Question

Determine the empirical formula of florhydral, a fragrance molecule with a fresh, natural aroma often used in fabric detergents. It is composed of \(86.65 \% \mathrm{C}, 4.48 \% \mathrm{H}\), and \(8.88 \% \mathrm{O}\).

Short answer

The empirical formula of florhydral is C\textsubscript{13}H\textsubscript{8}O.

Step-by-step solution

Convert Percentages to Grams

Assume a 100 g sample of florhydral. This means the composition will be 86.65 g of Carbon (C), 4.48 g of Hydrogen (H), and 8.88 g of Oxygen (O).

Convert Grams to Moles

Use the molar mass of each element to convert the mass to moles.Molar Mass:- Carbon (C): 12.01 g/mol- Hydrogen (H): 1.01 g/mol- Oxygen (O): 16.00 g/molCalculations:o Carbon: \[ \frac{86.65 \text{ g}}{12.01 \text{ g/mol}} = 7.22 \text{ moles} \]o Hydrogen: \[ \frac{4.48 \text{ g}}{1.01 \text{ g/mol}} = 4.44 \text{ moles} \]o Oxygen: \[ \frac{8.88 \text{ g}}{16.00 \text{ g/mol}} = 0.555 \text{ moles} \]

Determine the Simplest Whole Number Ratio

Divide the number of moles of each element by the smallest number of moles calculated to determine the simplest ratio.o Carbon: \[ \frac{7.22}{0.555} = 13.0 \]o Hydrogen: \[ \frac{4.44}{0.555} = 8.0 \]o Oxygen: \[ \frac{0.555}{0.555} = 1.0 \]This gives the ratio C\textsubscript{13}H\textsubscript{8}O\textsubscript{1}.

Key concepts

stoichiometry

Stoichiometry is a core concept in chemistry that involves calculating the quantities of reactants and products involved in chemical reactions. It is like a recipe for chemistry; using balanced equations, it tells you how much of each ingredient is needed and how much product will be made. In empirical formula calculations, stoichiometry helps establish the relationships between the elements within a compound.
In our problem with florhydral, stoichiometry helps us figure out the simplest formula by relating the masses and moles of Carbon (C), Hydrogen (H), and Oxygen (O). This ensures we determine the correct proportion of each element in the compound.

percent composition

Percent composition tells us the percentage of each element in a compound by mass. It's like finding out what fraction of a whole pie each type of ingredient is. In our florhydral problem, we start by converting each percentage of C, H, and O into grams. Assuming a 100g sample makes this simplification easy:

• Carbon (C): 86.65g
• Hydrogen (H): 4.48g
• Oxygen (O): 8.88g

This is a crucial first step to solving for the empirical formula since the percentages directly translate into the mass of each element within the sample.

mole conversion

Mole conversion changes mass into moles using the molar mass of each element (the mass of one mole of that element). This step is essential because moles provide a way to directly compare amounts of different elements, accounting for the fact that different atoms have different weights.
For florhydral, using molar masses:

• Carbon: \(\frac{86.65 \text{ g}}{12.01 \text{ g/mol}} = 7.22 \text{ moles}\)
• Hydrogen: \(\frac{4.48 \text{ g}}{1.01 \text{ g/mol}} = 4.44 \text{ moles}\)
• Oxygen: \(\frac{8.88 \text{ g}}{16.00 \text{ g/mol}} = 0.555 \text{ moles}\)

This step converts the masses into a comparable quantity: moles.

simplest ratio

The simplest ratio helps us express the relationship between the elements in their smallest whole number ratios. It's kind of like simplifying a fraction. To find this ratio, divide all mole values by the smallest number of moles calculated:

• Carbon: \(\frac{7.22}{0.555} = 13.0\)
• Hydrogen: \(\frac{4.44}{0.555} = 8.0\)
• Oxygen: \(\frac{0.555}{0.555} = 1.0\)

This results in the simplest whole number ratio of the elements, leading to our empirical formula: \( \text{C}_{13} \text{H}_8 \text{O}_1 \).

chemical formulas

Chemical formulas are shorthand representations of the types and numbers of atoms in a substance. The empirical formula is a type of chemical formula representing the simplest whole-number ratio of elements in a compound.
In the case of florhydral, after performing the calculations, we arrive at an empirical formula of \( \text{C}_{13} \text{H}_8 \text{O}_1 \). This formula reveals the simplest ratio of carbon, hydrogen, and oxygen atoms within the fragrance molecule, providing essential insights into its composition and aiding in further chemical analysis or synthesis.