Finding mass percent of barium ions by gravimetric analysis with BaSO4

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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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Discussion

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.
 
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.