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.

Step 1: Calculate the Initial Moles of Sulfuric Acid

First, we calculate the moles of sulfuric acid used initially. Since the molarity of sulfuric acid is given as \(0.1 \, \text{M}\) and its volume is \(100 \, \text{mL}\), we can calculate the moles using the formula:\[ \text{Moles of } H_2SO_4 = 0.1 \, \text{M} \times 0.1 \, \text{L} \]This gives us \(0.01 \text{ moles of } H_2SO_4\).

Step 2: Calculate the Moles of NaOH Used

Next, we use the given data that \(20 \, \text{mL}\) of \(0.5 \, \text{M}\) sodium hydroxide is used for neutralization. Calculate the moles of NaOH:\[ \text{Moles of } NaOH = 0.5 \, \text{M} \times 0.02 \, \text{L} \]This results in \(0.01 \text{ moles of } NaOH\).

Step 3: Calculate the Excess Moles of Acid

Since \(0.01\) moles of NaOH neutralize \(0.01\) moles of the sulfuric acid, there are \(0.01\) moles of sulfuric acid that were in excess. Therefore, the moles of sulfuric acid initially subtracted by the moles of NaOH used gives us the moles of sulfuric acid that reacted with ammonia, which is \(0.01 - 0.01 = 0\) moles reacted with ammonia.

Step 4: Identify Moles of Ammonia Produced

We observe that all the sulfuric acid was neutralized by the ammonia, meaning \(0.01\) moles of ammonia reacted. Therefore, the ammonia evolved corresponds to the moles of nitrogen in the organic compound.

Step 5: Compare with the Options

From the given options, calculate the moles of nitrogen for each compound based on their molar mass and nitrogen content, then match this with \(0.01\) moles.- Benzamide (molar mass: 121.14 g/mol, contains 1 nitrogen): \(M_{benzamide} = \frac{0.3}{121.14} \approx 0.0025\) mol.- Acetamide (molar mass: 59.07 g/mol, contains 1 nitrogen): \(M_{acetamide} = \frac{0.3}{59.07} \approx 0.0051\) mol.- Thiourea (molar mass: 76.12 g/mol, contains 2 nitrogens): \(M_{thiourea} = \frac{0.3}{76.12} \times 2 \approx 0.0079\) mol.- Urea (molar mass: 60.06 g/mol, contains 2 nitrogens): \(M_{urea} = \frac{0.3}{60.06} \times 2 \approx 0.01\) mol.The calculations show that urea produces universally \(0.01\) moles of nitrogen as required.

Key Concepts

Molecular mass calculation

Calculating molecular mass is an essential skill in chemistry. It involves adding together the atomic masses of all atoms present in a molecule. This is important because it allows us to convert between mass and moles for a given substance, crucial for quantitative chemistry. Each element has its own atomic mass, which can be found on the periodic table. The formula for calculating the molecular mass is as follows:

• Identify each element in the molecule and count the number of atoms of each element.
• Find the atomic mass for each element from the periodic table. For example, carbon is about 12 amu (atomic mass units).
• Multiply the atomic mass of each element by the number of atoms of that element in the molecule, then add all these values together to get the total molecular mass.

For example, to calculate the molecular mass of urea (CH_4N_2O1), you would add up the contributions from carbon, hydrogen, nitrogen, and oxygen based on their atomic masses: Carbon (1 atom  imes 12), Hydrogen (4 atoms  imes 1), Nitrogen (2 atoms  imes 14), and Oxygen (1 atom  imes 16). This results in a total molecular mass of approximately 60.06 g/mol.
Understanding this allows us to calculate moles, which is pivotal for solving many chemical problems, especially when reacting appropriate amounts of substances together.

Stoichiometry in chemical reactions

Stoichiometry is the math behind chemistry. It refers to the calculation of reactants and products in chemical reactions. Practicing stoichiometry ensures that students understand how chemical quantities relate to each other in a balanced reaction. When given a chemical process, stoichiometry helps in predicting how much of each reactant is necessary to produce a desired amount of product. This involves several steps:

• Writing and balancing the chemical equation for the reaction. This step ensures that the number of atoms on both the reactant and product sides are equal.
• Using coefficients from the balanced equation to find the mole ratio between reactants and products. This helps in determining the relative amounts of each substance involved in the reaction.
• Using the relationship between moles and molecular mass to convert given mass into moles and vice versa.

For instance, in the exercise involving the compound urea, stoichiometry is key to understanding the balance of nitrogen atoms between reactants and products. By knowing the molar mass of urea and calculating that its decomposition releases two nitrogen atoms, we can figure out that it produces 0.01 moles of nitrogen from an initial sample of 0.3 grams. This clear relationship is visualized using stoichiometry.

Neutralization reactions

In chemistry, neutralization reactions involve the reaction of acids and bases to form water and salt. This concept is integral in laboratory procedures and real-world chemical applications. During a neutralization reaction, the acid donates protons ( ext{H}^+) while the base offers hydroxide ions ( ext{OH}^-), resulting in water. It's important for students to comprehend this because these reactions are fundamental to many processes, including titration.

• A typical example outlined in the exercise is the neutralization of sulfuric acid and sodium hydroxide.
• The balanced reaction equation helps in determining the stoichiometric relationship: 2 imes ext{NaOH} + ext{H}_2 ext{SO}_4 ightarrow ext{Na}_2 ext{SO}_4 + 2 ext{H}_2 ext{O}.
• By understanding this chemical reaction, one can calculate how much of each reactant is needed to reach complete neutralization.

In the given problem, the excess sulfuric acid that wasn't neutralized by ammonia was completely reacted with sodium hydroxide, revealing the initial amount of ammonia produced. Neutralization reactions like these exemplify dynamic and practical chemical interactions, illustrating the importance of balancing reactions and predicting reactant-product quantities.