Problem 91 - Chemistry: An Atoms-Focused Approach

Question

Eugenol is one of the compounds responsible for the flavor of cloves. A 111 mg sample of eugenol is dissolved in \(1.00 \mathrm{g}\) of chloroform \(\left(K_{\mathrm{b}}=3.63^{\circ} \mathrm{C} / \mathrm{m}\right)\) increasing the boiling point of chloroform by \(2.45^{\circ} \mathrm{C}\) Calculate eugenol's molar mass. Eugenol is \(73.17 \% \mathrm{C}\) \(7.32 \% \mathrm{H},\) and \(19.51 \% \mathrm{O}\) by mass. What is the molecular formula of eugenol?

Step-by-Step Solution

Answer

Answer: The molecular formula of eugenol is C10H12O2.

Step 1: Calculate the molality of the solution

To calculate the molality, we will use the boiling point elevation formula: \(\Delta T_b=K_b \cdot m\). We are given the boiling point elevation (\(\Delta T_b = 2.45^{\circ}\mathrm{C}\)) and the boiling point elevation constant for chloroform (\(K_b = 3.63^{\circ}\mathrm{C}/\mathrm{m}\)). Thus, we can solve for molality (m): \(m = \frac{\Delta T_b}{K_b} = \frac{2.45}{3.63} \approx 0.6751\, \mathrm{mol/kg}\)

Step 2: Calculate the moles of eugenol

The molality of the solution is moles of solute (eugenol) per kilogram of solvent (chloroform). We can use this information to find the moles of eugenol: \(\text{number of moles} = \text{molality} \cdot \text{mass of solvent }(Kg) = 0.6751\, \mathrm{mol/kg} \cdot 0.001\, \mathrm{kg} \approx 6.751 \times 10^{-4}\, \mathrm{mol}\)

Step 3: Calculate the molar mass of eugenol

Now we can calculate the molar mass of eugenol using the moles and mass of the eugenol sample: \(\text{molar mass} = \frac{\text{mass of eugenol}}{\text{moles}} = \frac{111\, \mathrm{mg}}{6.751\times10^{-4}\, \mathrm{mol}} \approx 164.33\, \mathrm{g/mol}\)

Step 4: Calculate the empirical formula

We are given the mass percentages of the elements in eugenol: C (73.17%), H (7.32%), and O (19.51%). Let's assume 100 g of eugenol is present, so we will have 73.17 g of C, 7.32 g of H, and 19.51 g of O. We can convert these masses to moles using the molar masses of the elements: - Moles of C: \(\frac{73.17\, \mathrm{g}}{12.01\, \mathrm{g/mol}} = 6.090\, \mathrm{mol}\) - Moles of H: \(\frac{7.32\, \mathrm{g}}{1.01\, \mathrm{g/mol}} = 7.248\, \mathrm{mol}\) - Moles of O: \(\frac{19.51\, \mathrm{g}}{16.00\, \mathrm{g/mol}} = 1.219\, \mathrm{mol}\) Now divide all the moles by the smallest mole value to get the ratio: - Ratio of C: \(\frac{6.090}{1.219} \approx 5\) - Ratio of H: \(\frac{7.248}{1.219} \approx 6\) - Ratio of O: \(\frac{1.219}{1.219} = 1\) The empirical formula is C5H6O.

Step 5: Calculate the molecular formula

We can now compare the empirical formula mass to the molar mass of eugenol to determine if the empirical formula is the same as the molecular formula: Empirical formula mass: \((5 \times 12.01) + (6 \times 1.01) + (1 \times 16.00) = 82.09\, \mathrm{g/mol}\) We can now find the ratio between the molar mass and the empirical formula mass: \(n = \frac{164.33\, \mathrm{g/mol}}{82.09\, \mathrm{g/mol}} \approx 2\) Since the ratio is approximately 2, the molecular formula is twice the empirical formula: (C5H6O)2 = C10H12O2. The molecular formula of eugenol is C10H12O2.

Key Concepts

Boiling Point Elevation

When a solute is added to a solvent, the boiling point of the solution usually increases; this is known as boiling point elevation. It's a colligative property, meaning it depends on the number of particles in a solution rather than the type of particles. We calculate it using the formula: \[ \Delta T_b = K_b \cdot m \]where:

• \(\Delta T_b\) is the boiling point elevation,
• \(K_b\) is the ebullioscopic constant of the solvent, and
• \(m\) is the molality of the solution.

In this exercise, knowing the change in boiling temperature and the constant \(K_b\), we can determine the molality, which tells us the concentration of the eugenol solute in chloroform. Understanding this concept is crucial as it illustrates how the presence of solute particles alters the properties of the solvent.

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of atoms of each element in the compound. This is determined using the mass percentages of each element in the compound. To find the empirical formula, follow these steps:

• Convert the percentage of each element to grams (assuming 100 g of the compound for simplicity).
• Convert grams to moles using the molar mass of each element.
• Divide all mole values by the smallest number of moles to find the ratio of elements.

For eugenol, the empirical formula calculation shows a ratio of \(C: H: O\) as \(5:6:1\), resulting in the formula \(\text{C}_5\text{H}_6\text{O}\). Knowing how to derive the empirical formula helps us understand the basic atomic composition of a compound.

Molecular Formula

The molecular formula indicates the actual number of atoms of each element in a molecule of a compound. It can sometimes be the same as the empirical formula or a multiple of it. To determine the molecular formula, we need the empirical formula and the molar mass of the compound. The steps are:

• Calculate the mass of the empirical formula.
• Divide the molar mass of the compound by the empirical formula mass to find the multiple (n).
• Multiply the subscripts in the empirical formula by this multiple to get the molecular formula.

In the case of eugenol, the molecular formula \(\text{C}_{10}\text{H}_{12}\text{O}_2\) is derived from doubling the empirical unit \(\text{C}_5\text{H}_6\text{O}\) as the molar mass is roughly twice that of the empirical formula mass.

Elemental Composition

The elemental composition of a compound gives the percentage by mass of each element within it. It's essential for figuring out both the empirical and molecular formulas. The steps involve:

• Taking the percentage by mass of each element (e.g., C is 73.17% in eugenol).
• Assuming a certain mass of the compound (often 100 g for easier calculation).
• Converting these percentages into moles by dividing by each element's molar mass.

These conversions help make the ratios needed to establish both empirical and molecular formulas. For eugenol, understanding its elemental composition was vital to determining that it contains primarily carbon, along with hydrogen and oxygen, illustrating the proportions of these atoms in its structure.