The ammonia evolved from the treatment of 0.30\ g of an organic compound for the estimation of nitrogen was pa

Question

The ammonia evolved from the treatment of 0.30\ g of an organic compound for the estimation of nitrogen was passed in 100\ mL of 0.1\ M sulphuric acid. The excess of acid required 20\ mL of 0.5\ M sodium hydroxide solution for complete neutralization. The organic compound is:

Accepted Answer

Initial milliequivalents of H_2SO_4 = 100 × 0.1 × 2 = 20\ meq . Milliequivalents of NaOH used for excess acid = 20 × 0.5 × 1 = 10\ meq . Milliequivalents of H_2SO_4 reacted with NH_3 = 20 - 10 = 10\ meq . Milli-moles of NH_3 = 10\ mmol (since NH_3 reacts with 1 equivalent). Mass of Nitrogen = 10 × 10^-3 × 14 = 0.14\ g . \% N = 0.14/0.30 × 100 ≈ 46.6\% . Urea ( NH_2CONH_2 ) has a molar mass of 60 and contains 28\ g of Nitrogen. \% N in urea = 28/60 × 100 = 46.6\% . Thus, the compound is urea.

The ammonia evolved from the treatment of $0.30 g$ of an organic compound for the estimation of nitrogen was passed in $100 mL$ of $0.1 M$ sulphuric acid. The excess of acid required $20 mL$ of $0.5 M$ sodium hydroxide solution for complete neutralization. The organic compound is:

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The ammonia evolved from the treatment of 0.30 g of an organic compound for the estimation of nitrogen was passed in 100 mL of 0.1 M sulphuric acid. The excess of acid required 20 mL of 0.5 M sodium hydroxide solution for complete neutralization. The organic compound is: