Does anyone have the results for the sulfate fertiliser experiment. Also can you tell me how to calculate the percentage. Btw we didn't do the titration method using EDTA etc.
Does anyone have the results for the sulfate fertiliser experiment. Also can you tell me how to calculate the percentage. Btw we didn't do the titration method using EDTA etc.
Hey, i think the results are irrelevant for this experiements, because in the exam they will likely give you a sample and ask you to find the contend in that.. but anywho.. heres the calculations you can use
Ammonium sulfate if the main constitunet in most lawn fertilisers.
Ba + SO4 ---> Baso4 but the {n of moles Baso4= n moles so4}
convert mass of Baso4 to moles.... n= m/M
as 1:1 ratio
convert moles of sulfate to grams m= nM
express this as a % of total weight.. ie... .0005/1.5 x 100
Let's say you used 1.00 grams of ground and accurately weighedfertiliser.
Then after grounding you dissolve with HCl, precipitate it with BaCl2 then filter it out and finally you have 1.8 grams.
The formulas involed to find the percentage of sulfate in a sample of fertiliser is:
m(SO42-) = m(BaSO4) x M(SO42-) / M(BaSO4)
% m(SO42-) = m(SO42-) x 100 / m(fertiliser)
In words:
(mass of sulfate precipitate) = (mass of barium sulfate precipitate) x (molecular weight of sulfate) / (molecular weight of barium suflate)
(percentage mass of suflate precipitate) = (mass of sulfate) x 100 / (mass of fertiliser)
Answer:
m(SO42-) = 1.8 x 96.06 / 233.37 = 0.74
% mass SO42- = 0.74 x 100 / 1.00 = 74.1%
The question was modified from a previous HSC but you should NOT get a value this high because of the poor reliability as asked in a subsequent question.
Acid should be used to help dissolve the fertiliser and the precipitate should be filtered several times.