Problem 63 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\(5 \mathrm{~mL}\) of a gas containing only carbon and hydrogen were mixed with an excess of oxygen \((30 \mathrm{~mL})\) and the mixture exploded by means of an electric spark. After the explosion, the volume of the mixed gases remaining was \(25 \mathrm{~mL}\). On adding a concentrated solution of potassium hydroxide, the volume further diminished to \(15 \mathrm{~mL}\) of the residual gas being pure oxygen. All volumes have been reduced to N. T.P. Calculate the molecular formula of the hydrocarbon gas. \([1979]\)

Step-by-Step Solution

Answer

The molecular formula of the hydrocarbon is C2H2.

Step 1: Determine Initial Volumes at NTP

Initially, we have a hydrocarbon gas with 5 mL volume, mixed with 30 mL of oxygen. After the explosion, the total volume becomes 25 mL.

Step 2: Volume Analysis after Reaction

Post-explosion, the gases reduced to 25 mL due to the combustion of the hydrocarbon. Since the oxygen was in excess and remaining inactive, the drop indicates the volume used up by the carbon dioxide and water vapor.

Step 3: CO2 Volume Calculation

The introduction of KOH allows us to determine 10 mL of gas as carbon dioxide since it reacts with KOH, dissolving to disappear from the measurement, leading to 15 mL of remaining gas. Thus, volume of CO2 = 10 mL.

Step 4: Determine Moles of CO2

Since 10 mL of CO2 is formed, it suggests all carbon from the hydrocarbon has combusted to CO2. We know volume and moles are directly proportional at NTP, and therefore we assume that CO2 comes entirely from carbon atoms of the hydrocarbon.

Step 5: Calculate Initial Volumes Used

Since 15 mL of the original 30 mL oxygen remains and 10 mL is used to form CO2, 5 mL of oxygen must have contributed to forming water (as O atoms), indicating 5 mL of H2O formed.

Step 6: Determine Hydrocarbon Molecular Formula

For a hydrocarbon CxHy combusting completely to CO2 and H2O, we solve:\[\text{C}_x\text{H}_y + (x+y/4)\text{O}_2 \rightarrow x\text{CO}_2 + y/2\text{H}_2\text{O}\]Using the 5 mL O2 reacting equals the O2 for combustion:CO2 and H2O equilibria indicate formula is \[\text{C}_2\text{H}_2\].

Key Concepts

Molecular Formula Calculation

To deduce the molecular formula of the hydrocarbon based on the given experiment, changes in the volume of gases before and after combustion are crucial. Initially, students must understand that the molecular formula represents the number of each type of atom in a single molecule of the compound.

The provided scenario features a hydrocarbon with 5 mL volume mixing with excess oxygen. Once the mixture reacts, it produces carbon dioxide and water vapor. By carefully analyzing the changes in gas volumes pre- and post-reactive explosion, we can infer the amounts of carbon dioxide and water produced, thus giving insight into the hydrocarbon's molecular composition.

• Volume reductions imply specific gases have reacted or been absorbed.
• 10 mL of CO2 indicates the presence of carbon from the hydrocarbon.
• 5 mL used with water shows involvement of hydrogen.

Using stoichiometry, we calculate the ratio of carbon to hydrogen, revealing that the molecular formula is \(\text{C}_2\text{H}_2\). This understanding of direct volume to molecular conversion is fundamental in such exercises.

Gas Volume Analysis

Gas volume changes signify the consumption or production of gases in reactions. In the experiment, analyzing these shifts reveals insights into the combustion process. Initially, there's a combination of hydrocarbon gas with oxygen—forming a 35 mL reactive mixture. After the explosion, the remaining gas measures 25 mL. This implies that some gases have reacted, primarily the hydrocarbon with oxygen, forming the products carbon dioxide and water vapor. Let's break it down:

• Overall reduction specifies reaction completeness and stoichiometric relationships.
• The 10 mL volume drop, after adding KOH, helps identify the carbon dioxide amount, since KOH captures carbon dioxide gas.
• Remaining oxygen, as noted, narrows down the amounts used versus excess.

Such analysis is key in deducing reaction pathways in gas reactions, assisting students to visually and numerically track how each component contributes to the final measurement changes.

Chemical Stoichiometry

Chemical stoichiometry involves calculating the quantitative relationships of reactants and products in a chemical reaction. In the case of hydrocarbon combustion, stoichiometry lets students map out the required oxygen and resultant carbon dioxide and water molecules. In this exercise, recognize that the ratio of initial substances to final products stems directly from the balanced equation of the combustion:

• The hydrocarbon takes in oxygen proportionally to the number of carbon and hydrogen atoms.
• The disappearance of 10 mL due to KOH absorption indicates the exact volume of CO2 formed.
• Remaining oxygen not tied to reaction reveals unreacted volume, simplifying equation balancing.

Practicing chemical stoichiometry with gas reactions, such as this, grants a hands-on understanding of how to derive molecular formulas from real reaction data. This linking of volumetric data to stoichiometric coefficients is fundamental in mastering chemistry calculations.