Problem 42 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

Consider the reaction sequence from \(\mathbf{P}\) to \(\mathbf{Q}\) shown below. The overall yield of the major product \(\mathbf{Q}\) from \(\mathbf{P}\) is \(75 \%\). What is the amount in grams of \(\mathbf{Q}\) obtained from \(9.3 \mathrm{~mL}\) of \(\mathbf{P} ?\) (Use density of \(\mathbf{P}=1.00 \mathrm{~g}\) \(\mathrm{mL}^{-1} ;\) Molar mass of \(\mathrm{C}=12.0, \mathrm{H}=1.0, \mathrm{O}=16.0\) and \(\mathrm{N}=14.0 \mathrm{~g} \mathrm{~mol}^{-}\) ')

Step-by-Step Solution

Answer

The actual amount of Q obtained is 6.975 g.

Step 1: Calculate Mass of P

To find the mass of \( P \), use the formula for density: \( \text{mass} = \text{density} \times \text{volume} \). From the problem, the density of \( P \) is \( 1.00 \, \mathrm{g/mL} \) and the volume is \( 9.3 \, \mathrm{mL} \). Thus, the mass of \( P \) is \( 9.3 \, \mathrm{g} \).

Step 2: Determine Theoretical Yield

Assuming 100% conversion, the theoretical yield would be equal to the mass of \( P \) before any losses, which is \( 9.3 \, \mathrm{g} \).

Step 3: Calculate Actual Yield of Q

Given the overall yield of the major product \( Q \) from \( P \) is \( 75\% \), calculate the actual yield using: \( \mathrm{Actual \ yield} = \mathrm{Theoretical \ yield} \times \frac{\text{percentage yield}}{100} \). Thus, \( \mathrm{Actual \ yield} = 9.3 \, \mathrm{g} \times 0.75 = 6.975 \, \mathrm{g} \).

Step 4: Conclusion: Actual Amount of Q

The actual amount of \( Q \) obtained from \( 9.3 \, \mathrm{mL} \) of \( P \) at a \( 75\% \) yield is \( 6.975 \, \mathrm{g} \).

Key Concepts

Understanding Reaction Yield

In chemistry, reaction yield is an important concept that relates to how much product is formed from a reaction compared to the theoretical amount. The concept of theoretical yield assumes that all of the starting material (reactants) converts perfectly to the product, without any waste. However, in reality, this is often not the case. There are many reasons why the actual yield may be less than the theoretical yield, including incomplete reactions, side reactions, or practical issues such as product loss during transfer.
To calculate the reaction yield percentage, you use the formula:
\[ \text{Reaction Yield ( ext{Percentage})} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \% \]
Knowing the percentage yield is valuable, as it tells you the efficiency of the reaction under given conditions. In the provided exercise, the reaction yield from P to Q is 75%. This means only 75% of the theoretical yield is achieved in practice.

Steps in Mass Calculation

Mass calculation often serves as the foundation for understanding quantities in chemical reactions. In our example problem, we start by calculating the mass of substance P using the formula:

• Density \( \times \) Volume = Mass

Given the density of P is 1.00 g/mL and its volume is 9.3 mL, the mass of P is found to be:
\[ \text{Mass of } P = 1.00 \, \mathrm{g/mL} \times 9.3 \, \mathrm{mL} = 9.3 \, \mathrm{g} \]
This mass calculation is critical as it serves as the starting point for determining the theoretical yield. It is always essential to calculate the mass accurately to ensure the subsequent calculations for yield and conversion are correct. Any error at this stage could carry forward, affecting the accuracy of results.

Conversion of Volume to Mass in Chemistry

Conversion of volume to mass is a frequently encountered task in chemical calculations, especially when dealing with liquids. As seen in the example, understanding how to use density to convert between these two properties is important. Density acts as a bridge in this conversion.
The relationship is given by:

• Mass = Density \( \times \) Volume

This simple multiplication helps convert the volume of a liquid into a tangible mass, which is more easily used for further chemical calculations such as determining the number of moles or reaction yields.
In our example, knowing the density of P allowed us to convert its measured volume into mass (9.3 grams). This step is fundamental in stoichiometry where precise mass and moles are necessary to predict the products of a reaction accurately. Therefore, mastering the volume-to-mass conversion is a key skill in chemical sciences.