now strongly heated till the whole of the organic compound is burnt up. The flow of oxygen is continued to drive CO2 and water vapours completely to the U-tubes. The apparatus is cooled and the U-tubes are weighed separately.
Observed and Calculations.
1. Mass of organic compound taken = w.g.
2. Mass of water produced = x g (Increase in mass of CaCk tube)
3. Mass of carbon dioxide produced = y g. (Increase in mass of KOH tube)
To determine % of carbon
Molar mass of CO2 = 44g mol-1
Now, 44g of CO2 = contains 12 g of C.
∴ y g of CO2 will contain of \(\frac{12y}{44}\)g of C.
This amount of carbon was present in w. g. of the substance
∴ %C = \(\frac{12y}{44}\) x \(\frac{100}{w}\)
To determine % of Hydrogen
Molar mass of water = 18 g mol-1
Now 18g of H2O contains 2 g of H2
∴ x g of H2O will contain \(\frac{2x}{18}\)g of H2
This amount of hydrogen was present in weight of substance.
∴ %H2 = \(\frac{2x}{18}\) x \(\frac{100}{w}\)
