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Organic Chemistry – Some Basic Principles And Techniques

Question
CBSEENCH11007632

What do you understand by the estimation of elements? Discuss breifly a method for the estimation of carbon and hydrogen in an organic compound.

Solution

Estimation of elements. Estimation of elements means to determine the percentage composition of each element present in the organic compound.
Estimation of carbon and hydrogen (Liebig’s method): A known weight of the organic compound is strongly heated with an excess of dry copper oxide in an atmosphere of dry and pure oxygen or air. Carbon of the compound is oxidised to carbon dioxide and hydrogen to water.
straight C plus 2 CuO space rightwards arrow space space CO subscript 2 upwards arrow space plus space 2 Cu
2 straight H space plus space CuO space rightwards arrow space space straight H subscript 2 straight O space upwards arrow space plus space Cu
Water vapours and carbon dioxide are bubbled through weighed U-tube containing anhydrous calcium chloride and weighed potash bulbs containing a strong solution of potassium hydroxide. The increase in mass of U-tube gives the mass of water formed and increase in mass of potash bulbs gives the mass of carbon dioxide formed. Knowing the masses of water and carbon dioxide formed, the percentage of hydrogen and carbon can be calculated.

Calculations:
Let the mass of organic compound taken  = Wg
Let the mass of water formed = a g
(increase in mass of anhydrous CaCl2 U- tube)
Let the mass of carbon dioxide formed  = bg
(increase in mass of potash bulbs)
(i) Percentage of hydrogen. 
                                  straight H subscript 2 straight O space identical to space space 2 straight H
18 g of water contains hydrogen  = 2g
therefore space space space straight a space straight g space of space water space contains space hydrogen space space equals space 2 over 18 space cross times space straight a space straight g
Hence comma space percentage space of space hydrogen space equals space 2 over 18 cross times straight a over straight W cross times 100
therefore space space Percentage space of space hydrogen
space equals space 2 over 18 cross times space fraction numerator Mass space of space straight H subscript 2 straight O space formed over denominator Mass space of space organic space compound end fraction cross times 100
(ii) Percentage of carbon.
                 CO subscript 2 space identical to space straight C
44 g of carbon dioxide contains carbon = 12 g
therefore space straight b space straight g space of space carbon space dioxide space contains space carbon space equals space 12 over 44 cross times space straight b space straight g
therefore space space percentage space of space carbon space equals space 12 over 44 cross times straight b over straight W cross times 100
Thus percentage of carbon
           equals space 12 over 44 cross times fraction numerator Mass space of space CO subscript 2 space formed over denominator Mass space of space organic space compound end fraction cross times 100