A115-mg sample of eugenol, the compound responsible for the odor of cloves, was placed in an evacuated fl ask with c05thepropertiesofgases. indd page 183 6/18/12 4: 40 pm user-f393 /users/user-f393/desktop 184 chapter 5 the properties of gases a volume of 500.0 ml at 280.0 c. the pressure that eugenol exerted in the fl ask under those conditions was found to be 48.3 torr. in a combustion experiment, 18.8 mg of eugenol burned to give 50.0 mg of carbon dioxide and 12.4 mg of water. what is the molecular formula of eugenol?

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A 115-mg sample of eugenol, the compound responsible for the odor of cloves, was placed in an evacuated flask with a volume of 500.0 mL at 280.0°C. The pressure that eugenol exerted in the flask under those conditions was found to be 48.3 Torr. In a combustion experiment, 18.8 mg of eugenol burned to give 50.0 mg of carbon dioxide and 12.4 mg of water. What is the molecular formula of eugenol?

C₁₀H₁₂O₂

Explanation:

First, let's find the molar mass of the eugenol by the ideal gas law:

PV = nRT

Where P is the pressure, V is volume (0.5 L), n is the number of moles, R is the gas constant (62.3637 torr.L/mol.K), and T is the temperature (280°C + 273 = 553 K).

n = PV/RT

n = (48.3*0.5)/(62.3637*553)

n = 7.00x10⁻⁴ mol

The molar mass (M) is the mass (0.115 g) divided by the number of moles:

M = 0.115/7.00x10⁻⁴

M = 164 g/mol

The burn of eugenol produced only carbon dioxide and water, so the compound is formed only by carbon, oxygen, and hydrogen, and can be represented as CxHyOz. The combustion reaction is:

CxHyOz + O₂ → CO₂ + H₂O

All the carbon in the compound must form CO₂ and all the hydrogen must form H₂O, so the number of moles of them can be determined knowing that the molar mass of CO₂ is 44 g/mol, and the molar mass of water is 18 g/mol.

nCO₂: 0.05g/44g/mol = 1.136x10⁻³ mol

nC = nCO₂ = 1.136x10⁻³ mol

nH₂O = 0.0124g/18g/mol = 6.889x10 ⁻⁴ mol

nH = 2*nH₂O = 1.378x10⁻³ mol

These are the number of moles of the elements at eugenol, so their masses can be calculated, knowing that the molar mass of C is 12g/mol, and of H is 1g/mol:

mC = 12*1.136x10⁻³ = 0.013632 g = 13.632 mg

mH = 1*1.378x10⁻³ = 1.378x10⁻³ g = 1.378 mg

mO = 18.8 - (13.632 + 1.378) = 3.79 mg

The molar mass of O is 16 g/mol, so:

nO = 0.00379g/16g/mol = 2.36875x10⁻⁴ mol

Dividing the number of moles for the smallest (2.36875x10⁻⁴), we can find the empirical formula:

C: 1.136x10⁻³/2.36875x10⁻⁴ ≅ 5

H: 1.378x10⁻³/2.36875x10⁻⁴ ≅6

O: 2.36875x10⁻⁴/2.36875x10⁻⁴ = 1

The empirical formula is C₅H₆O, which has molar mass:

5*12 + 6*1 + 16 = 82 g/mol

Then:

n(C₅H₆O) = 164 g/mol

n*82 = 164

n = 2

The molecular formula is:

C₁₀H₁₂O₂