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 \(0.984\) 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.

Short answer

The molecular formula of cyclopropane is \(\mathrm{C_3H_6}\). Compared to Argon, cyclopropane is expected to deviate more from the ideal gas behavior at moderately high pressures and room temperature due to its larger size and molecular complexity.

Step-by-step solution

Determine the empirical formula from the given mass percentages

To find the empirical formula of cyclopropane, which is the simplest whole number ratio of atoms in the molecule, we can assume a 100 g sample and determine the moles of Carbon and Hydrogen. For 100 g sample: - 85.7 g of Carbon (\(\mathrm{C}\)) - 14.3 g of Hydrogen (\(\mathrm{H}\)) Molar mass of \(\mathrm{C}\) = 12.01 g/mol Molar mass of \(\mathrm{H}\) = 1.008 g/mol Moles of \(\mathrm{C}\) = \(\frac{85.7}{12.01} \approx 7.14\) moles Moles of \(\mathrm{H}\) = \(\frac{14.3}{1.008} \approx 14.2\) moles \(\frac{7.14}{7.14} \mathrm{C} \mathrm{H}\frac{14.2}{7.14}\) \(= 1\mathrm{C}2\mathrm{H}\) The empirical formula is \(\mathrm{CH_2}\).

Use the Ideal Gas Law to find the molar mass

With the given sample conditions, we can now use the Ideal Gas Law to find the molar mass of cyclopropane. The Ideal Gas Law Equation: \(PV = nRT\) where P = pressure = \(0.984 \mathrm{~atm}\) V = volume = \(1.00 \mathrm{~L}\) T = temperature = \((50.0 + 273.15) = 323.15 \mathrm{K}\) R = gas constant = \(0.08206 \frac{\mathrm{L}\cdot {\mathrm{atm}}}{{\mathrm{mol}}\cdot {\mathrm{K}}}\) First, we need to find the moles (n) of cyclopropane in the 1-liter sample: \(n =\frac{PV}{RT} =\frac{(0.984 \times 1.00)}{(0.08206 \times 323.15)} \approx 0.03733 \; \text{moles}\) Next, use the moles of cyclopropane and the given sample mass (1.56 g) to find the molar mass: Molar mass = \(\frac{\text{sample mass (g)}}{\text{moles}} =\frac{1.56}{0.03733} \approx 41.78 \; \text{g/mol}\)

Determine the molecular formula from the empirical formula and molar mass

To find the molecular formula of cyclopropane, we will divide the molar mass by the mass of the empirical formula. Mass of empirical formula (\(\mathrm{CH_2}\)): \(12.01 + (2 \times 1.008) = 14.03 \; \text{g/mol}\) Divide the molar mass by the empirical formula mass to find the multiplier: Multiplier = \(\frac{41.78}{14.03} \approx 3\) Molecular formula = (Empirical formula) × Multiplier = \(\mathrm{(CH_2)_3}\) = \(\mathrm{C_3H_6}\) So, the molecular formula of cyclopropane is \(\mathrm{C_3H_6}\).

Analyzing the deviation from ideal-gas behavior and comparison with Ar

Cyclopropane (\(\mathrm{C_3H_6}\)) is a heavier and more complex molecule compared to Argon (\(\mathrm{Ar}\)). The Ideal Gas Law assumes that gas molecules are point particles with no volume and no interactions between them. However, this is not the case with complex molecules like cyclopropane. In real situations, molecules have volume and there are both attractive and repulsive forces between them. At moderately high pressures and room temperature, these forces can cause deviations from ideal gas behavior. Compared to Argon, we can expect cyclopropane to deviate more from the ideal gas behavior due to its higher molecular complexity and larger size. In conclusion, the molecular formula of cyclopropane is \(\mathrm{C_3H_6}\), and we can expect its behavior to deviate more from the ideal gas behavior than that of Argon, particularly at moderately high pressures and room temperature, due to its larger size and molecular complexity.

Key concepts

Understanding the Molecular Formula

The molecular formula of a compound represents the actual number of each type of atom present in a molecule. Unlike the empirical formula, which only provides the simplest ratio of elements, the molecular formula offers precise atomic composition. In the case of cyclopropane, its molecular formula is found to be \( \mathrm{C_3H_6} \). This formula was obtained using known conditions and measurements such as pressure, volume, and temperature together with the empirical findings. It tells us that a single molecule of cyclopropane consists of 3 carbon atoms and 6 hydrogen atoms.

Applying the Ideal Gas Law

The Ideal Gas Law is a crucial equation in chemistry, used to relate measurable properties of an ideal gas. The formula is \( PV = nRT \), where:

• \( P \) = pressure
• \( V \) = volume
• \( n \) = number of moles
• \( R \) = ideal gas constant
• \( T \) = temperature

In the task of identifying cyclopropane's formula, this law helps estimate the number of moles under given conditions. Knowing the moles allows students to calculate its molar mass. Applying this knowledge bridged the gap between empirical findings and an accurate molecular formula. It's important to remember that the Ideal Gas Law assumes no interaction between molecules, which is not perfect for real gases like cyclopropane, particularly under high pressure or low temperature.

Dissecting the Empirical Formula

The empirical formula provides the simplest integer ratio of elements in a compound. For cyclopropane, this formula is \( \mathrm{CH_2} \). It suggests that for every carbon atom, there are two hydrogen atoms. How is it determined? By calculating the moles of each element from their mass percentages. Based on the given problem, we observed remains of carbon at approximately 85.7% and hydrogen at 14.3% mass contribution. Working through ratios, the derived simplest form was \( \mathrm{CH_2} \), which served as the foundation for further analysis towards deriving the molecular formula.

Exploring Chemical Composition

Chemical composition involves the types and amounts of elements that make up a compound. Cyclopropane consists predominantly of carbon and hydrogen, which contribute to its properties as a flammable gas used in anesthetics. With a chemical breakdown of 85.7% carbon and 14.3% hydrogen by mass, knowing these proportions is crucial for understanding its reactions and behaviors. Information derived from the chemical composition is vital not only for academic exercises but also for industrial and medical applications. This insight allows researchers to predict characteristics, behaviors under different conditions, and potential uses of a compound like cyclopropane.