The Student Room Group
1.0g sulphur = 1/32 moles =0.03125 moles

S + NH3 --> H2S + NxSy (no stoichoimetry)

H2S occupies 420cm3 = 420/22499 moles = 0.01875

relationship between moles S and moles H2S = 0.03125/0.01875 =1.67 = 5/3

so for every 5 moles S there are 3 moles H2S created

5S + NH3 --> 3H2S + NxSy


5S + 2NH3 --> 3H2S + N2S2

so the solid has the formula N2S2
number of mols initial S = mass/Ar = 1/32 = 0.03125
no. mol H2S = 0.420/22.4 = 0.01875
no. mol S remaining = 0.03125-0.01875 = 0.0125
no. mol H used from NH3 = 0.01875*2 = 0.0375
no. mol NH3 reacted = 0.0375/3 = 0.0125

now theres a "proper" way of working out empirical formulae, but we can just see that the no. mol NH3 reacted = no. mol S remaining.

So that means the ratio of N:s-smilie: in the remaining solid is 1:1 (since we arent told that any Nitrogen gas escapes etc) so the empirical forumla is NS...

I hope thats all right...


edit: charco you bastard, you beat me - but we got the same result it seems. Different method though, i cant quite follow it.
Reply 3
u didnt get same answers

rock u got NS
charco got N2S2

which one is righ6t?
rock worked out the empirical formula while I worked out the actual formula - so they're both right (in a sense)
Deduce the empirical fomula of the solid?

charco - are you sure youre able to work out the actual formula from the data provided? Ill look at your method again but I cant understand it

EDIT: it seems to make sense now charco - Ive never seen that process before.

qwert001 - if the question wants the empirical formula, u should probably use the lowest ratio which is 1:1
Reply 6
oh thanks

just woundering in a exam if u do a method which is "strange" but u get the right answer u wont lose marks or anything wud u?
check past papers and their answers. Ofton there are marks for working, so u may have to show certain key steps. Sometimes if theres not many marks available, it might just give u full marks for a correct answer (usually in 2 mark Qs, where you could 1 for working even if your answer was wrong)

but check your board's past papers
you are quite corrrect - I assumed that the compound NS is highly improbable as nitrogen would have to have an oxidation state of -II and sulphur +II. (Nitrogen is more electronegative than sulphur and would have to have a negative oxidation state, nitrogen's only common negative state is -III, but sulphur never forms a +III state)

However, a compound containing two nitrogens and two sulphurs would leave more room for possibilies.
However, a compound containing two nitrogens and two sulphurs would leave more room for possibilies.

that may be so (I dont know), but still the *empirical* ratio is properly stated as 1:1, rather than multiples of that. Its important for the thread starter because in an exam, if they want the empirical formula, they may not accept anything other than the lowest ratio
RE you're quite righty - I'm agreeing with you!
well i disagree!
I'll tell my Dad - he's bigger than yours
lol who neg repped me for this thread with "crap at chem"???

was it you charco.. o_0
It wasn't me guv, I didn't do it - I was in bed at home - ask me mum...
... besides I don't know how to!

while we're here, how do you go about getting 4 warnings and a ban? seems pretty harsh!
ive been banned twice actually, im so cool

the warnings are temporary, they come and go (if u get over 15 at once u get banned I think)

basically for looking at mods the wrong way, and not including enough :hugs:'s and :tsr:'s in my posts