Problem 39 - Chemistry and Chemical Reactivity

Question

A halothane-oxygen mixture \(\left(\mathrm{C}_{2} \mathrm{HBrClF}_{3}+\mathrm{O}_{2}\right)\) can be used as an anesthetic. A tank containing such a mixture has the following partial pressures: \(P\) (halothane) \(=170 \mathrm{mm} \mathrm{Hg}\) and \(P\left(\mathrm{O}_{2}\right)=570 \mathrm{mm} \mathrm{Hg}\) (a) What is the ratio of the number of moles of halothane to the number of moles of \(\mathrm{O}_{2} ?\) (b) If the tank contains \(160 \mathrm{g}\) of \(\mathrm{O}_{2}\), what mass of \(\mathrm{C}_{2} \mathrm{HBrClF}_{3}\) is present?

Step-by-Step Solution

Answer

(a) Mole ratio \( \approx 0.298.\) (b) About 294.1 g of halothane.

Step 1: Understanding Partial Pressure and Mole Ratio

To find the ratio of moles, we use Dalton's Law which states that the partial pressure of a gas in a mixture is proportional to its mole fraction. Thus, the number of moles ratio can be determined by the ratio of their partial pressures. Given that the partial pressure of halothane is 170 mm Hg and oxygen is 570 mm Hg, we find the mole ratio: \( \frac{n_{\text{halothane}}}{n_{\text{O}_2}} = \frac{P_{\text{halothane}}}{P_{\text{O}_2}} = \frac{170}{570} \approx 0.298.\)

Step 2: Calculate Moles of Oxygen

To find out how many moles of oxygen are in 160 g, use the molar mass of \(\text{O}_2\), which is 32.00 g/mol. \( n_{\text{O}_2} = \frac{160 \text{ g}}{32.00 \text{ g/mol}} = 5.00 \text{ moles}.\)

Step 3: Determine Moles of Halothane

Using the mole ratio from Step 1, \( \frac{n_{\text{halothane}}}{n_{\text{O}_2}} = 0.298,\) and \(n_{\text{O}_2} = 5.00\text{ moles},\) calculate \(n_{\text{halothane}}:\) \( n_{\text{halothane}} = 0.298 \times 5.00 \approx 1.49 \text{ moles}.\)

Step 4: Calculate the Mass of Halothane

To find the mass of halothane, use its molar mass \(\text{C}_2 \text{HBrClF}_3,\) which is approximately 197.38 g/mol. Multiply the number of moles by the molar mass:\( m_{\text{halothane}} = 1.49 \text{ moles} \times 197.38 \text{ g/mol} \approx 294.1 \text{ g}.\)

Key Concepts

Mole Ratio

The concept of mole ratio is fundamental to understanding gas mixtures and reactions. It helps convey the relationship between different substances in a mixture.
The mole ratio is derived using Dalton's Law of Partial Pressures, which allows us to relate a gas's partial pressure to its proportional concentration in the mixture. Dalton's Law states that the partial pressure of each gas in a mixture is directly proportional to the number of moles of the gas. Therefore, by comparing the partial pressures, one can determine the mole ratio.
For instance, consider a gas mixture with partial pressures: Halothane = 170 mm Hg and Oxygen = 570 mm Hg. Using the formula:

• Mole ratio = \(\frac{P_{\text{halothane}}}{P_{\text{O}_2}}\)

Plugging in the values:

• Mole ratio = \(\frac{170}{570} \approx 0.298\)

This signifies that for every mole of oxygen, there are approximately 0.298 moles of halothane. By understanding and calculating these ratios, one can predict the outcomes of reactions or the behavior of the mixture.

Molar Mass Calculation

Calculating molar mass is a crucial skill when dealing with chemical quantities, conversions, and stoichiometry. To determine molar mass, sum the atomic masses of all atoms within a molecule.
Take the example of oxygen, \(\text{O}_2\). The atomic mass of oxygen is approximately 16.00 g/mol. Since \(\text{O}_2\) contains two oxygen atoms, the molar mass becomes:

• Molar mass of \(\text{O}_2 = 16.00 \times 2 = 32.00 \text{ g/mol}\)

Using this molar mass, if we have 160 grams of \(\text{O}_2\), the number of moles is calculated by dividing the mass by its molar mass:

• \(n_{\text{O}_2} = \frac{160 \text{ g}}{32.00 \text{ g/mol}} = 5.00\text{ moles}\)

Molar mass calculations become even more essential when dealing with complex molecules like halothane \(\text{C}_2 \text{HBrClF}_3\), where each atomic mass contributes to the overall weight. For halothane, the molar mass can be calculated individually, summing the masses of the constituent atoms, leading to an approximate molar mass of 197.38 g/mol. This value is then used in further calculations involving mass and moles.

Anesthetic Mixture

Anesthetic mixtures typically involve the combination of gases or compounds that result in a physiological effect, such as sedation or unconsciousness.
In the context of halothane-oxygen mixtures, halothane acts as the anesthetic agent while oxygen is the carrier gas. The combination ensures that the patient remains sedated and receives essential life-sustaining oxygen. This makes calculating the proportions of each component vital for efficiency and safety.
Given the partial pressures of halothane (170 mm Hg) and oxygen (570 mm Hg) in the exercise, providing the correct mole ratio, as established, is crucial in optimizing the anesthetic effect while maintaining oxygenation. Therefore, understanding how to manipulate and calculate these mixtures ensures that anesthetic procedures are both effective and safe for patients. It is crucial to comprehend these mixtures when preparing for real-life applications in medical settings.