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 \(\mathrm{C}_{\mathrm{x}} \mathrm{H}_{y}\). The ratio \(\underline{\mathrm{y}}\) is \(\mathrm{x}\) _______.

Step-by-Step Solution

Answer

The ratio \(\underline{\mathrm{y}}/\mathrm{x}\) is 1.

Step 1: Understand the Problem

We need to determine the formula of a hydrocarbon based on its combustion with oxygen, producing carbon dioxide and water. The problem gives us the volumes of the gases involved in the reaction, and our goal is to find the ratio of hydrogen to carbon atoms in the hydrocarbon formula, represented as \(\underline{\mathrm{y}} / \mathrm{x}\).

Step 2: Write the Combustion Equation

The combustion of a hydrocarbon \( \mathrm{C}_x \mathrm{H}_y \) is given by:\[ \mathrm{C}_x \mathrm{H}_y + O_2 \rightarrow x \mathrm{CO}_2 + \frac{y}{2} \mathrm{H}_2O \].We know that 15 mL of the hydrocarbon requires 45 mL of oxygen and produces 30 mL of carbon dioxide.

Step 3: Apply Volume Ratios from the Reaction

Given that 15 mL of the hydrocarbon produces 30 mL of \( \mathrm{CO}_2 \), and using Avogadro's Law, which states that equal volumes of gases at the same conditions contain equal numbers of molecules, we conclude \(x = 2\) because it produces twice the amount of \( \mathrm{CO}_2 \) compared to the volume of the hydrocarbon consumed.

Step 4: Analyze the Oxygen Requirement

45 mL of O\(_2\) is consumed. We know from the equation \( \mathrm{C}_x \mathrm{H}_y + O_2 \rightarrow x \mathrm{CO}_2 + \frac{y}{2} \mathrm{H}_2O \) that each carbon atom requires one O\(_2\) for complete combustion to form \( \mathrm{CO}_2 \), and hydrogen requires \( \frac{y}{2} \) O\(_2\) to form water.

Step 5: Solve the Equation

The total oxygen used is given as 45 mL, fulfilling both the carbon to \( \mathrm{CO}_2 \) and hydrogen to \( \mathrm{H}_2O \) reactions. Therefore,\[ 2\cdot15 + \frac{y}{2} \cdot15 = 45 \]. Solving gives:\[ 30 + \frac{15y}{2} = 45 \] \[ \frac{15y}{2} = 15 \] \[ y = 2 \].

Step 6: Write the Formula and the Ratio

The hydrocarbon formula is found to be \( \mathrm{C}_2\mathrm{H}_2 \), which gives the ratio \( y/x = 2/2 = 1 \).

Key Concepts

Hydrocarbon formula determination

Determining the formula of a hydrocarbon involves understanding how it reacts with oxygen in a combustion process to produce carbon dioxide and water. A hydrocarbon is a compound made only of carbon and hydrogen atoms. The formula of a hydrocarbon can be represented as \( \mathrm{C}_x \mathrm{H}_y \), where \( x \) stands for the number of carbon atoms, and \( y \) for the number of hydrogen atoms in the molecule.

To identify these numbers, we apply the principles of stoichiometry and volume analysis based on a combustion reaction. The key step is to utilize the volumes of gases involved in the reaction. According to Avogadro's law, equal volumes of gases at the same conditions contain equal numbers of molecules. This means we can directly compare the volumes provided in the problem to understand the ratios of molecules reacting or produced. By analyzing the given information and applying these principles, you can find clear relationships between the volumes of the hydrocarbon used, and the products formed as well as the reactants consumed per the balanced chemical equation. This helps in concluding that the formula of such a hydrocarbon is derived from the carbon and hydrogen atom stoichiometry shown in the combustion equation.

Volume ratios in reactions

Volume ratios play a crucial role in analyzing chemical reactions involving gases, especially combustion reactions. In the exercise scenario, we examined the combustion of a hydrocarbon, producing carbon dioxide and water, with given volumes of the hydrocarbon, oxygen, and carbon dioxide.

Using Avogadro's Law, we can deduce that since 15 mL of the hydrocarbon produced 30 mL of carbon dioxide, this indicates a molar ratio. The hydrocarbon produces twice the volume of carbon dioxide, meaning each molecule of the hydrocarbon has two carbon atoms, as one mole of carbon dioxide comes from one mole of carbon atom. This helps us identify \( x = 2 \) in the formula \( \mathrm{C}_x \mathrm{H}_y \). The next step is to determine how much oxygen was required, which, through stoichiometry, gives us further insight into the composition of the hydrocarbon.

By exploring the relationships between these gas volumes, we can conclude with confidence which elements and in what ratios participate in the reaction, helping you to deduce the molecular composition of the reacting hydrocarbon.

Balancing chemical equations

Balancing chemical equations is essential in setting a clear relationship between reactants and products in a chemical reaction. It ensures that the number of atoms for each element is equal on both sides of the equation. In the case of analyzing hydrocarbons, this balancing act helps relate known gas volumes to unknown molecular formulas.

When balancing the combustion equation \( \mathrm{C}_x \mathrm{H}_y + O_2 \rightarrow x \mathrm{CO}_2 + \frac{y}{2} \mathrm{H}_2O \), we matched the given volume relationships to elements participating in the reaction. The coefficients obtained through balancing are not just numbers; they indicate the exact ratios of molecules reacting and forming. When we determined that 45 mL of oxygen are used,

• This amount verified through balancing is in harmony with the quantities of CO\(_2\) and water that should theoretically result from combustion given the balanced equation.
• Balancing the equation verified the results, as each carbon atom requires one oxygen molecule to form CO\(_2\), effectively utilizing 30 mL of oxygen in total verification.

The balance and stoichiometry thus confirmed that \( y = 2\) in \(\mathrm{C}_2 \mathrm{H}_2\), supporting the final ratio \( y/x = 1 \). Effective balancing illustrates the broader principles of conserving mass and energy in chemical reactions.