Sulfate content in lawn fertilizer (1 Viewer)

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Hi

Heres the question:

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

a) Calculate the percentage by mass of sulfate in the fertilizer.
b) Evaluate the reliability of the experimental precedure used.

Thanks a lot!!
 

CM_Tutor

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(a)

% SO<sub>4</sub><sup>2-</sup> in BaSO<sub>4</sub> = [M(SO<sub>4</sub><sup>2-</sup>) / M(BaSO<sub>4</sub>)] * (100 / 1)
= [(32.07 + 4 * 16.00) / (137.3 + 32.07 + 4 * 16.00)] * (100 / 1)
= 41.16 ... %

So, m(SO<sub>4</sub><sup>2-</sup> in orig sample) = 41.16 ... % * 1.8 = 0.7409 ... g

So, % sulfate in fertiliser = (0.7409 ... / 1.0) * (100 / 1) = 74 % (2 sig fig)

(b) You tell me... :)
 

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Hi

I did the exact same thing.....but we cannot say that excess BaCl2 was added for complete ppt of BaSO4.... i think we need to take into account the concentration and volume of BaCl2

Thanks !!
 

CM_Tutor

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I would check that BaCl<sub>2</sub> was in excess as part of the discussion of (b), but even if it is limiting, you still will get the same answer for sulfate content, but with the problem that the answer is unreliable as there was unprecipitated sulfate left behind in the solution.
 

Steven12

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dude, you got that from the 2003 HSC paper didnt you?

all I did was finding the percentage if SO4 in 1 mol of BaSo4,
and times that percentage to 1.8 g(amount so4)

then divide it over 1 g and you get the percentage.
i think that is the right way. i think..
 

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yeah i did get it from the 2003 HSC paper.

for part b) i just said that excess BaCl2 should be added in order for all the sulfate to be precipitated. Also, the experiment should be repeated a number of times, until consistent values are obtained and then averaged to get a more reliable answer.

Would that get 3 marks?

Thanks!
 

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