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ri0t · Aug 8, 2009, 04:34 PM

An extremely useful colourless, volatile liquid compound is usually synthesised by the reaction of sulfur dioxide with phosphorus pentachloride, or by reaction of sulfur dichloride

When 0.1508 g of this compound , which comprises 13.45% oxygen by mass, was slowly added to 50 ml of water an acidic mixture was formed, which reacted with excess silver nitrate solution to form a white precipitate. After filtration, washing and drying this precipitate was found to have a mass of 0.6334g. This white precipitate turned grey upon exposure to sunlight

this is what I done so far:
the white precipitate is Silver chloride (% composition of chlorine = 24.74%)
therfore the amount of Cl => 24.74% x 0.3634 = 0.08989

therefore the % of Cl in the compound => 0.08989/0.1508 x 100 = 59.61 %

I am stuck here and do not know how to work out the empirical formula

Unknown008 · Aug 9, 2009, 05:17 AM

You need to know what other element you have in the compound. I've never met such a reaction, but I'm pretty sure the other element is sulfur. Can you now find the empirical formula?

To do this, make three columns, with headings Oxygen, Chlorine and Sulfur.
Under oxygen, write its percentage composition, and do the same for the others.
Write the mass of each element under the percentages.
Divide the percentages by the masses.
Finally, with the values you obtained, divide them by the smallest value you just obtained from the three.

The final number gives you the number of elements present in the compound in its simplest ratio.

Hope it helped!

Post the empirical formula you obtained.

ri0t · Aug 10, 2009, 01:41 AM

Thank you for that,

I got an answer of SOCl_2

Unknown008 · Aug 10, 2009, 09:09 AM

Ok, I'll do it.

1. Put the percentages under their corresponding elements. (Got 12.94 from 100% - (13.45%+59.61%))
2. Put their respective atomic mass uderneath.
3. Divide each percentage by atomic mass.
4. Divide each resulting value by the smallest of the three just obtained.
5. That means there is 1 S, 2O and 4 Cl, or $SO_2Cl_4$

$S\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Cl \\12.94\,\,\,\,\,\,13.45\,\,\,\,\,\,\,59.61 \\ 32\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,16\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,35.5 \\ 0.41\,\,\,\,\,\,\,\,\,0.841\,\,\,\,\,\,\,\,1.679 \\ 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4$

ri0t · Aug 10, 2009, 01:54 PM

Wait Sorry, isn't the mass of Chlorine 35.453

Unknown008 · Aug 11, 2009, 06:35 AM

Ah, now, you got me!I made a typo, that should be 35.5 instead of 55.5 lol. I'll go and correct that. Thanks for pointing that out! :)