Problem 88 - A Complete Resource Book in Chemistry for JEE Main

Question

An organic compound containing carbon, hydrogen and oxygen contains \(52.2 \%\) carbon and \(13.04 \%\) hydrogen. Vapour density of the compound is 23 . Its molecular formula will be (a) \(\mathrm{C}_{4} \mathrm{H}_{\mathrm{g}} \mathrm{O}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{s}} \mathrm{O}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\) (d) \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}\)

Step-by-Step Solution

Answer

The molecular formula is (c) \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \).

Step 1: Calculate Percentage of Oxygen

Since the percentages of elements in a compound must add up to 100%, we calculate the percentage of oxygen by subtracting the given percentages of carbon and hydrogen from 100%: \( 100\% - (52.2\% + 13.04\%) = 34.76\% \) oxygen.

Step 2: Calculate Empirical Formula

Convert the percentages to grams (assuming 100 g of the compound):- Carbon: 52.2 g- Hydrogen: 13.04 g- Oxygen: 34.76 gConvert each to moles using their atomic weights (C = 12, H = 1, O = 16):- Moles of C = \( \frac{52.2}{12} = 4.35 \)- Moles of H = \( \frac{13.04}{1} = 13.04 \)- Moles of O = \( \frac{34.76}{16} = 2.17 \)Divide by the smallest number of moles (2.17):- C: \( \frac{4.35}{2.17} \approx 2 \)- H: \( \frac{13.04}{2.17} \approx 6 \)- O: \( \frac{2.17}{2.17} = 1 \)Thus, the empirical formula is \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \).

Step 3: Calculate Molecular Formula Using Vapour Density

Vapour density, which is half of the molar mass, is given as 23. Therefore, the molar mass of the compound is \( 23 \times 2 = 46 \).Calculate the molar mass of the empirical formula \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \):\( 2 \times 12 + 6 \times 1 + 16 = 46 \).Since the calculated molar mass equals the empirical formula molar mass, the empirical formula is also the molecular formula: \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \).

Step 4: Match with Given Options

Compare the molecular formula derived, \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \), with the given options:- (a) \( \mathrm{C}_4 \mathrm{H}_9 \mathrm{O} \)- (b) \( \mathrm{C}_3 \mathrm{H}_8 \mathrm{O} \)- (c) \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \)- (d) \( \mathrm{C}_4 \mathrm{H}_{10} \mathrm{O} \)The correct answer is (c) \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \).

Key Concepts

Vapour Density

Vapour density is a crucial concept in chemistry that denotes the density of a vapor in comparison to that of hydrogen. It's an important parameter when determining the molecular mass of a compound. The vapour density is numerically half of the molar mass of the compound. For instance, if you know that the vapour density of a compound is 23, you can easily calculate the molar mass by multiplying the vapour density by 2. This results in a molar mass of 46 for the given compound.
Understanding vapour density helps chemists verify the consistency of an empirical formula with the correct molecular formula. By matching the calculated molar mass with known empirical data, scientists can ascertain the exact composition of a compound. In our previous solution, the empirical formula and molecular formula were shown to match, confirming the compound's chemical structure as \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \). Such comparisons are an essential part of problem-solving in organic chemistry.

Organic Chemistry

Organic chemistry focuses on compounds mainly containing carbon and hydrogen, often accompanied by elements like oxygen, nitrogen, or sulfur. Carbon's versatility in forming various bonds enables the immense diversity of organic compounds. In our exercise, we dealt with an organic compound composed of carbon, hydrogen, and oxygen.
In organic chemistry, determining the molecular formula from an empirical formula involves a series of calculations. This procedure not only helps identify the compound but also provides insights into its possible reactions and properties. Such exercises are common in organic chemistry courses and require a solid understanding of the relationships between elemental percentages, molecular weight, and structural composition.
By knowing how to calculate these, students can better predict and understand the behavior of organic compounds in various chemical reactions. This foundational knowledge is vital for anyone pursuing a career in chemistry or related scientific fields.

Molar Mass Calculation

Molar mass calculation is an essential skill for chemists and plays a critical role in identifying compounds. The molar mass is calculated by summing the atomic masses of all atoms present in a molecule. In exercises like ours, these calculations showed that the molar mass of the empirical formula \( \mathrm{C}_2 \mathrm{H}_6 \mathrm{O} \) corresponds with the molecular formula.
Here's how it works: calculate the molar mass of each element (e.g., carbon, hydrogen, oxygen), and multiply each by the number of atoms in the empirical formula. By adding them together, you'll find that \( 2 \times 12 + 6 \times 1 + 16 = 46 \). This verified the given molar mass of the compound, confirming our earlier steps in determining the molecular formula.
Correctly calculating molar mass is crucial for predicting the physical and chemical properties of substances, understanding reactions, and when synthesizing new compounds. Regular practice with such calculations enhances problem-solving skills in laboratory settings and theoretical chemistry.