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Chemistry help - How much Iron in a compound

If someone could help with with this i'd be so grateful...
The questions are:

•

Calculate how much Iron must have been in the mass of hydrated Iron(II) ammonium sulfate that you used.

•

Calculate the percentage of iron in the hydrated Iron(II) ammonium sulfate crystals.


This is the first task we've been set and I'm not sure if this is something simple that I should already know- please help me out asap if you can. Thanks :smile:
Reply 1
Avatar for excelcisor
excelcisor
For the first bit just work out the number of moles of Iron by using the balanced equation or work it out. Then work out the mass by doing moles times the RFM (55.8) and that should be it. For the next bit do the mass of iron over the mass of the ammonium compound x100 or moles x100
Hope this helps :smile:
Reply 2
Avatar for Srf123
Srf123
OP
Original post by excelcisor
For the first bit just work out the number of moles of Iron by using the balanced equation or work it out. Then work out the mass by doing moles times the RFM (55.8) and that should be it. For the next bit do the mass of iron over the mass of the ammonium compound x100 or moles x100
Hope this helps :smile:


Thankyou- I still think I've gone wrong. We used 5g of iron ammonium sulfate so to find the moles I did 5/55.845 ??
(edited 10 years ago)
Original post by Srf123
Thankyou- I still think I've gone wrong. We used 5g of iron ammonium sulfate so to find the moles I did 5/55.845 ??


In your experiment you have probably titrated the iron(II) using manganate(VII) ions...

.. you use the titration value to work out the moles of iron and hence the mass of iron in the 5g of ammonium iron(II) sulphate that you originally weighed out.
Reply 4
Avatar for excelcisor
excelcisor
Original post by Srf123
Thankyou- I still think I've gone wrong. We used 5g of iron ammonium sulfate so to find the moles I did 5/55.845 ??


It will be 5/the Rfm of the ammonium sulphate so fe (55.8)+ NH3 (17) + SO4 (96.1) should come out as something like 0.03 moles I think. I think that is correct
Reply 5
Avatar for Srf123
Srf123
OP
Original post by excelcisor
It will be 5/the Rfm of the ammonium sulphate so fe (55.8)+ NH3 (17) + SO4 (96.1) should come out as something like 0.03 moles I think. I think that is correct


Thankyou so much for the help, really appreciate it:smile:
Reply 6
Avatar for excelcisor
excelcisor
No problem do you know if you got it correct? :smile:
Reply 7
Avatar for Srf123
Srf123
OP
Original post by excelcisor
No problem do you know if you got it correct? :smile:

No not yet its due tomorrow

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