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?

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.

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

To find the mass of barium ions in the sample, we can use the stoichiometry of the reaction between sodium sulfate and barium compound.

1. Find the molar mass of BaSO4:
- Ba: 1 atom x 137.33 g/mol = 137.33 g/mol
- S: 1 atom x 32.06 g/mol = 32.06 g/mol
- O4: 4 atoms x 16.00 g/mol = 64.00 g/mol
- Total mass: 137.33 g/mol + 32.06 g/mol + 64.00 g/mol = 233.39 g/mol

2. Set up a stoichiometric ratio using the molar mass of BaSO4:
- 233.39 g BaSO4 / 1 mol BaSO4

3. Convert the given mass of the precipitate (513 mg BaSO4) to moles of BaSO4:
- 513 mg BaSO4 x (1 g / 1000 mg) x (1 mol BaSO4 / 233.39 g BaSO4) = 0.0022 mol BaSO4

4. Since the stoichiometric ratio is 1:1 between Ba and BaSO4, the moles of Ba ions are also 0.0022 mol.

5. Convert the moles of Ba ions to grams:
- 0.0022 mol Ba x 137.33 g/mol = 0.30 g Ba

Therefore, there are 0.30 grams of barium ions in the 458-mg sample of the barium compound.

To find the mass percentage of barium in the compound:

6. Calculate the mass percentage of barium:
- Mass percentage = (mass of Ba / mass of sample) x 100%
- Mass percentage = (0.30 g Ba / 0.458 g sample) x 100% = 65.50%

The mass percentage of barium in the compound is 65.50%.

To find out the mass of barium ion in the sample, we need to use stoichiometry and the mass of the precipitate formed.

First, we need to calculate the molar mass of BaSO4.
- The molar mass of barium (Ba) is 137.33 g/mol.
- The molar mass of sulfur (S) is 32.07 g/mol.
- The molar mass of oxygen (O) is 16.00 g/mol.
Therefore, the molar mass of BaSO4 is 137.33 + 32.07 + (4 * 16.00) = 233.38 g/mol.

Next, we convert the mass of the precipitate (513 mg) to moles of BaSO4:
513 mg BaSO4 * (1 g / 1000 mg) * (1 mol BaSO4 / 233.38 g BaSO4) = 0.00220 mol BaSO4

Since there is an excess of sodium sulfate, all the barium ion will combine with sulfate to form BaSO4. This means that the moles of BaSO4 formed are also equal to the moles of barium ion in the sample.

Now we can find the moles of barium ion by using the stoichiometry of the reaction:
1 mol BaSO4 = 1 mol Ba2+
Therefore, the sample contains 0.00220 mol of barium ion.

To calculate the mass of barium ion in the sample, multiply the moles of barium ion by the molar mass of barium:
0.00220 mol Ba2+ * 137.33 g Ba2+/mol = 0.302 g Ba2+

Therefore, there are 0.302 grams of barium ion in the 458-mg sample of the barium compound.

To find the mass percentage of barium in the compound, divide the mass of barium ion by the mass of the sample and multiply by 100:
(0.302 g Ba2+ / 458 mg sample) * 100 = 66.0%

The mass percentage of barium in the compound is 66.0%.