A forensic chemist is given a white solid that is suspected of being pure cocaine $\left(\mathrm{C}_{17} \mathrm{H}_{21} \mathrm{NO}_{4}, \text { molar mass }=303.35 \mathrm{g} / \mathrm{mol}\right)\( She dissolves \)1.22 \pm 0.01 \mathrm{g}\( of the solid in \)15.60 \pm 0.01 \mathrm{g}$ benzene. The freezing point is lowered by \(1.32 \pm 0.04^{\circ} \mathrm{C}\) a. What is the molar mass of the substance? Assuming that the percent uncertainty in the calculated molar mass is the same as the percent uncertainty in the temperature change, calculate the uncertainty in the molar mass. b. Could the chemist unequivocally state that the substance is cocaine? For example, is the uncertainty small enough to distinguish cocaine from codeine $\left(\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}, \text { molar }\right.\( mass \)=299.36 \mathrm{g} / \mathrm{mol}$ )? c. Assuming that the absolute uncertainties in the measurements of temperature and mass remain unchanged, how could the chemist improve the precision of her results?

Step by step solution

01

Write down the freezing point depression equation

The freezing point depression equation is given by: ∆T = Kf * m where ∆T is the change in freezing point, Kf is the cryoscopic constant (5.12 °C·kg/mol for benzene), and m is the molality of the solution. The molality is given by: m = (moles of solute) / (mass of solvent in kg) Since we need to calculate the molar mass of the substance, let's first express the moles of solute in terms of molar mass: moles of solute = (mass of solute) / (molar mass) Now, the molality can be written as: m = (mass of solute) / ((molar mass) * (mass of solvent in kg))

02

Find the molality of the solution

Solve the freezing point depression equation for molality: m = ∆T / Kf m = (1.32 ± 0.04 °C) / (5.12 °C·kg/mol) = 0.258 ± 0.0078 mol/kg

03

Determine the molar mass

Using the expression for molality from step 1, we can now find the molar mass (M): M = (mass of solute) / (m * (mass of solvent in kg)) M = (1.22 ± 0.01 g) / (0.258 ± 0.0078 mol/kg * 0.0156 ± 0.0001 kg) M ≈ 305.43 g/mol

04

Calculate the uncertainty of molar mass

The percent uncertainty of the calculated molar mass can be found using the percent uncertainty of the temperature change: Percent uncertainty of ∆T = (0.04 °C) / (1.32 °C) * 100% ≈ 3.03% Now, apply the same percent uncertainty to the calculated molar mass: Uncertainty in molar mass = (3.03%) * (305.43 g/mol) ≈ 9.25 g/mol The molar mass of the substance is approximately 305.43 ± 9.25 g/mol.

05

Compare the molar mass and uncertainty with cocaine and codeine

Using the calculated molar mass of the substance (305.43 ± 9.25 g/mol), we cannot say for certain that the substance is cocaine as the molar mass of codeine (299.36 g/mol) also falls within the range of uncertainty.

06

Improving the precision of the experiment

To improve the precision of the results, the chemist could: 1. Utilize a more accurate measuring device for the mass and temperature measurements to reduce their uncertainties. 2. Perform several trials of the experiment and calculate the average molar mass. 3. Employ a method with less environmental factors influencing the results, such as using a solution with a lower freezing point depression constant.

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