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

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

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

Based on the calculations you provided, it seems like you have approached the problem correctly. However, there might be a small arithmetic mistake in your final calculation. Let's go through the steps to check for any errors.

1. Determine the empirical formula:
- C: 39.56% / 12.01 g/mol = 3.293 mol
- H: 7.75% / 1.008 g/mol = 7.6885 mol
- O: 52.70% / 16.00 g/mol = 3.2937 mol

Divide all the values by the smallest number (3.293) to get whole number ratios:
C: 3.293 / 3.293 ≈ 1
H: 7.6885 / 3.293 ≈ 2.33 ≈ 2
O: 3.2937 / 3.293 ≈ 1

Therefore, the empirical formula is CH2O.

2. Calculate the molar mass of the empirical formula:
Molecular weight of C = 12.01 g/mol
Molecular weight of H = 1.008 g/mol
Molecular weight of O = 16.00 g/mol

Molar mass of CH2O = (12.01 * 1) + (1.008 * 2) + (16.00 * 1) = 30.03 g/mol

Based on these calculations, CH2O has a formula weight of 30.03 g/mol.

3. Calculate the number of moles of the carbohydrate:
(78.49 - 78.29) / 1.23 = 0.1626 mol/kg
n = 0.1626 mol/kg * 0.375 kg = 0.0609 mol

4. Calculate the molar mass of the carbohydrate in grams:
molar mass = mass / moles = 11.3 g / 0.0609 mol ≈ 185.46 g/mol

Now, check if there was an arithmetic error in the last calculation. I obtained 185.46 g/mol, which is slightly different from the value you found (185.32 g/mol). This discrepancy may be the reason why the homework program marked your answer as incorrect.

Try using the corrected molar mass (185.46 g/mol) and divide it by the empirical formula weight (30.03 g/mol) to obtain the final ratio:

185.46 g/mol / 30.03 g/mol ≈ 6.175 ≈ 6

So, the molecular formula of the carbohydrate should be C6H12O6.

Make the necessary adjustment and redo the calculation with the updated molar mass value. Hopefully, this will give you the correct answer according to the homework program.