# Chemical Forums

## Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on November 12, 2012, 10:12:08 AM

Title: Problem of the week - 12/11/2012
Post by: Borek on November 12, 2012, 10:12:08 AM
The boiling point of a substance A is 7°C. A is unstable, and if left turns into a compound B, which sublimates at 135°C. When heated B decomposes into a solid C (melting point 160°C), a liquid D (boiling point 67°C) and a gas E. The liquid D is not stable, and if left turns into a solid F (melting point 175°C).

Using table of molar and mass fractions determine formulas of all compounds, write all reaction equations, tell which compounds are ionic and which are covalent.
Title: Re: Problem of the week - 12/11/2012
Post by: Dan on November 12, 2012, 05:55:27 PM
That's a cool question. A nice demonstration of the chemistry of these compounds - I am somewhat amazed that I could recall it from the recesses of my brain. Hasn't been all that long though.
Title: Re: Problem of the week - 12/11/2012
Post by: CrazyAssasin on November 13, 2012, 01:23:25 PM
A,B - PCl2F3
C - PCl5
D - PCl4F
E - PF5
F - PCl4F
While speaking about bonding type all of them are covalent.
Title: Re: Problem of the week - 12/11/2012
Post by: Dan on November 14, 2012, 02:52:30 AM
While speaking about bonding type all of them are covalent.

What's the difference between A/B and D/F then?
Title: Re: Problem of the week - 12/11/2012
Post by: Borek on November 15, 2012, 04:27:35 PM
Determining the formulas wasn't hard, but apparently further details are harder... Although some simple logic should help.
Title: Re: Problem of the week - 12/11/2012
Post by: Big-Daddy on November 17, 2012, 05:08:32 PM
The boiling point of a substance A is 7°C. A is unstable, and if left turns into a compound B, which sublimates at 135°C. When heated B decomposes into a solid C (melting point 160°C), a liquid D (boiling point 67°C) and a gas E. The liquid D is not stable, and if left turns into a solid F (melting point 175°C).

Using table of molar and mass fractions determine formulas of all compounds, write all reaction equations, tell which compounds are ionic and which are covalent.

The dashes indicate 0% content, correct? (Not "withheld knowledge")
Title: Re: Problem of the week - 12/11/2012
Post by: Borek on November 17, 2012, 05:09:50 PM
Yes.
Title: Re: Problem of the week - 12/11/2012
Post by: Borek on November 19, 2012, 11:50:48 AM
Technically there is still no correct answer...
Title: Re: Problem of the week - 12/11/2012
Post by: CrazyAssasin on November 19, 2012, 01:04:39 PM
So, I think that B and F should be ionic compounds, because their intermolecular forces increases, while all the others are covalent.
Title: Re: Problem of the week - 12/11/2012
Post by: Borek on November 19, 2012, 02:32:57 PM
Care to list ions?

These are really interesting compounds, quite surprising.
Title: Re: Problem of the week - 12/11/2012
Post by: CrazyAssasin on November 19, 2012, 03:07:20 PM
I give up, I can't understand how those compounds turn into ionic
Title: Re: Problem of the week - 12/11/2012
Post by: Dan on November 19, 2012, 06:10:48 PM
Hint:

Title: Re: Problem of the week - 12/11/2012
Post by: DrCMS on November 20, 2012, 04:15:15 AM
Hint:

Very nicely hinted Dan - the correct answer without being obvious.
Title: Re: Problem of the week - 12/11/2012
Post by: Rutherford on November 26, 2012, 12:42:10 PM
I hope that it is okay that I ask something again:
When determining the molar masses, I used compound C. I used the molar shares to get the formula XY5. Using mass shares, I got two equations, which are identical (x is the molar mass of X and y is of Y):
x/(x+5y)=0.1487
5y/(x+5y)=0.8513
Solving one of them: x=0.873y. When I put 35.5g/mol for y I get 31g/mol for x, but this is done on assumption. Can I calculate x and y without assumpting the molar masses?

