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Organic Chemistry – Some Basic Principles And Techniques

Question
CBSEENCH11007637

In a Duma's nitrogen estimation method, 0.3 g of an organic compound gave 50 mL of nitrogen collected at 300K and 715 mm pressure. Calculate the percentage of nitrogen in the compound. (Aqueous tension of water at 300 K is 15 mm).

Solution

Mass of organic compound = 0.3 g
(i) To calculate the volume of nitrogen at N.T.P.
(Given)                                (At N.T.P.)
straight V subscript 1 space equals space 50 space mL space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space straight V subscript 2 space equals space ?
straight P subscript 1 space equals space 715 minus 15 space equals space 700 space mm space space space space space space space space space space space straight P subscript 2 space equals space 760 space mm
straight T subscript 1 space equals space 300 space straight k space space space space space space space space space space space space space space space space space space space space space 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 space we space have comma space
space space space space straight V subscript 2 space equals space fraction numerator straight P subscript 1 straight V subscript 1 straight T subscript 2 over denominator straight T subscript 1 cross times straight P subscript 2 end fraction space equals space fraction numerator 700 space cross times space 50 space cross times space 273 over denominator 300 space cross times space 760 end fraction
space space space space space space space space space equals space 41.90 space mL

(ii) To calculate the percentage of nitrogen:
Now 22400 mL of nitrogen at N.T.P. weighs  = 28 g.
therefore space 41.90 space mL space of space nitrogen space at space straight N. straight T. straight P. space weighs
space space space space space space equals space 28 over 22400 cross times 41.90 space straight g
therefore space space Percentage space of space nitrogen
space space space space space space space space space equals space 28 over 22400 cross times fraction numerator 41.90 over denominator 0.3 end fraction cross times 100 space equals space 17.46