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.