Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. \((\mathbf{a})\) If $1.56 \mathrm{~g}\( of cyclopropane has a volume of \)1.00 \mathrm{~L}$ at 99.7 \(\mathrm{kPa}\) and \(50.0^{\circ} \mathrm{C}\), what is the molecular formula of cyclopropane? (b) Judging from its molecular formula, would you expect cyclopropane to deviate more or less than Ar from ideal-gas behavior at moderately high pressures and room temperature? Explain. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, \(\mathrm{CH}_{4} ?\)

Given the percentage composition by mass as \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\), we can find the moles of each element by assuming we have a 100 g sample of the gas. \(85.7 \, \mathrm{g \, C * \frac{1 \, mol \, C}{12.01 \, g \, C}} \approx 7.14 \, \mathrm{mol \, C}\) \(14.3 \, \mathrm{g \, H * \frac{1 \, mol \, H}{1.008 \, g \, H}} \approx 14.2 \, \mathrm{mol \, H}\) Now, divide the moles of both elements by the minimum value. \(\frac{7.14 \, \mathrm{mol \, C}}{7.14} \approx 1 \, \mathrm{mol \, C}\) \(\frac{14.2 \, \mathrm{mol \, H}}{7.14} \approx 2 \, \mathrm{mol \, H}\) The empirical formula of cyclopropane is CH2.

We are given a sample of the gas with mass 1.56 g and volume 1.00 L under the conditions 99.7 kPa and 50.0°C. Use the Ideal Gas Law to determine the molar mass of the gas. \(PV = nRT \) \[n = \frac{PV}{RT} \] Here, \(P = 99.7 \, \mathrm{kPa}\), \(R = 8.314 \, \mathrm{\frac{J}{mol \cdot K}}\), \(T = 50.0 + 273.15 = 323.15 \, \mathrm{K}\), and \(V = 1.00 \, \mathrm{L} = 1000 \, \mathrm{mL}\). Convert pressure from kPa to Pa. \(P = 99.7 \, \mathrm{kPa} * 1000 \, \mathrm{\frac{Pa}{kPa}} = 99700 \, \mathrm{Pa}\) Now, plug in the values and calculate the number of moles: \(n = \frac{(99700 \, \mathrm{Pa})(1.00 \times 10^{-3} \, \mathrm{m^3})}{(8.314 \, \mathrm{\frac{J}{mol \cdot K}})(323.15 \, \mathrm{K})} \approx 0.0373 \, \mathrm{mol}\) Next, find the molar mass by dividing the mass by the moles: \(Molar \, Mass = \frac{1.56 \, \mathrm{g}}{0.0373 \, \mathrm{mol}} \approx 41.8 \, \mathrm{\frac{g}{mol}}\)

The empirical formula of cyclopropane is CH2, which has a molar mass of approximately 14.02 g/mol. Divide the molar mass of the compound by the empirical formula's molar mass and round to the nearest whole number: \(\frac{41.8 \, \mathrm{\frac{g}{mol}}}{14.02 \, \mathrm{\frac{g}{mol}}} \approx 3\) The molecular formula of cyclopropane is C3H6, which is obtained by multiplying the empirical formula by the factor 3.

Cyclopropane's molecular formula, C3H6, suggests that it is a more complex molecule than a noble gas like argon. As a result, it is more likely to have stronger intermolecular forces and therefore deviate more from ideal gas behavior, particularly at moderately high pressures and room temperature.

The effusion rate of a gas is inversely proportional to the square root of its molar mass. To compare the effusion rates of cyclopropane and methane, we can use the following relation: \(\frac{Rate_{C3H6}}{Rate_{CH4}} = \sqrt{\frac{Molar \, Mass_{CH4}}{Molar \, Mass_{C3H6}}}\) \(Molar \, Mass_{CH4} = 16.04 \, \mathrm{\frac{g}{mol}}\) \(\frac{Rate_{C3H6}}{Rate_{CH4}} = \sqrt{\frac{16.04 \, \mathrm{\frac{g}{mol}}}{41.8 \, \mathrm{\frac{g}{mol}}}} \approx 0.62\) Since the rate is less than 1, cyclopropane would effuse more slowly than methane.