Problem 74 - A Complete Resource Book in Chemistry for JEE Main

Question

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 \(0.35 \mathrm{~g}\) of silver chloride. The percentage of chlorine in the compound is (a) \(36.6 \%\) (b) \(45.3 \%\) (c) \(34.6 \%\) (d) \(54.8 \%\)

Step-by-Step Solution

Answer

The percentage of chlorine in the compound is 34.6% (option c).

Step 1: Write the Reaction Equation

The organic compound containing chlorine reacts under the described conditions to form silver chloride. The reaction can be written as: \[ \text{Compound} + \text{AgNO}_3 + \text{HNO}_3 \rightarrow \text{AgCl} + \text{other products} \]

Step 2: Calculate the Moles of AgCl

The molar mass of silver chloride (AgCl) is calculated by adding the atomic masses of silver (Ag) and chlorine (Cl): \[ M_{\text{AgCl}} = 107.87 + 35.45 = 143.32 \text{ g/mol} \] With 0.35 grams of AgCl produced, we can find the moles of AgCl: \[ n_{\text{AgCl}} = \frac{0.35}{143.32} \approx 0.00244 \text{ moles} \]

Step 3: Determine Moles of Chlorine

Each mole of AgCl contains one mole of chlorine. Thus, the moles of Cl in the sample is also approximately 0.00244 moles.

Step 4: Calculate Mass of Chlorine

The mass of chlorine can be computed using its molar mass (35.45 g/mol): \[ m_{\text{Cl}} = n_{\text{Cl}} \times M_{\text{Cl}} = 0.00244 \times 35.45 \approx 0.0865 \text{ g} \]

Step 5: Calculate Percentage of Chlorine

The percentage of chlorine in the original compound (0.25 g) is calculated by: \[ \text{Percentage of Cl} = \left( \frac{0.0865}{0.25} \right) \times 100 \approx 34.6\% \]

Step 6: Final Answer Selection

From the given options, the closest percentage is 34.6% which corresponds to option (c).

Key Concepts

Percentage Composition

Percentage composition is a useful concept in chemistry that shows the relative amount of each element in a compound. It is expressed in terms of percentage by mass and helps in identifying the proportions of elements in a given sample. To find the percentage composition of a certain element in a compound, you need the mass of the element in the sample and the total mass of the sample itself.
To calculate the percentage composition of chlorine in an unknown compound, we use the formula:

• First, calculate the mass of the element (chlorine in this case).
• Then divide the mass of the chlorine by the total mass of the compound sample.
• Multiply the result by 100 to convert it to a percentage.

This measurement is crucial as it provides insight into the chemical makeup of a compound, which informs other analyses like the molecular structure or when synthesizing new compounds.

Molar Mass Calculation

Molar mass is an essential concept in chemistry that refers to the mass of one mole of a substance. It allows us to convert between the mass of a sample and the number of moles of the substance. This conversion is vital in stoichiometry and helps in understanding how substances interact in chemical reactions.
To calculate the molar mass, sum the atomic masses of all atoms in a molecule, as found on the periodic table.

For this example, we determine the molar mass of silver chloride (AgCl) by adding the atomic weight of silver (approx. 107.87 g/mol) and chlorine (approx. 35.45 g/mol):

• Molar Mass of AgCl = 107.87 + 35.45 = 143.32 g/mol.

Once the molar mass is known, it is possible to calculate the number of moles present in any given mass of the compound, which is crucial for further percentage composition calculations or reaction yield assessments.

Chemical Reaction Equation

A chemical reaction equation is a symbolic representation of a chemical reaction where reactants are transformed into products. These equations allow chemists to understand and predict the outcomes of chemical reactions. They show the conservation of mass, meaning the number of atoms of each element is the same on both sides of the equation.
Writing a chemical equation involves:

• Identifying the reactants and products of the reaction.
• Writing their chemical formulas.
• Balancing the equation so that the number of atoms for each element is the same on both sides.

In the given example, the reaction between an organic compound containing chlorine, silver nitrate (\( \text{AgNO}_3 \) and nitric acid \( \text{HNO}_3 \), produces silver chloride (\( \text{AgCl} \)) and other products. The equation needs to clearly demonstrate the stoichiometric relationships between reactants and products, which allow for calculations like those to determine the amount of chlorine in the starting compound, beautifully illustrating the workspace of analytical chemistry.