[Borek]

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.

[Dan]

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.

[CrazyAssasin]

A,B - PCl2F3
C - PCl5
D - PCl4F
E - PF5
F - PCl4F
While speaking about bonding type all of them are covalent.

[Dan]

While speaking about bonding type all of them are covalent.

What's the difference between A/B and D/F then?

[Borek]

Determining the formulas wasn't hard, but apparently further details are harder... Although some simple logic should help.

[Big-Daddy]

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")

[Borek]

Yes.

[Borek]

Technically there is still no correct answer...

[CrazyAssasin]

So, I think that B and F should be ionic compounds, because their intermolecular forces increases, while all the others are covalent.

[Borek]

Care to list ions?

These are really interesting compounds, quite surprising.

[CrazyAssasin]

I give up, I can't understand how those compounds turn into ionic

[Dan]

Hint:

[DrCMS]

Hint:

Very nicely hinted Dan - the correct answer without being obvious.

[Rutherford]

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.

[Borek]

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.