Question

A "canned heat" product used to warm buffet dishes consists of a homogeneous mixture of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and paraffin, which 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 to produce \(1.07 \mathrm{kPa}\) of ethanol vapor pressure at \(35^{\circ} \mathrm{C}\) ? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is \(13.3 \mathrm{kPa}\).

Short answer

The mass of ethanol to be added to 620 kg of paraffin to produce 1.07 kPa of ethanol vapor pressure at 35°C is approximately 7,389.02 g.

Step-by-step solution

Understanding Raoult's Law

Raoult's law states that the partial pressure of a component in a mixture is equal to the mole fraction of that component multiplied by the vapor pressure of the pure component. Mathematically, it can be represented as: \( P_A = x_A P_A^* \) where, \(P_A\) is the partial pressure of component A in the mixture, \(x_A\) is the mole fraction of component A in the mixture, and \(P_A^*\) is the vapor pressure of pure component A.

Determine the mole fraction of ethanol

In the given problem, the vapor pressure of ethanol in the mixture is 1.07 kPa, and the vapor pressure of pure ethanol is 13.3 kPa. We can use Raoult's law to determine the mole fraction of ethanol in the mixture: \(1.07 \mathrm{kPa} = x_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \times 13.3 \mathrm{kPa} \) Solving for \( x_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \): \( x_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = 0.0805 \)

Calculate the moles of paraffin and ethanol

To find the mass of ethanol that should be added to the mixture, we need to determine the moles of paraffin and ethanol. The mass of paraffin is given as 620 kg, and the average molecular formula of paraffin is C24H50. First, let's find the molar mass of paraffin: \( \mathrm{MM}_{\mathrm{C}_{24}\mathrm{H}_{50}} = 24 \times 12.01 + 50 \times 1.01 = 338.34 \hspace{2mm} \mathrm{g/mol} \) Now, we can find the moles of paraffin: \(\mathrm{moles}_{\mathrm{C}_{24}\mathrm{H}_{50}} = \frac{620,000 \hspace{2mm} \mathrm{g}}{338.34 \hspace{2mm} \mathrm{g/mol}} = 1833.49 \hspace{2mm} \mathrm{mol} \) Now, let's find the moles of ethanol using the mole fraction of ethanol: \( \frac{\mathrm{moles}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}}}{\mathrm{moles}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} + \mathrm{moles}_{\mathrm{C}_{24}\mathrm{H}_{50}}} = 0.0805 \) Solving for the moles of ethanol: \( \mathrm{moles}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = 0.0805 \times \mathrm{moles}_{\mathrm{C}_{24}\mathrm{H}_{50}} / (1 - 0.0805) = 160.41 \hspace{2mm} \mathrm{mol} \)

Calculate the mass of ethanol to be added

Finally, we can find the mass of ethanol to be added by using the molar mass of ethanol, which is: \( \mathrm{MM}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = 2 \times 12.01 + 6 \times 1.01 + 1 \times 16 = 46.07 \hspace{2mm} \mathrm{g/mol} \) The mass of ethanol to be added is: \( \mathrm{mass}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = \mathrm{moles}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \times \mathrm{MM}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} \) \( \mathrm{mass}_{\mathrm{C}_{2}\mathrm{H}_{5}\mathrm{OH}} = 160.41 \hspace{2mm} \mathrm{mol} \times 46.07 \hspace{2mm} \mathrm{g/mol} = 7,389.02 \hspace{2mm} \mathrm{g} \) So, 7,389.02 g of ethanol should be added to 620 kg of the paraffin to produce 1.07 kPa of ethanol vapor pressure at 35°C.

Key concepts

Vapor Pressure

Vapor pressure is a crucial concept in understanding how substances evaporate and behave in mixtures. It refers to the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. In simpler terms, it's the "push" that a liquid can exert when some of it turns into gas. The higher the vapor pressure, the more volatile the substance is, meaning it evaporates more easily.

In the exercise, we are interested in the vapor pressure of ethanol as part of a mixture. Pure ethanol has a vapor pressure of 13.3 kPa at 35°C. However, when mixed with paraffin, the vapor pressure decreases because the ethanol's presence in the vapor is partially diluted by the other component. The resulting vapor pressure is 1.07 kPa. This change in pressure within a mixture is what we analyze using Raoult's Law.

Mole Fraction

Mole fraction is a way to express the concentration of a component in a mixture. It describes how much of one substance is present compared to the total amount of all components.

• It is represented as a ratio: for a component A in a mixture, the mole fraction \( x_A \) is calculated as the moles of A divided by the total moles in the mixture.
• Mole fraction is a dimensionless quantity, meaning it has no units.

In the given problem, Raoult's Law helps us calculate the mole fraction of ethanol in the mixture. Using the vapor pressures (1.07 kPa for the mixture and 13.3 kPa for pure ethanol), we use the formula:\[ 1.07 \text{ kPa} = x_{\text{C}_2\text{H}_5\text{OH}} \times 13.3 \text{ kPa} \]Solving this gives us a mole fraction \( x_{\text{C}_2\text{H}_5\text{OH}} = 0.0805 \). This result means that in this mixture, 8.05% of the moles are ethanol.

Mixture Composition

When discussing mixture composition, we're looking at how different substances are combined and the resulting properties. In this problem, we have a homogeneous mixture of ethanol and paraffin. Homogeneous means the mixture is uniform throughout; you can't easily distinguish one component from another by looking at it.

• Ethanol, with its formula \( \text{C}_2\text{H}_5\text{OH} \), is a common alcohol in chemical applications.
• Paraffin, here represented by \( \text{C}_{24}\text{H}_{50} \), is a non-volatile, wax-like hydrocarbon.

Their combination within the mixture alters both physical and chemical characteristics, including things like boiling point and, importantly for Raoult's Law, vapor pressure. The target vapor pressure of 1.07 kPa at 35°C implies a specific balance between ethanol and paraffin. This balance impacts how much ethanol is needed to reach equilibrium in the mixture composition.

Molecular Weight

Understanding molecular weight is key to converting mole quantities into mass. Molecular weight (also known as molar mass) is the sum of the atomic weights of all atoms in a molecule and is expressed in grams per mole (g/mol).

• For paraffin (\(\text{C}_{24}\text{H}_{50}\)), the calculated molar mass is 338.34 g/mol.
• For ethanol (\(\text{C}_2\text{H}_5\text{OH}\)), it is 46.07 g/mol.

In solving the exercise, we first find the number of moles of paraffin using its molecular weight and given mass. Once the moles of paraffin are known, we use the mole fraction of ethanol to solve for the necessary moles of ethanol in the mixture. Finally, multiplying the moles of ethanol by its molecular weight provides the mass needed for the mixture. This link between moles and mass is vital for understanding how much of each component is needed to achieve any desired effects in the mixture.