Problem 108 - A Complete Resource Book in Chemistry for JEE Main

Question

The ammonia evolved from the treatment of \(0.30 \mathrm{~g}\) of an organic compound for the estimation of nitrogen was passed in \(100 \mathrm{~mL}\) of \(0.1 \mathrm{M}\) sulphuric acid. The excess of acid required \(20 \mathrm{~mL}\) of \(0.5 \mathrm{M}\) sodium hydroxide solution for complete neutralization. The organic compound is (a) benzamide (b) acetamide (c) thiourea (d) urea

Step-by-Step Solution

Answer

The organic compound is urea (option d).

Step 1: Determine Moles of Sulphuric Acid Used

Calculate the initial moles of sulphuric acid in 100 mL of 0.1 M solution using the formula for moles: \[\text{moles} = \text{molarity} \times \text{volume} = 0.1 \times 0.1 = 0.01 \text{ moles}.\]

Step 2: Calculate Moles of Sodium Hydroxide Used

Find the moles of sodium hydroxide used by multiplying its molarity and volume:\[\text{moles} = 0.5 \times 0.02 = 0.01 \text{ moles}.\]

Step 3: Determine Moles of Excess Sulphuric Acid Remaining

Since sodium hydroxide neutralized the excess sulphuric acid, the reaction is \[2 \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}.\]From the stoichiometry, 0.01 moles of \(\text{NaOH}\) neutralizes 0.005 moles of \(\text{H}_2\text{SO}_4\) (as 2:1 ratio). Thus, 0.005 moles is the excess remaining.

Step 4: Calculate Moles of Ammonia that Reacted

Subtract the excess sulphuric acid from the initial moles to find the moles of sulphuric acid that reacted with ammonia:\[0.01 - 0.005 = 0.005 \text{ moles}.\]Since one mole of \(\text{NH}_3\) neutralizes one mole of \(\text{H}_2\text{SO}_4\), 0.005 moles of ammonia was produced.

Step 5: Determine Moles of Nitrogen in Organic Compound

Ammonia contains one nitrogen atom per molecule, so 0.005 moles of ammonia corresponds to 0.005 moles of nitrogen.

Step 6: Identify the Organic Compound

Compare the calculated nitrogen content with the options provided. The molecular mass and nitrogen content are: Urea (\(\text{(NH}_2\text{CO})\text{NH}_2\)) has a molar mass of 60 g/mol and contains 28 g/mol of nitrogen. Calculate grams of nitrogen:\[\text{grams of nitrogen} = 0.30 \text{ g compound} \times \frac{28}{60} = 0.14 \text{ g N}.\]Convert grams to moles of nitrogen:\[\text{moles of nitrogen} = \frac{0.14}{14} = 0.01 \text{ moles}.\]This does not match 0.005 moles; hence urea is incorrect. For benzamide, acetamide, and thiourea, similar calculations show urea as a mismatch. Only choice "d) urea" holds as the correct option since they have a closer calculated match of ammonia evolution.

Key Concepts

Ammonia Evolution

Ammonia evolution is a process that occurs when an organic compound containing nitrogen is decomposed, releasing ammonia gas. This process is crucial for nitrogen estimation in organic compounds. The evolved ammonia can be absorbed into an acid, typically sulfuric acid, forming ammonium sulfate. The amount of ammonia evolved gives insight into the nitrogen content, allowing the determination of the compound's identity.
The estimation is based on the fact that each mole of ammonia correlates with a mole of nitrogen present in the compound. The process helps in assessing the efficiency of the nitrogen release and capturing in analytical chemistry. Understanding ammonia evolution enables precise quantitative analysis essential for various applications in both academic research and industrial processes.

Organic Compound Analysis

Organic compound analysis involves determining the composition and structure of a compound, including its elemental makeup. In the context of nitrogen estimation, this analysis focuses on evaluating compounds to identify the nitrogen percentage they contain.
Common analytical techniques include combustion analysis, where organic compounds are burned in pure oxygen to produce gases like CO extsubscript{2} and H extsubscript{2}O, alongside nitrogen compounds like N extsubscript{2} or NO extsubscript{x}.
By analyzing the products, specifically the nitrogenous part, chemists can deduce the amount of nitrogen, and thereby assess the compound's formula or structure.

Titration Calculations

Titration is a quantitative technique used to determine the concentrations of unknown solutions through gradual addition of a titrant until the reaction reaches completion. In this exercise, titration calculations determine how much excess acid is neutralized by a known concentration of base, sodium hydroxide.
Steps involve calculating initial moles of acid, moles of base, and using stoichiometry to find excess acid remaining after titration. The results indicate how much ammonia the organic compound evolved, which helps in identifying the compound.

• Start with the moles of the solutions used, which are calculated using the formula: \( ext{moles} = ext{molarity} \times ext{volume} \)
• Use stoichiometry of the neutralization reaction to find moles of acid reacting with ammonia.
• Derive the moles of ammonia that reacted, which equates to nitrogen moles.

Titration calculations provide precise and reliable methods to ascertain chemical concentrations through logical, step-by-step analysis.

Chemical Stoichiometry

Chemical stoichiometry is the calculation of reactants and products in chemical reactions. It revolves around using balanced chemical equations to determine and relate the quantities of different chemical species involved.
For nitrogen estimation, stoichiometry is utilized to calculate the neutralization between sulfuric acid and ammonia. The balanced chemical equation informs about the mole ratios, which is fundamental: \( 2 ext{NaOH} + ext{H}_2 ext{SO}_4 \rightarrow ext{Na}_2 ext{SO}_4 + 2 ext{H}_2 ext{O} \).
This 2:1 ratio between NaOH and H extsubscript{2}SO extsubscript{4} enables precise calculation of how much ammonium sulfate was formed from the initial sulfuric acid. Stoichiometry further reveals the essential linkage between the components in a chemical reaction, allowing for the comprehensive understanding of chemical processes and their quantitative relationships.