Problem 67 - Chemistry: Principles and Reactions

Question

Pure benzene boils at \(80.10^{\circ} \mathrm{C}\) and has a boiling point constant, \(k_{\mathrm{b}}\), of \(2.53^{\circ} \mathrm{C} / \mathrm{m} .\) A sample of benzene is contaminated by naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8}\). The boiling point of the contaminated sample is \(81.20^{\circ} \mathrm{C}\). How pure is the sample? (Express your answer as mass percent of benzene.)

Step-by-Step Solution

Answer

The mass percent of benzene in the contaminated sample is approximately 91.88%.

Step 1: Calculate the change in boiling point (ΔT)

To find the change in boiling point, we can subtract the boiling point of pure benzene from the boiling point of the contaminated sample: ΔT = \(81.20^\circ C - 80.10^\circ C = 1.10^\circ C\)

Step 2: Calculate the molality of the solution

To find the molality, use the boiling point elevation formula and solve for molality: molality = ΔT / \(k_b\) molality = \(1.10^\circ C / 2.53^\circ C/ \mathrm{m} = 0.435 \mathrm{m}\)

Step 3: Molecular weights of benzene and naphthalene

The molecular weight of benzene (C6H6) is \(6(12.01) + 6(1.01) = 78.12 \ g/mol\). The molecular weight of naphthalene (C10H8) is \(10(12.01) + 8(1.01) = 128.18 \ g/mol\).

Step 4: Let x be the mass naphthalene in grams

We have the molality of the solution (0.435). Molality is defined as moles of solute (naphthalene) per kilogram of solvent (benzene). If we let x be the mass of naphthalene in grams, we have: molality = moles of naphthalene / mass of benzene in kg 0.435 = \([(x \ g) / (128.18 \ g/mol)] / [(1000 - x) \ g * (1 \ kg / 1000 \ g)]\)

Step 5: Solve for x - mass of naphthalene

Multiply both sides by (1000 - x) and simplify: 0.435(1000 - x) = (x) / (128.18) 435 - 0.435x = x / 128.18 0.435x + x / 128.18 = 435 x(0.435 + 1 / 128.18) = 435 x = 435 / (0.435 + 1 / 128.18) x = 81.16

Step 6: Calculate the mass percent of benzene

Finally, we can find the mass percent of benzene in the contaminated sample as follows: Mass percent of benzene = (mass of benzene / total mass) * 100 = \([(1000 - 81.16) \ g / 1000 \ g] * 100\) = \(918.84/1000 * 100\) = \(91.88 \%\) The contaminated sample of benzene is approximately \(91.88\%\) pure.

Key Concepts

Understanding Molality

Molality is an essential concept in chemistry, especially when studying solutions. It refers to the number of moles of solute per kilogram of solvent. This is different from molarity, which is moles of solute per liter of solution. In the context of boiling point elevation, molality is crucial because it directly connects to the change in boiling point.

When you have a solute dissolved in a solvent, like naphthalene in benzene, the boiling point of the solution can change. The formula for boiling point elevation involves molality:

• Boiling Point Elevation, ΔT = i \( \times \) \( k_b \times \) molality

Here, \( i \) is the van't Hoff factor, which is 1 for naphthalene since it doesn’t dissociate. Using this, you can solve for molality if you know the boiling point change and \( k_b \).

For example, in our problem, we found the change in boiling point, ΔT, to be \(1.10^{\circ} C\) and \( k_b = 2.53^{\circ}C/m \). By rearranging the formula and dividing ΔT by \( k_b \), we determine the solution's molality to be approximately \(0.435\ m\). This tells us how concentrated our naphthalene solute is in the benzene solvent.

What is Mass Percent?

Mass percent is a way of expressing the concentration of a component in a mixture. It's given as the mass of the solute divided by the total mass of the solution, multiplied by 100, giving the percentage by mass.

In practical terms, it helps you understand the proportion of substances in a mixture. For example, determining how much benzene is in a contaminated sample compared to impurities can tell you the sample's purity.

• Mass Percent = \( \frac{\text{mass of component}}{\text{total mass}} \times 100\)

In our problem, after calculating the mass of naphthalene in the mixture, we found the mass of benzene. By subtracting the mass of naphthalene from the total 1000 g sample, and then using the mass percent formula, we calculated that the benzene's purity was approximately \(91.88\%\). This means that 91.88% of the solution's mass is made up of benzene, and the rest is naphthalene contaminant.

Using Molecular Weights

Molecular weight (or molar mass) is crucial when dealing with compounds in solutions because it allows you to convert between grams and moles. It’s particularly important in calculations involving molality, where you need to know how many moles of a solute you have.

To find molecular weight, add together the atomic weights of all the atoms in a molecule. For example:

• Benzene (\(C_6H_6\)): \(6 \times 12.01 + 6 \times 1.01 = 78.12\ g/mol\)
• Naphthalene (\(C_{10}H_8\)): \(10 \times 12.01 + 8 \times 1.01 = 128.18\ g/mol\)

These weights help us calculate the molality in our solution. Knowing the molecular weight of naphthalene allowed us to convert its mass into moles. Thus, we could determine the molality and then calculate other parameters like mass percent.

This understanding is crucial in real-world applications, like quality control or chemical analysis, where the purity of a substance often matters greatly.