you add excess sodium sulfate to a solution of a soluble barium compound in order to precipitate all of the barium ion as barium sulfate, BaSO4. Suppose a 458 mg sample of the barium compound is dissolved to create the solution. Then when the sodium is added, 513 mg of barium sulfate precipitates.

a) how many milligrams of barium ion are in the solution, and therefore in the original barium compound sample?

To find the amount of barium ion in the solution, we can use the information given about the precipitated barium sulfate and the reaction that took place.

We know that barium sulfate, BaSO4, forms when barium ions, Ba2+, react with sulfate ions, SO42-. The balanced chemical equation for this reaction is:

Ba2+ + SO42- -> BaSO4

According to the question, 513 mg of barium sulfate (BaSO4) precipitates when excess sodium sulfate is added to the solution. From the balanced chemical equation, we know that one mole of BaSO4 contains one mole of Ba2+. So, we need to determine the number of moles of BaSO4 formed.

To find the number of moles of BaSO4, we can use its molar mass. The molar mass of BaSO4 can be calculated by summing the atomic masses of each element:

Ba: 137.33 g/mol
S: 32.06 g/mol
O: 16.00 g/mol (oxygen has a subscript of 4, so we multiply by 4)

Molar mass of BaSO4 = (137.33 g/mol) + (32.06 g/mol) + (16.00 g/mol * 4)
= 233.39 g/mol

Now, we can calculate the number of moles of BaSO4 using the given mass (513 mg) and its molar mass:

Number of moles = Mass / Molar mass
= 513 mg / (233.39 g/mol)

To convert mg to g, divide by 1000:

Number of moles = (513 mg / 1000) g / (233.39 g/mol)

Simplifying, we have:

Number of moles of BaSO4 = 0.0022 mol

Since BaSO4 and Ba2+ are in a 1:1 ratio according to the balanced chemical equation, the number of moles of Ba2+ is also 0.0022 mol.

Finally, to find the mass of barium ion (Ba2+), we can multiply the number of moles of Ba2+ by its atomic mass:

Mass of Ba2+ = Number of moles * Atomic mass of Ba
= 0.0022 mol * 137.33 g/mol

Thus, the mass of barium ion in the solution, and hence in the original barium compound sample, is:

Mass of Ba2+ = 0.3009 g = 300.9 mg

Therefore, there are 300.9 milligrams of barium ion in the solution and in the original barium compound sample.

mg Ba in original sample = mg BaSO4 x (atomic mass Ba/molar mass BaSO4)= ? mg Ba.

% Ba in the sample = (? mg Ba from above/458 mg sample)*100 = ??