Question

A carbohydrate contains 39.56% C, 7.75% H, and 52.70% O by mass. A solution of 11.3 g of this carbohydrate in 375 g of ethanol has a boiling point of 78.49 deg C. The boiling point of ethanol is 78.29 deg C, and Kb for ethanol is 1.23 deg C*kg/mol. What is the molecular formula of this carbohydrate?


So this is what I have so far:

C: 39.56/12.01 = 3.2939 mol = 3.2939/3.2937 = 1.00

H: 7.75/1.008 = 7.6885 mol = 7.6885/3.2937 = 2.33 = 2

O: 52.70/16 = 3.2937 mol = 3.2937/3.2937 = 1

So the empirical formula is CH2O, and the formula mass is 30.026 g/mol

Then,

(78.48-78.29)/1.23 = 0.1626016 mol/kg
n=0.1626016 mol/kg*375*(1kg/1000g) = 0.0609756 mol
M=11.3g/0.0609756 mol = 185.32 g/mol
185.32/30.026 = 6.172 = 6 formula units per molecule

Meaning the molecular formula should be C6H12O6. But the homework program I'm using says this is incorrect. Can you help me see where I might have gone wrong? Thanks

Answer 1

Nevermind, I got it.

I totally forgot about multiplying empirical formulas to the nearest whole number. It's actually: C3H7O3

Which makes the formula mass 91.086

Which makes the molecular formula C6H14O6

Answer 2

Frankly, I don't understand your reasoning that goes from the wrong answer to the right answer; however, your final answer is correct.