Questions: Suppose you mix 15 g of halothane vapor with 23.5 g oxygen gas. If the total pressure of the mixture is 855 mmHg, what are the partial pressures of each gas?

Solution Steps

First, we need to determine the molar mass of each gas.

• Haloethane (C\(_2\)H\(_3\)ClF\(_3\)):

  • Carbon (C): \(2 \times 12.01 \, \text{g/mol} = 24.02 \, \text{g/mol}\)
  • Hydrogen (H): \(3 \times 1.008 \, \text{g/mol} = 3.024 \, \text{g/mol}\)
  • Chlorine (Cl): \(1 \times 35.45 \, \text{g/mol} = 35.45 \, \text{g/mol}\)
  • Fluorine (F): \(3 \times 19.00 \, \text{g/mol} = 57.00 \, \text{g/mol}\)

  Total molar mass of haloethane = \(24.02 + 3.024 + 35.45 + 57.00 = 119.494 \, \text{g/mol}\)

• Oxygen (O\(_2\)):

  • Oxygen (O): \(2 \times 16.00 \, \text{g/mol} = 32.00 \, \text{g/mol}\)

Using the formula \(n = \frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass:

• Haloethane: \[ n_{\text{haloethane}} = \frac{15 \, \text{g}}{119.494 \, \text{g/mol}} = 0.1255 \, \text{mol} \]

• Oxygen: \[ n_{\text{oxygen}} = \frac{23.5 \, \text{g}}{32.00 \, \text{g/mol}} = 0.7344 \, \text{mol} \]

Add the moles of each gas to find the total number of moles in the mixture:

\[ n_{\text{total}} = n_{\text{haloethane}} + n_{\text{oxygen}} = 0.1255 + 0.7344 = 0.8599 \, \text{mol} \]

Using Dalton's Law of Partial Pressures, the partial pressure of each gas is given by:

\[ P_i = \left(\frac{n_i}{n_{\text{total}}}\right) \times P_{\text{total}} \]

where \(P_{\text{total}} = 855 \, \text{mmHg}\).

• Partial Pressure of Haloethane: \[ P_{\text{haloethane}} = \left(\frac{0.1255}{0.8599}\right) \times 855 = 124.8 \, \text{mmHg} \]

• Partial Pressure of Oxygen: \[ P_{\text{oxygen}} = \left(\frac{0.7344}{0.8599}\right) \times 855 = 730.2 \, \text{mmHg} \]

Final Answer