0.29 g of an organic compound were analysed by Liebig's method. The increase in the mass of U-tube and the potash bulbs at the end of the experiment were found to be 0.27 g and 0.66 g respectively. Calculate the percentage of carbon and hydrogen in it.

Explanation

Mass of organic compound= 0.29g. Mass of `H_(2)O` formed= Increase in the mass of U-tube= 0.27g
Mass of `CO_(2)` formed= Increase in the mass of potash bulb= 0.66g
Percentage of Carbon: The mass of Carbon present in 0.66g of `CO_(2)= (12)/(44) xx 0.66= 0.18g`
Therefore, 0.29g of the given compound contains 0.18g of Carbon.
Hence, %C `= (0.18)/(0.29) xx 100= 62.1%`.
Percentage of Hydrogen: The mass of Hydrogen present in 0.27g of `H_(2)O = (2)/(18) xx 0.27= 0.03g`
Therefore, 0.29g of the given compound contains 0.03g of Hydrogen.
Hence, % `H= (0.03)/(0.29)xx 100= 11.1%`