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Offline Rutherford

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Cyclic trimers
« on: February 20, 2013, 12:13:57 PM »
The interaction of thionyl chloride and sodium azide at –30°С gives colorless crystals Х, containing 36.4 wt.% of Cl. The crystals consist of cyclic trimers. Find the composition of X and give the reaction equation. Draw two stereoisomers of Х.

I assumed that there is 1 chlorine atoms, therefore: 35.5:M=36.4:100, M=97.5g/mol. For this molar mass, the formula SOClN would fit. The reaction would be: SOCl2+NaN3 :rarrow: NaCl+N2+SOClN. Now, I can't determine the two stereoisomers of this compound. I need help with this.

Offline Dan

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Re: Cyclic trimers
« Reply #1 on: February 20, 2013, 01:13:04 PM »
You were told it is trimers - i.e. the empirical formula is SOClN, so what is the molecular formula of the trimer.
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Offline Rutherford

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Re: Cyclic trimers
« Reply #2 on: February 20, 2013, 01:19:58 PM »
It is attached, but I can't see the other stereoisomer  ???.

Offline Dan

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Re: Cyclic trimers
« Reply #3 on: February 20, 2013, 01:22:29 PM »
Google "trimer" - what does it mean?
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Offline Rutherford

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Re: Cyclic trimers
« Reply #4 on: February 20, 2013, 01:31:28 PM »
Shame. Okay, I attached the new structure.

Are the isomers:
1st=when all Cl atoms are above or under the plane
2nd=when two are under or above, and the last one is dirrected in the opposite direction?

Offline Dan

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Re: Cyclic trimers
« Reply #5 on: February 20, 2013, 01:38:35 PM »
Yes, well done - it's called sulfanuric chloride trimer.
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Offline Rutherford

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Re: Cyclic trimers
« Reply #6 on: February 20, 2013, 01:54:39 PM »
Okay, but now it gets complicate:
A colorless liquid Y was prepared by a reaction between X and antimony(III) fluoride. Addition of 1.00 g of Y to the excess of barium acetate aqueous solution gave the precipitate with the mass of 3.96 g. Determine the chemical formula of Y, draw its structure and write the reaction equation.

I don't know what to assume when doing the calculation. Is 3.96-1=2.96g the mass of barium added?

Offline Rutherford

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Re: Cyclic trimers
« Reply #7 on: February 20, 2013, 02:28:11 PM »
I think now that in the reaction with the fluoride, the chlorine atoms are replaced by fluorine. Still the very big mass increase confuses me. Can't find the reaction of Y with barium acetate ???.

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Re: Cyclic trimers
« Reply #8 on: February 20, 2013, 04:17:18 PM »
I guess hydrolysis, then think about which ions in the resulting soup will precipitate barium...

Edit: Yes, that works if you assume all the barium salts are completely insoluble.

First, consider what you get from hydrolysis of (SOFN)3, then think about insoluble barium salts.
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Offline Rutherford

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Re: Cyclic trimers
« Reply #9 on: February 21, 2013, 08:23:38 AM »
Maybe: (SON)33-. Now with positively charged barium, this salt should be made: Ba3[(SON)3]2, but the calculation won't show a correct result.

What is made from hydrolysis then? Are it SO42-, NO2- and F-, but then the number of oxygen atoms is very high  ?

Offline Dan

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Re: Cyclic trimers
« Reply #10 on: February 21, 2013, 09:31:23 AM »
What is made from hydrolysis then? Are it SO42-, NO2- and F-, but then the number of oxygen atoms is very high  ?

You're close. H2SO4 and HF will be two of the hydrolysis products, but NO2- is not correct.

Look at your starting material (SOFN)3 - what is the oxidation state of each atom. Hydrolysis is not a redox reacion.
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Offline Rutherford

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Re: Cyclic trimers
« Reply #11 on: February 21, 2013, 09:37:45 AM »
Okay, N is -3. Is NH3 produced?

Offline AWK

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Re: Cyclic trimers
« Reply #12 on: February 21, 2013, 09:53:28 AM »
May be BaF2 is a precipitate
AWK

Offline Rutherford

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Re: Cyclic trimers
« Reply #13 on: February 21, 2013, 10:38:46 AM »
Assumed that the precipitates are BaSO4 and BaF2, that way I got their mass together to be 3.95g, while in the problem it is given as 3.96g. The difference is very small, so it is probably correct.

I will try now the last part of the problem.

Offline Dan

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Re: Cyclic trimers
« Reply #14 on: February 21, 2013, 11:29:51 AM »
Assumed that the precipitates are BaSO4 and BaF2, that way I got their mass together to be 3.95g, while in the problem it is given as 3.96g. The difference is very small, so it is probably correct.

Yes, this is what I think is happening too.

(SOFN)3 + 9H2O + (4.5)Ba(OAc)2  :rarrow: 3BaSO4 + (1.5)BaF2 + 3NH4OAc + 6AcOH

Your 3.95 is probably the result of a rounding error. Here is my calculation:

1.00 g (SOFN)3 => 1000/243.2 = 4.112 mmol

Therefore we expect: 4.112*1.5 = 6.168 mmol BaF2 => 6.168*175.3 = 1081 mg

And also: 4.112*3 = 12.336 mmol BaSO4 => 12.336*233.4 = 2879 mg

Total = 1081 + 2879 = 3960 mg

=> 3.96 g
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