Problem 72 - Chemistry An Atoms First Approach

Question

A solution contains \(3.75 \mathrm{g}\) of a nonvolatile pure hydrocarbon in \(95 \mathrm{g}\) acetone. The boiling points of pure acetone and the solution are \(55.95^{\circ} \mathrm{C}\) and \(56.50^{\circ} \mathrm{C},\) respectively. The molal boilingpoint constant of acetone is \(1.71^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} .\) What is the molar mass of the hydrocarbon?

Step-by-Step Solution

Answer

The molar mass of the hydrocarbon is approximately 122.7 g/mol.

Step 1: Calculate the boiling point elevation

Boiling point elevation (ΔT_b) is the difference between the boiling point of the solution and the boiling point of the pure solvent. In this case, the solvent is acetone. ΔT_b = T_solution - T_solvent = 56.50°C - 55.95°C = 0.55°C

Step 2: Calculate the molality of the hydrocarbon

We can use the formula ΔT_b = Kb * molality to find the molality of the hydrocarbon in the solution. Molality = ΔT_b / Kb = 0.55°C / 1.71°C·kg/mol = 0.3216 mol/kg

Step 3: Calculate the moles of hydrocarbon in the solution

Since molality represents the moles of solute per kilogram of solvent, we can calculate the moles of hydrocarbon, knowing the mass of acetone in the solution: moles_hydrocarbon = molality * mass_of_acetone / 1000 = 0.3216 mol/kg * 95g * (1 kg/1000g) = 0.03054 mol

Step 4: Calculate the mass of hydrocarbon per mole

We are given the mass of the hydrocarbon in the solution, so we can calculate the mass of hydrocarbon per mole: mass_per_mole = mass_of_hydrocarbon / moles_hydrocarbon = 3.75g / 0.03054 mol = 122.7 g/mol

Step 5: Determine the molar mass of the hydrocarbon

The molar mass of the hydrocarbon is the mass per mole calculated in the previous step. Molar mass of hydrocarbon = 122.7 g/mol The molar mass of the hydrocarbon is approximately 122.7 g/mol.

Key Concepts

Molar Mass

Molar mass is a fundamental concept in chemistry that refers to the mass of a given substance (element or compound) divided by the amount of substance. It is expressed in grams per mole (g/mol), representing the weight of one mole of molecules or atoms of a material. Calculating molar mass is essential for understanding how substances interact in chemical reactions, as it determines the proportions in which different materials combine.

In the context of the problem, finding the molar mass involves determining the mass of the hydrocarbon present and dividing it by the number of moles. Since we have the mass of the solute as 3.75 g and moles calculated in step 4 as 0.03054 mol, we divide these values to find the molar mass to be about 122.7 g/mol. Understanding molar mass is crucial for comprehending the physical and chemical properties of substances.

• Helps quantify chemical reactions
• Assists in converting moles to grams
• Essential for calculating concentrations

Molality

Molality ( μ ) is another way to express the concentration of a solution, defined as the number of moles of solute per kilogram of solvent. Unlike molarity, molality is not dependent on temperature since it relies solely on the mass for its calculation rather than volume, which can change with temperature.

In the given problem, molality is calculated using the boiling point elevation formula, ΔT_b = K_b × molality, where the molal boiling point constant ( K_b ) is a unique value for each solvent. For acetone, K_b is 1.71°C·kg/mol. Given the boiling point elevation, ΔT_b, as 0.55°C, we find that the molality of the hydrocarbon is 0.3216 mol/kg. This concept illustrates how changes in boiling point can help determine concentration values of solutes in a solution.

• Independent of temperature
• Utilizes mass to measure concentration
• Useful for boiling and freezing point calculations

Hydrocarbon

Hydrocarbons are organic compounds composed entirely of hydrogen and carbon atoms. They are the simplest type of organic compound and serve as the backbone for more complex molecules in organic chemistry. Hydrocarbons are classified based on the types of bonds between carbon atoms:

1. Alkanes (single bonds)
2. Alkenes (double bonds)
3. Alkynes (triple bonds)

In this exercise, the hydrocarbon is a nonvolatile solute in acetone, affecting the solution's boiling point, which allows us to calculate properties like molality and molar mass. Understanding hydrocarbons is vital because they:

• Form the basis for all organic chemistry
• Are major components of fuels and plastics
• Vary significantly in their physical and chemical properties based on structure

Acetone

Acetone is a simple ketone with the molecular formula C_3H_6O. It is a clear, volatile, flammable liquid commonly used as a solvent in cleaning and nail polish removers. Its properties as a solvent make it particularly important in both industrial and laboratory settings.

Acetone's ability to dissolve a wide range of substances, and its high volatility, mean that when a solute like a hydrocarbon is added, significant changes can occur in properties like the boiling point. In this particular situation, acetone aids in determining the molar mass and molality of a solute due to its known boiling point elevation constant. Acetone's significance is highlighted by its role in scientific calculations involving solution properties, emphasizing its practical applications in chemistry research and industry.

• Highly effective solvent
• Commonly used in industry
• Useful in calculating changes in solution properties