Question

Chloromethane, \(\mathrm{CH}_{3} \mathrm{Cl}\), arises from microbial fermentation and is found throughout the environment. It is also produced industrially and is used in the manufacture of various chemicals and has been used as a topical anesthetic. Calculate how much energy is required to convert \(92.5 \mathrm{~g}\) liquid to a vapor at its boiling point, \(-24.09^{\circ} \mathrm{C}\). (The heat of vaporization of \(\mathrm{CH}_{3} \mathrm{Cl}\) is \(21.40 \mathrm{~kJ} / \mathrm{mol}\).)

Short answer

The energy required is 39.2 kJ.

Step-by-step solution

Calculate moles of Chloromethane

First, determine the molar mass of chloromethane (\(\text{CH}_3 \text{Cl}\)). The atomic masses are: C = 12.01 \(\text{g/mol}\), H = 1.01 \(\text{g/mol}\), and Cl = 35.45 \(\text{g/mol}\). Adding these gives the molar mass: \(12.01 + (3\times 1.01) + 35.45 = 50.49\ \text{g/mol}\). Now, calculate the moles of \(92.5\ \text{g}\) of \(\text{CH}_3 \text{Cl}\) by \(\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{92.5}{50.49} \approx 1.832\ \text{mol}\).

Apply heat of vaporization

Now, use the heat of vaporization to find the energy required to vaporize the given mass of chloromethane. The formula for energy is \( \text{energy} = \text{moles} \times \text{heat of vaporization} \). Plug in the values: \(1.832\ \text{mol} \times 21.40\ \text{kJ/mol}\ = 39.1928\ \text{kJ}\).

Round off the answer

Finally, provide the energy required by rounding to a suitable number of significant figures. Given that the significant figures in the problem (92.5 and 21.40) suggest three significant figures, the energy required is approximately \(39.2\ \text{kJ}\).

Key concepts

Chloromethane

Chloromethane, also known as methyl chloride, is a chemical compound with the formula \(\text{CH}_3\text{Cl}\). It is a colorless, flammable gas with a faint sweet odor. Chloromethane occurs naturally, being released by some types of plankton, fungi, and other plants, and is also produced through microbial fermentation in the environment. Industrially, it is synthesized and used in the production of various chemicals.In addition to its uses in manufacturing, chloromethane has applications as a local anesthetic. Characteristics such as its molecular structure, consisting of one carbon bonded to three hydrogen atoms and a chlorine atom, contribute to its chemical properties and the way it interacts with other substances. The relatively low molar mass of chloromethane makes it efficient to transport when used industrially.

Boiling Point

The boiling point of a substance is the temperature at which its liquid form turns into vapor. For chloromethane, this temperature is exceptionally low, at \(-24.09^{\circ} \mathrm{C}\). This means that chloromethane transitions from liquid to gas at temperatures well below freezing, indicating it has weak intermolecular forces compared to other substances that require higher temperatures to boil.Boiling points give insight into a substance's volatility. Substances like chloromethane with low boiling points are quite volatile and can easily evaporate when exposed to air at room temperature. When considering industrial applications, this property must be considered in storage and handling protocols to prevent unwanted losses through evaporation or accidental releases.

Energy Calculation

The energy calculation for vaporizing chloromethane involves understanding the concept of heat of vaporization. This is the amount of heat energy required to turn one mole of a liquid into vapor without changing its temperature. For chloromethane, this is \(21.40 \mathrm{~kJ/mol}\).To find out how much total energy is needed to vaporize a given mass of chloromethane, you first convert the mass into moles. This conversion requires knowing the molar mass, which for chloromethane is \(50.49\ \text{g/mol}\). By dividing the sample mass by the molar mass, you determine the number of moles.Once you have the number of moles, you multiply it by the heat of vaporization to get the total energy. For \(92.5\ \text{g}\) of chloromethane, this procedure results in approximately \(39.2\ \text{kJ}\) of energy needed, illustrating how much energy is involved in transformations at the molecular level.