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# Chem Synoptic watch

1. Ok, this is a question from the Chemistry Edexcel June 2002 paper:

A fertilser is known to contain ammonium sulphate (NH4)2SO2, as the only ammonium salt. A sample of this fertiliser weighing 3.80g was dissolved in water and the volume made to 250cm^3. To 25cm^3 portions of this solution about 5cm^3 (an excess) of aqueous methanal was added. The following reaction took place:

4(NH4)^+ + 6HCHO ------> C6H12N4 + 4(H)^+ +6H2)

The liberated acid was titrateddirectly with 0.1 mol.dm^-3 aqueous NaOH. The average volume required was 28 cm^3. Calculate the percentage of ammonium sulphate in the fertiliser.

Thanks everyone
2. A fertilser is known to contain ammonium sulphate (NH4)2SO2, as the only ammonium salt. A sample of this fertiliser weighing 3.80g was dissolved in water and the volume made to 250cm^3. To 25cm^3 portions of this solution about 5cm^3 (an excess) of aqueous methanal was added. The following reaction took place:

4(NH4)^+ + 6HCHO ------> C6H12N4 + 4(H)^+ +6H2)

The liberated acid was titrateddirectly with 0.1 mol.dm^-3 aqueous NaOH. The average volume required was 28 cm^3. Calculate the percentage of ammonium sulphate in the fertiliser.

Moles NaOH used = (0.1/1000) * 28 = 0.0028 moles
Therefore moles H+ ions in 25cm^3 = 0.0028 moles
Therefore moles NH4 in 25cm^3 = 0.0028 moles
Therefore moles NH4 in 250cm^3 = 0.028 moles
Therefore moles (NH4)2SO2 in 250cm^3 = 0.014moles
Molecular mass (NH4)2SO2 = 100
Mass (NH4)2SO2 in 3.80g sample = 1.4g
Percentage by mass = (1.4/3.8)*100 = 36.8%
3. (Original post by JeremyG)
Moles NaOH used = (0.1/1000) * 28 = 0.0028 moles
Therefore moles H+ ions in 25cm^3 = 0.0028 moles
Therefore moles NH4 in 25cm^3 = 0.0028 moles
Therefore moles NH4 in 250cm^3 = 0.028 moles
Therefore moles (NH4)2SO2 in 250cm^3 = 0.014moles
Molecular mass (NH4)2SO2 = 100
Mass (NH4)2SO2 in 3.80g sample = 1.4g
Percentage by mass = (1.4/3.8)*100 = 36.8%
Firstly, i don't really understand the bold bit. Why do you multiply by 2?

Secondly, isn't the Mr of ammonium sulphate 132?
4. (Original post by Silly Sally)
Firstly, i don't really understand the bold bit. Why do you multiply by 2?

Secondly, isn't the Mr of ammonium sulphate 132?
It said: "Therefore moles NH4 in 250cm^3 = 0.028 moles
Therefore moles (NH4)2SO2 in 250cm^3 = 0.014moles"

This is divided by 2, not multiplied by 2.
It is divided because there are 2 moles of NH4 in every mole of (NH4)2SO2

rosie
5. (Original post by Silly Sally)
Firstly, i don't really understand the bold bit. Why do you multiply by 2?

Secondly, isn't the Mr of ammonium sulphate 132?
Yeah Ammonium sulphate is 132. Someone has typed (NH4)2SO2 instead of (NH4)2SO4. The rest of the method is correct though:

No. of moles of (NH4)2SO4 in 250cm^3 = 0.014moles
Mass of (NH4)2SO4 = 1.848

therefore %purity=48.6% (which I know is correct answer as did this question earlier!)

Hey Sally, how come you didn't look at the markscheme Mrs P gave us? I thought it explained it quite well, although admitedly briefly!
6. Dang, sorry for the SO2 typo
7. (Original post by crana)
This is divided by 2, not multiplied by 2.
It is divided because there are 2 moles of NH4 in every mole of (NH4)2SO2

rosie
Sorry i meant divide by 2 - but beacuse it is (NH4)2SO4 i thought it would be multiply by 2???
8. Sorry for the error.
Wasn't thinking and should have realised you meant (NH4)2S04 not (NH4)2SO2!

as there are two moles of ammonia for every one mole of ammonium sulphate you need to divide the number of moles of ammonia to get the number of moles of ammonium sulphate.

Good luck on Tuesday!
9. OHHHH!!! So we divide by 2 as there are 4 moles of NH4 ^+ in the original equation right??? I am so slow today sorry!!!
10. (Original post by JeremyG)
A fertilser is known to contain ammonium sulphate (NH4)2SO2, as the only ammonium salt. A sample of this fertiliser weighing 3.80g was dissolved in water and the volume made to 250cm^3. To 25cm^3 portions of this solution about 5cm^3 (an excess) of aqueous methanal was added. The following reaction took place:

4(NH4)^+ + 6HCHO ------> C6H12N4 + 4(H)^+ +6H2)

The liberated acid was titrateddirectly with 0.1 mol.dm^-3 aqueous NaOH. The average volume required was 28 cm^3. Calculate the percentage of ammonium sulphate in the fertiliser.

Moles NaOH used in titration = (0.1/1000) * 28 = 0.0028 moles
NaOH and H+ react one to one:
Therefore moles H+ ions in 25cm^3 = 0.0028 moles

In the equation 4(NH4)^+ + 6HCHO ------> C6H12N4 + 4(H)^+ +6H2)
there is one mole of H+ ions for every one mole of NH4+ ions
Therefore moles NH4 in 25cm^3 = 0.0028 moles
We need to multipy by ten to get no. moles in 250cm^3
Therefore moles NH4 in 250cm^3 = 0.028 moles
Ammonium sulphate dissolves in water to give:
(NH4)2SO4 ----> 2NH4+ + SO42-
So we need to divide the number of moles of NH4 in the sample by two to get the number of moles of ammonium sulphate originally used.
Therefore moles (NH4)2SO4 in 250cm^3 = 0.014moles
Molecular mass (NH4)2SO4 = 132
Mass (NH4)2SO2 in 3.80g sample = 1.848g
Percentage by mass = (1.848/3.8)*100 = 48.6%
Make sense?
11. (Original post by JeremyG)
Make sense?
YEAH!!! Thanks!!!

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