[JayDee]

Sulfur in the form of sulfate is another important plant nutrient and is there- fore found in many fertiliser preparations.
As sulfates easily form precipitates with suitable metal cations, a gravi- metric method based on this feature is a common way to analyse for sulfate. The usual metal ion used to form such a precipitate is barium, owing to the very low solubility of barium sulfate.
The equation for this reaction is: Ba2+(aq) + SO4 2-(aq) -----> BaSO4(s)

METHOD 1. Grind up some lawn food in a mortar and pestle, then accurately weigh
out about 1 g.
2. Add this to about 100 mL of distilled water and dissolve as much as possible.
3. Filter the solution into a 400 mL (or larger) beaker. Wash the residue in the filter paper with several portions of distilled water.
4. Add about 5 mL of 1 mol L␣1 hydrochloric acid to the filtrate followed by sufficient distilled water to give a total volume of 200 mL.
5. Heat to near boiling and then add about 15 mL of the barium chloride solution, stirring throughout.
6. Allow the precipitate to settle and then add a few drops of the barium chloride solution. If a precipitate forms, add a further 2 mL of the barium chloride solution. Repeat this step until no further precipitate forms.
7. Allow the mixture to stand for as long as possible before filtering. 8. Weigh a clean and dry suction funnel, together with a piece of filter paper. 9. Carefully filter the supernatant liquid and then the precipitate. Scrape
difficult-to-remove traces of precipitate into the funnel. Rinse the beaker a few times with water and add these washings to the filter paper.
10. Wash the precipitate with several portions of warm water followed by methylated spirits.
11. Dry the crucible and contents in an oven at 110°C overnight.
12. Weigh when dry and calculate the mass of barium sulfate produced.
13. Use your results to calculate the percentage of sulfur in the lawn food tested.

Okay, the buchner funnels and filter paper were not working very well so alot of the percipitate was not filtered.
The expected value is 18.60%
but I got 9.85%

Mass of BaSO4 produced =  .827g

so.. I did...

n(BaSO4) = m/M = .827/269.4 = 0.003069785 mol
:n(SO4) = nxM = .003069785 x 32.06  (just the molar mass of Sulfur) = 0.294883547g
:%(SO4) = 0.294883547/1.00 x 100
= 9.85%

is that right?

[cliverlong]

<< snip >>
n(BaSO4) = m/M = .827/269.4 = 0.003069785 mol
<< snip >>

where does 269.4 come from? (I have checked several times and I don't end up with that value)

Clive

[JayDee]

ooopps it should be 233.4
is the method right though?
let me do it with this and see if it works

[JayDee]

Okay, now I got 11.36%

is the method right?

Ba2+ + SO4 2- -----> BaSO4(s)

n(BaSO4) = .827/233.4
= 0.003543273 mol

n(SO4 2-) = n(BaSO4)
= 0.003543273 mol

m(SO4) = nxM = 0.003543273 x 32.06
= 0.113597344g
= 0.113597344/1.00 x 100
11.36%

[cliverlong]

Okay, now I got 11.36%

is the method right?

I haven't time to check it all but looks ok

Barium is the heavy atom so a low-ish percentage "feels" right

[JayDee]

Yay!
I do not think there is a very big difference between the expected value and mine especially as there were so many random errors throughout the practical.
But if you have some time, could you please let me know if the method is being carried out correctly.
Thank you!!