Question

Cyclopropane mixed in the proper ratio with oxygen can be used as an anesthetic. At \(755 \mathrm{~mm} \mathrm{Hg}\) and \(25^{\circ} \mathrm{C}\), it has a density of \(1.71 \mathrm{~g} / \mathrm{L}\). (a) What is the molar mass of cyclopropane? (b) Cyclopropane is made up of \(85.7 \% \mathrm{C}\) and \(14.3 \% \mathrm{H}\). What is the molecular formula of cyclopropane?

Short answer

Answer: The molecular formula of cyclopropane is C3H6.

Step-by-step solution

Determine the molar volume of the gas mixture at the given conditions

First, we need to calculate the molar volume of the gas under the given conditions. We will use the ideal gas law to achieve this: \(PV=nRT\) where: P = pressure = 755 mm Hg * (1 atm/760 mmHg) = 0.9934 atm V = volume (in L) n = number of moles R = ideal gas constant = 0.0821 L*atm/(mol*K) T = temperature = 25°C + 273.15 = 298.15 K Rearranging the ideal gas equation for molar volume (V/n), it becomes: \(\frac{V}{n} =\frac{RT}{P}\) Now, we can plug in the values and calculate the molar volume: \(\frac{V}{n} = \frac{(0.0821)(298.15)}{0.9934} = 24.47 \mathrm{L/mol}\)

Calculate the molar mass of cyclopropane using the given density

Using the density and molar volume that was calculated in Step 1, we can determine the molar mass of cyclopropane. To do this, use the formula: \( ρ = \frac{mass}{volume} = \frac{n * molar~mass}{molar~volume}\) Rearranging the formula to solve for the molar mass: \( molar~mass = \frac{ρ*molar~volume}{n}\) Now plug in the known values: \(molar~mass = \frac{1.71 * 24.47}{1} = 41.83 \mathrm{g/mol}\)

Calculate the empirical formula of cyclopropane using the given percentages

To find the empirical formula, we can start by assuming 100g of the molecule, and then calculate the moles of carbon and hydrogen present: moles of C: \(\frac{85.7g}{12.01 \mathrm{~g/mol}} = 7.14\mathrm{~mol}\) (using the molar mass of Carbon = 12.01 g/mol) moles of H: \(\frac{14.3g}{1.01 \mathrm{~g/mol}} = 14.16\mathrm{~mol}\) (using the molar mass of Hydrogen = 1.01 g/mol) Now, divide each molar amount by the smallest value to find the ratio of the elements: \(\frac{7.14}{7.14} : \frac{14.16}{7.14} = 1 : 1.99 \approx 1 : 2\) Thus, the empirical formula is CH2.

Determine the molecular formula using the molar mass and empirical formula mass

First, calculate the empirical formula mass: Empirical formula mass = (12.01 g/mol C) * 1 + (1.01 g/mol H) * 2 = 14.03 g/mol Now, find the ratio between the molar mass and empirical formula mass: Ratio = \(\frac{molar~mass}{empirical~formula~mass} = \frac{41.83}{14.03} = 2.98 \approx 3\) Finally, determine the molecular formula by multiplying the empirical formula by the ratio: Molecular formula = CH2 * 3 = C3H6 So, the molecular formula of cyclopropane is C3H6.

Key concepts

Understanding the Ideal Gas Law

The ideal gas law is a critical concept for anyone studying chemistry or physics, as it describes the behavior of an 'ideal' gas under various conditions. It is represented by the equation PV=nRT, where P stands for pressure, V for volume, n for the amount of substance in moles, R for the universal gas constant, and T for temperature in Kelvin.

To use the ideal gas law effectively, it's essential to ensure that the units used are consistent. For pressure, atmospheres (atm) are often preferred, and the volume must be in liters (L) to align with the gas constant R, which has units of L*atm/(mol*K). When dealing with temperature, always use Kelvin, as it is the SI unit for thermodynamic temperature scale.

In the context of calculating the molar mass of cyclopropane, the ideal gas law helps to determine the molar volume at which one mole of gas occupies under given pressure and temperature. This foundational understanding is the stepping stone towards finding the molar mass, as the density is connected to molar volume through the equation \( \rho = \frac{mass}{volume} \).

Empirical and Molecular Formula Determination

The distinction between empirical and molecular formulas is fundamental in chemistry. An empirical formula represents the simplest whole number ratio of the elements in a compound, while the molecular formula shows the actual number of atoms of each element in a molecule. To determine these formulas, the percentage composition of each element must be known.

Taking the example of cyclopropane, the empirical formula is derived by converting percentages into moles, based on the assumption of a 100g sample. Then, the moles of each element are divided by the smallest number of moles to achieve the simplest ratio. For many compounds, the empirical formula is a building block for the molecular formula. To bridge the gap from the empirical formula to the molecular formula, the molar mass (obtained from molar volume and density via the ideal gas law) is divided by the mass of the empirical formula unit. The resulting ratio is used to multiply the subscripts in the empirical formula to obtain the molecular formula, giving a true representation of the substance's composition.

Molar Mass Calculation from Gas Density

The molar mass of a substance is a fundamental physical property defined as the mass of one mole of that substance. It is a key piece of information for many chemical calculations. In the case of gases, molar mass can be determined from the density using the ideal gas law. By manipulating the law to express n/V (moles per volume), it becomes possible to solve for the molar mass when density (mass per volume) is given.

In the solution for the molar mass of cyclopropane, density (\( \rho \)) is utilized along with the calculated molar volume under the specified conditions of temperature and pressure. The formula \( molar~mass = \frac{\rho*molar~volume}{n} \) directly relates density to molar mass, simplifying the process of calculation. The determined molar mass not only provides insight into the molecular weight of the gas but also serves as a stepping stone to finding the molecular formula. By combining the understanding of molar mass, empirical formula, and molecular formula, one gains a comprehensive overview of the compound's identity in terms of its compositional structure.