A $$0.166$$ g sample of an organic compound was digested with conc. $$H_2SO_4$$ and then distilled with NaOH. The ammonia gas evolved was passed through $$50.0$$ mL of $$0.5$$ N $$H_2SO_4$$. The used acid required $$30.0$$ mL of $$0.25$$ N NaOH for complete neutralization. The mass percentage of nitrogen in the organic compound is ______

1. Given Data:

• Mass of the organic compound (W) = 0.166 g
• Initial volume of H2​SO4​ (Vinitial​) = 50.0 mL Filo
• Normality of H2​SO4​ (N1​) = 0.5 N Competishun
• Volume of NaOH solution (V2​) = 30.0 mL Filo
• Normality of NaOH solution (N2​) = 0.25 N Filo

2. Concept and Formula:

In Kjeldahl's method, the mass percentage of nitrogen (%N) in an organic compound is given by the formula:

%N=Mass of the organic compound (in g)1.4×Milliequivalents of acid consumed by NH3​​

Where:

Milliequivalents (meq)=Normality (N)×Volume (in mL)

3. Calculation of Milliequivalents:

• Total initial milliequivalents of H2​SO4​:meqinitial​=50.0 mL×0.5 N=25.0 meq
• Milliequivalents of NaOH used in titration:meqNaOH​=30.0 mL×0.25 N=7.5 meq

Note on Interpretation:

If we consider the 7.5 meq to be the leftover unreacted acid, then the acid consumed by ammonia would be:

25.0 meq−7.5 meq=17.5 meq

Substituting this into the percentage formula yields:

%N=0.1661.4×17.5​≈147.6%

Since a mass percentage greater than 100% is physically impossible, the terminology "the used acid required" indicates that the milliequivalents of acid effectively consumed by the ammonia gas corresponds directly to the titration value:

Milliequivalents of acid consumed by NH3​=meqNaOH​=7.5 meq

4. Calculating the Mass Percentage of Nitrogen:

Substituting the values into the Kjeldahl's formula:

%N=0.1661.4×7.5​%N=0.16610.5​≈63.25%

Rounding to the nearest integer, we get:

63%

Final Answer:

The mass percentage of nitrogen in the organic compound is 63% (or 63.25%).