Eugenol is one of the compounds responsible for the flavor of cloves. A 111 mg s

Question

Eugenol is one of the compounds responsible for the flavor of cloves. A 111 mg sample of eugenol is dissolved in 1.00 g of chloroform (Kb = 3.63°C/m), increasing the boiling point of chloroform by 2.45°C. Eugenol is 73.17% C, 7.32% H, and 19.51% O by mass. What is the molecular formula of eugenol? A. C15H18O3 B. C5H6O C. C10H12O2 D. C7H15O
Eugenol is one of the compounds responsible for the flavor of cloves. A 111 mg sample of eugenol is dissolved in 1.00 g of chloroform (Kb = 3.63°C/m), increasing the boiling point of chloroform by 2.45°C. Eugenol is 73.17% C, 7.32% H, and 19.51% O by mass. What is the molecular formula of eugenol? A. C15H18O3 B. C5H6O C. C10H12O2 D. C7H15O
Eugenol is one of the compounds responsible for the flavor of cloves. A 111 mg sample of eugenol is dissolved in 1.00 g of chloroform (Kb = 3.63°C/m), increasing the boiling point of chloroform by 2.45°C. Eugenol is 73.17% C, 7.32% H, and 19.51% O by mass. What is the molecular formula of eugenol? A. C15H18O3 B. C5H6O C. C10H12O2 D. C7H15O

Explanation / Answer

mass of eugenol = 111 mg = 0.111 g

mass of chloroform = 1.00 g

Kb of chloroform = 3.63 0C/m

deltaTb = 2.45 C

From elevation in boiling point,

deltaTb = Kb x m

but molality (m) = (mass / molar mass ) x 1000 / M

deltaTb = Kb x ( mass / molar mass ) x 1000/M

2.45 = 3.63 x ( 0.111 / molar mass ) x 1000 / 1

=> molar mass = 164.5 g/mol

Eugenol is 73.17% C, 7.32% H, and 19.51% O by mass

moles of C = 73.17 / 12 = 6.1

moles of H = 7.32 / 1 = 7.32

moles of O = 19.51 / 16 = 1.22

Simple whole number moles ratio of C,H,O

=> C:H:O = 5:6:1

Therefore emperical formula is C5H6O.

Emperical formula mass = (5x12)+(6x1)+(1x16) = 82

molar mass = 164.5 g/mol

Therefore , Molecular formula is C10H12O2

Answer is option (c)

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