Percentage by mass (1 Viewer)

[tennille]

Hey, could somebody help me with this question. When we were determining the amount of sulfate in fertiliser, we used titration and not hte precipitatyion reaction, so i'm not sure how to do this.

This was a question in last years HSC paper.

A student carried out an investigation to analyse the sulfate content of lawn fertiliser. The student weighed out 1.0g of fertiliser and dissolved it in water. 50mL of 0.25 mol/L barium chloride solution was then added. A white precipitate of barium sulfate formed, which weighed 1.8g.

a. Calculate the percentage by mass of sulfate in the fertiliser.

b. Evaluate hte reliability of the experimental procedure used.

 

[danif]

heya i hope this helps:

a) Mass of barium sulfate = 1.8g
% mass of sulfate in the barium sulfate=
32.1+16x4 (i.e. MM of sulfate)/ 32.1 + 16x4 + 137.3 (i.e. MM of barium sulfate)
= 96.1/233.4
therefore amount of sulfate in 1.8g of precipitate = 96.1/233.4 X 1.8
=0.7411...g
therefore the % sulfate in the fertilizer = 0.741/1
=74.113...
=74.1% (1d.p.)

b) The student did not use a very reliable procedure for a number of reasons.
- the experiement was not repeated.
- the Barium sulfate crystals are very fine and need to be filtered using sintered glass filter, there was no indication of this being done.
- not everything in the fertilizer would have been dissolved with just water. A strong acid such as HCl should have been used. it should have then been filtered, removing any solids.
- the student used a very small amount, therefore the % error would be greater.
.... etc. etc.

but i do have a question, when we did this experiement in class we used something to make the barium sulfate coagulate so it wasn't so fine... does anyone know what this could be? (that could be another bad experimental procedure)

 

[grimreaper]

heya i hope this helps:

a) Mass of barium sulfate = 1.8g
% mass of sulfate in the barium sulfate=
32.1+16x4 (i.e. MM of sulfate)/ 32.1 + 16x4 + 137.3 (i.e. MM of barium sulfate)
= 96.1/233.4
therefore amount of sulfate in 1.8g of precipitate = 96.1/233.4 X 1.8
=0.7411...g
therefore the % sulfate in the fertilizer = 0.741/1
=74.113...
=74.1% (1d.p.)

b) The student did not use a very reliable procedure for a number of reasons.
- the experiement was not repeated.
- the Barium sulfate crystals are very fine and need to be filtered using sintered glass filter, there was no indication of this being done.
- not everything in the fertilizer would have been dissolved with just water. A strong acid such as HCl should have been used. it should have then been filtered, removing any solids.
- the student used a very small amount, therefore the % error would be greater.
.... etc. etc.

but i do have a question, when we did this experiement in class we used something to make the barium sulfate coagulate so it wasn't so fine... does anyone know what this could be? (that could be another bad experimental procedure)

One way of coagulating it is "digesting" it ie heating it

 

[tennille]

Thank you so much for that.

 

[hayleynicole4]

My prac method has an extra step that i dont see the point of (barium chloride was used to precipitate sulphate ions instead of barium carbonate). The step is just before precipitate is weighed:
Collect the last drops of filtrate in a 100 mL beaker and test for the presence of chloride ions by adding a few drops of silver nitrate solution to the filtrate. If the solution becomes cloudy, wash the precipitate with a further 10 mL of warm water and repeat the test.
Help?!

 

[xiao1985]

I think the step you mentioned is to further remove any prescense of barium chloride, which might be trapped in the precipitate.

Something non-related: Wow, old thread. *tsk *tsk @ tennile. Good o' HSC eh.

 

[crammy90]

My prac method has an extra step that i dont see the point of (barium chloride was used to precipitate sulphate ions instead of barium carbonate).

CL- are mainly soluble so by adding it instead of carbonate (most insoluble) you are reducing the risk of getting other precipitates which would be included in the calculation.
You could do this before the precipitate is weighed to precipitate the BaSO4 out and the cl dissociates to form solubles which are removed. If BaCO3 was used the CO3 could dissociate and could react with stuff if there (i.e. anything not in group 1 or NH4+ salts) to form insolubles which would be takin into our naive calcuations to be baso4.
i think thats what it means