percentage error of the determination resulting from the presence of the Na2SO4
2 answers
Complete problem : in a gravimetric determination of sulfur as BaSO4, 0.8863 g of the ignited precipiate is found to contain 7.71 mg of coprecipitated Na2SO4.
What was the weight of the sample used initially? Did you calculate % S. If so, what was the value?
You can calculate that 7.71 mg = 0.00771 g Na2SO4 converts to 0.0071 g x (molar mass BaSO4/molar mass Na2SO4) = 0.00771 x (233.34/142?) = ? and subtract from 0.8863. You can find the error in that easy enough. You need to redo the molar mass Na2SO4 to a better figures. I just remembered the whole numbers. Post your work if you get stuck.
You can calculate that 7.71 mg = 0.00771 g Na2SO4 converts to 0.0071 g x (molar mass BaSO4/molar mass Na2SO4) = 0.00771 x (233.34/142?) = ? and subtract from 0.8863. You can find the error in that easy enough. You need to redo the molar mass Na2SO4 to a better figures. I just remembered the whole numbers. Post your work if you get stuck.