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Organic Chemistry – Some Basic Principles And Techniques

Question
CBSEENCH11007636

How is nitrogen estimated in a given organic compound by Duma's method?

Solution
Duma’s method: It is used for the estimation of nitrogen in all nitrogenous compounds. It consists of heating a known mass of the organic compound with an excess of copper oxide and copper in an atmosphere of carbon dioxide at about 975 K. The carbon and hydrogen are oxidised to carbon dioxide and water respectively while nitrogen is set free. If any oxide of nitrogen is produced during the process, it is reduced back to free nitrogen by passing over heated copper gauze. The nitrogen gas is collected.

over a concentrated solution of potassium hydroxide taken in a nitrometer and its volume is measured. From the volume of nitrogen collected, the percentage of nitrogen can be calculated.
Calculations
Let the mass of the organic compound taken = W g
Volume of nitrogen collected = V mL
  Atmospheric pressure  = P mm
  Room temperature straight t degree straight C = (273 + t)K
  Aqueous tension at straight t degree straight C = p mm
(i) To calculate the volume of nitrogen gas at NTP.
          straight P subscript 1 space equals space left parenthesis straight P space minus straight p right parenthesis space mm space space space space space space straight P subscript 2 space equals space 760 space mm
straight V subscript 1 space equals space straight V space ml space space space space space space space space space space space space space space space space straight V subscript 1 space equals space ?
straight T subscript 1 space equals left parenthesis 273 space plus space straight t right parenthesis straight K space space space space space space space space space space straight T subscript 2 space equals space 273 space straight K
Applying the general gas equation
         fraction numerator straight P subscript 1 straight V subscript 1 over denominator straight T subscript 1 end fraction space equals space fraction numerator straight P subscript 2 straight V subscript 2 over denominator straight T subscript 2 end fraction comma space we space have
Volume at N.T.P. (V2) = fraction numerator straight P subscript 1 straight V subscript 1 straight T subscript 2 over denominator straight T subscript 1 straight P subscript 2 end fraction
                 equals space fraction numerator left parenthesis straight P minus straight p right parenthesis space cross times space straight V space cross times space 273 over denominator left parenthesis 273 space plus straight t right parenthesis 760 end fraction space equals space straight V apostrophe space mL space left parenthesis say right parenthesis
(ii) To calculate the percentage of nitrogen:
 22400 mL of nitrogen gas at N.T.P. weighs  = 28 g
therefore  V' mL of nitrogen gas at N.T.P. weighs
                                 equals 28 over 22400 cross times straight V apostrophe space straight g
Hence percentage of nitrogen
                             equals space 28 over 22400 cross times fraction numerator straight V apostrophe over denominator straight W end fraction cross times 100
Percentage of nitrogen
             equals space fraction numerator Mass space of space nitrogen space at space straight N. straight T. straight P. over denominator Mass space of space organic space compound end fraction cross times 100