Problem 179 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(300 \mathrm{~K}\) and \(1 \mathrm{~atm}, 15 \mathrm{~mL}\) a gaseous hydrocarbon requires \(375 \mathrm{~mL}\) air containing \(20 \% \mathrm{O}_{2}\) by volume for complete combustion. After comustion the gases occupy \(330 \mathrm{~mL}\). Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is: (a) \(\mathrm{C}_{3} \mathrm{H}_{8}\) (b) \(\mathrm{C}_{4} \mathrm{H}_{8}\) (c) \(\mathrm{C}_{4} \mathrm{H}_{10}\) (d) \(\mathrm{C}_{3} \mathrm{H}_{6}\)

Step-by-Step Solution

Answer

The formula of the hydrocarbon is \(\mathrm{C}_3\mathrm{H}_8\).

Step 1: Understanding the Problem

We have a hydrocarbon undergoing complete combustion in the presence of oxygen. This reaction produces carbon dioxide and potentially water vapor, but we assume water forms as a liquid. The key is to find the hydrocarbon's formula given the volume changes in gaseous products and reactants.

Step 2: Determine Oxygen Utilized

Since the air is 20% oxygen, the 375 mL of air contains \(0.2 \times 375 = 75\) mL of \(\mathrm{O}_2\). Thus, 75 mL of oxygen was used for combustion.

Step 3: Analyze the Gaseous Volume Change

The combustion of the hydrocarbon results in a decrease in gas volume from 375 mL (total gaseous reactants) to 330 mL (total gaseous products), indicating the hydrocarbon absorbed some oxygen volume. Initially, there is 390 mL of gas (375 mL air + 15 mL hydrocarbon), decreasing to 330 mL, confirming that water is accounted as a liquid.

Step 4: Derive Hydrocarbon Product Volumes

The 15 mL hydrocarbon reacted to entirely consume 75 mL of \(\mathrm{O}_2\). The reduction of 45 mL suggests 45 mL (from 75 mL of \(\mathrm{O}_2\) and 15 mL hydrocarbon) of volume was converted to liquid form (water), leaving 30 mL as carbon dioxide (i.e. 30 mL gas after combustion from hydrocarbon).

Step 5: Equate to Hydrocarbon Formulas

Since 30 mL of \(\mathrm{CO}_2\) is produced, deduce the carbon atoms in the hydrocarbon as 3 (15 mL hydrocarbon yields 30 mL \(\mathrm{CO}_2\), thus the conversion factor is \(\mathrm{nCO}_2=2n_{\text{hydrocarbon}}\)). A reasonable hydrocarbon formula with 3 carbons is \(\mathrm{C}_3\). Therefore, the original hydrocarbon is either \(\mathrm{C}_3\mathrm{H}_8\) or \(\mathrm{C}_3\mathrm{H}_6\) which can be validated against hydrogen water formation.

Step 6: Balance Hydrogen and Water Formation

Consider \(\mathrm{C}_3\mathrm{H}_8\) (propane): Complete combustion forms carbon dioxide and water. For 3 carbons, \(\mathrm{C}_3\), to yield 15 mL of water at standard, 8 hydrogens ensure water formation, confirming \(\mathrm{C}_3\mathrm{H}_8\) is viable, \(\mathrm{C}_3\mathrm{H}_6\) would not match water outcome under identical proportions (i.e., incorrect hydrogen accounting).

Step 7: Conclusion: Identify Hydrocarbon Formula

Matching combustion products to typical hydrocarbon combustion indicates that the correct formula for the hydrocarbon leading to these specific outcomes is \(\mathrm{C}_3\mathrm{H}_8\), found by occlusion testing and volume adjustments.

Key Concepts

Hydrocarbon Combustion

Hydrocarbon combustion is a fundamental chemical reaction in which a hydrocarbon reacts with oxygen to produce carbon dioxide and water. This process releases energy, often in the form of heat and light. A complete combustion requires a sufficient amount of oxygen to ensure that all the carbon and hydrogen in the hydrocarbon fully oxidize. In the exercise we are looking at, the hydrocarbon combusts completely.Hydrocarbons can vary in structure and complexity. The simplest, methane (CH₄), combines with oxygen (O₂) to form carbon dioxide (CO₂) and water (H₂O) as follows:\[CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O\]This equation shows the stoichiometry in combustion reactions. For hydrocarbons with higher numbers of carbon, such as propane (C₃H₈), the combustion equation would be:\[C_3H_8 + 5O_2 \rightarrow 3CO_2 + 4H_2O\]This illustrates that the complete combustion of propane results in three molecules of carbon dioxide and four molecules of water vapor. Understanding the combustion reaction is crucial for deducing the empirical formula of hydrocarbons through stoichiometry and volume changes.

Gas Volume Changes

Gas volume changes are a vital aspect of combustion reactions, especially when dealing with gases. During a combustion reaction, the volume of gases may change due to the consumption of reactants and production of new gaseous products. This principle is utilized in the provided exercise to help calculate the hydrocarbon's formula. Initially, we have 375 mL of air mixed with 15 mL of hydrocarbon, which totals 390 mL of gases pre-combustion. After the reaction, the total gas volume reduces to 330 mL. This decrease is because some of the gaseous reactants (O₂ in this case) are fully consumed, and some products (water) may be in liquid form, not affecting the gaseous volume. By analyzing the gaseous volumes, differentials between the volumes of reactants and products can help identify the precise stoichiometric ratios involved. In the example, the reduction from 375 mL (of air including 75 mL of O₂) to 330 mL (post-combustion) implies a significant change in gaseous volume due to combustion, indicating potential liquid water formation and complete consumption of oxygen by the hydrocarbon.

Chemical Stoichiometry

Chemical stoichiometry refers to the calculation of reactants and products in chemical reactions, which is fundamental for balancing chemical equations. In this exercise, stoichiometry helps deduce the formula of a hydrocarbon from the volumes of gaseous compounds involved. The reaction of any hydrocarbon with oxygen follows a stoichiometric pattern where the amounts of reactants and products are proportional. Given the volume of oxygen and hydrocarbon, one can determine the stoichiometry of the reaction. For instance, if 75 mL of O₂ is used to combust 15 mL of a hydrocarbon, and we end up with 30 mL of CO₂, as seen, the stochiometric calculation involves finding the balance point that matches these volumes with typical formulas of hydrocarbons. The balance point is determined by recognizing that for each mole of carbon in the hydrocarbon, we can expect a mole of CO₂. Applying this to the mL to mL ratio helps deduce whether the hydrocarbon is a simple compound like propane (C₃H₈) through its match with the produced carbon dioxide and inferred water quantities. Hence, the stoichiometric analysis of volume changes guides us to the hydrocarbon’s formula in the exercise.