Question

An anaesthetic can be prepared by mixing gaseous cyclopropane \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right.\) molecular weight \(\left.=42 \mathrm{~g} / \mathrm{mole}\right)\), and oxygen \(\left(\mathrm{O}_{2}\right.\), molecular weight \(=32 \mathrm{~g} /\) mole \() .\) If a gas cylinder is prepared with cyclopropane at a partial pressure of 170 torr and oxygen at a partial pressure of 570 torr, calculate the ratio of the number of moles of cyclopropane to that of oxygen \(\left[\left(\mathrm{n}_{(\mathrm{C}) 3(\mathrm{H}) 6}\right) /\left(\mathrm{n}_{(\mathrm{H}) 2 \mathrm{O}}\right)\right]\)

Short answer

The ratio of moles of cyclopropane to oxygen in the gas cylinder is approximately 0.2982.

Step-by-step solution

Recall the Ideal Gas Law formula

The Ideal Gas Law formula is given by \(PV = nRT\), where P represents pressure, V represents volume, n represents the number of moles, R represents the ideal gas constant, and T represents temperature.

Find the moles of each gas

Using the Ideal Gas Law formula, we will be solving for the moles of each gas, given their respective partial pressures and assuming the volume and temperature are the same for both gases. \(PV = n_cRT_c\) and \(PV = n_oRT_o\), where n_c and n_o indicate the moles of cyclopropane and oxygen, respectively, while RT_c and RT_o denote the product of R and T. The ratio of moles can be calculated as follows: \(\frac{n_c}{n_o} = \frac{P_cV}{RT_c} \div \frac{P_oV}{RT_o}\)

Simplify the equation

Simplify the ratio equation by canceling out the common terms V, R, and T: \(\frac{n_c}{n_o} = \frac{P_c}{P_o}\)

Plug in the given values

Now, using the given partial pressures for cyclopropane (170 torr) and oxygen (570 torr), plug them into the ratio equation: \(\frac{n_c}{n_o} = \frac{170 \ \text{torr}}{570 \ \text{torr}}\)

Calculate the ratio

Divide the partial pressures to get the ratio of moles: \(\frac{n_c}{n_o} = 0.2982\) The ratio of moles of cyclopropane to oxygen in the gas cylinder is approximately 0.2982.

Key concepts

Partial Pressure

Partial pressure refers to the pressure that a single gas in a mixture would exert if it occupied the entire volume alone at a given temperature. It is a crucial concept in understanding gas mixtures. When dealing with gases like cyclopropane and oxygen in a cylinder, each gas contributes to the total pressure of the mixture with its partial pressure.
A helpful way to think about this is to imagine the gas molecules bouncing around inside a container. Each type of gas bumping against the container walls generates a pressure independent of the others. The sum of all these partial pressures gives the total pressure in the cylinder. For this exercise, we're given that cyclopropane exerts a partial pressure of 170 torr, while oxygen exerts a partial pressure of 570 torr.

Moles of Gases

In chemistry, the concept of moles allows us to count particles like atoms or molecules by weighing them. A mole is a convenient number (Avogadro's number, roughly \(6.022 \times 10^{23}\)) that gives us a way to express amounts of a substance. With gases, we often relate moles directly to volume, temperature, and pressure in equations like the Ideal Gas Law: \(PV = nRT\).
In this problem, although we don't directly use the volume or temperature, knowing the pressure helps us calculate moles when other conditions are constant. In our case, it is given that both cyclopropane and oxygen are under specific pressures, and assuming similar temperature and volume conditions, we can find the ratio of their moles based on these pressures alone.

Ratio Calculation

Ratio calculations are common in chemistry to compare quantities like moles of different substances under similar conditions. It helps in understanding how much of one substance is present relative to another.
In this exercise, we leverage a simplified form of the Ideal Gas Law equation to find the mole ratio of cyclopropane to oxygen. The equation \(\frac{n_c}{n_o} = \frac{P_c}{P_o}\) derives from our earlier acknowledgment that total volume, temperature, and the ideal gas constant have been canceled out. Thus, we're left to compare the partial pressures directly. Given the pressures of 170 torr for cyclopropane and 570 torr for oxygen, the ratio works out to about 0.2982.

Chemical Calculation

Chemical calculation involves using established mathematical relationships and chemical formulas to ascertain unknown quantities such as moles or concentration. These calculations are foundational in chemistry and allow us to predict and adjust the outcomes of chemical reactions and mixtures.
For this problem, we've employed a straightforward chemical calculation rooted in the Ideal Gas Law to find the ratio of moles between the gases. Starting with known values of partial pressures and simplifying the equation, we solve for the mole ratio. Such calculations not only help in theoretical problems like these but are routinely used in practical applications to prepare precise mixtures of gases in industries and laboratories.