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) \(\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\) (b) \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\) (c) \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\) (d) \(\mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}\)

Step-by-Step Solution

Answer

The molecular formula is \(C_2H_6O\) (option c).

Step 1: Determine the Percentage of Oxygen

Since the compound contains only carbon, hydrogen, and oxygen, and we know the percentages of carbon and hydrogen, we can find the percentage of oxygen by subtracting the sum of the percentages of carbon and hydrogen from 100%.\[\%\text{Oxygen} = 100\% - 52.2\% - 13.04\% = 34.76\%\]

Step 2: Calculate the Empirical Formula

To find the empirical formula, we assume 100 grams of the compound are present, which gives us 52.2 g of carbon, 13.04 g of hydrogen, and 34.76 g of oxygen. We divide these masses by their respective atomic masses to get the mole ratios:- Moles of Carbon: \(52.2 / 12.01 \approx 4.35\)- Moles of Hydrogen: \(13.04 / 1.008 \approx 12.94\)- Moles of Oxygen: \(34.76 / 16.00 \approx 2.17\)Next, divide each value by the smallest number of moles, 2.17, to get a ratio:- Carbon: \(4.35 / 2.17 \approx 2\)- Hydrogen: \(12.94 / 2.17 \approx 6\)- Oxygen: \(2.17 / 2.17 = 1\)The empirical formula is \(C_2H_6O\).

Step 3: Determine the Molecular Formula

Next, we need to calculate the molecular formula using the vapour density. The vapour density is given as 23, which means the molar mass of the compound is \(23 \times 2 = 46\text{ g/mol}\).Calculate the molar mass of the empirical formula \(C_2H_6O\): - \(C_2\): \(12.01 \times 2 = 24.02\)- \(H_6\): \(1.008 \times 6 = 6.048\)- \(O\): \(16.00\)Adding these, the molar mass of the empirical formula is: \[24.02 + 6.048 + 16.00 = 46.068\, \text{g/mol}\]Since the calculated molar mass (46.068 g/mol) matches the given molar mass (46 g/mol), the molecular formula is the same as the empirical formula, \(C_2H_6O\).

Step 4: Verify the Answer

Compare the molecular formula \(C_2H_6O\) to the options provided:(a) \(C_4H_8O\)(b) \(C_3H_8O\)(c) \(C_2H_6O\)(d) \(C_5H_{10}O\)The correct molecular formula matches with option (c), \(C_2H_6O\).

Key Concepts

Empirical Formula Calculation

Empirical formula calculation is often the first step in determining the identity of an unknown compound, especially when the compound's composition is known in terms of the percentage of each element. In this example, the compound contains carbon, hydrogen, and oxygen. To calculate the empirical formula, we perform the following steps:
First, assume a sample size, which is often 100 g, to simplify calculations with percentage compositions. This assumption allows us to directly express the percentages as masses in grams. Given here are 52.2 g of carbon, 13.04 g of hydrogen, and, by subtraction, 34.76 g of oxygen.
Next, convert these masses into moles by dividing by the respective atomic masses: 12.01 g/mol for carbon, 1.008 g/mol for hydrogen, and 16.00 g/mol for oxygen. This provides the mole ratios for each element. In this case, you get approximately 4.35 moles of carbon, 12.94 moles of hydrogen, and 2.17 moles of oxygen.
Finally, to find the simplest whole-number ratio, often called the stoichiometric ratio, divide each mole value by the smallest number of moles present, which here is 2.17. The adjusted mole ratio in this example results in an empirical formula of \( C_2H_6O \).

• Tip: Always remember to convert mass to moles when dealing with percentage composition in empirical formula calculations.

Vapour Density

Vapour density plays a critical role in determining the molecular formula from the empirical formula. Vapour density is defined as the mass of a certain volume of a gas or vapor compared to the mass of an equal volume of hydrogen under identical conditions. The key relation to remember here is that the molar mass of the compound is twice the vapour density.
In our problem, the vapour density of the compound is given as 23. To find the molar mass, multiply the vapour density by 2, resulting in a molar mass of 46 g/mol.

• This is an essential link between vapour density and molar mass: \( \text{Molar Mass} = \text{Vapour Density} \times 2 \).

With the calculated molar mass, you can confirm it against the molar mass of the empirical formula, validating the molecular formula.

Percentage Composition

Percentage composition provides a snapshot of the elemental makeup of a compound, allowing us to measure how much of each element is present in terms of percentage by mass. This is crucial for empirical formula calculation, as we saw in the previous step.
For example, if a compound is known to contain 52.2% carbon and 13.04% hydrogen, you can easily find the remaining percentage attributed to oxygen by subtracting the sum of the known percentages from 100%: \[ 100 ext{%} - 52.2 ext{%} - 13.04 ext{%} = 34.76 ext{%} \]
Such information aids in converting mass percentages directly into masses when you assume a total mass of 100 g for the compound. This simplification directly leads to finding the number of moles for each element and, subsequently, the empirical formula.

Organic Compound Analysis

Organic compound analysis involves studying the composition and molecular structure of organic molecules. The information gathered from experiments, such as percentage composition analysis, can determine both the empirical and molecular formulas. This is fundamental in identifying unknown compounds.
Through organic compound analysis, like that in our exercise, we can determine essential parts of a molecule's identity, such as:

• Identifying elements present in the compound based on compositional data.
• Determining the empirical formula through mass-to-mole conversions.
• Using vapour density to establish the compound's molar mass.

Critical in organic chemistry, these techniques allow scientists to conduct structural elucidation—transforming raw data into meaningful chemical information that identifies compounds within organic mixtures.