a compound containing only boron, nitrogen and hydrogen was found to be 40.3% B, 52.2%N, and 7.5% H by mass. If 3.301g of this compound is dissolved in 50.00g of benzene, the solution produced freezes at 1.30 degrees celsius. If Kf(freezing point depression constant) for benzene is 5.12 degrees celsius per metre and the freezing point of benzene is 5.48 degrees celsius, what is the molecular weight of this compound?

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chem - DrBob222, Friday, November 7, 2008 at 4:24pm 
You can use the percents to determine the empirical formula. 
Take 100 g sample which will give you 
40.3 g C, 52.2 g N and 7.5 g H. Now divide each by the atomic mass of C, N, and H respectively, find the ratio in small whole numbers and you have the empirical formula. 
Now use delta T = kf*molality to determine molality. Finally, knowing molality = # mols/kg solvent, you can determine the molar mass and from that the molecular formula and from that the molar mass. 
Post your work if you get stuck.

ok i was able to find the ratio to be 1:1:2 
and i also found the molality from the equation delta T=Kf*molality to be 1.07/m. I get lost about what to do after

DrBob222 · Answer

The ratio of B1N1H2 is correct.
The molality is not.
delta T = Kf*m (by the way, Kf is 5.12 degrees C/molal [not metre].)
(5.48-1.30)=5.12*m.
molality = 0.816 but you should confirm this.
molality = # mols/kg solvent.
Solve for mols.
0.816 = # mols/0.050 = 0.04
# mols = g/molar mass
0.04 = 3.301/molar mass
molar mass approximately 81. These calculations need to be redone for I have estimated and rounded here and there. Also, it is worth noting here that molar masses determined this way are not exact, usually within 5 or so. Therefore, there must be a better method BUT this gives an approximate number to work with.
Now we go through the B,N,H think which you have done and obtained B1N1H2.
Lets do the empircal formula mass. It is
10.8 + 14 + 2 = 26.8.
So how many formula masses must be in the molecular formula. That would be 81/26.8 = 3.02, we round that to 3.000 and say the molecular formula is EXACTLY (BNH2)3 = B3N3H6. Now look up the atomic masses of B, N, and H, multiply by the subscripts and add them together to obtain the molar mass of the compound.
Summary of what we've done.
Freezing point work gives an approximate molar mass.
B,N,H analysis gives an exact EMPIRICAL formula.
We divide the approximate molar mass by the exact empirical formula mass to get a number, round that to a whole number, and multiply that whole number by each of the subscripts in the empircal formula to obtain the molecular formula. Then look up atomic masses on periodic table and determine the exact molar mass. Check my work. Check for typos. Redo the calculations to make sure the numbers are correct.