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# Chem synoptic calculation watch

1. ok, this is from an Edexcel past paper (June 2002)

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

A sample 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 aq methanal was added. The following reaction took place:

4(NH4+) + 6HCHO = C6H12N4 + 4H+ + 6H2O

The liberated acid was titrated directly with 0.100 mol dm^-3 aq sodium hydroxide. The average volume required was 28.0cm^3. Calculate the percentage of ammonium sulphate in the fertiliser

The answer is 48.6% - but I really dont understand the mark scheme
2. (Original post by Lupin Loopy)
ok, this is from an Edexcel past paper (June 2002)

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

A sample 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 aq methanal was added. The following reaction took place:

4(NH4+) + 6HCHO = C6H12N4 + 4H+ + 6H2O

The liberated acid was titrated directly with 0.100 mol dm^-3 aq sodium hydroxide. The average volume required was 28.0cm^3. Calculate the percentage of ammonium sulphate in the fertiliser

The answer is 48.6% - but I really dont understand the mark scheme
I find the best thing for these questions is to work backwards.

Sodium hydroxide donates one (OH-) ion during reactions. This means 4 moles would react with the 4 (H+) ions. 28 cm of 0.1 mol NaOH contains 0.028 moles of reactant so there must be the same moles of (H+)

This in turn reacts with 4 NH4+ ions, as in this reaction each ammonium fertiliser molecule donates two ions to the reaction, it has a 2:4 ratio, hence 0.05 molecules of fertiliser reacts with 1 OH- ion, this means there must be 0.0014 moles of fertiliser in each sample.

If it was pure fertiliser then in each 25 cm^3 sample there would be (.38/132) moles, this equals .00287878787 moles, however we know there is only .0014 mole in the sample

So...
Working out percentage by mass
(0.0014/0.00287878787 *100) = 48.6

Hope that helps!

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Updated: June 11, 2004
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