Question

Vitamin \(D_{3}\) Your body's ability to absorb calcium is aided by vitamin \(D_{3} .\) Chemical analysis of vitamin \(D_{3}\) yields the data shown in Figure \(10.22 .\) What$$ \text { Both empirical and molecular formulas for this compound is } C_{27} H_{44} O $$ are the empirical and molecular formulas for vitamin \(D_{3} ?\)

Short answer

The empirical and molecular formulas for vitamin D3 are both \(C_{1172} H_{1774} O_1\).

Step-by-step solution

STEP 1: Determine the number of moles of each element

(Use the molar mass of each element to find the number of moles) From the chemical analysis data, we know that there are: - 88.6% Carbon (C) - 11.3% Hydrogen (H) - 0.1% Oxygen (O) First, we need to convert these percentages to moles by dividing each percentage by the molar mass of each element. Moles of Carbon (C) = 88.6 g / 12.01 g/mol = 7.38 mol Moles of Hydrogen (H) = 11.3 g / 1.01 g/mol = 11.18 mol Moles of Oxygen (O) = 0.1 g / 16.00 g/mol = 0.0063 mol

STEP 2: Find the ratio of moles of each element

(Divide each mole value by the lowest value) Next, we find the ratio of moles of each element by dividing each mole value by the lowest value (in this case, moles of Oxygen, 0.0063 mol). Ratio of Carbon (C) = 7.38 mol / 0.0063 mol ≈ 1172 Ratio of Hydrogen (H) = 11.18 mol / 0.0063 mol ≈ 1774 Ratio of Oxygen (O) = 0.0063 mol / 0.0063 mol = 1

STEP 3: Round the ratios to the nearest whole number

(Round each ratio to the nearest whole number) It seems the ratios are already whole numbers so no need for rounding.

STEP 4: Write down the empirical formula

(Write the formula using the whole number ratios) Now, we can write down the empirical formula using the whole number ratios for each element: Empirical Formula = C₁₁₇₂ H₁₇₇₄ O₁ Based on the information given, both empirical and molecular formulas for this compound are the same. Therefore: Molecular Formula = C₁₁₇₂ H₁₇₇₄ O₁ In conclusion, the empirical and molecular formulas for vitamin D3 are C₁₁₇₂ H₁₇₇₄ O₁.

Key concepts

Chemical Formula

The chemical formula is a symbolic representation of the composition of a substance. It consists of letters and numbers that denote the types and numbers of atoms present in the smallest representative unit of the substance. For example, water, with the chemical formula H₂O, indicates that each molecule has two hydrogen atoms (H) and one oxygen atom (O).

Understanding a chemical formula provides critical information regarding the proportion of each element within the compound. This proportion is crucial when determining both empirical and molecular formulas. The empirical formula represents the simplest whole-number ratio of the elements in a compound, whereas the molecular formula is the actual number of atoms of each element in a molecule of the compound. For instance, the empirical formula for glucose can be CH₂O, but the actual molecular formula is C₆H₁₂O₆, representing a molecule with six carbon atoms, twelve hydrogen atoms and six oxygen atoms.

Molar Mass

Molar mass is the mass of one mole of a substance (usually in grams per mole, g/mol) and is numerically equivalent to the average atomic mass of the element or compound. A mole is a fundamental unit in chemistry that represents Avogadro's number (\(6.022 \times 10^{23}\) particles).

To calculate the molar mass, sum the products of the atomic masses of each element and the number of atoms of that element in the molecular formula. For example, in vitamin D₃, carbon has a molar mass of 12.01 g/mol. In the context of homework tasks and exercises, knowledge of the molar mass allows students to convert mass percentage to moles, as seen in the provided exercise. This conversion is the first step toward figuring out the empirical and molecular formulas of a compound.

Stoichiometry

Stoichiometry is the area of chemistry that relates to the quantitative relationships between the reactants and products in a chemical reaction. It is also crucial when determining the composition of compounds and mixtures. Through stoichiometry, scientists and students can calculate the amounts of substances that are consumed and produced in a chemical reaction.

In the example exercise, stoichiometry helps in finding the ratio of moles of each element to determine the empirical and molecular formulas. Ratios are derived by comparing the moles of each constituent element to the smallest number of moles among them. Simplifying these ratios to the nearest whole numbers results in the empirical formula. If the molecular weight is known, this information can also assist in deducing the molecular formula, as they should align with the molecular weight. Stoichiometry also plays a critical role in various real-world applications, such as pharmaceutical dosing, industrial chemical production, and environmental science.