Question

Determine the empirical formula of jasvanol, a fragrance molecule with a floral aromatic jasmine note with a hint of fruit. It is composed of \(81.02 \% \mathrm{C}, 11.79 \% \mathrm{H}\), and \(7.20 \% \mathrm{O}\).

Short answer

The empirical formula for jasvanol is \( \text{C}_{15}\text{H}_{26}\text{O}_{1} \).

Step-by-step solution

- Define Molar Masses

Determine the molar mass of each element present in jasvanol: Carbon (C) has a molar mass of 12.01 g/mol, Hydrogen (H) has a molar mass of 1.008 g/mol, and Oxygen (O) has a molar mass of 16.00 g/mol.

- Convert Percentages to Masses

Assume you have 100 grams of jasvanol. This simplifies the math as the percentages can directly translate to masses: Carbon: 81.02 g Hydrogen: 11.79 g Oxygen: 7.20 g.

- Convert Masses to Moles

Use the molar mass of each element to convert the masses to moles: Moles of Carbon: \[ \frac{81.02}{12.01} = 6.75 \text{ moles} \] Moles of Hydrogen: \[ \frac{11.79}{1.008} = 11.70 \text{ moles} \] Moles of Oxygen: \[ \frac{7.20}{16.00} = 0.45 \text{ moles} \]

- Find the Simplest Ratio

Determine the simplest whole number ratio by dividing the moles of each element by the smallest number of moles, which is 0.45 in this case: \( \text{Ratio of C:} \frac{6.75}{0.45} = 15 \) \( \text{Ratio of H:} \frac{11.70}{0.45} = 26 \) \( \text{Ratio of O:} \frac{0.45}{0.45} = 1 \)

- Write the Empirical Formula

Using the simplest whole number ratio, write the empirical formula for jasvanol: \( \text{C}_{15}\text{H}_{26}\text{O}_{1} \).

Key concepts

Molar Masses

To determine an empirical formula, you first need to know the molar masses of the elements involved. The molar mass of an element is the mass of one mole of that element. It is typically measured in grams per mole (g/mol) and can be found on the periodic table.
In the case of jasvanol, the molar masses are:

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

Using these values accurately is essential because they play a critical role in converting mass to moles in a later step.

Percent Composition

Percent composition refers to the percentage by mass of each element present in a compound. It essentially tells you how much of each element is in a compound based on its total mass.
If you have a compound like jasvanol with 81.02% Carbon, 11.79% Hydrogen, and 7.20% Oxygen, you can assume you have 100 grams of the compound for simplicity.
This means the mass of each element directly correlates with its percentage, simplifying calculations:

• Carbon: 81.02 grams
• Hydrogen: 11.79 grams
• Oxygen: 7.20 grams

Assuming a 100-gram sample helps streamline the process of converting these masses to moles.

Mole Conversion

Converting mass to moles is one of the crucial steps to determine the empirical formula. One mole of any element contains Avogadro's number of atoms, often denoted as \text{N}_{A} \[6.022 \times 10^{23}\]. Moles are calculated using the formula:
\[ \text{moles} = \frac{ \text{mass}}{ \text{molar mass} }\]
For jasvanol:

• For Carbon: \[ \frac{81.02}{12.01} = 6.75 \text{ moles} \]
• For Hydrogen: \[ \frac{11.79}{1.008} = 11.70 \text{ moles} \]
• For Oxygen: \[ \frac{7.20}{16.00} = 0.45 \text{ moles} \]

Converting to moles allows us to find out how many atoms of each element are present in the compound.

Simplest Ratio

To determine the empirical formula, we need the simplest whole number ratio of the moles of each element. This step involves dividing the moles of each element by the smallest number of moles in the set.
For jasvanol, with 0.45 moles of Oxygen being the smallest:

• Ratio of Carbon: \[ \frac{6.75}{0.45} = 15 \]
• Ratio of Hydrogen: \[ \frac{11.70}{0.45} = 26 \]
• Ratio of Oxygen: \[ \frac{0.45}{0.45} = 1 \]

These ratios tell us that for every Oxygen atom, there are 15 Carbon atoms and 26 Hydrogen atoms. Hence, the empirical formula for jasvanol is \( \text{C}_{15} \text{H}_{26} \text{O}_{1} \).