Problem 99 - Chemistry The Central Science

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 \(\mathrm{C}_{24} \mathrm{H}_{50}\). 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.

Step-by-Step Solution

Answer

Approximately 6.9 kg of ethanol (C2H5OH) should be added to 620 kg of paraffin to formulate the mixture with an 8 torr vapor pressure of ethanol at 35 °C.

Step 1: 1. Recall Raoult's Law formula

Raoult's Law states: \[P_A = x_A P^*_A\] where \(P_A\) is the vapor pressure of component A over the mixture, \(x_A\) is the mole fraction of component A in the mixture, and \(P^*_A\) is the vapor pressure of the pure component A. In our case, A is ethanol (C2H5OH).

Step 2: 2. Rearrange the formula to find the mole fraction of ethanol

We need to find the mole fraction of ethanol in the mixture, so we'll rearrange Raoult's Law formula: \[x_A = \frac{P_A}{ P^*_A}\]

Step 3: 3. Calculate the mole fraction of ethanol

We are given the vapor pressure of ethanol over the mixture (\(P_A\)) to be 8 torr and the vapor pressure of pure ethanol (\(P^*_A\)) to be 100 torr. Substituting the values into the rearranged formula, we get: \[x_A = \frac{8 \, \text{torr}}{100 \, \text{torr}} = 0.08\]

Step 4: 4. Relate the mole fraction to mass and mole

We know that mole fraction (\(x_A\)) can be represented as \[x_A = \frac{\text{moles of } A}{\text{moles of } A + \text{moles of} \, B}\] where B is paraffin (\(C_{24}H_{50}\)).

Step 5: 5. Convert mass to moles

In order to use the mole fraction equation, we'll need to convert the mass of ethanol and paraffin into moles. We'll denote ethanol's mass as \(m_{C_2H_5OH}\) and paraffin's mass as \(m_{C_{24}H_{50}}\). We are given the mass of paraffin in the mixture (\(m_{C_{24}H_{50}}\)) as 620 kg, and we need to find the mass of ethanol (\(m_{C_2H_5OH}\)). The molecular weights are: \[M_{C_2H_5OH} = 46 \, \frac{g}{\text{mole}}\] and \[M_{C_{24}H_{50}} = 338 \, \frac{g}{\text{mole}}\] Converting the mass of paraffin to grams and moles, we get: \[n_{C_{24}H_{50}} = \frac{620 \, \text{kg}}{1} \times \frac{1000 \, g}{1 \, \text{kg}} \times \frac{1 \, \text{mole}}{338 \, g} = 1834.32 \, \text{moles}\]

Step 6: 6. Calculate the moles of ethanol

Using the mole fraction equation from step 4 and the mole fraction of ethanol from step 3, we can now solve for the moles of ethanol: \[0.08 = \frac{n_{C_2H_5OH}}{n_{C_2H_5OH} + 1834.32}\] Rearranging for \(n_{C_2H_5OH}\) and solving, we get: \[n_{C_2H_5OH} = 149.6 \, \text{moles}\]

Step 7: 7. Convert moles of ethanol to mass

Using the molecular weight of ethanol (\(M_{C_2H_5OH}\)) from step 5, we can convert the number of moles of ethanol back to mass: \[m_{C_2H_5OH} = 149.6 \, \text{moles} \times \frac{46\, g}{1\, \text{mole}} = 6881.6 \, g\]

Step 8: 8. Conclusion

6881.6 g, or approximately 6.9 kg of ethanol (C2H5OH), should be added to 620 kg of paraffin to formulate the mixture with an 8 torr vapor pressure of ethanol at 35 °C.

Key Concepts

Vapor Pressure

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a specific temperature. In simpler terms, it describes how much a liquid tends to evaporate. Liquids with high vapor pressures evaporate more easily. For example, at 35°C, pure ethanol has a vapor pressure of 100 torr, meaning it readily evaporates under these conditions.

Raoult's Law helps us understand how the vapor pressure of a component behaves when it is part of a mixture. According to the law, the vapor pressure of a component in a solution (\(P_A\)) is proportional to the vapor pressure of the pure component (\(P^*_A\)) times its mole fraction in the mixture. Raoult's Law is expressed as:

• \(P_A = x_A P^*_A\)

Using this, one can calculate how the presence of other substances influences the vapor pressure of a component in a mixture.

Mole Fraction

The mole fraction is a way of expressing the concentration of a component in a mixture. It is defined as the ratio of the moles of one component to the total moles in the mixture. In our example exercise, the mole fraction of ethanol (\(x_A\)) tells us how much of the mixture is ethanol.

To find the mole fraction, use this formula:

• \(x_A = \frac{\text{moles of component A}}{\text{total moles in the mixture}}\)

By rearranging Raoult's Law and inserting known values (vapor pressures), the mole fraction of ethanol was determined to be 0.08. This means ethanol makes up 8% of the moles in the mixture. Understanding mole fraction allows us to see how much of each substance is present compared to the entire mixture.

Mixture Formulation

Mixture formulation is the process of combining substances in precise ratios to achieve desired properties, such as a specific vapor pressure. Here, the goal is to reach an 8 torr vapor pressure for ethanol in a mixture with paraffin.

This involves calculating the amount of each component needed, using:

• Molecular weights to convert mass to moles
• The mole fraction to allocate the right quantities of each substance

Specifically, to find the required mass of ethanol, we first determined the moles using the given mole fraction of 0.08. By converting these moles back to mass using ethanol's molecular weight, it was found that approximately 6.9 kg of ethanol should be combined with 620 kg of paraffin to create the desired mixture. This systematic approach ensures the mixture will behave as intended in applications like maintaining a specific vapor pressure for fuel use.