Problem 22 - Chemistry and Chemical Reactivity

Question

Chloromethane, \(\mathrm{CH}_{3} \mathrm{Cl}\), arises from the oceans and from microbial fermentation and is found throughout the environment. It is used in the manufacture of various chemicals and has been used as a topical anesthetic. What quantity of heat must be absorbed to convert \(92.5 \mathrm{g}\) of liquid to a vapor at its boiling point, \(-24.09^{\circ} \mathrm{C} ?\) The heat of vaporization of \(\mathrm{CH}_{3} \mathrm{Cl}\) is \(21.40 \mathrm{kJ} / \mathrm{mol}\).

Step-by-Step Solution

Answer

39.21 kJ of heat is required.

Step 1: Determine the Molar Mass of CH3Cl

To find the molar mass of chloromethane, \[ \text{CH}_3\text{Cl} \] we add the atomic masses of all atoms in the molecule:- Carbon (C): \(12.01 \text{ g/mol}\)- Hydrogen (H): \(1.01 \text{ g/mol}\) for each hydrogen, and there are 3 hydrogens.- Chlorine (Cl): \(35.45 \text{ g/mol}\)Thus, the molar mass of CH₃Cl is:\[ 12.01 + (3 \times 1.01) + 35.45 = 50.49 \text{ g/mol} \]

Step 2: Convert Mass to Moles

We need to convert the given mass of chloromethane to moles using the molar mass:\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{92.5 \text{ g}}{50.49 \text{ g/mol}} \]Calculate:\[ n \approx \frac{92.5}{50.49} \approx 1.832 \text{ mol} \]

Step 3: Calculate Heat Required for Vaporization

We use the number of moles and the heat of vaporization to find the total heat required:\[ q = n \times \Delta H_{vap} = 1.832 \text{ mol} \times 21.40 \text{ kJ/mol} \]Calculate:\[ q \approx 39.21 \text{ kJ} \]

Step 4: Final Answer

The heat required to convert 92.5 g of liquid chloromethane to vapor at its boiling point is approximately 39.21 kJ.

Key Concepts

Molar Mass Calculation

Understanding molar mass is essential when working with chemical substances. It represents the mass of one mole of a substance and is expressed in grams per mole (g/mol). To find the molar mass of a compound like chloromethane (CH₃Cl), follow these simple steps:

• Identify the atomic masses of each atom in the molecule from the periodic table.
• Carbon (C) has an atomic mass of 12.01 g/mol.
• Hydrogen (H) has an atomic mass of 1.01 g/mol. Since there are three hydrogen atoms, you multiply this by 3.
• Chlorine (Cl) has an atomic mass of 35.45 g/mol.

Add them all together:\[ 12.01 + (3 \times 1.01) + 35.45 = 50.49 \text{ g/mol} \]This calculation tells us that one mole of chloromethane weighs 50.49 grams.

Conversion from Mass to Moles

After determining the molar mass, converting mass to moles is a vital step in many calculations. Moles are a standard unit in chemistry that allows us to relate mass to the number of molecules involved in a reaction.To convert a given mass to moles, use the formula:\[ n = \frac{\text{mass}}{\text{molar mass}} \]Where:

• \( n \) is the number of moles
• \( \text{mass} \) is the given mass of the substance
• \( \text{molar mass} \) is the molar mass calculated previously

In our example, with 92.5 g of chloromethane and a molar mass of 50.49 g/mol, the calculation is:\[ n \approx \frac{92.5}{50.49} \approx 1.832 \text{ mol} \]So, we have approximately 1.832 moles of chloromethane.

Boiling Point

The boiling point is a crucial property that helps us understand when a substance changes from liquid to gas. It occurs when the vapor pressure of the liquid equals the surrounding atmospheric pressure. For chloromethane, this transition occurs at \(-24.09^{\circ} \text{C}\).Knowing the boiling point is essential when calculating the energy required for vaporization. At the boiling point, the heat absorbed by the liquid is used to overcome intermolecular forces without changing temperature, known as the heat of vaporization.For chloromethane:

• The heat of vaporization is \(21.40 \text{kJ/mol}\).

Using this, you can calculate the energy needed to convert 1.832 moles of liquid chloromethane to vapor. The formula is:\[ q = n \times \Delta H_{vap} = 1.832 \text{ mol} \times 21.40 \text{ kJ/mol} \]Thus, approximately \(39.21 \text{kJ}\) of heat must be absorbed for the phase change.