Problem 94 - Chemistry The Central Science

Question

The koala dines exclusively on eucalyptus leaves. Its digestive system detoxifies the eucalyptus oil, a poison to other animals. The chief constituent in eucalyptus oil is a substance called eucalyptol, which contains \(77.87 \% \mathrm{C}, 11.76 \% \mathrm{H},\) and the remainder O. (a) What is the empirical formula for this substance? (b) A mass spectrum of eucalyptol shows a peak at about 154 amu. What is the molecular formula of the substance?

Step-by-Step Solution

Answer

The empirical formula for eucalyptol is \(C_{10}H_{18}O\), and since the molar mass from the mass spectrum is approximately equal to the molar mass of the empirical formula, the molecular formula is also \(C_{10}H_{18}O\).

Step 1: Calculate the moles of each element in eucalyptol

Given the percentage composition of C, H, and O, we can assume we have 100 g of eucalyptol. This will make the calculations easier since the percentage composition can be treated as the mass of each element. a) Calculate moles of Carbon (C): Percentage composition of Carbon (C) = 77.87 % Mass of Carbon in 100 g eucalyptol = 77.87 g Molar mass of Carbon (C) = 12.01 g/mol Moles of Carbon (C) = (mass)/(molar mass) = (77.87 g)/(12.01 g/mol) = 6.49 mol b) Calculate moles of Hydrogen (H): Percentage composition of Hydrogen (H) = 11.76 % Mass of Hydrogen in 100 g eucalyptol = 11.76 g Molar mass of Hydrogen (H) = 1.008 g/mol Moles of Hydrogen (H) = (mass)/(molar mass) = (11.76 g)/(1.008 g/mol) = 11.67 mol c) Calculate moles of Oxygen (O): Percentage composition of Oxygen (O) = 100 - 77.87 - 11.76 = 10.37 % Mass of Oxygen in 100 g eucalyptol = 10.37 g Molar mass of Oxygen (O) = 16.00 g/mol Moles of Oxygen (O) = (mass)/(molar mass) = (10.37 g)/(16.00 g/mol) = 0.65 mol

Step 2: Determine the empirical formula for eucalyptol

Now that we know the moles of each element, we need to find the simplest whole number ratio for each element. Divide each mole value by the smallest mole value: Moles of C: 6.49/0.65 = 10 Moles of H: 11.67/0.65 = 18 Moles of O: 0.65/0.65 = 1 This results in a simple whole number ratio of C: 10, H: 18, and O: 1. Therefore, the empirical formula for eucalyptol is C10H18O1 or C10H18O.

Step 3: Find the molecular formula of eucalyptol

To find the molecular formula, we need to determine how many empirical formula units are present in a molecule of eucalyptol. (a) Calculate the molar mass of the empirical formula, C10H18O: Molar mass of C10H18O = (10 x Molar mass of C) + (18 x Molar mass of H) + (1 x Molar mass of O) = (10 x 12.01g/mol) + (18 x 1.008 g/mol) + (1 x 16.00 g/mol) = 154.26 g/mol (b) Use the molar mass in the mass spectrum to find the molecular formula: Given mass from the mass spectrum = 154 amu The ratio (n) = (Given mass from the mass spectrum)/(Molar mass of the empirical formula) n = 154 amu / 154.26 g/mol ≈ 1 Since n is approximately 1, the molecular formula (C10H18O) is the same as the empirical formula. The molecular formula of eucalyptol is C10H18O.

Key Concepts

Stoichiometry

Stoichiometry is the study of quantitative relationships between the amounts of reactants used and products formed by a chemical reaction. It's based on the conservation of mass where the total mass of the reactants equals the total mass of the products.

For students diving into stoichiometry, it's important to understand the concept of the 'mole', which is a unit for counting particles, just like a 'dozen' for eggs. A mole (mol) represents Avogadro's number, which is roughly equal to \(6.022 \times 10^{23}\) entities of the substance.

To ace stoichiometry problems, grasp these steps:

• Determine the molar mass of each reactant and product.
• Convert mass of reactants to moles using their molar mass.
• Use the balanced chemical equation to set up mole ratios.
• Convert moles back to grams if necessary to find the mass of the desired product.

Stoichiometry is crucial to understand because it explains how ingredients combine in the perfect proportions to form the desired compounds without waste—just like following a precise recipe in cooking.

Percent Composition

Percent composition is a way of expressing the relative amount of each element within a compound. To determine the percent composition, one must know the mass of each element in a mole of the compound and the molar mass of the entire compound.

The formula for percent composition is:\[ \text{Percent Composition} = \left( \frac{\text{mass of element}}{\text{molar mass of compound}} \right) \times 100\% \]
This concept is used in the initial steps when determining the empirical formula of a compound. The empirical formula represents the simplest whole-number ratio of the different atoms in the substance and is determined by converting the percent composition of each element to moles, and then simplifying these ratios into the smallest whole numbers.

In practical terms, calculating the percent composition also helps chemists in analyzing samples, checking the purity of substances, and formulating products with exact specifications.

Molar Mass

The molar mass is the mass of one mole of a substance (usually in grams per mole, g/mol) and is the bridge between the atomic level and the macroscopic level. It's a sum of the atomic weights of all the atoms in a chemical formula.

To find the molar mass, simply multiply the number of atoms of each element in the molecular formula by the atomic mass of that element from the periodic table, and then add all of these values together:\[ \text{Molar Mass} = \sum (\text{Number of atoms of element} \times \text{Atomic weight of element}) \]
Understanding the molar mass allows you to convert between the mass of a substance and the number of moles, a crucial step in stoichiometry. It's also essential for calculating concentrations and for converting between the mass of a substance and the volume of gas at Standard Temperature and Pressure (STP).

For example, if a molecule has a peak at 154 amu in a mass spectrum, it means the mass of one mole of this molecule is 154 grams. This information can tell us how many molecules or atoms are in a specific sample by simply weighing it.