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Organic Chemistry – Some Basic Principles And Techniques

Question
CBSEENCH11007638

How is nitrogen estimated in a given organic compound by Kjeldahl’s method?

Solution
Kjeldahl’s method: It is employed largely in the analysis of food stuff and fertilisers. It is preferred to Duma’s method. It cannot be applied to the compounds which contain nitrogen atom linked either to oxygen or to other nitrogen atoms.

It consists in heating a known weight of a nitrogenous compound with concentrated sulphuric acid and a little K2SO4 and anhydrous CuSO(or mercury). Potassium sulphate raises the boiling point of H2SO4 and CuSO4 acts as a catalyst. Nitrogen present in the compound is quantitatively converted into ammonium sulphate. The resultant liquid is heated with concentrated sodium hydroxide. The ammonia gas thus liberated is absorbed in a known volume of an excess of a standard solution of acid. The acid left unused is titrated against a standard solution of alkali. Knowing the volume of acid used to neutralise ammonia, the volume of ammonia evolved and finally the percentage of nitrogen can be calculated.
Reactions involved are:
left parenthesis straight i right parenthesis space stack Organic space compound with left parenthesis straight C comma space straight H comma space straight N right parenthesis below space rightwards arrow from Heat to straight H subscript 2 SO subscript 4 of space CO subscript 2 space plus space straight H subscript 2 straight O space plus space left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4
left parenthesis ii right parenthesis space left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4 space plus space 2 NaOH space rightwards arrow with Distillation on top space 2 NH subscript 3 space plus space Na subscript 2 SO subscript 4 space plus space 2 straight H subscript 2 straight O
left parenthesis iii right parenthesis space 2 NH subscript 3 space plus space stack straight H subscript 2 SO subscript 4 with Standard space acid below space rightwards arrow space space space left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4
left parenthesis iv right parenthesis space 2 NaOH space plus space straight H subscript 2 SO subscript 4 space rightwards arrow space space space stack Na subscript 2 SO subscript 4 space plus space 2 straight H subscript 2 straight O with left parenthesis Titration space of space excess space acid right parenthesis below
Calculations:
 Let the mass of organic compound taken = Wg
  Volume of standard straight H subscript 2 SO subscript 4 taken
                                       = V mL of  1 M solution
   After the absorption of ammonia.
  Volume of alkali required for the excess acid
                                 space equals space straight V subscript 1 mL space of space 1 space straight M space solution
space space space straight V subscript 1 mL space of space 1 straight M space NaOH space equals space straight V subscript 1 divided by straight V subscript 2 space mL space of space 1 MH subscript 2 SO subscript 4
therefore space space Volume space of space straight H subscript 2 SO subscript 4 space used space for space neutralising space ammonia equals space
space space space space space space space space space open parentheses straight V subscript 1 space minus space straight V subscript 2 over 2 close parentheses space mL space of space 1 thin space straight M space solution
         
     equals space 2 open parentheses straight V minus straight V subscript 1 over 2 close parentheses space mL space of space 1 thin space straight M thin space NH subscript 3 space solution
Also, 1000 mL of 1 M NH3 solution contains 14g nitrogen
therefore space space space 2 open parentheses straight V minus straight V subscript 1 over 2 close parentheses mL space of space 1 straight M space NH subscript 3 space solution space would space contain
space space space space space space space space equals 14 over 1000 cross times 2 open parentheses straight V minus straight V subscript 1 over 2 close parentheses space cross times space straight M space straight g space of space nitrogen
therefore space space space space Percentage space of space nitrogen
space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 14 space cross times space straight M space cross times space 2 left parenthesis straight V minus straight V subscript 1 divided by 2 right parenthesis over denominator 1000 end fraction cross times 100 over straight W
space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 1.4 space cross times straight M space cross times 2 space left parenthesis straight V minus straight V subscript 1 divided by 2 right parenthesis over denominator straight W end fraction space space space space space space space