Asked by Anna
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?
Answers
Answered by
DrBob222
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.
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.
Answered by
Anna
Thanks for walkin me through that one =]. It helped a lot with the other problems like that too.