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1. A fertiliser is known to contain ammonium sulphate (NH4)2SO4 as the only ammonium salt

a) a sample weighing 3.80g was dissolved in water and the volume made to 250 cm3. to 25.0 cm3 portions of solution 5cm3 (Xcess) of methanal added

4NH4+ + 6hcho----->C6H12N4+ 4H+ + 6h20

liberated acid titrated with 0.100 moldm-3 aq.NaOH. Average volume needed is 28.0cm3.

Calculate Ammonium sulphate in fertiliser!

(jun02)
2. (Original post by littleone)
A fertiliser is known to contain ammonium sulphate (NH4)2SO4 as the only ammonium salt

a) a sample weighing 3.80g was dissolved in water and the volume made to 250 cm3. to 25.0 cm3 portions of solution 5cm3 (Xcess) of methanal added

4NH4+ + 6hcho----->C6H12N4+ 4H+ + 6h20

liberated acid titrated with 0.100 moldm-3 aq.NaOH. Average volume needed is 28.0cm3.

Calculate Ammonium sulphate in fertiliser!

(jun02)
3. I think it is 48.6% by mass.

from equation, moles acid : moles ammonium moles is 1:1,

therefore moles NH4+ = (28 * 0.1) / 1000 = 0.0028

moles NH4+ in whole solution = 0.0028 * 10 = 0.028

ammonium sulphate is (NH4)2SO4, so moles ammonium sulphate is 0.028 / 2 = 0.014

Rmm ammonium sulphate = (14 + 4) * 2 + 32 + 16 * 4 = 132

mass ammonium sulphate = 132 * 0.014 = 1.848g

% by mass = (1.848/3.8) * 100 = 48.6% (3 s.f.)

I only just worked that out quickly, so please correct me if I'm wrong. Do you have the actual answer?
4. i got 97.3%......it is prob rong as i didnt use all the info in the qn
5. moles acid = moles alkali (1:1 ratio)

moles acid = conc x volume = 0.1 x 0.028 = 0.0028

moles NH4+ = 0.0028 (1:1 ratio)

Above is the moles in 25cm3

moles NH4+ = 0.028 (moles in 250cm3)

mass ammonium sulphate = moles x Mr = 0.028 x 132 = 3.70g (to 3 s.f.)

Percentage ammounium sulphate in fertiliser = 3.7 / 3.8 x 100 = 97.3%

that's what i got
6. yep my thinking process was the same as revenged
7. I go with Revenge...as got tha same answer....me thinx
8. actually, i think i should have divided the moles of NH4 by two because each mole of ammonia sulphate has 2 moles of NH4 in
9. (Original post by mr_shoe_uk)
ammonium sulphate is (NH4)2SO42, so moles ammonium sulphate is 0.028 / 2 = 0.014
I don't want to say I'm right before we actually know yet, but I think you guys missed out this step. If you see above for the rest of my working, I think it's the same.
10. Here is the mark scheme, the answer was 48.6%.
Attached Images

11. DAMN....lol rushed. it...lol well have to take time nxt time
12. Yay! :P

Shall we do a sort of thing where whoever gets it right first finds the next question?

So... Should I find another practise question?
13. yeah! something organic would be good
14. good idea
15. (Original post by mr_shoe_uk)
I don't want to say I'm right before we actually know yet, but I think you guys missed out this step. If you see above for the rest of my working, I think it's the same.
ye i was trying to figure out y u divided by 2 cos u sed formula for ammonium sulphate was (NH4)2SO42.........instead of (NH4)2SO4. so i was gettin baffed

well dun for right answer tho
16. Ok, just a little one, and not a calculation (this wasn't meant to be only maths based questions was it?). Also, I don't have the mark-scheme, but I think I know the scoring points anyway, as we went through this paper in class.

Jan 2004, 4.d:

(i) Give an equation for the reaction of the Grignard reagent methylmagnesium bromide, CH3MgBr, with water.

(ii) State TWO precautions that you would have to take to prevent water from being able to react with a Grignard reagent.

P.S. Thank you KY1986, I edited my workings so that it should make sense to anyone who might want to see how to do it in the future
17. i cant answer that question cuz it's not on AQA syllabus
18. Oh, sorry, if I ask another thing that isn't on AQA, just tell me. Someone else who does Edexcel can try and answer that one..

I'll make up another one.. let's see..

Describe and explain what will be seen when ammonia is added till in excess to a solution of hydrated Cu2+ ions.
19. I believe that the [Cu(H2O)6]2+ is first hydrolysed to Cu(H2O)4(OH)2, so therefore there is a blue precipitate formed from a blue solution.

Then there is ligand substitution occurring with excess NH3 resulting in [Cu(H
2O)2(NH3)4]2+. Therefore this forms a deep blue solution.
20. (Original post by junners)
I believe that the [Cu(H2O)6]2+ is first hydrolysed to Cu(H2O)4(OH)2, so therefore there is a blue precipitate formed from a blue solution.

Then there is ligand substitution occurring with excess NH3 resulting in [Cu(H
2O)2(NH3)4]2+. Therefore this forms a deep blue solution.
Almost, the reaction with hyrdoxide ions is a deprotonation. Otherwise correct though

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