OCR Chemistry: Trends and Patterns module

jseldis

No, it wasn't. At least i'm almost 99% sure it wasn't :biggrin:

Even if it was you'd definately get ECF.

Was i right on the Electronic configuration one?

Ok after raking my memory I have remembered what it was:
2FeSO4 -> SO2 + SO3 + Fe2O3

Therefore there is a 2:1 ratio of FeSO4 : the gas. To work the moles of FeSO4 were twice that of the gas. Which is also equal to moles of FeSO4.7H2O.

Now just work out the number of moles of FeSO4.7H2O that there would be in the pure sample. Divide these two and multiply by 100 to get a percentage which was about 76/77 !

I cant remember the given masses or volume of gas but if someone remembers post them and I will give the proper solution!

Reply 241

Nazmoth

Good news I had values stored on calculator- question was:
2FeSO4 -> SO2 + SO3 + Fe2O3
92.5cm3 of so2 were formed when 2.784g of FeSO4

Experimental number of moles of FeSO4.7H2O:
. calculate moles of so2- (92.5)/(24x1000)=3.85x10^-3mol
. ratio feso4: so2 = 2:1 => 7.71x10^-3mol of FeSO4 or FeSO4.7H2O

Number of moles of FeSO4.7H2O in 2.784g of pure
. moles=mass/mr=2.784/277.9=0.0100mol

Percentage purity=(7.71x10^-3/0.01)x100 = 76.9%

Let me know if I have done anything wrong or if you agree!

Reply 242

Otley

Nazmoth

Good news I had values stored on calculator- question was:
2FeSO4 -> SO2 + SO3 + Fe2O3
92.5cm3 of so2 were formed when 2.784g of FeSO4

Experimental number of moles of FeSO4.7H2O:
. calculate moles of so2- (92.5)/(24x1000)=3.85x10^-3mol
. ratio feso4: so2 = 2:1 => 7.71x10^-3mol of FeSO4 or FeSO4.7H2O

Number of moles of FeSO4.7H2O in 2.784g of pure
. moles=mass/mr=2.784/277.9=0.0100mol

Percentage purity=(7.71x10^-3/0.01)x100 = 76.9%

Let me know if I have done anything wrong or if you agree!

I got 77.0% because of different rounding. Could be I made the rounding error, hopefully it won't matter too much.

Reply 243

Benhancox

sounds gud 2 me
got a horrible feeling i put dwn 86.9% dwn by mistake, coz dats wot i saved in my calc
but i did the exact same method and 76.86 rings a bell 4 sum reason

Reply 244

arpeggio

12

Anyone know how many marks you'd lose if you didn't do anything about the 7H2O? I did everything else, just used FeSO4 rather than FeSO4.7H2O for some reason.

Reply 245

mc999

what did everyone put for the last question where you had to use complexes of copper only?

Reply 246

Nazmoth

Benhancox

sounds gud 2 me
got a horrible feeling i put dwn 86.9% dwn by mistake, coz dats wot i saved in my calc
but i did the exact same method and 76.86 rings a bell 4 sum reason

Good to see I wasn't the only one who did it this way! A few people earlier in this thread nearly convinced me that I had done it wrong!
thanks for reassuring me!

Reply 247

Otley

mc999

what did everyone put for the last question where you had to use complexes of copper only?

I explained co-ordinate bonding, defined a ligand, said that complex ions are usually tetrahedral or octohedral and drew [Cu(H2O)6]2+ including bond angle of 90 degrees. I then explained ligand substitution and did:
[Cu(H2O)6]2+ + 4Cl- -----------> [CuCl4]2- + 6H2O.
Then I said [Cu(H2O)6]2+ = light blue, [CuCl4]2- = yellow.

Reply 248

Jarve99

3

I did the NH3 one, instead. Hopefully that's ok. I think I got nearly every mark on the last question, just got quite a few earlier on wrong. My enthalpy value thing was negative, which I've heard is wrong.

Reply 249

1mhward

1

Hey, just wondering how many marks you think ill get for the copper coimplexes question if i wrote everythning write, but put Cu+ insetead of 2+?? Will I just be penalised 1? or for everything were i wrote it

Reply 250

Otley

1mhward

Hey, just wondering how many marks you think ill get for the copper coimplexes question if i wrote everythning write, but put Cu+ insetead of 2+?? Will I just be penalised 1? or for everything were i wrote it

It depends where you used it. If it was in the ligand substitution then you'd loose a mark. If it was also in the structure or whatever then you might loosr another mark too.