Achemist 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 441-mg sample of the barium compound if a solution of the sample gave 403 mg baso4 precipitate? what is the mass percentage of barium in the compound?

Answer : The mass percentage of barium in the compound is, 53.8 %

Explanation : Given,

Mass of barium compound = 441 mg

Mass of barium sulfate = 403 mg = 0.403 g       (1 mg = 0.001 g)

The balanced chemical reaction will be:

First we have to calculate the moles of

Molar mass of = 233.38 g/mole

Now we have to calculate the moles of barium ion.

From the balanced chemical reaction, we conclude that

As, 1 mole of barium sulfate produced from 1 mole of barium ion

So, 0.001727 mole of barium sulfate produced from 0.001727 mole of barium ion

Now we have to calculate the mass of barium ion.

Molar mass of barium = 137.3 g/mol

Now we convert the mass of barium ion from gram to mg.

Conversion used : (1 g = 1000 mg)

Mass of barium ion = 0.2371 g = 237.1 mg

Now we have to calculate the mass percentage of barium in the compound.

Mass percent of barium =

Thus, the mass percentage of barium in the compound is, 53.8 %

the volume increases