Problem 46 - A Complete Resource Book in Chemistry for JEE Main

Question

A dibasic acid containing \(\mathrm{C}, \mathrm{H}\) and \(\mathrm{O}\) was found to contain \(\mathrm{C}=26.7 \%\) and \(\mathrm{H}=2.2 \%\). The vapour density of diethyl ester of the acid was found to be \(73 .\) What is the molecular formula of the acid? (a) \(\mathrm{CH}_{2} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{2} \mathrm{O}_{4}\) (c) \(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{O}_{4}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{O}_{4}\)

Step-by-Step Solution

Answer

The molecular formula of the acid is \( C_2H_2O_4 \) (option b).

Step 1: Calculate the percentage of Oxygen

The dibasic acid contains only carbon (C), hydrogen (H), and oxygen (O). Start by calculating the percentage of O. Given the percentages of C and H, the percentage of O can be calculated as: \( \text{O} = 100\% - \text{C} - \text{H} = 100\% - 26.7\% - 2.2\% = 71.1\% \).

Step 2: Determine the moles of each element

Calculate the moles of C, H, and O by assuming a 100g sample.- Moles of C: \( \frac{26.7}{12} \approx 2.225 \)- Moles of H: \( \frac{2.2}{1} = 2.2 \)- Moles of O: \( \frac{71.1}{16} \approx 4.444 \)

Step 3: Find the simplest mole ratio

Divide each value by the smallest number of moles to find the empirical formula:- C: \( \frac{2.225}{2.2} \approx 1.011 \to 1 \)- H: \( \frac{2.2}{2.2} = 1 \)- O: \( \frac{4.444}{2.2} \approx 2.02 \to 2 \)The empirical formula is \( \text{CHO}_2 \).

Step 4: Calculate the molar mass of the diethyl ester

The vapor density of the diethyl ester is given as 73. Since vapor density \( D = \frac{M}{2} \), the molar mass \( M \) is \( 73 \times 2 = 146 \).

Step 5: Determine the molar mass of the original acid

The general form for a diethyl ester of a dibasic acid is \( (C_nH_{2n+1}O)_2R \). If R is \( CHO_2 \), the ester formula becomes \( C_4H_{10}O_4 \). Check if this fits the empirical formula of the acid:

Step 6: Find the molecular formula of the acid

If the ester has the molar mass of 146, splitting this mass into two molecules results in the molar mass of the acid equal to \( \frac{146}{2} = 73 \), which points to an acid with formula \( C_2H_2O_4 \) matching the option (b).

Key Concepts

Empirical Formula Calculation

The empirical formula represents the simplest whole number ratio of the elements in a compound. To derive it, start by knowing the percentage composition of each element. For a dibasic acid containing carbon, hydrogen, and oxygen, we've identified:

• 26.7% carbon
• 2.2% hydrogen
• The remainder as oxygen

By calculating 100% minus the sum of carbon and hydrogen, we find oxygen's percentage. Next, convert these percentages to moles by assuming a 100g sample. Divide the mass percentages by the atomic masses (C=12, H=1, O=16) to obtain moles:

• Moles of carbon: approximately 2.225
• Moles of hydrogen: exactly 2.2
• Moles of oxygen: approximately 4.444

To finalize the empirical formula, divide all mole figures by the smallest mole quantity identified, typically used for comparison. This produces a simple formula: CHO₂, indicating that every atom of carbon and hydrogen is accompanied by two oxygen atoms within the compound.

Vapor Density

Vapor density is a critical concept used for determining the molecular weight of gases and volatile compounds. Generalized as the mass of a substance compared to that of hydrogen, vapor density provides a hint towards the molar mass of a compound through the formula: \[ D = \frac{M}{2} \] where \( D \) is the vapor density and \( M \) represents the molecular mass of the compound. Given that the dibasic acid's diethyl ester has a vapor density of 73, multiplying this value by 2 gives the molar mass of the ester:\[ M = 73 \times 2 = 146 \]This calculated molar mass is instrumental in deducing the molecular formula of the parent acid by evaluating how this mass correlates to the formula obtained from empirical calculations.

Percentage Composition

Percentage composition refers to the derivative percentage by weight of each element within a compound. It's an essential step for identifying molecular formulas, especially in complex organic compounds like dibasic acids. Consider our problem, where:

• Carbon (C) accounts for 26.7%
• Hydrogen (H) represents 2.2%
• Oxygen (O) constitutes 71.1%

Each percentage is pivotal for converting into moles, thus crafting the empirical formula. These calculated amounts reflect the exact number found in each significant portion of the acidic compound. By grasping these elemental weights, one smoothly transitions into identifying not only empirical ratios but potentially the full molecular formula.

Dibasic Acid Analysis

Dibasic acids, characterized by the presence of two replaceable hydrogens, can form esters and salts indicative of their molecular structure. Deeply analyzing the original acid is crucial after gauging its components through empirical formulas and related molecular evidence.To unravel the molecular formula, consider the specific ester concept. The known determination of the diethyl ester sheds light on the acid through its vapor density findings, leading to molar mass comparisons. With the diethyl ester, we derived its approximate molecular structure: \[(C_nH_{2n+1}O)_2R: C_4H_{10}O_4\]When split, the ester structure is consistent with the predicted formula: \[C_2H_2O_4\]This conforms to having a matching molar mass of 73 for the acid, as validated by the empirical and molecular deductions. Analyzing such esters and vapor attributes is invaluable for tracing any dibasic acid's true identity in practical applications, be it in labs or industrial processes.