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 compound's molecular formula is (c) \( \mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O} \).

Step 1: Calculate the Mass of Oxygen

Begin by calculating the percentage of oxygen in the compound. Since the sum of all percentages must equal 100%, we have: \[ \text{Percentage of O} = 100\% - 52.2\% - 13.04\% = 34.76\% \]

Step 2: Assume a Sample Mass and Calculate Moles

Assume a sample mass of 100 grams for convenience. This gives us 52.2 grams of carbon, 13.04 grams of hydrogen, and 34.76 grams of oxygen in the sample. Calculate the moles for each element: \[ \text{Moles of C} = \frac{52.2}{12} = 4.35 \]\[ \text{Moles of H} = \frac{13.04}{1} = 13.04 \]\[ \text{Moles of O} = \frac{34.76}{16} = 2.17 \]

Step 3: Determine the Empirical Formula

To find the empirical formula, divide all moles by the smallest number of moles, which is 2.17: \[ \text{Ratio of C} = \frac{4.35}{2.17} \approx 2 \]\[ \text{Ratio of H} = \frac{13.04}{2.17} \approx 6 \]\[ \text{Ratio of O} = \frac{2.17}{2.17} = 1 \]Thus, the empirical formula is \( \text{C}_2\text{H}_6\text{O} \).

Step 4: Calculate the Empirical Formula Mass

Find the mass of the empirical formula \(\text{C}_2\text{H}_6\text{O}\) by adding the atomic masses:\[ \text{Molar Mass} = (2 \times 12) + (6 \times 1) + (1 \times 16) = 46 \text{ grams/mol} \]

Step 5: Determine the Molecular Formula Using Vapour Density

Given that the vapour density is 23, the molar mass of the molecular formula is double that (since vapour density is half the molar mass):\[ \text{Molar Mass} = 23 \times 2 = 46 \text{ grams/mol} \]Since this is equal to the empirical formula mass, the empirical formula is also the molecular formula. Thus, the molecular formula is \(\text{C}_2\text{H}_6\text{O} \).

Key Concepts

Empirical Formula Calculation

To determine the empirical formula of a compound, you start by knowing the percentages of each constituent element. In our exercise, we know the percentages of carbon and hydrogen, so we can easily deduce the percentage of oxygen by subtracting the sum of carbon and hydrogen percentages from 100%. This gives us the complete set of percentages:

• 52.2% Carbon
• 13.04% Hydrogen
• 34.76% Oxygen (since 100% - 52.2% - 13.04% = 34.76%)

Next, you translate these mass percentages into moles, using an assumed sample mass, for example, 100 grams. This simplifies your calculations, as the mass percentage will be equal to the mass in grams. For each element, divide the mass by its atomic mass to find the number of moles. This lets you understand the relative number of atoms. Dividing each by the smallest value among them further streamlines it, providing you the simplest whole number ratio of atoms in the compound. This results in the empirical formula, reflecting the simplest whole-number relationship of atoms in the molecule, which for this exercise is \({\text{C}_2\text{H}_6\text{O}}\).

Vapour Density

Vapour Density is a crucial concept in determining the molecular formula of a compound. Defined as the mass of a certain volume of a gas compared to the mass of an equal volume of hydrogen at the same conditions, it offers insights into the molecular weight. The vapour density is directly related to the molar mass of a compound, being half of it. To calculate the molar mass using vapour density, use:

• \( \text{Molar Mass} = 2 \times \text{Vapour Density} \)

In this case, given a vapour density of 23, the molar mass is \(46 \text{ grams/mol}\). Recognizing the equality between the empirical formula mass and this molar mass allows us to conclude that the empirical formula is also the molecular formula for the compound.

Stoichiometry

Stoichiometry is the heart of balancing chemical reactions and observing the conservation of mass. It involves using balanced chemical equations to calculate the amounts of reactants and products involved in reactions. In the exercise, stoichiometry helps relate the moles of each element calculated to form the empirical formula. It's the process of translating mass into moles and finding the proportional distribution of atoms around in a molecule. The transition from grams to moles is made through stoichiometric coefficients in balanced equations that allow predictions on product yield or necessary reactant quantities needed, given the initial conditions. Essentially, stoichiometry governs the compilation of chemical elements inside a particular molecular structure.

Chemical Composition Analysis

Understanding the chemical composition of a compound is critical in molecular formula determination. This analysis involves deciphering the percentages of each element within a compound, providing insight into its structure. When analyzing chemical composition, tools such as mass spectrometry or chromatography might be used in practical settings, but in theoretical exercises, it's the calculation that shows the exact percentages. From there, the calculation of empirical formulas is made possible as you see in our exercise where we determined percentages from the compound, followed by calculating the element masses as part of our sample. Analyzing chemical composition is a paramount skill in chemistry, enabling the classification of unknown substances, guiding synthesis, and increasing understanding of molecular structures. It ensures that scientists and students alike can categorize and replicate substances for a wide array of applications.