sulfate content calculation problem (1 Viewer)

with-chu · May 26, 2010

from conquering chemistry pate 224, question 31

how do you calculate the percentage of sulfate?? I took mass of ppt as 0.727 after averaging 3 closest values and got mass of sulfate. How do you calculate the percentage when all you're given is the volume of the original sample (50mL)?

DNETTZ · May 26, 2010

The problem please. We dont all use your text book, or sleep with it under our beds

with-chu · May 26, 2010

right. sorry!

To determine the sulfate concentration in a liquid fertiliser, an analyst first diluted 50 mL of the solution to 250 mL, pipetted out 25 mL of the dilute solution, added 200 mL water to it then precipitated the sulfate with barium nitrate solution. The ppt was filtered, washed, dried to constant mass. In repeated experiments, the mass of ppt was 0.728g, 0.773g, 0.722g and 0.732g. As accurately as you can, calculate percentage of sulfate in the original fertiliser solution. Justify the procedure you followed.

brenton1987 · May 26, 2010

with-chu said:
To determine the sulfate concentration in a liquid fertiliser, an analyst first diluted 50 mL of the solution to 250 mL, pipetted out 25 mL of the dilute solution, added 200 mL water to it then precipitated the sulfate with barium nitrate solution. The ppt was filtered, washed, dried to constant mass. In repeated experiments, the mass of ppt was 0.728g, 0.773g, 0.722g and 0.732g. As accurately as you can, calculate percentage of sulfate in the original fertiliser solution. Justify the procedure you followed.

m₁ = 0.728 g
m₂ = 0.773 g
m₃ = 0.722 g
m₄ = 0.732 g
m_a = 0.73875 g

0.73875 g * 10 = 7.3875 g
7.3875 g / 50 g * 100 % = 14.775 %

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Adding water to the 50 mL sample does not change the amount of sulfate. An aliquot of one tenth the volume is removed so logically one tenth of the sulfate is removed. The sulfate precipitate is one tenth the mass of sulfate in the original which is why the mass is multiplied by 10. Assume that the density of the fertiliser is 1 g^.mL^-1

ibbi00 · May 26, 2010

$n=\frac{m}{M} \\\ \\\n= \frac{.72733333333 }{233.38} \\\ \\\n= 3.116519539 \times 10^{-3} \\\ \\\ n(BaSO_{4}) = n(SO_{4}) = 3.116519539 \times 10^{-3} \\\ \\\ m(SO_{4}) =M(SO_{4}) \times n(SO_{4}) \\\ m= 96.07 \times 3.116519539 \times 10^{-3} \\\ m= 0.29940403219$

I know I'm correct up until this step. I'm still trying to get my head around the dilutions they done to see what you divide by.

p.s Final answer should come down to 5.988%

EDIT: 0.773g is an outlier and is therefore not used when averaging the weight.

sulfate content calculation problem (1 Viewer)

with-chu

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DNETTZ

Camp-italist Fatcat

with-chu

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brenton1987

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ibbi00

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