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.
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.
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.
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?
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.