Question

A compound with empirical formula \(\mathrm{CH}_{2} \mathrm{O}\) has a vapour density of \(30 .\) Its molecular formula is (a) \(\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}_{3}\) (b) \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) (c) \(\mathrm{C}_{2}^{3} \mathrm{H}_{2} \mathrm{O}_{2}^{3}\) (d) \(\mathrm{C}_{6}^{2} \mathrm{H}_{12} \mathrm{O}_{6}^{-}\)

Short answer

The molecular formula is \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\).

Step-by-step solution

Calculate Molar Mass using Vapour Density

The vapour density of a compound is half its molar mass. Given that the vapour density is 30, we can find the molar mass of the compound by multiplying the vapour density by 2. So the molar mass is \(30 \times 2 = 60\, \text{g/mol}\).

Find the Molar Mass of the Empirical Formula

The empirical formula given is \( \text{CH}_2 \text{O} \). Calculate the molar mass: carbon (C) is \(12\, \text{g/mol}\), hydrogen (H) is \(1\, \text{g/mol}\) and oxygen (O) is \(16\, \text{g/mol}\). Thus, the molar mass of \( \text{CH}_2 \text{O} \) is \(12 + 2 \times 1 + 16 = 30\, \text{g/mol}\).

Determine the Molecular Formula

The molecular formula is a multiple of the empirical formula. Divide the molar mass of the compound (60 g/mol) by the molar mass of the empirical formula (30 g/mol), which gives \(60 / 30 = 2\). Multiply each subscript in the empirical formula by this multiple to get the molecular formula: \(\text{C}_{2} \text{H}_{4} \text{O}_{2}\).

Key concepts

Understanding Empirical Formula

In chemistry, the empirical formula is a way to express the simplest ratio of elements in a compound. It's crucial because it provides a snapshot of the proportionate amounts of atoms, without detailing the actual number of atoms in a molecule.

To find the empirical formula:

• Identify the elements in your compound.
• Measure or estimate the ratio of these elements by number of moles, not by mass.
• Simplify these ratios to the smallest whole numbers possible.

For example, in the compound with empirical formula \(\text{CH}_2 \text{O}\), there is 1 Carbon atom for every 2 Hydrogen and 1 Oxygen atom. Despite the simplicity, the empirical formula is essential for understanding the basic stoichiometry of a compound.

Keep in mind, while the empirical formula provides a basic framework, it does not reveal the complete structure or molecular composition. This leads us to determine the molecular formula.

Vapour Density and Its Role

Vapour density is an important concept when determining the molecular formula of a gas. Vapour density is defined as the mass of a specific volume of a substance compared to the mass of the same volume of hydrogen.

Why is this important?

• It's directly related to molar mass, a key trait for molecular formula determination.
• The vapour density of a compound is always half its molar mass.

In the exercise, with a vapour density of 30, we can find the molar mass by multiplying the vapour density by 2. Thus, the molar mass comes out to be \(30 \times 2 = 60 \, \text{g/mol}\).

Understanding and calculating vapour density makes it possible to deduce the molecular formula, leading to deeper insights into the compound's structure.

Grasping Molar Mass Calculation

Molar mass calculation is fundamental to bridge the gap between empirical and molecular formulas. The molar mass is the weight of one mole of a chemical compound, showing in grams per mole \(\text{g/mol}\).

Let's break it down further:

• Identify individual elements of the empirical formula.
• Use known atomic masses: for carbon, it is 12 \(\text{g/mol}\), hydrogen 1 \(\text{g/mol}\), and oxygen 16 \(\text{g/mol}\).
• Sum up these values for the atoms in the empirical formula. For \(\text{CH}_2 \text{O}\), it's \(12 + 2 \times 1 + 16 = 30 \, \text{g/mol}\).

Now, compare the empirical formula's molar mass with the molar mass derived from the vapour density. The discrepancy suggests how many times the empirical formula needs to be multiplied to mirror the molecular formula.

Divide the molar mass of the compound by this empirical molar mass. In our example, you see \(60 / 30 = 2\), indicating the empirical formula needs doubling. The final molecular formula then mirrors the compound's true nature as \(\text{C}_{2} \text{H}_{4} \text{O}_{2}\).