Problem 50 - Chemistry The Central Science

Question

The density of toluene \(\left(\mathrm{C}_{7} \mathrm{H}_{8}\right)\) is \(0.867 \mathrm{g} / \mathrm{mL},\) and the density of thiophene \(\left(\mathrm{C}_{4} \mathrm{H}_{4} \mathrm{S}\right)\) is 1.065 \(\mathrm{g} / \mathrm{mL}\) . A solution is made by dissolving 8.10 \(\mathrm{g}\) of thiophene in 250.0 \(\mathrm{mL}\) of toluene.(a) Calculate the molefraction of thiophene in the solution. (b) Calculate the molality of thiophene in the solution. (c) Assuming that the volumes of the solute and solvent are additive, what is the molarity of thiophene in the solution?

Step-by-Step Solution

Answer

(a) The mole fraction of thiophene in the solution is 0.039. (b) The molality of thiophene in the solution is 0.443 mol/kg. (c) The molarity of thiophene in the solution, assuming that the volumes of the solute and solvent are additive, is 0.373 M.

Step 1: Calculate moles of thiophene

The given mass of thiophene is 8.10 grams. To calculate the moles of thiophene, we have to use the molar mass of thiophene, which is: C4H4S = 4(12.01) + 4(1.008) + 32.07 = 84.14 g/mol. Now, we can find the moles of thiophene: Moles of thiophene = Mass / Molar mass Moles of thiophene = 8.10 g / 84.14 g/mol = 0.096 mol

Step 2: Calculate moles of toluene

The mass of toluene can be calculated using its density and volume. Mass of toluene = Density x Volume Mass of toluene = 0.867 g/mL x 250 mL = 216.75 g Now, we will calculate moles of toluene using the molar mass of toluene: C7H8 = 7(12.01) + 8(1.008) = 92.14 g/mol. Moles of toluene = Mass / Molar mass Moles of toluene = 216.75 g / 92.14 g/mol = 2.352 mol

Step 3: Calculate mole fraction of thiophene

Mole fraction of thiophene = Moles of thiophene / (Moles of thiophene + Moles of toluene) Mole fraction of thiophene = 0.096 / (0.096 + 2.352) = 0.039

Step 4: Calculate molality of thiophene

Molality (m) = Moles of solute / Mass of solvent (in kg) Mass of toluene in kg = 216.75 g / 1000 = 0.21675 kg Molality of thiophene = 0.096 mol / 0.21675 kg = 0.443 mol/kg

Step 5: Calculate molarity of thiophene

To calculate the molarity, we need the volume of the solution in liters. The solution volume can be found by adding the volumes of thiophene and toluene. Volume of thiophene = Mass / Density = 8.10 g / 1.065 g/mL = 7.606 mL Volume of solution = Volume of thiophene + Volume of toluene = 7.606 mL + 250 mL = 257.606 mL Convert the volume to liters: 257.606 mL / 1000 = 0.257606 L Molarity (M) = Moles of solute / Volume of solution (in L) Molarity of thiophene = 0.096 mol / 0.257606 L = 0.373 M To summarize: (a) Mole fraction of thiophene in the solution: 0.039 (b) Molality of thiophene in the solution: 0.443 mol/kg (c) Molarity of thiophene in the solution: 0.373 M

Key Concepts

Mole Fraction

Understanding the mole fraction of a component within a mixture is essential to grasp solution concentration calculations. Mole fraction, often symbolized as \(X\), is a dimensionless quantity that provides a ratio of the number of moles of a particular substance to the total number of moles in the mixture. To put it simply, if you envision a large party where every guest represents a molecule, the mole fraction would tell you the proportion of guests who are wearing a red hat (your substance of interest) compared to all the hats in the room.

Let's apply this to our problem with thiophene and toluene. After determining the moles of each substance, the mole fraction of thiophene is the number of moles of thiophene divided by the sum of the moles of thiophene and the moles of toluene. This ratio reveals how much of the mixture's composition is made up by thiophene. In our scenario, the calculation yielded a mole fraction of 0.039, indicating that thiophene makes up about 3.9% of the total moles in the solution.

Molality

Molality is a measure of the solute concentration in a solution but, unlike molarity, it is not affected by changes in temperature. It's expressed as moles of solute per kilogram of solvent, symbolized by \(m\). Imagine you’re creating a special tea blend. You are careful to measure a fixed amount of tea leaves (the solute) for a certain amount of water (the solvent), regardless of whether it expands slightly when heated. This fixed ratio ensures a consistent flavor each time.

For our exercise, once we have the mass of toluene in kilograms and the moles of thiophene, we calculate the molality. It is the ratio of the moles of thiophene to the mass of the toluene solvent—yielding a molality of 0.443 moles per kilogram of toluene. This tells us how concentrated the thiophene is within the toluene solvent.

Molarity

Molarity is one of the most common measures of solution concentration. Represented by \(M\), molarity is the number of moles of solute per liter of solution. Consider a chef making a broth; for every liter of water added, a precise number of bouillon cubes is dissolved to maintain a specific flavor intensity. This is akin to molarity, which speaks to how 'flavorful' the solution is — that is, how many molecules of solute are present in a given volume.

In our case, after computing the total volume of the solution by adding the volumes of both thiophene and toluene, and adjusting it to liters, we find the molarity by dividing the moles of thiophene by this volume. The resultant molarity of thiophene in this solution is 0.373 M, telling us that for every liter of the solution, there are 0.373 moles of thiophene.

Solute and Solvent

The key players in any solution are the solute and the solvent. The solute is the substance that is dissolved, while the solvent is the component that does the dissolving. Think of making lemonade: the sugar is the solute that gets dissolved into the water, the solvent. A good rule of thumb is that there is usually more solvent than solute in a solution.

In our textbook example, the solute is thiophene and the solvent is toluene. The solute, thiophene, is present in a smaller amount and is distributed within the solvent, toluene, which constitutes the majority of the solution’s volume. This dynamic determines the final characteristics of the solution, such as molarity and molality, and the calculation of these factors hinges on accurately knowing the amounts of each component in the mix.