Aqa chem 4/ chem 5 june 2016 thread

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Can someone pls explain question 4b from the june 2010 paper of chem4. Thanks (:

Paper : http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-QP-JUN10.PDF

Mark scheme : http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-MS-JUN10.PDF

So the 5 peaks on the M/Z are
Carbon-12
Hydrogen-1
Oxygen-16
Chlorine-35
Chlorine-37

And then the for the most abundant peak
1) get an average abundance for Cl-35 and Cl-37

So (37x25+35x75)/100

Which gives you 35.5

2)most abundant peak is the molecular ion
So basically add up the Mr

12*12 + 1*4 + 2*16 + 4*35.5 = 322.0

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Reply 241

Gezza_O'Brien

Original post by Super199

Yh if you don't mind, I've not seen a qs like this before

ImageUploadedByStudent Room1465816662.937155.jpg

I find it helps to write out the actual chemical equations so you can see what's going on

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Reply 242

sockface

can someone explain when you times by two and when you divide by two for a diprotic and dibasic molecule

Reply 243

bat_man

Original post by Gezza_O'Brien

Right, so i get the 5 peaks now. That makes sense, but still with the MR, i got 322 as well but the answer is 320.. :/

Reply 244

bat_man

Original post by Gezza_O'Brien

Oh wait nvm, it says 322.0 as well, i didnt read the mark scheme fully. thanks!!

Reply 245

Gezza_O'Brien

Original post by bat_man

Right, so i get the 5 peaks now. That makes sense, but still with the MR, i got 322 as well but the answer is 320.. :/

It says 320.0 or 322.0, the reason people get 320.0 is because they took the average of Cl as 35 because they rounded the decimal down. Don't worry 322.0 is the correct (and more accurate) answer

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Reply 246

hopingmedicinae

Original post by bat_man

The molecular formula of TCDD is C12H4O2Cl4Chlorine exists as two isotopes 35Cl (75%) and 37Cl (25%).Deduce the number of molecular ion peaks in the mass spectrum of TCDD andcalculate the m/z value of the most abundant molecular ion peak.

So we're looking at Cl which has two Isotopes 35,(75%) and 37 (25%)

So it Could be the Mr of C12H4O2 + 35 + 35 + 35+35
Or 37 + 37 + 37 +37
Or 37+37+35+35
Or 37+35+35+35
Or 35+37+37+37

Could be any combination of isotopes, 5 peaks formed
M/z of most abundant ion peak (35 is 75% abundant so) you would add Mr of C12H14O2 to (4X35) which gives you 320.0

Hope that helps!

(edited 7 years ago)

Reply 247

Sniperdon227

http://filestore.aqa.org.uk/subjects/AQA-CHEM4-QP-JUN13.PDF
http://filestore.aqa.org.uk/subjects/AQA-CHEM4-W-MS-JUN13.PDF

Q3e) stuck on the final bit worked out which was in excess, it was H2SO4, then worked out the conc of H+ (0.014928x1000/46) is this = to 2[H+] or [H]+ im confused is it equal to 2[H+] since weve already multiplied the moles by 2???

Reply 248

hopingmedicinae

Original post by RedDevil1997

Can someone please explain how this has 11 peaks in its Carbon 13 nmr?!?
Thanks in advance

Hope this helps! By the way, number 8 is weirdly back in the benzene ring, I forgot to mark it so had to go back to it!

13454121_462897353919763_167680737_n.jpg46.6KB

Reply 249

Parallex

Original post by Sniperdon227

There's 0.02904 moles of H+ (2x the moles of H2SO4) and 0.014112 moles of OH-. Excess moles of H+ = 0.014928. Concentration of H+ = 0.014928/(46*10^-3). -log(10) of this = 0.49 to 2 dp.

Reply 250

Super199

Original post by Gezza_O'Brien

I find it helps to write out the actual chemical equations so you can see what's going on

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hmm. So HY decreases so you take away.
Y- increases so you add?

Reply 251

Sniperdon227

Original post by Parallex

There's 0.02904 moles of H+ (2x the moles of H2SO4) and 0.014112 moles of OH-. Excess moles of H+ = 0.014928. Concentration of H+ = 0.014928/(46*10^-3). -log(10) of this = 0.49 to 2 dp.

yeah but its diprotic so does multiplying the moles by two in previous steps make 2[H+]

Reply 252

Parallex

Original post by Sniperdon227

yeah but its diprotic so does multiplying the moles by two in previous steps make 2[H+]

Yes.

Reply 253

shiney101

http://filestore.aqa.org.uk/subjects/AQA-CHM4-W-QP-JUN08.PDF

Can someone help me with q1di?

Reply 254

hopingmedicinae

Original post by Sniperdon227

Nope it's equal to H+, you only multliply the concentration by 2 when working out moles of H2S04 because it's got 2H+ in it

So xs moles you just do over total volume to get new [H+], then do -log[H+]

Reply 255

Gezza_O'Brien

Original post by Super199

hmm. So HY decreases so you take away.
Y- increases so you add?

Yeah exactly. So if you were to add, say, HCl instead of NaOH, then there is more H+ ions in the equilibrium that need to be used, so the equilibrium shifts to the left (to use up H+) so there is less Y- and more HY, so you take the moles of HCl from Y- and you add them to HY, then put the new moles into Kc blah blah blah, does that make sense? Tbf it's pretty confusing but you've just got to think about it in terms of what's happening to the equilibrium

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Reply 256

Sniperdon227

Original post by hopingmedicinae

yeah but that 'new [H+]' must be equal to 2[H+] surely

Reply 257

Parallex

Original post by Sniperdon227

yeah but that 'new [H+]' must be equal to 2[H+] surely

What do you mean? H2SO4 fully dissociates so the moles of H+ is 2x the moles of H2SO4. That's all it is.

Reply 258

Sniperdon227

Original post by Parallex

What do you mean? H2SO4 fully dissociates so the moles of H+ is 2x the moles of H2SO4. That's all it is.

In that question you multiply the moles of H2SO4 by two as you must since its diprotic, we find out that its in excess, by x moles, we divide xmoles by total volume this gives [H+] (or 2[H+]) -log the conc of H+ to give you your pH, whenever working out the pH of diprotic acids you must multiply by two?

Reply 259

Parallex

Original post by Sniperdon227

Yeah. You're looking at how much excess there is after neutralisation. If it's a strong acid then you have to multiply by 2 because it will fully dissociate and 2H+ is available for neutralisation.