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May 4, 2016
Posted by **Desz** on Tuesday, August 6, 2013 at 9:26pm.

- Chemistry -
**bobpursley**, Tuesday, August 6, 2013 at 9:47pmI dont 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. - Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 9:57pmNo. Not even close. If you will post your work I will find the error. I suspect you didn't obtain the correct molar mass of 92.

- Chemistry -
**Desz**, Tuesday, August 6, 2013 at 9:57pmI'm getting a little confused..would you mind telling me how you would solve this problem? Thank you.

- To Bob P---Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 10:00pmoops! 0.52 should be 186

- Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 10:01pmThat's 1.86

- To Desz--Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 10:03pmPost your work and we can help you through it.

- Chemistry -
**Desz**, Tuesday, August 6, 2013 at 10:06pmI actually did find 92 for my molar mass, but for the molar mass of the molecular formula I found 1666.67 and so I did 1666.67/92 which equals to 18. I then multiplied my empirical formula, C3H8O3 by 18 and got the answer C54H144O54.I found the molar mass of the molecular formula by finding molality: .0894=m x 1.86 and I got m=.0480645612. I then multiplied .04806 m with .025 kg to get moles and I got moles=.00120. To get molar mass, I did 2g/.00120moles and got 1666.67.

- Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 10:12pmOK. Everything you did with the freezing point data is correct and the molar mass actually is 92 from that data. Where you went wrong is in the 2.00 g. That is the mass of the CHO compound.

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. - Chemistry -
**Desz**, Tuesday, August 6, 2013 at 10:17pmI actually did all that already.. Do you want me to post how I did it? And for the 2.00 g, don't I need the mass of the CHO compound and divide it by the moles to get the MM of the molecular formula? And then I do MM of molecular/MM of empirical to get the multiplying factor to find the molecular formula? Thank you so much for all the help!

- Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 10:20pmIf you did all of that already, what did you find for the empirical formula? I don't see that anywhere in your respone and that's the key to the problem.

- Chemistry -
**Desz**, Tuesday, August 6, 2013 at 10:22pmI found C3H8O3for my empirical formula, that's where I got the 92.

- Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 10:32pmI think you have the solution in the palm of your hand and you don't recognize it. No, that isn't where you got the 92. You obtained the 92 from the freezing point data. That gives you the molar mass of 92. Then the empirical formula is C3H8O3 from the combustion data.

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? - Chemistry -
**Desz**, Tuesday, August 6, 2013 at 10:37pmOhhhh I get it now.. Thank you so much!!! One last question, how would you calculate the 92 from the freezing point data? I understand everything now, but just not that one point.

- Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 10:53pmTwo points here.

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. - Chemistry -
**Desz**, Tuesday, August 6, 2013 at 11:00pmOhhhhhh I understand now.. I thought that I did the freezing point wrong since you said I got the wrong answer.. But I understand now. THANK YOU SO MUCH! Sorry for taking up so much of your time!

- Chemistry -
**DrBob222**, Tuesday, August 6, 2013 at 11:07pmGo back and read the thread. I said your freezing point data was correct. Don't worry about the time. What's important is that you understand it. We get paid the same whether we spend 1 min or 1 hour. All of us on this board are volunteers.