Analysis of a compound indicates that it is 49.02% carbon, 2.743% hydrogen, and 48.23% chlorine by mass. A solution is prepared by dissolving 3.150 grams of the compound in 25.00 grams of benzene, C6H6. Benzene has a normal freezing point of 5.50degreeC and the solution freezes at 1.12degreeC. The molal freezing point constant,kf, for benzene is 5.12C/molal.

Question

1.Find the empirical formula of this compound.
2.Using the freezing point data calculate the molar mass of the compound.
3. Calculate the mole fraction of benzene in the solution.
4. The vapor pressure of benzene at 35degreeC is 150.0mmHg. Calculate the vapor pressure of benzene over the solution described in this problem at 35degreeC.

Answer 1

Part 1.
Take a 100 g sample. That will give you
49.02 g C, 2.743 g H, and 48.23 g Cl.
Convert grams to moles.
49.02/12.01 = ??
2.743/1.008 = ??
48.23/35.45 = ??
Find the mole ratio of the elements to each other. The easy way to do this is to divide the smallest number by itself (thus assuring that will be 1.00), then divide the other two numbers by the same small number. Round to whole numbers. That will give you the empirical formula.

part 2.
delta T = Kf*molality
solve for molality.
m = moles/kg solvent.
Solve for moles.
moles = grams/molar mass
solve for molar mass.

part 3.
mole fraction benzene = moles benzene/total moles. You have grams benzene which can be convert to moles and you have moles of the unknown compound.

part 4.
Pbenzene = Xbenzene*Pnormal benzene.
post your work if you get stuck.

Answer 2

1) C=3 ;H=2 and Cl=1
so the empirical formula of the compound is C3H2Cl.

I seem not to get the others clear.

Answer 3

C=3 ;H=2 and Cl=1. Therefore the empirical formula of the compound is C3H2Cl

The others don't seem clear to me