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. (c) Would cyclopropane effuse through a pinhole faster or more slowly than methane, CH. \(_{4} ?\)

Short answer

(a) The molecular formula of cyclopropane is C3H6. (b) Cyclopropane would deviate more from ideal gas behavior than Argon at moderately high pressures and room temperature due to its larger and more complex molecules with stronger intermolecular forces. (c) Cyclopropane will effuse more slowly than methane, as its effusion rate ratio is less than 1.

Step-by-step solution

Find the moles of C and H in 100 g of the compound

Given, a mass composition of 85.7% C and 14.3% H in cyclopropane. In 100 g of the compound, we have: 85.7 g of Carbon (C) and 14.3 g of Hydrogen (H). Now, find the moles of each element in 100 g of the compound: (For Carbon, the molar mass is 12.01 g/mol, and for Hydrogen, the molar mass is 1.008 g/mol.) Moles of Carbon (nC) = (85.7 g) / (12.01 g/mol) = 7.14 mol Moles of Hydrogen (nH) = (14.3 g) / (1.008 g/mol) = 14.2 mol

Find the empirical formula and molar mass of the empirical formula

Divide the moles of each element by the smallest number to get the empirical formula. Empirical Formula: C(7.14/7.14)H(14.2/7.14) = CH2 Now, calculate the molar mass of the empirical formula: Molar mass of CH2 = (12.01 g/mol + 2 * 1.008 g/mol) = 14.03 g/mol

Use the ideal gas equation to find the moles of gas

Given, 1.56 g of cyclopropane has a volume of 1.00 L at 0.984 atm and 50.0°C. Apply the ideal gas equation, 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. Temperature in Kelvin: T = 50.0 + 273.15 = 323.15 K Gas constant R = 0.0821 L atm / (mol K) Rearranging the equation to solve for n (moles): n = PV / (RT) Substitute the known values: n = (0.984 atm)(1.00 L) / (0.0821 L atm/mol K)(323.15 K) n = 0.0367 mol

Determine molecular formula and molar mass of cyclopropane

Calculate the molar mass using the given 1.56 g of cyclopropane and the moles found in step 3: Molar mass of cyclopropane = (1.56 g) / (0.0367 mol) = 42.5 g/mol Divide the molar mass of cyclopropane by the molar mass of the empirical formula to find the whole number multiplier: Multiplier = (42.5 g/mol) / (14.03 g/mol) ≈ 3 Multiply the empirical formula (CH2) by 3 to get the molecular formula of cyclopropane: Molecular Formula = C3H6 (a) The molecular formula of cyclopropane is C3H6.

Analyze cyclopropane's deviation from ideal gas behavior

(b) Cyclopropane (C3H6) consists of larger and more complex molecules with stronger intermolecular forces compared to Argon (Ar), which is a noble gas with weak intermolecular forces. As a result, at moderately high pressures and room temperature, cyclopropane would deviate more from ideal gas behavior than Argon.

Compare the effusion rate of cyclopropane to methane

(c) 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. To compare the effusion rate of cyclopropane (C3H6) and methane (CH4), we need their molar masses. C3H6 (Cyclopropane) Molar Mass = 42.5 g/mol CH4 (Methane) Molar Mass = 12.01 g/mol + 4 * 1.008 g/mol = 16.04 g/mol Effusion Rate Ratio = (Effusion Rate of Cyclopropane) / (Effusion Rate of Methane) = sqrt(Molar Mass of Methane) / sqrt(Molar Mass of Cyclopropane) = sqrt(16.04 g/mol) / sqrt(42.5 g/mol) = 0.617 As the ratio is less than 1, cyclopropane will effuse more slowly than methane.

Key concepts

Empirical Formula Calculation

At the heart of understanding chemical compositions, the empirical formula serves as a fundamental stepping stone. It represents the simplest whole-number ratio of the elements in a compound. For students grappling with this concept, the process begins with the percentage composition by mass of each element, which is then used to calculate the moles of the individual elements. Once you have the moles, it's a game of comparison: each element's mole count is divided by the smallest mole count among them. The resulting numbers are translated into the empirical formula through rounding to the nearest whole number. Our example with cyclopropane, which consists of 85.7% Carbon and 14.3% Hydrogen by mass, led to the empirical formula CH₂.

Understanding this process is crucial, as it lays the groundwork for the subsequent determination of the molecular formula. The empirical formula gives the smallest, simplest view of what's inside, but it's the molecular formula that conveys the actual number of atoms in each molecule of a compound; cyclopropane's being C₃H₆. This jump from empirical to molecular formula hinges on the ratio between the compound's empirical formula mass and its actual molar mass.

Ideal Gas Law

The ideal gas law is a powerful tool in the arsenal of chemistry and physics, relating the pressure, volume, temperature, and number of moles of a gas with the timeless equation PV = nRT. Often, for students, the most challenging aspect is understanding each component and how they affect one another. It's crucial to note that R, the gas constant, acts as the linchpin in this relationship, ensuring that the units of P (pressure), V (volume), T (temperature), and n (moles) are compatible.

When we tackle problems like determining the molecular formula of cyclopropane, the ideal gas law comes into play to find the number of moles of gas under specific conditions. In our case, with known mass, pressure, temperature, and volume, the moles of cyclopropane are calculated, which is the key to unlock the door towards its molecular formula. Remember to always convert temperature to Kelvin and pay attention to the units for pressure and volume to ensure they're consistent with the value of R you are using.

Graham's Law of Effusion

Velocity enters the picture when we talk about Graham's law of effusion – a principle that deals with the speed at which gases escape through a tiny opening into a vacuum. For students, envisioning effusion can be simplified by likening it to letting air out of a balloon through a needle's pinhole: the rate at which it escapes is determined by the gas's molar mass - lighter gases zip out faster than heavier ones. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass, mathematically expressed as the rate of effusion being directly proportional to \(\sqrt{1/M}\), where M represents the molar mass.

Comparing two gases, like cyclopropane and methane in the given exercise, involves calculating the square root of the inverse ratio of their molar masses. This approach means that the lighter methane gas effuses quicker than the heavier cyclopropane. This principle not only helps us predict effusion rates but also reinforces concepts of molar mass and gas behavior.