Problem 118 - Chemistry The Central Science In Si Units

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. \((\mathbf{a})\) 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, \(\mathrm{CH}_{4} ?\)

Step-by-Step Solution

Answer

Cyclopropane's molecular formula is \( \mathrm{C}_3\mathrm{H}_6 \). It deviates more from ideal-gas behavior than Ar and effuses slower than methane.

Step 1: Find Molar Mass of Cyclopropane

Given the mass of cyclopropane is 1.56 g and its volume is 1.00 L at 99.7 kPa and 50.0°C, we use the ideal gas equation to find the molar mass. First, convert the temperature to Kelvin: \( T = 50.0 + 273.15 = 323.15 \) K. Use the ideal gas law: \( PV = nRT \), where \( P = 99.7 \) kPa (or 0.997 atm), \( V = 1.00 \) L, and \( R = 0.08206 \) L atm / K mol. Solve for \( n \), the number of moles: \( n = \frac{PV}{RT} = \frac{(0.997)(1.00)}{(0.08206)(323.15)} \approx 0.0376 \) moles. The molar mass \( M \) is then \( \frac{1.56}{0.0376} \approx 41.5 \) g/mol.

Step 2: Determine Empirical Formula

Cyclopropane is composed of 85.7% carbon and 14.3% hydrogen. Assume 100 g of cyclopropane, then it contains 85.7 g of C and 14.3 g of H. Calculate moles of each: moles of C = \( \frac{85.7}{12.01} \approx 7.14 \), moles of H = \( \frac{14.3}{1.008} \approx 14.2 \). The ratio is \( \frac{7.14}{7.14} = 1 \) for C and \( \frac{14.2}{7.14} \approx 2 \) for H, giving an empirical formula of \( \mathrm{CH}_2 \).

Step 3: Find Molecular Formula

The molar mass of the empirical formula \( \mathrm{CH}_2 \) is approximately 14.0 g/mol. Divide the molar mass of cyclopropane (41.5 g/mol) by the empirical mass (14.0 g/mol): \( \frac{41.5}{14.0} \approx 3 \). Multiply the empirical formula by 3 to find the molecular formula: \( \mathrm{C}_3\mathrm{H}_6 \).

Step 4: Analyze Deviation from Ideal Gas Behavior

Cyclopropane, \( \mathrm{C}_3\mathrm{H}_6 \), is more complex than Argon, a noble gas. Due to its molecular structure and greater intermolecular forces, we expect it to deviate more from ideal-gas behavior than Argon, especially at moderately high pressures.

Step 5: Compare Effusion Rates

Use Graham's law of effusion, which states that effusion rates are inversely proportional to the square root of molar masses. The molar mass of methane \( \mathrm{CH}_4 \) is 16 g/mol. The ratio of effusion rates \( \frac{\text{Rate of cyclopropane}}{\text{Rate of methane}} = \sqrt{\frac{16}{41.5}} \approx 0.623 \). Thus, cyclopropane effuses slower than methane.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle connecting the pressure, volume, temperature, and amount of a gas in moles. It's expressed as:

• \( PV = nRT \)

Where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume the gas occupies.
• \( n \) is the number of moles.
• \( R \) is the universal gas constant (0.08206 L atm / K mol).
• \( T \) is the temperature in Kelvin.

To solve for the number of moles \( n \) of cyclopropane, we rearrange the formula:

• \( n = \frac{PV}{RT} \)

Starting with conversions is crucial. Convert the temperature from Celsius to Kelvin by adding 273.15 to give us 323.15 K. Pressure should be in atm, hence 99.7 kPa becomes 0.997 atm. After substituting into the equation, we find:

• \( n = \frac{(0.997)(1.00)}{(0.08206)(323.15)} \approx 0.0376 \text{ moles} \)

This number of moles can then be combined with the mass to determine molar mass, which in turn helps find the molecular formula.

Effusion Rates

Effusion describes the process of gas particles escaping through a tiny hole. The rate at which different gases effuse can be determined by Graham's Law, which tells us:

• \( \frac{\text{Rate of gas 1}}{\text{Rate of gas 2}} = \sqrt{\frac{M_2}{M_1}} \)

Where \( M_1 \) and \( M_2 \) are the molar masses of gas 1 and gas 2, respectively. This law highlights that lighter gases effuse more quickly. For cyclopropane (molar mass \( 41.5 \text{ g/mol} \)) and methane (molar mass \( 16 \text{ g/mol} \)), the comparison becomes:

• \( \frac{\text{Rate of cyclopropane}}{\text{Rate of methane}} = \sqrt{\frac{16}{41.5}} \approx 0.623 \)

This means that cyclopropane effuses slower than methane, due to its higher molar mass. This insight is valuable during laboratory experiments, where gas effusion rates may impact experimental results.

Empirical Formula Determination

The empirical formula of a compound represents the simplest whole number ratio of atoms of each element in the compound. To determine it, focus on mass percentages. For cyclopropane:

• 85.7% of the mass is carbon, and 14.3% is hydrogen.
• We assume a total mass of 100 g to ease calculations.
• Compute moles: carbon \( \frac{85.7}{12.01} \approx 7.14 \) and hydrogen \( \frac{14.3}{1.008} \approx 14.2 \).

To find the ratio:

• Carbon: \( \frac{7.14}{7.14} = 1 \)
• Hydrogen: \( \frac{14.2}{7.14} \approx 2 \)

Leading to an empirical formula of \( \text{CH}_2 \). The molecular formula is often a multiple of the empirical formula. With the molar mass, calculate the multiple:

• \( \text{Molar mass of } \text{CH}_2 = 14.0 \text{ g/mol} \)
• \( \text{Molar mass of cyclopropane} = 41.5 \text{ g/mol} \)
• \( \text{Ratio} = \frac{41.5}{14.0} \approx 3 \)

Thus, multiply the empirical formula by this ratio to find the molecular formula: \( \text{C}_3\text{H}_6 \). Understanding this process is essential in chemistry as it connects molecular structure with observable properties.