Question

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 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 1

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:

$Ba^{2+}(aq)+Na_2SO_4(aq)\rightarrow BaSO_4(s)+2Na^+(aq)$

First we have to calculate the moles of $BaSO_4$

$\text{Moles of }BaSO_4=\frac{\text{Mass of }BaSO_4}{\text{Molar mass of }BaSO_4}$

Molar mass of $BaSO_4$ = 233.38 g/mole

$\text{Moles of }BaSO_4=\frac{0.403g}{233.38g/mole}=0.001727mole$

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.

$\text{ Mass of }Ba^{2+}=\text{ Moles of }Ba^{2+}\times \text{ Molar mass of }Ba^{2+}$

Molar mass of barium = 137.3 g/mol

$\text{ Mass of }Ba^{2+}=(0.001727moles)\times (137.3g/mole)=0.2371g$

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 = $\frac{237.1mg}{441mg}\times 100=53.8\%$

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