# Question 3.19: Eugenol is an ingredient in several spices, including bay le...

Eugenol is an ingredient in several spices, including bay leaves and cloves (Figure 3.36). Eugenol contains carbon, hydrogen, and oxygen. Combustion of 21.80 mg of eugenol yields 58.5 mg of CO_{2} and 14.4 mg of H_{2}O. What is the empirical formula of eugenol? If the mass spectrum of eugenol shows a molecular ion at 164 amu, what is the molecular formula of eugenol?

**blue check mark**means that this solution has been answered and checked by an expert. This guarantees that the final answer is accurate.

Learn more on how we answer questions.

**Collect and Organize** We are given the masses of CO_{2} (58.5 mg) and H_{2}O (14.4 mg) produced by the combustion of 21.80 mg of eugenol and are asked to determine its empirical formula. We also know its molar mass, and we need to determine its molecular formula.

**Analyze** First we determine the number of moles of C and H in the CO_{2} and H_{2}O produced during combustion. These values are equal to the number of moles of C and H in the combusted sample. Multiplying these mole values by the atomic masses of C and H yields the mass of these elements in the original sample, which we then subtract from the total mass of the sample to obtain the mass of O in the sample. This mass can then be converted into moles of O in the sample:

The mole ratio of C:H:O is then calculated and reduced to a ratio of small whole numbers to obtain the empirical formula. We divide the mass of the empirical formula into the molar mass to obtain the multiplier, n, that allows us to convert the empirical formula to a molecular formula.

**Solve** The moles of C and H in the CO_{2} and H_{2}O collected during combustion are

58.5 \sout{mg CO_{2}} \times \frac{1 \sout{g CO_{2}}}{10^{3} \sout{mg CO_{2}}} \times \frac{1 \sout{mol CO_{2}}}{44.01 \sout{g CO_{2}}} \times \frac{1 mol C}{1 \sout{mol CO_{2}}} =1.329 \times 10^{-3} mol C

14.4 \sout{mg H_{2}O_2} \times \frac{1 \sout{g H_{2}O}}{10^{3} \sout{mg H_{2}O}} \times \frac{1 \sout{mol H_{2}O}}{18.02 \sout{g H_{2}O}} \times \frac{2 mol H}{1 \sout{mol H_{2}O}} =1.598 \times 10^{-3} mol H

The masses of C and H are

1.329 \times 10^{-3} \sout{mol C} \times \frac{12.01 g C}{1 \sout{mol C}} =1.596 \times 10^{-2} g C

1.598 \times 10^{-3} \sout{mol H} \times \frac{1.008 g H}{1 \sout{mol H}} =1.611 \times 10^{-3} g H

In order to more easily compare the masses of C and H to the original sample mass of 21.8 mg, let’s convert them into milligrams as well:

1.596 \times 10^{-2} \sout{g C} \times \frac{10^{3} mg C}{1 \sout{g C}}=15.96 mg C

1.611 \times 10^{-3} \sout{g H} \times \frac{10^{3} mg H}{1 \sout{g H}}=1.611 mg H

The sum of these two masses (15.96 mg C + 1.611 mg H = 17.57 mg) is less than the mass of the sample (21.8 mg). The difference must be the mass of oxygen in the sample:

Mass of oxygen = 21.80 mg-17.57 mg = 4.23 mg O

The number of moles of O atoms in the sample is

4.23 \times 10^{-3} \sout{g O} \times \frac{1 mol O}{16.00 \sout{g O}} =2.64 \times 10^{-4} mol O

The mole ratio of the three elements in the sample is

1.329 \times 10^{-3} mol C:1.598 \times 10^{-3} mol H:2.64 \times 10^{-4} mol O

Dividing through by the smallest value (2.64 \times 10^{-4} mol) gives a mole ratio of 5:6:1, making the empirical formula of the sample C_{5}H_{6}O. The empirical formula mass is

(5×12.01 g/mol + 6×1.008 g/mol + 16.00 g/mol) n=82.10 g/mol

The molecular ion identifies the molecular mass of eugenol as 164 amu, so, the molar mass of eugenol is 164 g/mol. Therefore the multiplier, n, to convert the empirical formula into a molecular formula is

n=\frac{molar mass}{empirical formula mass} =\frac{164 g/mol}{82.10 g/mol} =2

and the molecular formula is

(C_{5}H_{6}O)_{n} = (C_{5}H_{6}O)_{2} = C_{10}H_{12}O_{2}

**Think About It** Our answer makes sense because we have a simple whole-number ratio of the elements.