Answer to Question #116639 in Chemistry for HARU

Question

A student was tasked to perform gravimetric analysis of a soluble sulfate. His unknown sample weighed 0.7543 g. The sample underwent precipitation using BaCl2 and was digested for overnight. The precipitate was then filtered off to obtain white crystalline precipitate that was collected in an ash less filter paper. In performing constant weighing, he obtained a crucible mass that is 29.9442 g. After burning his samples inside the crucible, the obtained mass was 30.3375 g. Compute for the experimental mass (g) of SO3 in grams obtained by the student

Answer 1

In the reaction of sulphate anion with BaCl₂ the BaSO₄ precipitate is formed:

SO₄^2- + BaCl₂ $\rightarrow$ BaSO₄$\downarrow$ + 2Cl^-.

The mass of the precipitate BaSO₄ is the mass of the crucible with the burned precipitate minus the mass of the crucible obtained by constant weighing:

$m(\text{BaSO}_4) =30.3375 - 29.9442 = 0.3933$ g.

The number of the moles of BaSO₄ is its mass divided by its molar mass $M$ (233.39 g/mol):

$n(\text{BaSO}_4) = 0.3933/233.39 = 0.001685$ mol.

The number of the moles of sulphur in BaSO₄ and in SO₃ relate as:

$n(\text{BaSO}_4) = n(\text{SO}_3)$ .

Therefore, the experimental mass of SO₃ obtained by the student is the product of its number of the moles and the molar mass of SO₃ (80.06 g/mol):

$m(\text{SO}_3) = n·M = 0.001685·80.06 = 0.1349$ g.

Answer: the experimental mass of SO₃ in grams obtained by the student is 0.1349 g.