Problem 82 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound has \(\mathrm{C}\) and \(\mathrm{H}\) percentage in the ratio \(6: 1\) and \(\mathrm{C} \%\) and \(\mathrm{O} \%\) in the ratio \(3: 4\). The compound is (a) \(\mathrm{HCHO}\) (b) \(\mathrm{CH}_{3} \mathrm{OH}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) (d) \((\mathrm{COOH})_{2}\)

Step-by-Step Solution

Answer

The compound is (a) HCHO.

Step 1: Understand the Given Ratios

The problem gives us two ratios of percentages for the organic compound: \( C:H \) is 6:1 and \( C:O \) is 3:4. This means that in terms of percentage contribution to the compound's overall mass, for every 6 parts of carbon, there is 1 part of hydrogen, and for every 3 parts of carbon, there are 4 parts of oxygen.

Step 2: Relate Ratios to Atomic Mass

To find the empirical formula, consider the atomic masses: Carbon (C) is 12, Hydrogen (H) is 1, and Oxygen (O) is 16. The ratios are based on mass contribution; hence we can start determining the relative quantity by considering these atomic masses.

Step 3: Calculate Relative Amounts of Each Element

To find the empirical formula, we transform the ratios into the smallest whole numbers:1. **C:H Ratio:** If \( C:H = 6:1 \), using atomic masses, \( 6C = 6 \times 12 = 72 \) and \( H = 1 \times 1 = 1 \). Therefore, to balance, we need the ratio \( C:H = \frac{6}{72}:\frac{1}{1} = 6:1 \) remains the same.2. **C:O Ratio:** If \( C:O = 3:4 \), using atomic masses, \( 3C = 3 \times 12 = 36 \) and \( 4O = 4 \times 16 = 64 \). Therefore, the ratio \( C:O = \frac{3}{36}:\frac{4}{64} = 3:4 \) remains consistent.Based on these, adjust to maintain consistent ratios to atomic masses with the provided options.

Step 4: Compare to Compound Options

From the calculated information (ratios and chemical weights), compare with each compound given: - **Option (a) HCHO:** Carbon: 12, Hydrogen: 1, Oxygen: 16 -> Total 29 mass ratio. - **Option (b) CH₃OH:** Carbon: 12, Hydrogen: 3, Oxygen: 16 -> Total 31 mass ratio. - **Option (c) CH₃CH₂OH:** Carbon: 24, Hydrogen: 6, Oxygen: 16 -> Total 46 mass ratio. - **Option (d) (COOH)₂:** Carbon: 24, Hydrogen: 2, Oxygen: 32 -> Total 58 mass ratio. From this, (a) matches the empirical formula kinds calculated.

Step 5: Determine the Correct Compound

Option (a) HCHO matches the described ratio of C:H = 6:1 and C:O = 3:4 when considering equivalent ratios based on atomic masses. Thus, HCHO satisfies both given ratios when normalized back to the atomic level with numbers scaled appropriately for comparison based on each united element.

Key Concepts

Chemical Formula Calculation

Calculating the chemical formula of a compound involves understanding the proportions of each element present in the compound. This is done by determining the empirical formula, which is the simplest whole-number ratio of atoms in the compound. This process relies on the known atomic masses of the elements involved. 1. **Understanding Ratios**: When ratios of element percentages are given, such as the carbon to hydrogen ratio (C:H) of 6:1, and the carbon to oxygen ratio (C:O) of 3:4, it indicates the relative contribution by mass of each element to the compound. 2. **Using Atomic Masses**: To convert these ratios into a practical formula, we use atomic masses: Carbon (C) has an atomic mass of 12, Hydrogen (H) is 1, and Oxygen (O) is 16. These weights help us translate percentage ratios into actual amounts of substance. 3. **Deriving the Empirical Formula**: By applying the atomic weights to the given ratios, it's possible to determine the smallest whole numbers that can represent the relative amounts of the elements, thus forming the empirical formula. For example, the ratios given try to match with formulas like HCHO, which indeed satisfies C:H as 6:1 and C:O as 3:4 when analyzed by mass. The goal is to ensure the empirical formula reflects the precise proportionality of the elements involved, which lays the foundation for understanding the molecular structure of the compound.

Elemental Analysis

Elemental analysis is a critical technique used to determine the presence and percentage composition of elements within a compound. This is especially important in organic chemistry when trying to decipher complex molecular structures based on elemental compositions. - **Percentage Composition**: By analyzing the percentages of each element, such as how much percentage of a compound is carbon, hydrogen, or oxygen, we can derive meaningful relationships between them. In the given problem, the ratios provided a pathway to deduce the formula. - **Linking to Chemical Formulas**: The process involves converting mass ratios into mole ratios by using atomic masses. These are then compared to known potential formulas to find a match. For example, in the problem, this method aligned with HCHO as the correct formula because it met the percentage constraints dictated by the ratios. Elemental analysis thus acts as a bridge between raw compositional data and the structured, formulaic representation of the substance's chemical nature.

Organic Chemistry Concepts

Organic chemistry involves the study of carbon-containing compounds, which are not just limited to carbon and hydrogen, but may include oxygen, nitrogen, and other elements as well. Understanding these compounds often requires a combination of empirical data interpretation and structural analysis. - **Recognizing Organic Structures**: In the given exercise, students practice identifying potential organic structures from percentage data. This understanding helps in identifying functional groups and structural formulas. - **From Empirical to Molecular**: Initially, students determine the empirical formula, but often in organic chemistry, determining the molecular formula is the next step. This relates back to stoichiometry and understanding that the empirical formula is sometimes a simplified version of the molecule’s actual structure. - **Example Application**: In our problem, options like HCHO (formaldehyde) were considered based on the elemental analysis, illustrating how data interpretation in organic chemistry bridges the gap between theoretical ratios and practical chemical structures. In essence, mastering these concepts in organic chemistry aids in hypothesizing the nature of unknown compounds and understanding how carbon-based structures interact, forming the basis of myriad biological and synthetic processes.