Question

Cyclopropane, a gas used with oxygen as a general anesthetic, is composed of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\) by mass. (a) If \(1.56 \mathrm{~g}\) of cyclopropane has a volume of \(1.00 \mathrm{~L}\) at \(0.984 \mathrm{~atm}\) 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, \(\mathrm{CH}_{4}\) ?

Short answer

The molecular formula of cyclopropane is \(C_3H_6\). It is expected to deviate more from ideal-gas behavior than Argon at moderately high pressures and room temperature due to its more complex structure and stronger London dispersion forces. Cyclopropane would effuse more slowly than methane, as it has a larger molar mass.

Step-by-step solution

Determine the empirical formula

Divide the given mass percentages by their respective atomic masses: 85.7% C corresponds to \(\frac{85.7}{12.01} = 7.139\) moles of C, and 14.3% H corresponds to \(\frac{14.3}{1.01} = 14.158\) moles of H. Divide both values by the lowest (7.129) to get the ratio: \(C_1H_2\). This is the empirical formula.

Calculate the molar mass of cyclopropane using the Ideal Gas Law

The Ideal Gas Law is given by \(PV = nRT\), where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We know the mass of cyclopropane is 1.56 g, its volume is 1.00 L, pressure is 0.984 atm, and temperature is \(50.0 + 273.15 = 323.15\) K. Rearranging the Ideal Gas Law, we get the number of moles: \(n = \frac{PV}{RT}\). Plugging in the values, we get \(n = \frac{(0.984)(1.00)}{(0.0821)(323.15)} = 0.0366\) moles. Now, divide the mass of cyclopropane by the number of moles to find its molar mass: \(\frac{1.56 \mathrm{~g}}{0.0366 \mathrm{~mol}} = 42.62 \mathrm{~g/mol}\).

Determine the molecular formula of cyclopropane

Divide the molar mass of cyclopropane (42.62 g/mol) by the molar mass of the empirical formula (\(C_1H_2\), 14.03 g/mol): \(\frac{42.62}{14.03} = 3.04 \approx 3\). Multiply the empirical formula (\(C_1H_2\)) by 3 to get the molecular formula: \(C_3H_6\). This is the molecular formula of cyclopropane.

Analyzing deviation from ideal-gas behavior based on molecular formula

Cyclopropane, with its molecular formula \(C_3H_6\), has a more complex structure than noble gases like Argon. Deviations from ideal-gas behavior are more significant for molecules with larger size and stronger intermolecular forces. Since cyclopropane has additional London dispersion forces due to its more complex structure, it is expected to deviate more than Argon from the ideal-gas behavior at moderately high pressures and room temperature.

Compare effusion rates of cyclopropane and methane

According to Graham's Law of Effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass: \(Rate \propto \frac{1}{\sqrt{Molar~Mass}}\). Cyclopropane has a molar mass of 42.62 g/mol, while methane (\(\mathrm{CH}_{4}\)) has a molar mass of 16.04 g/mol. Comparing the rates of effusion, we can see that cyclopropane would effuse more slowly than methane, as it has a larger molar mass.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure, volume, and temperature of a gas, along with the number of moles present. The equation is written as \(PV = nRT\), where \(P\) denotes pressure, \(V\) is volume, \(n\) represents the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. This law assumes that gases behave ideally, meaning their particles do not interact except in elastic collisions and that they occupy no volume themselves.
To use the Ideal Gas Law, you must ensure that all units are consistent: pressure in atm, volume in liters, temperature in Kelvin, and the gas constant \(R\) as 0.0821 L·atm/mol·K.
When analyzing a sample like cyclopropane, you start by converting all measurements to the correct units. From there, rearrange the formula to solve for the unknown variable, such as the number of moles \(n\). This allows for the calculation of other properties, like molar mass, to aid in determining the molecular formula.

Empirical Formula Calculation

The empirical formula of a compound provides the simplest whole-number ratio of atoms within a molecule. To find it, you begin by converting the percentage composition of each element into moles using their atomic masses. For cyclopropane, with 85.7% carbon and 14.3% hydrogen, you convert these percentages to moles by dividing by the atomic masses of carbon (12.01 g/mol) and hydrogen (1.01 g/mol), respectively.
Once you have the mole ratios, you divide each by the smallest number of moles obtained to get the simplest ratio. For instance, cyclopropane gives a ratio of \(C_1H_2\), where carbon and hydrogen appear in a 1:2 ratio. This ratio may not represent the actual number of atoms in the molecule but helps in calculating the empirical formula. From this, using further data such as the molar mass, you can determine the molecular formula.

Graham's Law of Effusion

Graham's Law of Effusion describes how gases escape through a small opening. It states that the rate of effusion is inversely proportional to the square root of its molar mass: \(\text{Rate} \propto \frac{1}{\sqrt{\text{Molar Mass}}}\). This implies that lighter gases effuse faster than heavier ones.
For cyclopropane with a molar mass of 42.62 g/mol, comparing against methane (CH4) with a molar mass of 16.04 g/mol, cyclopropane effuses more slowly due to its greater mass. When calculating using Graham's Law, use the formula \(\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}}\) to find the relative rates of effusion between two gases. Understanding this law is crucial when predicting how quickly two different gases will escape or spread under similar conditions.

Deviation from Ideal Gas Behavior

Gases often deviate from ideal behavior, especially under high pressures or low temperatures. This is because the assumptions of the Ideal Gas Law—no intermolecular forces and negligible volume of gas particles—become less valid. Larger or more complex molecules tend to have more deviation.
Cyclopropane, with a larger and more complex molecular structure than a simple atom-like argon, is expected to deviate more. Its structure leads to stronger London dispersion forces, increasing interaction between molecules.
At times, knowing these deviations can help predict how gases will behave in different conditions. Cyclopropane, like many organic gases, shows these deviations at higher pressures and temperatures typical of real-world conditions.