Gravimetric Analysis Homework: Mass % of Barium Ions in Mixture

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SUMMARY

The discussion focuses on calculating the mass percent of barium ions in an impure mixture using gravimetric analysis. The participant analyzed 1.728g of the mixture and obtained 0.408g of dried barium sulfate (BaSO4). The molar mass of BaSO4 was calculated as 233.4 g/mol, leading to the determination of barium ions' mass as 0.0828g, resulting in a mass percent of 4.79%. The key takeaway is the 1:1 molar ratio between barium ions and barium sulfate, which simplifies the calculations.

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Homework Statement


An aqueous solution containing 1.728g of an impure mixture was analysed for barium ions by adding sufficient sulphuric acid solution to completely precipitate the barium ions as its sulphate. The precipitate was filtered, washed and dried to constant mass and finally weighed. What is the mass percent of barium ions in the mixture, if 0.408g of dried barium sulphate was obtained?


Homework Equations



I used this as the balanced equation but somehow it looks funny to me...

BaX + H2SO4 ------> BaSO4 + HX
1.728g 0.1408g

The Attempt at a Solution



no of mols BaSO4 produced = mass/Mr

Mr of BaSo4= (137.3)+ (32.1) + (16x4)
= 233.4

mols= 0.1408/233.4
=6.033x10^-4

BaX : BaSO4
1 : 1

mass Ba^2+ = molsx Mr
=(6.033x10^-4) x 137.3
=0.0828g

% of Ba^2+ present in mixture= (0.0828/1.728) x100
= 4.79%

Am i on the right track? Plz help!
 
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Hi leah3000! :smile:

You are very much on the right track!

Only one thing: you seem to have made a typo followed by the corresponding calculation errors when you wrote 0.1480 instead of 0.480.
 
oops. thanks for the help
 
The only thing that matters is that molar ratio between barium and barium sulfate is 1:1. This is enough to solve the problem.

Also note that while it is perfectly correct to convert everything to moles and back to masses, it is not necessary. See stoichiometric calculations using ratios for details.
 

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