Problem 86 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound contains \(49.3 \%\) carbon, \(6.84 \%\) hydrogen and its vapour density is 73. Molecular formula of the compound is (a) \(\mathrm{C}_{3} \mathrm{H}_{10} \mathrm{O}_{2}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{6} \mathrm{H}_{10} \mathrm{O}_{8}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}_{2}\)

Step-by-Step Solution

Answer

The molecular formula is C₆H₁₀O₄, correcting an initial confusion.

Step 1: Determine the Empirical Formula

To find the empirical formula, we first convert the percentage composition into grams. Assuming 100 grams of the compound, we have: - Carbon: 49.3 grams- Hydrogen: 6.84 grams- Oxygen (by difference): 100 - (49.3 + 6.84) = 43.86 grams.Next, we calculate the moles of each element:- Moles of Carbon = \( \frac{49.3}{12} = 4.108 \) moles- Moles of Hydrogen = \( \frac{6.84}{1} = 6.84 \) moles- Moles of Oxygen = \( \frac{43.86}{16} = 2.741 \) molesThen, we divide each by the smallest number of moles:- Carbon: \( \frac{4.108}{2.741} = 1.5 \)- Hydrogen: \( \frac{6.84}{2.741} = 2.5 \)- Oxygen: \( \frac{2.741}{2.741} = 1 \)The empirical formula proportions are approximately C₁.₅H₂.₅O₁, which are not whole numbers. Multiplying all by 2 to get whole numbers gives the empirical formula: C₃H₅O₂.

Step 2: Calculate the Empirical Formula Mass

Using the empirical formula C₃H₅O₂, we calculate the empirical formula mass: - Carbon: 3 × 12 = 36 - Hydrogen: 5 × 1 = 5 - Oxygen: 2 × 16 = 32 Adding them, we get: 36 + 5 + 32 = 73 g/mol.

Step 3: Calculate the Molecular Formula

The molecular formula is a multiple of the empirical formula C₃H₅O₂. Given the vapor density is 73, the molecular weight is: Molecular weight = 2 × 73 = 146 g/mol (since vapor density = molecular weight/2). Comparing the empirical formula mass (73 g/mol) to the molecular weight (146 g/mol), they differ by a factor of 2. Thus, the molecular formula is 2 times the empirical formula. Therefore, molecular formula = (C₃H₅O₂)₂ = C₆H₁₀O₄.

Key Concepts

Percentage Composition

The percentage composition of an element in a compound is a way to express how much of each element is present. Here, you convert these percentages to grams to find how much of each element there might be in a 100 gram sample of the compound.
This method simplifies calculations. For example, if a compound has a percentage composition of 49.3% carbon, 6.84% hydrogen, and the rest as oxygen, we can say:

• Carbon: 49.3 grams
• Hydrogen: 6.84 grams
• Oxygen: 43.86 grams (calculated by subtracting the sum of carbon and hydrogen percentages from 100)

This information is fundamental for moving on to calculate moles, as you need the mass of each element for that conversion.

Vapor Density

Vapor density helps in determining the molecular weight of a compound. It is defined as the mass of a certain volume of a gas compared to the mass of an equal volume of hydrogen at the same temperature and pressure.
To find the molecular weight from vapor density, we use the following relationship:
- Vapor density is half the molecular weight.
Thus, given a vapor density of 73, we can determine the molecular weight of the compound as:
\[ \text{Molecular weight} = 2 \times \text{Vapor density} \] Substituting the known vapor density value:
\[ \text{Molecular weight} = 2 \times 73 = 146 \, \text{g/mol} \]This molecular weight is crucial for finding the compound's molecular formula from its empirical formula.

Moles Calculation

Converting mass into moles is a key concept in chemistry. It allows us to understand how many atoms or molecules are present in a substance. To calculate moles, you use the formula:
\[\text{Moles} = \frac{\text{Grams of element}}{\text{Atomic mass of element}}\]Using the percentages converted to grams for each element in the compound, we calculate:

• Moles of Carbon = \( \frac{49.3}{12} = 4.108 \)
• Moles of Hydrogen = \( \frac{6.84}{1} = 6.84 \)
• Moles of Oxygen = \( \frac{43.86}{16} = 2.741 \)

Next, to find the simplest ratio of atoms in the compound (the empirical formula), divide each mole value by the smallest number of amongst them. In this case, oxygen’s moles (2.741) are the smallest, so we divide all the calculated moles by this value:

• Carbon: \( \frac{4.108}{2.741} \approx 1.5 \)
• Hydrogen: \( \frac{6.84}{2.741} \approx 2.5 \)
• Oxygen: \( \frac{2.741}{2.741} = 1 \)

Since chemical formulas must contain whole numbers, these values are scaled by multiplying them by 2 to get integers:

• Final empirical formula: C₃H₅O₂

The empirical formula is the first step in reaching the molecular formula, particularly when combined with vapor density calculations.