Question

Cortisol \((\mathscr{A}=362.47 \mathrm{~g} / \mathrm{mol})\) is a steroid hormone involved in protein synthesis. Medically, it has a major use in reducing inflammation from rheumatoid arthritis. Cortisol is \(69.6 \% \mathrm{C}\), \(8.34 \% \mathrm{H},\) and \(22.1 \% \mathrm{O}\) by mass. What is its molecular formula?

Short answer

The molecular formula is \( \text{C}_{20}\text{H}_{30}\text{O}_5 \).

Step-by-step solution

- Determine the mass of each element in 100 grams of cortisol

Assume you have 100 grams of cortisol. The mass of each element in 100 grams is:For Carbon (C): 69.6 gramsFor Hydrogen (H): 8.34 gramsFor Oxygen (O): 22.1 grams

- Calculate the number of moles of each element

Calculate the number of moles using the formula: \( \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \)For Carbon: \( \frac{69.6 \, g}{12.01 \, g/mol} = 5.8 \, mol \)For Hydrogen: \( \frac{8.34 \, g}{1.008 \, g/mol} = 8.28 \, mol \)For Oxygen: \( \frac{22.1 \, g}{16.00 \, g/mol} = 1.38 \, mol \)

- Determine the mole ratio

To find the empirical formula, divide the moles of each element by the smallest number of moles calculated:For Carbon: \( \frac{5.8}{1.38} = 4.20 \)For Hydrogen: \( \frac{8.28}{1.38} = 6.0 \)For Oxygen: \( \frac{1.38}{1.38} = 1.0 \)

- Write the empirical formula

The empirical formula based on the mole ratios found is approximately:\( \text{C}_4\text{H}_6\text{O}_1 \)

- Determine the molecular formula

Calculate the empirical formula mass:\( 4(12.01) + 6(1.008) + 16.00 = 74.08 \, g/mol \)Divide the molecular weight of cortisol by the empirical formula mass:\( \frac{362.47 \, g/mol}{74.08 \, g/mol} \approx 4.89 \)Since the ratio is approximately 5, multiply the subscripts in the empirical formula by 5:\( \text{Molecular formula} = \text{C}_{20}\text{H}_{30}\text{O}_5 \)

Key concepts

Empirical Formula

The empirical formula of a compound represents the simplest whole-number ratio of each element in the compound. It does not necessarily give the exact number of atoms, just the simplest ratio.

To find the empirical formula, you first convert the mass percentages to masses, assuming you have a 100-gram sample. This way, the percentage translates directly to grams.

Next, you convert the mass of each element to moles by dividing by their respective atomic masses.

Finally, to get the simplest ratio, divide all the mole quantities by the smallest number of moles calculated. In cases where you don't get whole numbers, you might need to multiply all ratios by a common factor to achieve whole numbers.

Mole Ratio

The mole ratio is fundamental in determining empirical and molecular formulas. It indicates the relative number of moles of each substance in a chemical reaction or compound.

After converting the masses to moles, finding the mole ratio involves dividing each element's mole value by the smallest number of moles calculated. This gives a ratio that can be used to derive the empirical formula.

For example, in cortisol, the ratio of 5.8 moles of carbon, 8.28 moles of hydrogen, and 1.38 moles of oxygen simplifies to roughly a 4:6:1 ratio when divided by the smallest mole quantity (1.38 moles of oxygen).

Elemental Composition

Elemental composition is the percentage by mass of each element in a compound. To determine the elemental composition, start with a known mass and proportion of each element.

In the example of cortisol, its composition is given as 69.6% carbon, 8.34% hydrogen, and 22.1% oxygen by mass. This step involves translating these percentages into actual masses for a given sample size, usually 100 grams.

Considering 100 grams simplifies calculations as the percentage values directly translate into grams, making it easier to determine mole ratios and subsequently empirical and molecular formulas.

Molecular Weight

Molecular weight, or molecular mass, is the sum of the atomic masses of all atoms in a molecule, measured in grams per mole (g/mol). It is crucial for going from empirical to molecular formulas.

You need the empirical formula's weight to determine how many empirical units fit into the actual molecular weight. In cortisol, the empirical formula weight is approximately 74.08 g/mol. The molecular weight is provided as 362.47 g/mol.

Dividing the molecular weight by the empirical formula mass gives a multiplier (in this case, about 4.89). This factor helps you scale up the empirical formula to the molecular formula, which more accurately reflects the compound's composition.

Stoichiometry

Stoichiometry deals with the quantitative relationships between reactants and products in a chemical reaction.

In terms of formula calculations, it connects the mass of different elements to their mole ratios and ultimately to the empirical and molecular formulas.

For example, converting 69.6 grams of carbon to moles using its molar mass, 8.34 grams of hydrogen, and 22.1 grams of oxygen similarly provides crucial ratios.

These ratios are essential in creating balanced chemical equations and understanding the mass relationships between components of a compound or reaction, allowing you to predict amounts of products or reactants needed.