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

We have 302 mg barium ions in the solution, this is 65.9 %

Explanation:

Step 1: Data given

Mass of sample of the barium compound = 458 mg = 0.458 grams

A solution of the sample gave 513 mg BaSO4 precipitate

Molar mass of BaSO4 = 233.4 g/mol

Molar mass of Ba = 137.3 g/mol

Gravimetric factor Ba/BaSO4=137.3/233.4=0.5883

Step 2: Calculate grams of barium ion

0.5883*513mg = 302mg or 0.302g

so there are 302mg Ba/458mg sol ution

%Ba=302mg Ba/458mg sol * 100 = 65.9%Ba

We have 302 mg barium ions in the solution, this is 65.9 %