Question

The sugar fructose contains 40.0%, 6.7% H, and 53.3% O by mass. A solution of 11.7g of fructose in 325g of ethanol has a boiling point of 78.59 degrees Celsius. The boiling point of ethanol is 78.35 degrees Celsius, and Kb for ethanol is 1.20 degrees Celsius/m. What is the molecular formula of fructose?

Answer 1

With the given data the first step is to calculate the empirical formula of fructose: You have to change mass % to moles assuming a sample of 100 g in this way:

atomic weights:

C= 12.01 g/mol

H= 1.01 g/mol

O = 16.00 g/mol

1) Change g to moles:

moles of C= 40/12.01g/mol = 3.33 moles C

moles of H = 6.7/1.01g/mol = 6.63 moles H

moles O = 53.3/16.00 g/mol = 3.33 moles O

We can write an empirical formula as : C_3.33 H_6.63 O_{3.33 .} Dividing by 3.33

CH₂O (empirical formula)

We have to calculate the molar mass of this formula:

(12.01 x 1) + ( 2x1.01) + 16 = 30.03 g/mol

The increase in the boiling point is given by:

ΔT_b = K_b · m where m = molality so

ΔT_b = K_b · moles fructose/mass ethanol = ΔT_b = K_b · (mass fructose/molar mass)/mass ethanol

solving for molar mass fructose:

Molar mass fructose = mass fructose xKb /ΔT_b x mass ethanol =

molar mass fructose = 11.7 g x 1.20 / ( 78.59-78-35) x0.325 = 180 g/mol remember to use mass of ethanol in Kg.

Dividing the molar mass of the substance by the molar mass of the molar mass of the empirical formula =

180/30.03 = 6

The molecular formula of fructose is 6 x (CH₂O) = C₆H₁₂O₆