Chemistry
posted by Desz on .
A compound of carbon, hydrogen, and oxygen was burned in oxygen, and 2.00g of the compound produced 2.868 g CO2 and 1.567 g H2O. In another experiment, 0.1107 g of the compound was dissolved in 25.0 g of water. This solution had a freezing point of 0.0894 degrees C. what is the molecular formula of the compound? I got C54H144O54. Is this answer correct?

I don't think so.
freezingptdepress=.52*(mass/molmass)/.0251107
molemass=.52/(.1107*.0251107*.0894)=2092
That is DIFFERENT than the mole mass of the formula you have. check my work. 
No. Not even close. If you will post your work I will find the error. I suspect you didn't obtain the correct molar mass of 92.

I'm getting a little confused..would you mind telling me how you would solve this problem? Thank you.

oops! 0.52 should be 186

That's 1.86

Post your work and we can help you through it.

I actually did find 92 for my molar mass, but for the molar mass of the molecular formula I found 1666.67 and so I did 1666.67/92 which equals to 18. I then multiplied my empirical formula, C3H8O3 by 18 and got the answer C54H144O54.I found the molar mass of the molecular formula by finding molality: .0894=m x 1.86 and I got m=.0480645612. I then multiplied .04806 m with .025 kg to get moles and I got moles=.00120. To get molar mass, I did 2g/.00120moles and got 1666.67.

OK. Everything you did with the freezing point data is correct and the molar mass actually is 92 from that data. Where you went wrong is in the 2.00 g. That is the mass of the CHO compound.
First convert grams CO2 to g C.
Convert grams H2O to grams H.
Then grams O = 2.00  g C  g H.
Convert g C to mols C.
Convert g H to mols H.
Convert g O to mols O.
Then find the ratio of C,H, and O to each other. That will give you the empirical formula, THEN you do the division bit to determine the number of units of the empirical formula there are in the molecular formula. 
I actually did all that already.. Do you want me to post how I did it? And for the 2.00 g, don't I need the mass of the CHO compound and divide it by the moles to get the MM of the molecular formula? And then I do MM of molecular/MM of empirical to get the multiplying factor to find the molecular formula? Thank you so much for all the help!

If you did all of that already, what did you find for the empirical formula? I don't see that anywhere in your respone and that's the key to the problem.

I found C3H8O3for my empirical formula, that's where I got the 92.

I think you have the solution in the palm of your hand and you don't recognize it. No, that isn't where you got the 92. You obtained the 92 from the freezing point data. That gives you the molar mass of 92. Then the empirical formula is C3H8O3 from the combustion data.
So the molecular formula is (C3H8O3)_{n} where n is the number of units of the empirical formula.So
n*empirical formula mass = molecular formula mass.
n*92 = 92
Therefore, n = 1 and the empirical formula is the same as the molecular formula. Got it? 
Ohhhh I get it now.. Thank you so much!!! One last question, how would you calculate the 92 from the freezing point data? I understand everything now, but just not that one point.

Two points here.
1. You told me how you did the freezing point data but I'll summarize it here.
dT = Kf*m
0.0894 = 1.86*m
m = about 0.048
m = mols/kg solvent so
mol = m x kg solvent = 0.048*0.025 = about 0.0012 (you can do these more accurately).
Then molar mass = g/mols = 0.1107/0.0012 = about 92.1 or so. Freezing point data almost never gives the EXACT molar mass so take these numbers as estimates. Then you do the n*92 = 92.1 and n = 1.01 which you round to the nearest whole number.
2. THEN you get the REAL molar mass, as you pointed out, from the C3H8O3. That freezing point data gives you the approximate molar mass, that allows you to know how many units of the empirical formula you have, you round that number off to a whole number, then calculate the molar mass from the molecular formula. Those approximate molar mass from freezing point data gives you a ball park figures for the molar mass. 
Ohhhhhh I understand now.. I thought that I did the freezing point wrong since you said I got the wrong answer.. But I understand now. THANK YOU SO MUCH! Sorry for taking up so much of your time!

Go back and read the thread. I said your freezing point data was correct. Don't worry about the time. What's important is that you understand it. We get paid the same whether we spend 1 min or 1 hour. All of us on this board are volunteers.