Problem 137 - A Complete Resource Book in Chemistry for JEE Main

Question

\(15 \mathrm{~mL}\) of gaseous hydrocarbon requires \(45 \mathrm{~mL}\) of oxygen for complete combustion which produces \(30 \mathrm{~mL}\) of \(\mathrm{CO}_{2}\) gas, measured under identical conditions. The formula of the hydrocarbon is \(\mathrm{C}_{x} \mathrm{H}_{y}\). The ratio \(\underline{\mathrm{y}}\) is X

Step-by-Step Solution

Answer

The value of \( y \) is 4.

Step 1: Write the Combustion Reaction Equation

The general reaction for the complete combustion of a hydrocarbon \( \mathrm{C}_x \mathrm{H}_y \) with oxygen \( \mathrm{O}_2 \) is:\[\mathrm{C}_x \mathrm{H}_y + \left( x + \frac{y}{4} \right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2 + \frac{y}{2} \mathrm{H}_2\mathrm{O}\]Given that the reaction produces \( 30 \; \mathrm{mL} \) of \( \mathrm{CO}_2 \), we know \( x = 2 \) since each \( \mathrm{CO}_2 \) molecule corresponds to 1 \( \mathrm{C} \) atom.

Step 2: Use the Given Volumes for Stoichiometry

We have \( 15 \; \mathrm{mL} \) of hydrocarbon reacting with \( 45 \; \mathrm{mL} \) of \( \mathrm{O}_2 \). The stoichiometry from the reaction equation requires:\[15 \; \mathrm{mL} \; \mathrm{C}_x\mathrm{H}_y + (15)\left(x + \frac{y}{4}\right) \; \mathrm{mL} \; \mathrm{O}_2 \]Plugging in \( x = 2 \), we have:\[15 \times \left(2 + \frac{y}{4}\right) = 45\]This simplifies to \( 30 + \frac{15y}{4} = 45 \).

Step 3: Solve for the Unknown y

Rearrange the equation \( 30 + \frac{15y}{4} = 45 \) to solve for \( y \):\[\frac{15y}{4} = 15\]Multiply both sides by 4:\[15y = 60\]Divide by 15 to solve for \( y \):\[y = 4\]Thus, the hydrocarbon has the formula \( \mathrm{C}_2\mathrm{H}_4 \).

Key Concepts

Stoichiometry

Stoichiometry is a key concept in chemistry that helps us understand the quantitative relationships in chemical reactions. It allows us to calculate how much of each reactant is needed and how much product is produced by using balanced chemical equations.

To understand stoichiometry, it's crucial to start with a correctly balanced equation like the combustion one here: \[\mathrm{C}_x \mathrm{H}_y + \left( x + \frac{y}{4} \right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2 + \frac{y}{2} \mathrm{H}_2\mathrm{O}\]

This equation tells us that one hydrocarbon molecule \(\mathrm{C}_x \mathrm{H}_y\) reacts with oxygen molecules \((\mathrm{O}_2)\) to form carbon dioxide \((\mathrm{CO}_2)\) and water \((\mathrm{H}_2\mathrm{O})\). By knowing the ratios from the balanced equation, we can find out the exact volumes or moles needed for the reaction.

For example, here we know \(15 \; \mathrm{mL}\) of hydrocarbon needs \(45 \; \mathrm{mL}\) of oxygen, which tells us about the mole ratio and allows us to calculate the values for \( x \) and \( y \) in the formulas.

Hydrocarbon Formulas

Hydrocarbons are organic compounds consisting entirely of hydrogen and carbon. The simplest of these are alkanes, which have the general formula \(\mathrm{C}_x \mathrm{H}_{2x+2}\), but in this exercise, we have a general formula \(\mathrm{C}_x \mathrm{H}_y\), allowing us to explore various electron configurations and bonding possibilities.

To find the specific formula of a hydrocarbon using given volumes, we first rely on the amounts of its combustion products. Here, knowing that \(30 \; \mathrm{mL}\) of \(\mathrm{CO}_2\) is produced helps us determine \(x\) because each \(\mathrm{CO}_2\) molecule represents one carbon atom. Therefore, if \(x = 2\), the formula starts as \(\mathrm{C}_2\).

Next, using the stoichiometric relations described above, we calculate \(y\). This involves solving: \[15 \times \left(2 + \frac{y}{4}\right) = 45\] By simplifying, we find \(y = 4\), leading to the hydrocarbon formula \(\mathrm{C}_2 \mathrm{H}_4\).

This shows how stoichiometry and understanding the structure of hydrocarbons guide us to determine precise molecular formulas.

Chemical Reaction Equations

Chemical reaction equations give us a clear way to describe what happens in a chemical reaction. They use symbols and formulas to succinctly represent the substances involved.

For combustion reactions, such as the one in this exercise, it’s essential to write a balanced equation that reflects the conservation of mass. This means having the same number of each type of atom on both sides of the equation. Consider again: \[\mathrm{C}_x \mathrm{H}_y + \left( x + \frac{y}{4} \right) \mathrm{O}_2 \rightarrow x \mathrm{CO}_2 + \frac{y}{2} \mathrm{H}_2\mathrm{O}\]

In this combustion equation, the hydrocarbon \(\mathrm{C}_x \mathrm{H}_y\) reacts with oxygen to produce carbon dioxide and water. We've balanced it using stoichiometric coefficients. Each part of the equation corresponds to a real amount of reactants and products, which can be calculated, as is done in this exercise.

Equations like these are the foundation of studying chemical reactions. They allow chemists to predict amounts and identify reactant or product requirements, leading to practical applications in fields like pharmaceuticals, energy production, and environmental science.