Question

Barium can be analyzed by precipitating it as BaSO4 and determining the mass of the precipitate. When a 0.269 g sample of a barium compound was treated with excess H2SO4, 0.0891 g of BaSO4 formed. What percentage of barium is in the compound?

Answer 1

Answer:

Percentage of Ba in the compound is 19.5%

Explanation:

According to formula of $BaSO_{4}$, 1 molecule of $BaSO_{4}$ contains 1 molecule of Ba.

So, 1 mol of $BaSO_{4}$ contains 1 mol of Ba.

Molar mass of $BaSO_{4}$ = 233.38 g/mol

Number of moles of a compound is ratio of mass to molar mass of the compound.

So, 0.0891 g of $BaSO_{4}$ = $\frac{0.0891}{233.38}moles$ of $BaSO_{4}$ = 0.000382 moles of $BaSO_{4}$

So, moles of Ba present in sample = 0.000382 moles

Molar mass of Ba = 137.33 g/mol

So, mass of Ba in sample = $(0.000382\times 137.33)g=0.0525g$

So, % of Ba in sample = $\frac{0.0525}{0.269}\times 100$% = 19.5%

Answer 2

The percentage of barium in the compound, according to the illustration, would be 19.49%

Stoichiometric calculations

From the equation of the reaction:

$Ba^{2+} + SO_4^{2-} ----------- > BaSO_4$

The mole ratio of Ba2+ to BaSO4 is 1:1

Mole of 0.0891 g BaSO4 = 0.0891/233.38 = 0.00038 moles

Equivalent mole of Ba2+ = 0.00038 moles

Mass of 0.00038 moles Ba2+ = 0.00038 x 137.327

                                                      = 0.052 g

Percentage of barium in the compound = 0.052/0.269

                                    = 19.49%

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