Titration calculation

Hi, I've been stuck on the last question on a sheet and can't find a round it:

"A fertiliser contains ammonium sulphate, (NH4)2SO4. A sample of 0.5g of fertiliser was warmed with sodium hydroxide solution. The ammonia evolved was absorbed in 100cm³ of 0.1 mol dm-³ of hydrochloric acid. The excess acid required 55.9cm³ of 0.1 mol dm-³ of sodium hydroxide for neutralisation.
Calculate the percentage of ammonium sulphate in the sample."

The question sounds quite ambiguous to me and I'm not sure whether it wants me to find the perecentage purity of ammonium sulphate or something else. Anyway, I've decided to split the question up into 3 equations (I'm not sure if this is the right step):

(NH4)2SO4 + 2NaOH > Na2SO4 + 2NH3 + 2H2O

NH3 + HCl > NH4Cl

NH3 + HCl > NaCl + H2O

I've been trying to find some sort of mass of ammonium sulphate apart from the 0.5g but I totally gave up after 3 hours. :-/

Some help will be much appreciated. Cheers

Reply 1

minnow2

You need to work backwards through it:

The final titration used 55.9cm³ of 0.1 mol dm-³ of sodium hydroxide and was excess HCl so the equation is:
HCl+NaOH --> NaCl +H2O
So there were 0.00559mol of NaOH present in that titration and therefore 0.00559mol of HCl. This HCl was left over when the ammonia was neutralised with excess HCl. The neutralisation involved 100cm3 of 0.1moldm-3 HCl. That is 0.01mol HCl. Since 0.00559mol was left over, 0.00441mol was used to neutralise the NH3.
From your equation:
NH3 + HCl > NH4Cl
They react in a 1:1 ratio, so there were 0.00441mol of NH3 present.
Given that the formula for the ammonium sulphate is (NH4)2SO4 the number of moles of ammonium sulphate will be half of the number of moles of NH3 evolved. that is 0.002205 moles.
The molar mass of ammonium sulphate is:132
Therefore the mass of ammonium sulphate was:0.29106g
The percentage in the sample mass 0.5g was therefore 58.212%
Round to 58%

Reply 2

Gabriel_sgs