Question

Eugenol, a chemical substance with the flavor of cloves, consists of \(73.19 \% \mathrm{C}, 19.49 \% \mathrm{O}\), and \(7.37 \% \mathrm{H}\). What is the empirical formula of eugenol?

Short answer

The empirical formula of Eugenol is \(C_5H_6O\).

Step-by-step solution

Convert Percentages to Grams

Since these are percentages by mass, assume a total sample mass of 100 g, so the mass of each element is the same as the given percentage. We will have 73.19 g of Carbon (C), 7.37 g of Hydrogen (H) and 19.49 g of Oxygen (O).

Convert Grams to Moles

The number of moles is calculated by dividing the mass of the element by its molar mass. The molar masses of C, H and O are approximately 12.01 g/mol, 1.008 g/mol and 16.00 g/mol respectively. So we get: \[\frac{73.19 g \mathrm{C}}{12.01 g/mol \mathrm{C}} \approx 6.09 moles \mathrm{C}\]\[\frac{7.37 g \mathrm{H}}{1.008 g/mol \mathrm{H}} \approx 7.31 moles \mathrm{H}\]\[\frac{19.49 g \mathrm{O}}{16.00 g/mol \mathrm{O}} \approx 1.22 moles \mathrm{O}\]

Determine Ratio of Moles

To find the simplest ratio, divide each number of moles by the smallest of these values which is 1.22. So we get: \[6.09 moles \mathrm{C} / 1.22 \approx 5 moles \mathrm{C}\]\[7.31 moles \mathrm{H} / 1.22 \approx 6 moles \mathrm{H}\]\[1.22 moles \mathrm{O} / 1.22 \approx 1 mole \mathrm{O}\]

Write Empirical Formula

The empirical formula is then written using these integers. Therefore Eugenol has an empirical formula of \(C_5H_6O\).

Key concepts

Chemical Composition

The chemical composition of a substance provides insight into what elements make up that substance and their relative proportions. For any compound, its chemical composition is often given in terms of the percentage by mass of each element present. This percentage is crucial as it informs us of the exact breakdown of the compound's makeup. In our example of eugenol, the given chemical composition is:

• 73.19% Carbon (C)
• 19.49% Oxygen (O)
• 7.37% Hydrogen (H)

This information lets us know that in every 100 grams of eugenol, these are the respective masses of carbon, hydrogen, and oxygen. From the chemical composition percentages, we can use them to further understand and determine the empirical formula of the compound, which tells us the simplest ratio of the elements within the compound.

Molar Masses

The concept of molar masses is fundamental when converting between grams and moles, which is essential in deriving the empirical formula of a compound. The molar mass of an element is the mass of one mole of its atoms, usually expressed in grams per mole (g/mol). For common elements involved in organic compounds like eugenol, their molar masses are approximately:

• Carbon (C): 12.01 g/mol
• Hydrogen (H): 1.008 g/mol
• Oxygen (O): 16.00 g/mol

In our exercise, these molar masses are used to convert the masses of each constituent element derived from the chemical composition, into moles. This conversion is necessary because chemical formulas and reactions are based on the number of moles, not masses. By dividing the mass of each element by its corresponding molar mass, you can calculate the number of moles, allowing us to find how each element contributes to the compound.

Percent Composition

Percent composition refers to the relative percentage by mass of each element in a compound. It plays a vital role in determining both chemical properties and the empirical formula of molecules. To find the empirical formula, you start by converting the percent composition into masses, typically under the assumption that you have a 100 g sample of the compound. This makes the conversion straightforward because the percentages directly translate to mass values in grams. Thus, with 100 grams of the sample, the percent composition numbers you have are exactly the number of grams for each element. Once you have the constituent masses, these are transformed into moles using the corresponding molar masses. In eugenol, these conversions lead to:

• Carbon: 73.19 g
• Hydrogen: 7.37 g
• Oxygen: 19.49 g

By taking the smallest number of moles calculated and using it as a divisor, you establish a mole ratio, which helps derive the empirical formula. This way, the percent composition not only serves as a guide for the mass of individual elements present but also as a stepping stone to the empirical formula, revealing the smallest whole number ratio in which atoms combine to form the compound.