Problem 79 - Chemistry The Central Science

Question

Adrenaline is the hormone that triggers the release of extra glucose molecules in times of stress or emergency. A solution of \(0.64 \mathrm{~g}\) of adrenaline in \(36.0 \mathrm{~g}\) of \(\mathrm{CCl}_{4}\) elevates the boiling point by \(0.49^{\circ} \mathrm{C}\). Is the molar mass of adrenaline calculated from the boiling-point elevation in agreement with the following structural formula?

Step-by-Step Solution

Answer

The molar mass of adrenaline is approximately 182.34 g/mol.

Step 1: Find the molality of the solution

To find the molality, we need the moles of the solute (adrenaline) and mass of the solvent (\(\mathrm{CCl}_{4}\)) in kg. Since we don't have the moles of adrenaline, let's represent it by "n". Also, we need to convert the mass of \(\mathrm{CCl}_{4}\) to kg: Mass of \(\mathrm{CCl}_{4}\) in kg = \(36.0 \mathrm{~g}\) * \(\frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} = 0.036 \mathrm{~kg}\) Molality (m) = \(\frac{n (moles\, of\, adrenaline)}{0.036\,\mathrm{kg\,of\,CCl}_{4}}\)

Step 2: Apply the boiling point elevation formula

According to the formula, ΔT = m * K_b. We substitute the given values and the molality from the previous step: \(0.49^{\circ} \mathrm{C} = \frac{n (moles\, of\, adrenaline)}{0.036\,\mathrm{kg\,of\,CCl}_{4}} * 5.03 \,\mathrm{K\, kg\, mol}^{-1}\)

Step 3: Find the moles of adrenaline

Solve the equation above for "n": \(n (moles\, of\, adrenaline) = \frac{0.49^{\circ} \mathrm{C} * 0.036\,\mathrm{kg\,of\,CCl}_{4}}{5.03 \,\mathrm{K\, kg\, mol}^{-1}} = 0.00351 \,\mathrm{mol}\)

Step 4: Calculate the molar mass of adrenaline

To find the molar mass of adrenaline, we will divide the mass in grams by the number of moles: Molar mass of adrenaline = \(\frac{0.64 \,\mathrm{g}}{0.00351\,\mathrm{mol}} = 182.34 \,\mathrm{g\,mol^{-1}}\) The approximate molar mass of adrenaline is 182.34 g/mol.

Key Concepts

Adrenaline

Adrenaline, also known as epinephrine, is a hormone that plays a crucial role in the body's "fight-or-flight" response. It is produced in the adrenal glands and helps the body react quickly during stressful or emergency situations. By releasing extra glucose into the bloodstream, it provides immediate energy needed for the body to respond to potential threats.

When studying chemical properties or reactions involving adrenaline, it's essential to consider its molecular structure. This molecule consists of specific elements that influence its chemical behavior, including how it interacts in various solutions. Understanding the molar mass and figuring out how it affects boiling point elevation can provide further insights into the intricate workings of this hormone in both biological and chemical environments.

Molar Mass Calculation

Calculating the molar mass of a compound like adrenaline involves determining the mass of one mole of its molecules. This is usually expressed in grams per mole (g/mol). To obtain the molar mass, you divide the total mass of the substance by the number of moles present.

In the exercise problem, adrenaline has a mass of 0.64 g, and the number of moles calculated was approximately 0.00351 mol. Using the formula:

• Molar mass (g/mol) = Mass (g) / Moles (mol)

gives us a molar mass of approximately 182.34 g/mol for adrenaline.
This value can then be compared with literature values for adrenaline to confirm its accuracy. Understanding how to perform molar mass calculations is key in chemistry, allowing us to verify chemical formulas and understand compound properties.

Molality

Molality is a measure of concentration that expresses the moles of solute per kilogram of solvent. It is particularly useful in colligative properties, such as boiling point elevation and freezing point depression, where changes in physical properties depend on the number of solute particles, not their identity.

To calculate molality in the context of the given problem, we use the formula:

• Molality (m) = Moles of solute (mol) / Mass of solvent (kg)

For adrenaline dissolved in carbon tetrachloride (CCl₄), we have a solute amount of 'n' moles and a solvent mass of 0.036 kg. By substituting these values into the boiling point elevation equation (\[ \Delta T = m \times K_b \]), where \( K_b \) is the ebullioscopic constant, we can find the molality needed to calculate boiling point elevation.

Understanding molality and its role in colligative properties helps demystify how solutions behave under various conditions, aiding in more accurate chemical predictions and applications.