Question

The empirical formula of a compound is \(\mathrm{CH}\). At \(200^{\circ} \mathrm{C}\) \(0.145 \mathrm{~g}\) of this compound occupies \(97.2 \mathrm{~mL}\) at a pressure of \(0.74 \mathrm{~atm}\). What is the molecular formula of the compound?

Short answer

The molecular formula of the compound is \( \text{C}_6\text{H}_6 \).

Step-by-step solution

Use the Ideal Gas Law

The ideal gas law is \( PV = nRT \). First, convert the volume to liters: \( 97.2 \text{ mL} = 0.0972 \text{ L} \). The temperature in Kelvin is \( 200^{\circ}C + 273.15 = 473.15 \text{ K} \). The universal gas constant \( R \) is \( 0.0821 \text{ L atm mol}^{-1} \text{ K}^{-1} \). Substitute the known values into the ideal gas law to find \( n \), the number of moles: \[ n = \frac{PV}{RT} = \frac{0.74 \times 0.0972}{0.0821 \times 473.15} \].

Calculate the Number of Moles

From the equation \( n = \frac{0.74 \times 0.0972}{0.0821 \times 473.15} \), calculate \( n \): \( n \approx 0.00185 \text{ moles} \).

Determine the Molar Mass

The molar mass \( M \) is given by the formula \( M = \frac{\text{mass}}{n} \). Substituting the mass of the compound (0.145 g) and number of moles (0.00185 moles) gives: \[ M = \frac{0.145}{0.00185} \approx 78.38 \text{ g/mol} \].

Find the Empirical Formula Mass

The empirical formula for the compound is \( \text{CH} \). The atomic mass of \( \text{C} \) is approximately 12 g/mol and \( \text{H} \) is approximately 1 g/mol, giving an empirical formula mass of \( 12 + 1 = 13 \text{ g/mol} \).

Calculate the Molecular Formula

The molecular formula is a multiple of the empirical formula. To find the multiple, divide the molar mass by the empirical formula mass: \[ \frac{78.38}{13} \approx 6 \]. Thus, the molecular formula is \( \text{C}_6\text{H}_6 \).

Key concepts

Empirical Formula

The empirical formula is a straightforward way to express the simplest integer ratio of elements in a compound. It does not give any information about the number of atoms but rather the ratio in which they combine.
For example, in the exercise, the empirical formula is given as \( \text{CH} \). This tells us that in the simplest form of this compound, the ratio of carbon to hydrogen is 1:1.
Empirical formulas are helpful when analyzing and comparing the elements present in compounds or when working with combustion analysis results.

• Empirical formula only provides the lowest ratio of elements.
• It differs from the molecular formula, which shows the exact number of each type of atom.

In practice, knowing the empirical formula allows chemists to establish baseline information about the substance and can serve as a starting point for further investigation into the compound's molecular structure.

Molecular Formula

The molecular formula reveals the actual number of atoms of each element in a molecule. While the empirical formula shows the simplest ratio, the molecular formula tells us precisely how many atoms are in each molecule of the substance.
In our problem, we established the empirical formula as \( \text{CH} \) with a molar mass calculation hinting that the molar mass is larger than that of one \( \text{CH} \) unit. Therefore, finding the molecular formula involves:

• Calculating or knowing the molar mass of the compound.
• Comparing this with the empirical formula mass.
• Determining the multiplier needed to convert the empirical formula into molecular form.

In this exercise, the molecular formula was found to be \( \text{C}_6\text{H}_6 \), indicating that each molecule contains six carbon atoms and six hydrogen atoms, thus representing the compound benzene.

Molar Mass

Molar mass is the mass of one mole of a substance, measured in grams per mole \( \text{g/mol} \). Calculating molar mass is crucial as it connects the microscopic scale of atoms to the macroscopic scale of grams.
To calculate it, you need the mass of the substance and the number of moles you've determined, such as with the ideal gas law.

• Molar mass is determined by the formula \( M = \frac{\text{mass}}{n} \).
• It serves as a conversion factor between the amount of substance and its mass.

The given exercise used a calculated molar mass of approximately 78.38 \( \text{g/mol} \), leading to the discovery of the compound's actual molecular formula. This calculated "weight" for a mole was used to confirm that the molecular formula needed a multiple adjustment from the empirical formula \( \text{CH} \) to \( \text{C}_6\text{H}_6 \), illustrating how molar mass aids in finding molecular composition.