[nortorius]

The question: A compound containing only boron, nitrogen, and hydrogen was found to br 40.3% boron, 52.2% nitrogen, and 7.5% hydrogen by mass. If 3.310 g of this compound is dissolved in 50.00 g of benzee, the solution produced freezes at 1.30 °C. If Kfinal for benzene is 5.12 °C m^-1 and the freezing point of pure beneze is 5.48 °C, what is the molecular weight of this compound?

My answer: First we find the mass that each element takes up in the compound..

(B) 3.301 x 0.403 = 1.33 g
(N) 3.301 x 0.522 = 1.72 g
(H) 3.301 x 0.075 = 0.25 g

than we find the amount of moles for each, resulting in..

(B) 0.123 mol
(N) 0.123 mol
(H) 0.25 mol

ntotal = 0.496 mol

finally, we use the equation M = m/n to get the molecular weight..

3.301 g / 0.496 mol = 6.66 g mol^-1

Is this right? It seems a little too easy. Thanks in advance for any help.

[Dan]

It's not right. You can tell easily because your molecular weight (6.66 g/mol) is less than that of B or N, so any compound with molecular weight 6.66 g/mol could not possibly contain any B or N.

Your error is to assume that adding up the no. of moles of each constituent atom gives the number of moles of the compound - it doesn't work like that.

eg. If I say that a sample of water has

2g H
16g O

and therefore:

2 mol H
1 mol O

it is incorrect to say we have 3 mol of water. We know water is H₂O, and so we have 1 mol water.

The ratio of the molar quantities B:N:H gives you the empirical formula (what is it?), but not the molecular formula (however, from the empirical formula there is one compound it is very likely to be, but it's not enough evidence).

Can you double check that it is 50g of benzene, because I think it should be 50mL?

To do this Q, you need to determine the concentration of the solution by using the data for the depression of freezing point. (you have been given the cryoscopic constant, Kfinal) I suggest Atkins' Pysical Chemistry (p176-7 in 7th edition) if you can't find how to do this in a book you have to hand.
Once you have the conc. you can determine the number of moles of the compound in 3.310g and from that follows the molecular weight.
Then using your empirical formula and molecular weight, you can suggest a molecular formula and a structure.

[nortorius]

Just rechecked, and it does say 50.00 g of benzene. But can't one make the assumption that 1 g = 1 mL, or does that only work for water because of its density?

Reading what you said, I have to use the equation deltaT_f = (K_f)(m)

so..

deltaT_f = 5.48 °C - 1.30 °C
deltaT_f = 4.18 °C

K_f = 5.12 °C m^-1

m is the molality of the solute (the moles of solute in one kilogram of solvent). Can I just solve the equation to find the unknown (m)? Thanks again.

[Dan]

Just rechecked, and it does say 50.00 g of benzene. But can't one make the assumption that 1 g = 1 mL, or does that only work for water because of its density?

OK, I didn't realise m in the equation highlighted below was in mol/kg solvent, I assumed it was mol dm^-3, my mistake, it should be g not mL.
The density of benzene is ~0.9 g/mL by the way, so it was throwing off my calculations when I was assuming m was in mol dm^-3.

Reading what you said, I have to use the equation deltaT_f = (K_f)(m)

m is the molality of the solute (the moles of solute in one kilogram of solvent). Can I just solve the equation to find the unknown (m)? Thanks again.

Yes

[nortorius]

deltaT_f = (K_f)(m)
(5.48 °C - 1.30 °C) = (5.12 °C m^-1)(m)
4.18 °C / 5.12 °C m^-1 = m
m = 0.8164 mol kg^-1

so now that I have the concentration I should be able to use C = n/V to solve for the amount of moles, and subsequently the molecular weight (would the volume be 50.00g or 53.301g?)

0.8164 mol kg^-1 = n/53.301 g
0.8164 kg^-1 / 0.053301 kg = n
n = 15.32 mol

... errr the molecular weight would be way off. What am I doing wrong?

[nortorius]

nevermind, got the correct answer (80.9) using the formula..

deltaT_f = (K)(m_solute (in grams)) / (M_solute)(m_solvent (in kilograms))