Question

A "canned heat" product used to warm chafing dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin that has an average formula of \(\mathrm{C}_{24} \mathrm{H}_{54}\). What mass of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\) should be added to \(620 \mathrm{~kg}\) of the paraffin in formulating the mixture if the vapor pressure of ethanol at \(35^{\circ} \mathrm{C}\) over the mixture is to be 8 torr? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.

Short answer

Approximately 6709.61 g (6.71 kg) of ethanol should be added to 620 kg of paraffin to get the desired pressure of 8 torr over the mixture.

Step-by-step solution

Calculate moles of paraffin and ethanol

First, let's find the moles of paraffin in 620 kg. The molecular weight of paraffin \(\mathrm{C}_{24} \mathrm{H}_{54}\) is: \((24 \times 12.01) + (54 \times 1.01) \approx 342.78 \,\mathrm{g/mol}\) Number of moles of paraffin = \(\frac{620 \times 10^3 \,\mathrm{g}}{342.78 \,\mathrm{g/mol}} \approx 1807.92 \,\mathrm{moles}\) Now, let's find the moles of ethanol \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\). The molecular weight of ethanol is: \((2 \times 12.01) + (6 \times 1.01) + (1 \times 16.00) \approx 46.07 \, \mathrm{g/mol}\) Let the moles of ethanol be \(n\). So, we have: $$ P_\mathrm{ethanol\,over\,mixture} = x_\mathrm{ethanol} \times P_\mathrm{pure\,ethanol} $$ Where \(8 \,\mathrm{torr} = x_\mathrm{ethanol} \times 100 \,\mathrm{torr}\). The mole fraction of ethanol (\(x_\mathrm{ethanol}\)) is given by: $$ x_\mathrm{ethanol} = \frac{n}{n + 1807.92} $$

Plug in values and solve for n

Next, plug the values into the equation and solve for n: $$ 8 \,\mathrm{torr} = \frac{n}{n + 1807.92} \times 100 \,\mathrm{torr} $$ Divide both sides by 100 torr and solve for n: $$ \frac{8}{100} = \frac{n}{n + 1807.92} $$ $$ n \approx 145.64 \,\mathrm{moles} $$

Calculate mass of ethanol

Finally, we calculate the mass of ethanol to be added by multiplying the moles with the molar mass of ethanol: Mass of ethanol = \(145.64 \,\mathrm{moles} \times 46.07 \,\mathrm{g/mol} \approx 6709.61 \,\mathrm{g}\) So, approximately 6709.61 g (6.71 kg) of ethanol should be added to 620 kg of paraffin to get the desired pressure of 8 torr over the mixture.

Key concepts

Mole Fraction Calculation

The mole fraction is a way to express the proportion of one component in a mixture relative to the total number of moles of all components. To calculate the mole fraction, you simply divide the number of moles of the component of interest by the total number of moles in the mixture. This ratio, which does not have units, is crucial in various applications in chemistry, such as determining the composition of solutions and analyzing colligative properties.

For instance, if we have a system where we're mixing ethanol with paraffin, we would calculate the mole fraction of ethanol as follows:
\[\begin{equation} x_\mathrm{ethanol} = \frac{n_\mathrm{ethanol}}{n_\mathrm{ethanol} + n_\mathrm{paraffin}}\end{equation}\]where:

• x_\mathrm{ethanol} is the mole fraction of ethanol,
• n_\mathrm{ethanol} is the number of moles of ethanol, and
• n_\mathrm{paraffin} is the number of moles of paraffin.

Understanding and calculating the mole fraction is essential because it is directly used in applying Raoult's Law, which relates to vapor pressure and colligative properties.

Colligative Properties

Colligative properties are the physical properties of solutions that depend on the number of solute particles, rather than the identity of the solute. These properties include boiling point elevation, freezing point depression, osmotic pressure, and vapor pressure lowering.

For students struggling with the concept of colligative properties, it is helpful to think about how the addition of a solute affects a solvent. When a non-volatile substance like salt is added to water, the boiling point of the mixture becomes higher than that of pure water because additional energy is required to allow the solvent molecules to escape from the solution. Similarly, the freezing point is lowered as the presence of the solute disrupts the formation of a solid structure in the freezing process. The key takeaway is that these changes in physical properties are dependent on the total number of solute particles and not what those particles are.
Vapor pressure lowering is particularly interesting. It occurs because the solute particles occupy space at the surface of the solution, which reduces the number of solvent molecules that can escape into the vapor phase. This means a solution will generally have a lower vapor pressure compared to the pure solvent. The precise relationship between mole fraction and vapor pressure is described by Raoult's Law, which brings us to our next section.

Raoult's Law

Raoult's Law is fundamental in understanding how vapor pressures of solutions behave. It states that the vapor pressure of each component in an ideal solution is directly proportional to the mole fraction of that component in the mixture. This allows us to express the vapor pressure of a solution as:\[\begin{equation} P_\mathrm{solution} = x_\mathrm{solute} \cdot P^\circ_\mathrm{solute}\end{equation}\]where:

• P_\mathrm{solution} is the vapor pressure of the solution,
• x_\mathrm{solute} is the mole fraction of the solute,
• P^\circ_\mathrm{solute} is the pure vapor pressure of the solute at that temperature.

In practice, Raoult's Law applies perfectly only to ideal solutions, where the forces between molecules are similar. For the step-by-step exercise at hand, the law allows us to calculate the mass of ethanol needed to achieve a specific vapor pressure over a mixture of ethanol and paraffin by considering the mole fraction of each substance.

By leveraging Raoult's Law in our calculations, we gain powerful insights into the behavior of solutions, enabling us to predict and control their vapor pressures through careful selection and measurement of components.