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%.

Answer 2

The percentage by mass of barium ion (Ba²⁺) in the unknown compound is 35.7%

We'll begin by writing the balanced equation for the reaction. This is given below:

Ba²⁺(aq) + Na₂SO₄(aq) —> BaSO₄(s) + 2Na⁺(aq)

Molar mass of Ba²⁺ = 137 g/mol

Mass of Ba²⁺ from the balanced equation = 1 × 137 = 137 g

Molar mass of BaSO₄ = 137 + 32 + (16×4)

= 233 g/mol

Mass of BaSO₄ from the balanced equation = 1 × 233 = 233 g

Thus,

From the balanced equation above,

137 g of Ba²⁺ reacted to produce 233 g of BaSO₄.

Next, we shall determine the mass of Ba²⁺ that reacted to produce 0.4105 g of BaSO₄. This can be obtained as follow:

From the balanced equation above,

137 g of Ba²⁺ reacted to produce 233 g of BaSO₄.

Therefore,

X g of Ba²⁺ will react to produce 0.4105 g of BaSO₄ i.e

X g of Ba²⁺ = $\frac{137 * 0.4105}{233}$

X g of Ba²⁺ = 0.2414 g

Finally, we shall determine the percentage of Ba²⁺ in the unknown compound. This can be obtained as follow:

Mass of Ba²⁺ = 0.2414 g

Mass of compound = 0.6760 g

Percentage of Ba²⁺ =?

$Percentage = \frac{mass of ion}{mass of compound} * 100\\\\= \frac{0.2414}{0.6760} * 100$

Percentage of Ba²⁺ = 35.7%

Therefore, the percentage by mass of barium ion (Ba²⁺) in the unknown compound is 35.7%

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