Problem 158 - Introductory Chemistry Atoms First

Question

The hormone estradiol contains carbon, hydrogen, and oxygen and has a molar mass of approximately \(272 \mathrm{~g} / \mathrm{mol}\). When \(1.15 \mathrm{~g}\) of estradiol is subjected to combustion analysis, the results yield \(79.26 \% \mathrm{C}\) and \(8.88 \% \mathrm{H}\). What is its molecular formula?

Step-by-Step Solution

Answer

The molecular formula for estradiol is \(C_9H_{12}O_{12}\).

Step 1: Calculate the moles of Carbon, Hydrogen, and Oxygen

First, we need to convert the percentages of each element to moles. We are given that estradiol contains 79.26% Carbon, 8.88% Hydrogen, and the rest is Oxygen. For a 1.15g sample of estradiol: - Carbon: \( \frac{79.26g}{12.01 g/mol(C)} = 6.60 mol \) - Hydrogen: \( \frac{8.88g}{1.01 g/mol(H)} = 8.79 mol \) - Oxygen: Since we know the total mass, we can subtract both the percentage of Carbon and Hydrogen to find the percentage of Oxygen. Then, we can use the mass of the sample to find the moles of Oxygen. Percentage of Oxygen: \( 100\% - 79.26\% - 8.88\% = 11.86\% \) Moles of Oxygen: \( \frac{11.86g}{16.00 g/mol(O)} = \frac{133.0g/mole(O)}{16.00 g/mol(O)} = 8.31 mol \)

Step 2: Determine the empirical formula

Now that we have the moles of each element, we can find the empirical formula by dividing each of the moles by the smallest number among them. Smallest number of moles: 6.60 moles (Carbon) Mole ratio: - Carbon: \( \frac{6.60}{6.60} = 1 \) - Hydrogen: \( \frac{8.79}{6.60} = 1.33 \) - Oxygen: \( \frac{8.31}{6.60} = 1.26 \) Since we can't have fractions in the empirical formula, we should round the numbers in the mole ratio to the nearest whole number. As both Hydrogen and Oxygen are close to 4/3 ratio, we will assume it as \( 1: 4/3: 4/3 \) and multiply each by 3, to get whole numbers: Empirical Formula: \(C_3 H_4 O_4\)

Step 3: Find the molecular formula

To find the molecular formula, we need to find the empirical formula's mass and compare it to the estradiol's molar mass (272 g/mol) to determine the multiplier (the number of empirical units in the molecular formula). Empirical Formula Mass: \( 3(12.01) + 4(1.01) + 4(16) = 36.03 + 4.04 + 64 = 104.07 \, g/mol \) Now we can find the multiplier (n): \( n = \frac{Molecular \, Mass}{Empirical \, Formula \, mass} = \frac{272 g/mol}{104.07 g/mol} = 2.61 \approx 3\) As the multiplier is approximately 3, we can multiply the empirical formula by 3 to find the molecular formula: Molecular Formula: \( C_{3 \times 3}H_{4 \times 3}O_{4 \times 3} = C_9H_{12}O_{12} \) Therefore, the molecular formula for estradiol is \(C_9H_{12}O_{12}\).

Key Concepts

Empirical and Molecular Formulas

Understanding the distinction between empirical and molecular formulas is fundamental in chemistry, especially when working with combustion analysis results. The empirical formula of a compound provides the simplest whole-number ratio of elements present, not the actual number of atoms like the molecular formula does.

For instance, the empirical formula for water is H₂O, which tells us that, proportionally, there are twice as many hydrogen atoms as oxygen atoms. However, the molecular formula conveys the actual number of atoms in a molecule; in the case of water, it’s the same as the empirical formula, but this isn't always the situation.

In the exercise, the empirical formula for estradiol was estimated initially, based on the mole ratios of each element. By simplifying the mole ratio to the lowest whole number terms, the empirical formula C₃H₄O₄ was deduced. To determine the actual number of atoms present in a molecule of estradiol, this empirical formula was then scaled up to match the molar mass of estradiol, resulting in the molecular formula C₉H₁₂O₁₂. Thus, the molecular formula indicates that each molecule of estradiol contains nine carbon atoms, twelve hydrogen atoms, and twelve oxygen atoms.

Molar Mass Calculation

The molar mass of a compound is equivalent to the mass of one mole of that compound, usually expressed in grams per mole (g/mol). It's a key concept in chemistry because it relates the mass of substances to the amount of substance (number of moles).

To calculate the molar mass of a compound, one must sum the atomic masses of all the atoms present in the formula. This is found on the periodic table where the atomic mass of each element—typically averaged across isotopes—is listed.

In our exercise, the empirical formula’s molar mass was calculated by summing the atomic masses of carbon (C), hydrogen (H), and oxygen (O) present in its empirical formula. This step is crucial as it establishes the relationship between the empirical formula and the molecular formula through the known molar mass of the compound, here given as 272 g/mol for estradiol. By comparing the molar mass of the empirical formula with the given molar mass of estradiol, we could determine the scaling factor to arrive at the correct molecular formula.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is used for calculations that involve the masses of reactants and products, the amount of product that can be formed from a given reactant, or the amount of reactant needed to produce a desired amount of product.

Within the context of our combustion analysis problem, stoichiometry enables us to determine the amount of carbon, hydrogen, and oxygen based on the given mass percentages. By converting these percentages to masses and then to moles, stoichiometry guides us to establish the appropriate mole ratios between the elements. These ratios are essential in determining the empirical formula, which, as previously mentioned, is then used along with molar mass information to find the molecular formula.

Stoichiometry is not only pivotal for solving this particular problem but is a cornerstone of chemical calculations, making it an indispensable tool for chemists and students alike to predict the outcomes of reactions quantitatively.