Question

A sample of 0.6760 g of an unknown compound containing barium ions (ba2+) is dissolved in water and treated with an excess of na2so4. if the mass of the baso4 precipitate formed is 0.4105 g, what is the percent by mass of ba in the original unknown compound?

Answer 1

Answer: 35.72 % of Barium ions will be present in the original unknown compound.

Explanation: The reaction of Barium ions and sodium sulfate is:

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

Here, Sodium sulfate is present in excess, Barium ions are the limiting reagent because it limits the formation of product.

Now, 1 mole of barium sulfate is produced by 1 mole of Barium ions.

Molar mass of Barium sulfate = 233.38 g/mol

Molar mass of Barium ions = 137.327 g/mol

233.38 g/mol of barium sulfate will be produced by 137.323 g/mol of Barium ions, so

0.4105 grams of barium sulfate will be produced by = $\frac{137.327g/mol}{233.38g/mol}\times 0.4105g$ of Barium ions

Mass of barium ions = 0.2415 grams

To calculate percentage by mass, we use the formula:

$\% mass=\frac{\text{Mass of solute (in grams)}}{\text{Total mass of the solution(in grams)}}\times 100$

Mass of the solution = 0.6760 grams

Putting the value in above equation, we get

$\% \text{ mass of }Ba^{2+}\text{ ions}=\frac{0.2415g}{0.6760g}\times 100$

% mass of Barium ions = 35.72%.