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 8 torr of ethanol vapor pressure at \(35^{\circ} \mathrm{C}\) ? The vapor pressure of pure ethanol at \(35^{\circ} \mathrm{C}\) is 100 torr.

Short answer

To produce 8 torr of ethanol vapor pressure at 35°C with a "canned heat" product, approximately 7.36 kg of ethanol should be added to 620 kg of paraffin.

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 product of its mole fraction and its vapor pressure in the pure state. Mathematically, this can be expressed as: \(P_i = X_i P^*_i\) where: - \(P_i\) is the partial vapor pressure of component i - \(X_i\) is the mole fraction of component i in the solution - \(P^*_i\) is the vapor pressure of the pure component i In this problem, we will apply Raoult's Law for ethanol in the mixture.

Calculate the mole fraction of ethanol

Given that the vapor pressure of the ethanol-paraffin mixture is 8 torr at 35°C, we can use Raoult's Law to calculate the mole fraction of ethanol: \(P_{ethanol} = X_{ethanol} P^*_{ethanol}\) We know that the vapor pressure of pure ethanol, \(P^*_{ethanol}\), at 35°C is 100 torr. We can rearrange the equation to find the mole fraction of ethanol: \(X_{ethanol} = \frac{P_{ethanol}}{P^*_{ethanol}} = \frac{8 \, \text{torr}}{100 \, \text{torr}} = 0.08\)

Calculate the moles of paraffin

Now we need to determine the number of moles of paraffin in the 620 kg of paraffin. The average molecular formula for paraffin is C24H50, with a molar mass of: \(M_{paraffin} = 24 \times (12.01 \, \text{g/mol}) + 50 \times (1.01 \, \text{g/mol}) = 338.74 \, \text{g/mol}\) We can convert the mass of paraffin (620 kg) into moles using the molar mass: n_{paraffin} = \(\frac{\text{mass of paraffin}}{\text{molar mass of paraffin}} = \frac{620,000 \, \text{g}}{338.74 \, \text{g/mol}} ≈ 1,831.32 \, \text{mol}\)

Calculate the moles of ethanol

Using the mole fraction of ethanol and the moles of paraffin, we can find the moles of ethanol in the mixture: \(\frac{n_{ethanol}}{n_{paraffin} + n_{ethanol}} = X_{ethanol}\) Rearrange the equation to find the moles of ethanol, \(n_{ethanol}\): \(n_{ethanol} = \frac{X_{ethanol} \times n_{paraffin}}{1 - X_{ethanol}} = \frac{0.08 \times 1,831.32 \, \text{mol}}{1 - 0.08} ≈ 159.67 \, \text{mol}\)

Calculate the mass of ethanol

Finally, we need to find the mass of ethanol that corresponds to the calculated moles. The molar mass of ethanol is: \(M_{ethanol} = 2 \times (12.01 \, \text{g/mol}) + 6 \times (1.01 \, \text{g/mol}) + 1 \times (16.00 \, \text{g/mol}) = 46.07 \, \text{g/mol}\) Using the moles of ethanol, we can find the mass of ethanol: mass_ethanol = \(n_{ethanol} \times M_{ethanol} = 159.67 \, \text{mol} \times 46.07 \, \text{g/mol} ≈ 7,359.96 \, \text{g}\) Therefore, the mass of ethanol that should be added to the 620 kg of paraffin to produce 8 torr ethanol vapor pressure at 35°C is approximately 7,359.96 g (7.36 kg).