Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Rutherford on February 20, 2013, 12:13:57 PM
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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.
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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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It is attached, but I can't see the other stereoisomer ???.
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Google "trimer" - what does it mean?
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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?
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Yes, well done - it's called sulfanuric chloride trimer.
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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?
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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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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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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 ?
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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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Okay, N is -3. Is NH3 produced?
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May be BaF2 is a precipitate
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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.
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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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Good, thanks for the help. There is one last question considering this problem:
One of the substances isoelectronic to Y transforms in the presence of water traces into polymer Z. 1.00 g of Z was dissolved in water and the resulting solution was added to the excess of barium acetate solution. The precipitate with the mass of 2.91 g was formed. Determine the formula of Z and draw its structure.
Isoelectronic - it could be a compound where an element from the ring is replaces with another element or group. I thought that the isoelectronic substance contains -OH groups instead of fluorine. The hydrolysis would be:
(SNOOH)3+6H2O :rarrow: 3H2SO4+3NH3
When barium is added, I got that the mass of the precipitate is 2.88g. Now the difference is slightly bigger, so I suspect if my solution is right. Is it? If yes what would be the formula of the polymer?
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I assumed now that 2 fluorine atoms are replaced with NH2 groups and one with OH group, and I got the correct mass. Still can't draw the formula of the polymer :-\.
EDIT: I got now a rush of ideas >:D, and I think that the molecule before polymerization is the one attached. When polymerization occurs, amide linkages are made. This could be it, only when hydrolysis ocured what was made from -CH3? C oxidation number is -2. Is ethene made?