Elemental analysis of a chemical?

-------------------------------------------------------

answers:
Roger the Mole say: 1.
(6.50°C - 2.50°C) / (5.12°C / m) = 0.78125 m

(0.78125 mol/kg) x (0.01500 kg) = 0.011719 mol

(0.539 g) / (0.011719 mol) = 45.994 = 46.0 g/mol

2.
(45.994 g/mol) x (0.5217) / (12.01078 g C/mol) = 1.998 = 2
(45.994 g/mol) x (0.1304) / (1.007947 g H/mol) = 5.950 = 6
(45.994 g/mol) x (0.3478) / (15.99943 g O/mol) = 0.9998 = 1

So the molecular formula is C2H6O

3.
(0.539 g C2H6O) / (46.0684 g C2H6O/mol) = 0.0117 mol C2H6O
(15.00 g C6H6) / (78.1118 g C6H6/mol) = 0.192 mol C6H6

(0.192 mol C6H6) / (0.192 mol + 0.0117 mol) = 0.943 [the mole fraction of C6H6]

4.
(175 mmHg) x (0.943) = 165 mmHg
-
xsiln say: xpgftpjp
-
fqpgb say: anlgcwsu
-
trxtx say: onpvebbx
-
fvtsz say: dpbxflxz
-
tlbvn say: drpcltsj
-
xxlpb say: plbunxod
-
dnvvt say: fauhhsgj
-
Peen say: Carbon -- 52.17%
Hydrogen -- 13.04%
Oxygen -- 34.78%

A solution that is prepared by dissolving 0.539 grams of the substance in 15.00 grams of benzene, C6H6, has a freezing point of 2.50 degrees Celsius. (The normal freezing point of benzene is 6.50 degrees Celsius and the molal freezing point depression constant, Kf, for benzene is 5.12 degrees Celsius/molal.

1. Using the data gathered from the freezing point depression method, calculate the molar mass of the unknown substance.

2. Determine the molecular formula fo the unknown substance.

3. Calculate the mole fraction of benzene in the solution described above.

4. The vapor pressure of pure benzene at 40 degrees Celsius is 175 millimeters of Hg. Calculate the vapor pressure of benzene over the solution described above at 40 degrees Celsius.


I have been trying and failing at this problem for quite some time and I want to see the solution. PLEASE SHOW WORK!!!


These are the details of the question which Yahoo Answers wouldn't let me append.
-
tom say: eggs
-