I know that the question isn't solved, yet, but I am not asking that part. The part I am asking here is solved by CrazyAssasin.
Title: Re: Problem of the week - 12/11/2012
Post by: Borek on November 26, 2012, 12:49:27 PM
Solving one of them: x=0.873y. When I put 35.5g/mol for y I get 31g/mol for x, but this is done on assumption. Can I calculate x and y without assumpting the molar masses?

No, you have to do it by trial and error.
Title: Re: Problem of the week - 12/11/2012
Post by: Rutherford on November 26, 2012, 01:48:42 PM
I got an idea here. It is obvious that A/B and D/F are isomers. As the compounds are trigonal bipyramidal it was hard to find the appropriate names of these isomers, they can't be cis/trans nor mer/fac. Then I found that stereoisomers of trigonal bipyramidal molecules can be axial and equatorial, so I believe that they much differ in stability and therefore in their chemical properties. The only problem is that I don't know which form is more stable and which is covalent and ionic.
Title: Re: Problem of the week - 12/11/2012
Post by: Dan on November 26, 2012, 02:15:00 PM
The trigonal bypyramidal compounds will always have 2 axial and 3 equatorial substituents, so you can't get isomers for XY5.

The trigonal bipyramidal compound could be represented as:
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Ft1.gstatic.com%2Fimages%3Fq%3Dtbn%3AANd9GcQsK41UmQuzTXhprZXr0mmVauJ0JpbI661D1sMz_XFOQhkFQOJ_xA&hash=ea49591391aba364399ceb205aa6fb8f)

Think about the polyhedra in the hint I posted.
Title: Re: Problem of the week - 12/11/2012
Post by: Rutherford on November 27, 2012, 09:08:10 AM
I didn't mean for XY5. I meant it for compounds A/B and D/F. There is more that one possible space arrangement for XY4Z.
I can't see how the tetrahedron and the octahedron are related to the geometry of these molecules, so I will leave this.
Title: Re: Problem of the week - 12/11/2012
Post by: Dan on November 27, 2012, 09:49:35 AM
I can't see how the tetrahedron and the octahedron are related to the geometry of these molecules, so I will leave this.

Hint: empirical vs molecular.
Title: Re: Problem of the week - 12/11/2012
Post by: Rutherford on December 08, 2012, 01:01:38 PM
As no one tried this and as I am really curios about the answer, I will try it again.
You said empirical vs molecular, and on the picture there were a tetrahedron and an octahedron. I came to the idea that the molecules combine, so that B and F make molecular crystal cells which is the reason why they are solids.
Title: Re: Problem of the week - 12/11/2012
Post by: Dan on December 10, 2012, 06:12:42 AM
Quote
I came to the idea that the molecules combine

You're so close now. How could XY5 (represented as a trigonal bipyramid) combine to form a tetrahedron and an octahedron?

Hint: How many vertices do these polyhedra have?
Title: Re: Problem of the week - 12/11/2012
Post by: Rutherford on December 10, 2012, 07:53:05 AM
Trigonal bipyramid has 5, tetrahedron 4 and octahedron 6. From two trigonal bipyramids, one tetrahedron and one octahedron can be formed. If  XY5 would have to make a tetrahedron, then it will lose one Y so it will become a cation (XY4+).  If XY5 would have to make a octahedron it has to accept a Y so it will become an anion (XY6-). These two can combine to make an ionic compound XY4*XY6 or X2Y10.
Title: Re: Problem of the week - 12/11/2012
Post by: Dan on December 10, 2012, 08:35:23 AM
Correct, phosphorous(V) halides can undergo autoionisation.

Se for example: http://en.wikipedia.org/wiki/Phosphorus_pentachloride#Structure

So, can you propose structures for B and F?
Title: Re: Problem of the week - 12/11/2012
Post by: Rutherford on December 10, 2012, 08:44:42 AM
Do they overlap?
Title: Re: Problem of the week - 12/11/2012
Post by: Dan on December 10, 2012, 12:21:30 PM
Do they overlap?