Help, coursework coming up next week (long post sorry). Watch

Deavsie
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Okay, basically we did a practical in school last week, but our teacher has a habit of just jumping into the experiment rather than explaining anything.

I think we're supposed to be working out the percentage of the mass of FeSO4 in lawn sand.

This is what I've got written down so far, so can someone just check and see if what I'm doing is right?

We put KMnO4 in a burette and titrated it into a mixture of lawn sand & H2SO4.
It's just the calculations I'm not sure about.
I've done half equations and ended up with this final equation:
\mathrm{MnO_4}^- + \mathrm{5Fe}^{2+} + \mathrm{8H}^+ \to \mathrm{Mn}^{2+} + 4H_2O + \mathrm{5Fe}^{3+}

So we put 50cm^3 of 0.02M KMnO4 in burette - aiming for 25 cm^3 as an endpoint.

I worked out the number of moles of KMnO4 = 0.0005
So the number of moles of Fe2+ = 0.0025 right?

Mass of Fe2+ / Fe2SO4 = moles X Mr,
= 0.0025 X 151.9
= 0.37975

I've been told to assume 6% of the mixture is iron sulphate.
So I need to weigh out around 6.33...g for the practical?

Now I get confused, I'm supposed to add H2SO4 to dissolve the lawn sand in,
because I have written down from the lesson,
"Number of moles of manganate = 0.0005 X 8 (H+)
= 0.004 (divide by 2, min volume needed = 2cm^3)
= 0.002."

Why have I done that last paragraph?!
Can someone explain?

Again sorry about the long post.
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charco
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I guess you are just tring to decide how much sulphuric acid to use. In reality, the potassium manganate (VII) is usually made up on large excess of acid so it doesn't really matter.
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ClaireHogben
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in reality you will dissolve in an excess of sulphuric acid, but in our coursework we had to work out the minimum volume of H2SO4 needed (ie to prove to the examiner that your method will work because you're using more than this).

this might be what you're looking for, so think about what calculations you might need (im not giving you all the answers obv ). Come back if you're stuck.

hope thats what you're looking for. I did this about 3 weeks ago with AQA
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Deavsie
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(Original post by ClaireHogben)
in reality you will dissolve in an excess of sulphuric acid, but in our coursework we had to work out the minimum volume of H2SO4 needed (ie to prove to the examiner that your method will work because you're using more than this).

this might be what you're looking for, so think about what calculations you might need (im not giving you all the answers obv ). Come back if you're stuck.

hope thats what you're looking for. I did this about 3 weeks ago with AQA
Yeah I think I am trying to work out the min volume needed,
and I think I get the rest of it,
but could you just explain why I need to multiply 0.0005 by 8,
and then divide by 2?
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ClaireHogben
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(Original post by Deavsie)
Yeah I think I am trying to work out the min volume needed,
and I think I get the rest of it,
but could you just explain why I need to multiply 0.0005 by 8,
and then divide by 2?
hmmmm i'll try....

MnO4- and H+ react in the ratio 1:8, so thats where the x8 bit comes from.
this gives you the moles of H+ needed

each mole of H2SO4 contains 2 H+s, so divide by 2

then re-arrange n = M.V to work out the volume of H2SO4 needed i would suggest.

(i'm assuming here that your answers are correct, i havent checked them)

hope this helps
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Deavsie
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(Original post by ClaireHogben)
hmmmm i'll try....

MnO4- and H+ react in the ratio 1:8, so thats where the x8 bit comes from.
this gives you the moles of H+ needed

each mole of H2SO4 contains 2 H+s, so divide by 2

then re-arrange n = M.V to work out the volume of H2SO4 needed i would suggest.

(i'm assuming here that your answers are correct, i havent checked them)

hope this helps
Arrrgh how did I not get that.
I got that H2SO4 was diprotic just completely forgot about the earlier equation :P

Thanks a lot +rep
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ClaireHogben
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no problem. i know how awful chemistry coursework is so im glad to help!

which reminds me i think i've got more of it next week too
why do we do it to ourselves? :rolleyes:
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Deavsie
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(Original post by ClaireHogben)
no problem. i know how awful chemistry coursework is so im glad to help!

which reminds me i think i've got more of it next week too
why do we do it to ourselves? :rolleyes:
Haha no idea.
Oh well only another 4 years of it to go for me :P
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ClaireHogben
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(Original post by Deavsie)
Haha no idea.
Oh well only another 4 years of it to go for me :P
same kind of. I'm doing biochem, which is mostly organic chemistry :eek:
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Deavsie
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(Original post by ClaireHogben)
same kind of. I'm doing biochem, which is mostly organic chemistry :eek:
The best bit IMO
And congrats on the Oxford offer
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Samera353
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Im currently doing the same coursework.

i am stuck on how to work out how much lawn sand to use?

Any help would be grately apreciated
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