Problem 71
Question
\(0.5 \mathrm{~g}\) of an organic compound in \(\mathrm{Kjeldahl}\) experiment needs \(29 \mathrm{ml}\) of \(\frac{\mathrm{N}}{5} \mathrm{H}_{2} \mathrm{SO}_{4}\) for complete neutralisation of \(\mathrm{NH}_{3}\). The \(\mathrm{N} \%\) is? (a) \(16.2\) (b) \(8.1\) (c) \(24.3\) (d) \(32.4\)
Step-by-Step Solution
Verified Answer
The nitrogen percentage is 16.2%, option (a).
1Step 1: Write Down Known Values and Formula
We begin by identifying the known values: the mass of the organic compound is 0.5 g, and the volume of the acid used is 29 ml. The acid concentration is \( \frac{N}{5} \). The formula for the percentage of nitrogen is given by \[ \text{N} \% = \frac{\text{Mass of Nitrogen}}{\text{Mass of Compound}} \times 100 \].
2Step 2: Calculate the Normality of Acid
The given normality of \( \mathrm{H}_{2}\mathrm{SO}_{4} \) is \( \frac{N}{5} \). This indicates the normality of the acid used in the reaction is \( 0.2 \text{ N} \).
3Step 3: Calculate the Amount of NH3 Neutralized
Given the volume of \( \mathrm{H}_{2}\mathrm{SO}_{4} \) used is 29 ml, and its normality is 0.2 N, convert volume to liters: \( V = 0.029 \text{ L} \). Calculate the moles of \( \mathrm{H}_{2}\mathrm{SO}_{4} \): \( = N \times V = 0.2 \times 0.029 = 0.0058 \text{ equiv} \). Assuming complete conversion, \( \text{moles of NH}_3 = \text{moles of H}_2\mathrm{SO}_4 = 0.0058 \text{ moles} \).
4Step 4: Calculate the Mass of Nitrogen
The reaction implies 1 mole of \( \mathrm{NH}_{3} \) gives 1 mole of \( \mathrm{N} \) and has a molar mass of \( 14 \text{ g/mol} \) for N. Therefore, the mass of nitrogen is \( 0.0058 \times 14 = 0.0812 \text{ g} \).
5Step 5: Calculate the Nitrogen Percentage
Using the formula, substitute the values calculated: \( \text{N} \% = \frac{0.0812}{0.5} \times 100 = 16.24\% \). We can approximate this to 16.2\%.
Key Concepts
Organic Compound AnalysisNitrogen Percentage CalculationAcid-Base Titration
Organic Compound Analysis
Organic compound analysis is a critical process to determine the composition of organic materials. This involves breaking down the compound into simpler substances to analyze the elements or functional groups present. The Kjeldahl method is a specific technique for analyzing nitrogen content within a compound. It involves digesting the organic material in sulfuric acid, converting nitrogen to ammonia, and then measuring this ammonia. This is crucial for applications in agriculture, food science, and environmental science to assess protein content and nitrogen levels. Students focusing on organic chemistry often see Kjeldahl as a classic method that combines both qualitative and quantitative aspects of chemical analysis.
Nitrogen Percentage Calculation
Calculating the nitrogen percentage in an organic compound is pivotal to understanding its composition. This involves a series of mathematical steps based on experimental data obtained from the Kjeldahl method.
- First, the mass of the compound tested is noted.
- The experiment then measures the nitrogen-containing ammonia produced, and with the aid of acid-base titration, determines the amount of nitrogen.
Acid-Base Titration
Acid-base titration is a fundamental chemical technique used to determine the concentration of a specific acid or base in a solution. In the context of the Kjeldahl method, titration plays an integral role. After the digestion phase, where nitrogen in the organic compound is converted to ammonia, this ammonia potentially contributes to a controlled acid-base reaction.
- A known volume and concentration of standard acid, like sulfuric acid, is used to neutralize the ammonia derived from the compound.
- The endpoint of this reaction is detected by a pH indicator or a pH meter.
Other exercises in this chapter
Problem 69
A compound with empirical formula \(\mathrm{CH}_{2} \mathrm{O}\) has a vapour density of 30 . Its molecular formula is (a) \(\mathrm{C}_{3} \mathrm{H}_{6} \math
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The identity of two different solid organic compounds having the same melting point can be best ascertained by determining their (a) dipole moment (b) solubilit
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When thiourea is heated with metallic sodium, the compound which cannot be formed is (a) \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) (b) \(\mathrm{Na}_{2} \mathrm{~S}\)
View solution Problem 74
When \(0.25 \mathrm{~g}\) of an organic compounds is heated with \(\mathrm{HNO}_{3}\) and \(\mathrm{AgNO}_{3}\) in a carius tube, it gives \(0.35 \mathrm{~g}\)
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