Fertiliser (1 Viewer)

horndog huang 2 · Jun 25, 2016

Hi all, i would like some to explain how to do this q and perhaps fertiliser questions in general. I just cant get my head around those. Thankd

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Suu · Jun 25, 2016

It's a gravimetric analysis question.
I find that this website has explained how to do these fertiliser questions pretty well:
http://www.easychem.com.au/monitoring-and-management/manufactured-products/sulfate-experiment

Basically, Barium Chloride reacts with Sulfate to form Barium Sulfate, which comes in the form of a precepitate.
From there, Mass of Sulfate/ Mass of Barium Sulfate * Amount of Barium Sulfate = Mass of Sulfate. From there, you can find the percentage of Sulfate in the Fertiliser.
Cheers

jorie · Jun 28, 2016

Jesus

ProdigyInspired · Jul 14, 2016

I just did an assignment on this and mainly used Shah's and other reports.

Here's how I set it out.
Consider how much sulphate SO4 is in Barium Sulfate, so mm(SO4)/mm(BaSO4) = ??%
Therefore the mass of SO4 is found through ??% x Mass (Barium Sulfate)
Then mass of SO4 / mass of fertiliser x 100 = percentage of sulphate in fertiliser.

My answer - 45.9% correct? Try it yourself.

$mass\quad of\quad Fertiliser\quad =\quad 2.00g\\ mass\quad of\quad precipitate\quad =\quad 2.23g\\ \begin{matrix} Percentage\quad of\quad S{ O }_{ 4 }\quad in\quad BaS{ O }_{ 4 }\quad =\quad \frac { 96.1 }{ 233.4 } \\ =\quad 41.1174\% \end{matrix}\\ \begin{matrix} Mass\quad of\quad { SO }_{ 4 }\quad in\quad BaS{ O }_{ 4 }\quad =\quad \%\quad S{ O }_{ 4 }\quad in\quad Ba{ SO }_{ 4 }\times \quad Mass\quad of\quad BaS{ O }_{ 4 } \\ =\quad 2.23\quad \times \quad 41.1174\% \\ =\quad 0.918179 \end{matrix}\\ \begin{matrix} Percentage\quad of\quad S{ O }_{ 4 }\quad in\quad Fertiliser\quad =\quad mass\quad of\quad S{ O }_{ 4 }\quad /\quad mass\quad of\quad Fertiliser\quad \times \quad 100\quad \\ =\quad 0.918179\quad /\quad 2.00\quad \times \quad 100 \\ =\quad 45.908955\% \end{matrix}$

Fertiliser (1 Viewer)

horndog huang 2

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Suu

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jorie

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ProdigyInspired

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