A chemist added an excess of sodium sulfate to a solution of a soluble barium compound to precipitate all of the barium ion as barium sulfate, BaSO4. How many grams of barium ion are in a 458-mg sample of the barium compound if a solution of the sample gave 513 mg BaSO4 precipitate? What is the mass percentage of barium in the compound?

Question

DrBob222 · Accepted Answer

513 mg BaSO4 = 0.513 g.
Convert to mols BaSO4.
Then convert mols BaSO4 to mols Ba ion.
Convert mols Ba ion to grams Ba ion.
That is how many g Ba ion that are in the sample.
Then %Ba ion = (mass Ba ion/wt of the sample g)*100
The weight of the sample is 0.458 g.
Check my thinking. Check my work.

Anna · Answer

Thanks for walkin me through that one =]. It helped a lot with the other problems like that too.