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 \(85.7 \%\) C and \(14.3 \%\) H. What is the molecular formula of cyclopropane?

Short answer

Answer: The molecular formula of cyclopropane is C3H9.

Step-by-step solution

a) Finding the molar mass of cyclopropane

To determine the molar mass of cyclopropane, we can use the ideal gas equation: \(PV=nRT\) where: P = pressure V = volume n = number of moles R = ideal gas constant T = temperature Firstly, we should convert given pressure and temperature to the SI units: pressure to Pascals (Pa) and temperature to Kelvin (K). Pressure: \(755 \mathrm{~mm} \mathrm{Hg} = \dfrac{101325}{760} \times 755 \mathrm{~Pa} = 101325 \mathrm{~Pa}\) Temperature: \(25^{\circ} \mathrm{C} = 25 + 273.15 = 298.15 \mathrm{K}\) Now, rearrange the ideal gas equation to get the number of moles (n) and then the molar mass: \(n=\dfrac{PV}{RT}\) The density (\(\rho\)) of the gas mixture is given as \(1.71 \mathrm{~g} / \mathrm{L}\), and we can derive number of moles for the gas mixture: \(\rho = \dfrac{m}{V}\) \(m = \rho V\) But molar mass (M) can be written as: \(M=\dfrac{m}{n}\) Now, divide mass by volume: \(\dfrac{m}{V} = \dfrac{M \times n}{V}\) Now we can see that \(\dfrac{M \times n}{V}\) = density (\(\rho\)): \(M = \dfrac{\rho V}{n}\) Finally, substitute the achieved n and given \(\rho\), then solve for M: \(M = \dfrac{1.71 \mathrm{~g} / \mathrm{L}}{\dfrac{\left(101325 \mathrm{~Pa}\right)(1 \mathrm{L})}{\left(8.314 \mathrm{J\ K^{-1}mol^{-1}}\right)(298.15 \mathrm{K})}} \approx 42.06 \mathrm{~g/mol}\) So, the molar mass of cyclopropane is approximately \(42.06 \mathrm{~g/mol}\).

b) Determining the molecular formula of cyclopropane

To find the molecular formula of cyclopropane, we will use the percentage composition of carbon and hydrogen in the compound: Carbon: \(85.7\%\) Hydrogen: \(14.3\%\) Since the molar mass of carbon (C) is \(12.01 \mathrm{~g/mol}\) and hydrogen (H) is \(1.01 \mathrm{~g/mol}\), we can find the ratio of moles of carbon and hydrogen in the compound by dividing their percentages by their molar masses. Moles of Carbon: \(\dfrac{0.857 \times M}{12.01 \mathrm{~g/mol}}\) Moles of Hydrogen: \(\dfrac{0.143 \times M}{1.01 \mathrm{~g/mol}}\) Now we'll find out the mole ratio between Carbon and Hydrogen: Mole Ratio (C : H) = \(\dfrac{\dfrac{0.857 \times M}{12.01 \mathrm{~g/mol}}}{\dfrac{0.143 \times M}{1.01 \mathrm{~g/mol}}}\) The molar ratio simplifies to approximately \(1 : 3\) Now we have determined the empirical formula of cyclopropane as CH3. To find the molecular formula, divide the molar mass of cyclopropane (42.06 g/mol) by the molar mass of the empirical formula (CH3, \(12.01 \mathrm{~g/mol} + 3 \times 1.01 \mathrm{~g/mol} \approx 15.04 \mathrm{~g/mol}\) ) and then multiply by the empirical formula: \(\mathrm{C}\mathrm{H}_{3} \times \dfrac{42.06 \mathrm{~}\mathrm{g/mol}}{15.04 \mathrm{~}\mathrm{g/mol}} =\mathrm{C}_{3}\mathrm{H}_{9}\) The molecular formula of cyclopropane is C3H9.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental part of understanding \({gases}\) and their behaviours under various conditions. This law combines several simple gas laws into one relationship which can be captured by the formula:

\(PV = nRT\)

where P stands for pressure, V represents volume, n is the number of moles of gas, R denotes the ideal, or universal, gas constant, and T is the absolute temperature in Kelvin. In determining the molar mass of cyclopropane, the ideal gas law was rearranged to solve for the number of moles (n). From there, the molar mass (M) was calculated using the given density and temperature, leading to the determination of cyclopropane's molar mass. This practical application demonstrates how the ideal gas law is integral to finding important chemical properties of substances under gaseous states.

Real Gases vs. Ideal Gases

It's important to remember that the ideal gas law is an approximation; it works well under many conditions but may not perfectly predict the behavior of real gases, especially at high pressures or low temperatures. For gases that closely follow the ideal behavior (such as noble gases under standard conditions), the law provides a very accurate description.

Molecular Formula Determination

Molecular formula determination involves calculating the actual numbers of atoms of each element present in a molecule. It gives a clear picture of the molecule's composition and is critical in fields such as \({chemistry}\) and \({pharmacology}\), where precise molecular information is necessary. In our example with cyclopropane, the molar mass of the compound was key to finding out how many times the empirical formula fit into the molecular formula. By dividing the molar mass of cyclopropane by the molar mass of the empirical formula, we obtained a factor that, when multiplied with the empirical formula, gave us the molecular formula C3H8. It's worth noting that there was a slight error in the original step by step solution, and the correct molecular formula is C3H8, not C3H9.

Significance in Drug Design

Molecular formula determination isn't just an academic exercise; it's crucial in designing and synthesizing new drugs, where even a single atom's difference can hugely affect a substance's biological activity.

Empirical Formula

An empirical formula represents the simplest whole-number ratio of atoms of each element in a compound. It's essentially the building block for a molecule, providing a simplified view of which elements are present, and in what ratio, without indicating the actual number of atoms found in the molecule. In our textbook problem, the ratio of carbon to hydrogen in cyclopropane was simplifed to about a 1:3 ratio, giving us the empirical formula CH3.

To calculate this, we divided the percent composition of each element by its atomic mass and simplified the resulting ratio to the lowest whole numbers. The empirical formula is particularly valuable in initial stages of chemical analysis and synthesis, allowing for quick assessment of composition before delving into more complex structural considerations.

Role in Identifying Unknown Substances

In a laboratory setting, determining an empirical formula is often a first step in the identification of unknown compounds, providing critical clues to the possible molecular structure of the material being studied.