Problem 104 - A Complete Resource Book in Chemistry for JEE Main

Question

\(0.1 \mathrm{~g}\) of an organic monobasic acid on complete combustion gave \(0.254 \mathrm{~g} \mathrm{CO}_{2}\) and \(0.0443 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\). For complete neutralisation \(0.122 \mathrm{~g}\) of the acid requires \(10 \mathrm{ml}\) of \(0.1 \mathrm{~m} \mathrm{KOH}\). The molecular formula of the acid is \(\mathrm{C}_{\mathrm{x}} \mathrm{H}_{6} \mathrm{O}_{2}\) the value of \({ }^{\prime} \mathrm{X}^{\prime}\) is?

Step-by-Step Solution

Answer

The value of \(x\) is 7.

Step 1: Determine the Number of Moles of CO2

First, calculate the number of moles of CO2 produced from the combustion. Using the formula, \[ \text{Number of moles of CO}_2 = \frac{\text{Mass of CO}_2}{\text{Molar mass of CO}_2} \] where the molar mass of CO2 is 44 g/mol. Substituting the given values, we have \[ \text{Number of moles of CO}_2 = \frac{0.254}{44} \approx 0.00577. \]

Step 2: Calculate Carbon Atoms in the Acid

Since each mole of CO2 contains one mole of carbon atoms, the number of moles of carbon in the original compound is the same as the number of moles of CO2, which is 0.00577 moles.

Step 3: Find Number of Moles of H2O

Calculate the number of moles of water produced, using the formula, \[ \text{Number of moles of H}_2\text{O} = \frac{\text{Mass of H}_2\text{O}}{\text{Molar mass of H}_2\text{O}} \] where the molar mass of H2O is 18 g/mol. So, \[ \text{Number of moles of H}_2\text{O} = \frac{0.0443}{18} \approx 0.00246. \]

Step 4: Calculate Hydrogen Atoms in the Acid

Since each mole of water contains two moles of H atoms, the number of moles of hydrogen atoms is \(2 \times 0.00246 = 0.00492\) moles.

Step 5: Determine the Valence factor of KOH

Recall that KOH neutralizes one equivalent of the acid. Since we have 0.1 molar KOH, we calculate the equivalents of KOH used as follows: \(\text{Equivalents of KOH} = 0.1 \times 0.01 = 0.001\) equivalents.

Step 6: Calculate the Molar Mass from Neutralization

Since 0.122 g of the acid is neutralized by 0.001 equivalents of KOH, the equivalent mass of the acid is \(\frac{0.122}{0.001} = 122\) g/equiv. As it is a monobasic acid, the molar mass is also 122 g/mol.

Step 7: Determine the Value of x in the Molecular Formula

The molecular formula is \(\text{C}_x\text{H}_6\text{O}_2\), so the molar mass is \(12x + 6 \times 1 + 2 \times 16 = 122\). Simplifying, \(12x + 6 + 32 = 122\) leads to \(12x = 84\). Thus, \(x = \frac{84}{12} = 7\).

Key Concepts

Combustion Analysis

In organic chemistry, combustion analysis is a key method used to identify the elemental composition of a compound. During combustion, a substance burns in oxygen, resulting in the formation of carbon dioxide and water among other products. By measuring the amounts of these products, we can infer the quantities of carbon and hydrogen in the original compound.

The process usually involves weighing the carbon dioxide and water produced when the compound is completely burned. By knowing the molar masses of these compounds, we can calculate the moles of carbon and hydrogen. For instance, if we know the amount of carbon dioxide generated, we can find out how much carbon was in the original compound. Similarly, by knowing the amount of water produced, we can determine how many hydrogen atoms were present.

• Each mole of CO extsubscript{2} corresponds to one mole of carbon atoms.
• Each mole of H extsubscript{2}O corresponds to two moles of hydrogen atoms.

Understanding combustion analysis is fundamental for deducing molecular formulas in organic chemistry.

Neutralization Reaction

Neutralization reactions occur when an acid and a base react to form water and a salt. In this educational exercise, we have a monobasic acid which reacts with potassium hydroxide (KOH), a strong base. This implies that one mole of acid reacts with one mole of base.

For example, if you know the concentration of the KOH solution and the volume used to neutralize a known amount of acid, you can determine the amount of acid that reacted. This is because the product formed contains equivalents that are specific to the acid's basicity. Neutralization reactions are helpful to determine the molar mass of acids and bases.

In the given problem, the amount of KOH used (measured in equivalents) was equated to the amount of acid present, using the fact that the acid is monobasic. This enabled us to deduce its molar mass. This step is essential as it links the experimental data to the molecular formula calculation.

Molecular Formula Calculation

Determining the molecular formula of an organic compound involves combining insights from combustion analysis and neutralization reactions. Let's break this process down into parts to fully comprehend it.

From combustion analysis, we identify the amount of carbon and hydrogen atoms in the sample. In our example, the moles of carbon and hydrogen derived from CO extsubscript{2} and H extsubscript{2}O give us partial information about the formula.

Then, by incorporating the molar mass information obtained from the neutralization step, we find the necessary details to finalize the molecular formula. The overall mass made up of carbon, hydrogen, and oxygen is equated to the molar mass deduced from the neutralization.

By solving the equation based on atomic masses (e.g., C, H, and O), we calculated the unknown components of the formula, such as the value of x in C extsubscript{x}H extsubscript{6}O extsubscript{2}. This involved simple algebraic calculations to ensure that the sum of all atomic masses matched the calculated molar mass. This step effectively bridges experimental data and theoretical molecular understanding.