Problem 6 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

The ratio of mass percent of \(\mathrm{C}\) and \(\mathrm{H}\) of an organic compound \(\left(\mathrm{C}_{\mathrm{X}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}\right)\) is \(6: 1\). If one molecule of the above compound \(\left(\mathrm{C}_{\mathrm{X}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}\right)\) contains half as much oxygen as required to burn one molecule of compound \(\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{Y}}\) completely to \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\). The empirical formula of compound \(\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{Y}} \mathrm{O}_{\mathrm{Z}}\) is : (a) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{4} \mathrm{O}_{2}\) (d) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{3}\)

Step-by-Step Solution

Answer

The empirical formula is \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}\) (option b).

Step 1: Understand the Ratio of Mass Percent

The problem states that the mass ratio of carbon (C) to hydrogen (H) in the compound is 6:1. This ratio is by mass, so it accounts for their atomic masses: carbon has an atomic mass of 12, and hydrogen has an atomic mass of 1. To find the mole ratio, we have to convert this mass ratio to a mole ratio.

Step 2: Convert Mass Ratio to Mole Ratio

Given the mass of 6 for C and 1 for H, divide the mass by their respective atomic masses to get moles: Carbon moles = \(\frac{6}{12} = 0.5 \), Hydrogen moles = \(\frac{1}{1} = 1 \). This simplifies to a mole ratio for C:H of 1:2.

Step 3: Analyze Oxygen Requirement

A complete combustion reaction for \( \text{CxHy} \) would require exactly \( \frac{y}{4} + x \) moles of O2, as per the stoichiometry of combustion reactions to form CO2 and H2O. The problem states that the oxygen content is half of what's needed for complete combustion.

Step 4: Determine the Oxygen Content

From the previous step, the required moles of O2 per mole of \( \text{CxHy} \) are \( \frac{y}{4} + x \). Since the compound contains only half this amount, the oxygen in \( \text{CxHyOz} \) should be \( \frac{y}{8} + \frac{x}{2} \).

Step 5: Find Possible Empirical Formulas

Potential formulas based on the given options are tested against the derived ratios: \( \text{C} \) and \( \text{H} \) ratio of 1:2 and for O content as depicted in the derived expression. Examine each option: (a) \( \text{C}_{3}\text{H}_{6}\text{O}_{3} \) does not match the simplification; (b) \( \text{C}_{2}\text{H}_{4}\text{O} \) matches both the centimeter and oxygen content derived; (c) \( \text{C}_{3}\text{H}_{4}\text{O}_{2} \) matches neither; (d) \( \text{C}_{2}\text{H}_{4}\text{O}_{3} \) misses the oxygen requirement.

Step 6: Verify Empirical Formula

By substitution, only \( \text{C}_{2}\text{H}_{4}\text{O} \) perfectly matches the derived stoichiometry, fulfilling both the requirement of mass ratio and the obvious criterion of oxygen required.

Key Concepts

Combustion Reaction

Combustion reactions are fundamental chemical processes where a substance combines rapidly with oxygen to release energy in the form of heat and light. During combustion, hydrocarbons, which are compounds made of hydrogen and carbon, typically react with oxygen to produce carbon dioxide (\(\mathrm{CO_2}\)) and water (\(\mathrm{H_2O}\)).
The general formula for the combustion reaction of a hydrocarbon \(\mathrm{C_x H_y}\) is:\[\mathrm{C_x H_y} + \left(\frac{y}{4}+x\right) \mathrm{O_2} \rightarrow x\ \mathrm{CO_2} + \frac{y}{2}\ \mathrm{H_2O}\]This equation illustrates the stoichiometry, or the exact proportions, needed for a complete combustion reaction.
In our exercise, we're analyzing a compound \(\mathrm{C_x H_y O_z}\), where it states that the compound has half the oxygen needed for complete combustion. This implies that we calculate the moles of oxygen necessary for combustion and then take half to find the actual oxygen content in the compound due to stoichiometric limitations.

Mole Ratio

The mole ratio is a key concept in stoichiometry that allows chemists to understand the quantitative relationships in a chemical reaction. It is derived from the coefficients of the balanced chemical equation and is crucial for calculating the amounts of reactants and products involved in the reaction.
For the problem at hand, the mass percent ratio of carbon to hydrogen is given as \(6:1\). To find the mole ratio based on this mass ratio:

• First, we take the mass for carbon as 6 and hydrogen as 1.
• Next, we convert these masses to moles using their atomic masses: 12 for carbon and 1 for hydrogen.
• Carbon moles are calculated as \(\frac{6}{12} = 0.5\) and hydrogen moles as \(\frac{1}{1} = 1\).

Thus, the mole ratio becomes \(1:2\) for carbon to hydrogen.
This ratio assists in determining the empirical formula of the compound by maintaining a proportional balance of atoms.

Mass Percent Ratio

Mass percent ratio indicates how much of each element is present by mass in a compound. It provides insight into the composition of chemical substances and is critical for deriving empirical formulas.
In the given problem, the mass percent ratio of carbon to hydrogen is specified as 6:1. This means carbon makes up a significantly larger portion of the compound by mass compared to hydrogen. Calculating the empirical formula from this ratio involves these key steps:

• Identify the mass ratio and convert to mole ratio by dividing by atomic masses of carbon (12) and hydrogen (1).
• The mole numbers determine how many atoms of each element are present in the simplest formula.

Understanding this ratio allows you to extrapolate the simplest whole number ratio of atoms - what we call the empirical formula. By matching this with possible molecular compositions (given in options in the exercise), the empirical formula that fulfills both the mass percent and oxygen condition (`as stated in the combustion analysis`) can be determined accurately.