Answer to Question #235461 in Chemistry for Dotun

Question

A sulfur containing compound is treated chemically to convert all its sulfur into barium sulfate. A 8.19 mg sample of the compound gave 5.46 mg barium sulfate.

What is the percentage of sulfur in the compound?

If there is one sulfur atom in the molecule, what is the molar mass of the compound ?

Answer 1

Solution:

The gravimetric factor (GF), represents the weight of analyte per unit weight of precipitate.

Analyte - S (sulfur); Precipitate - BaSO₄ (barium sulfate)

S → BaSO₄

The molar mass of S is 32.065 g/mol.

The molar mass of BaSO₄ is 233.38 g/mol.

Hence,

GF = (32.065 g S mol^-1 / 233.38 g BaSO₄ mol^-1) × (1/1) = 0.1374

GF = (Mass of analyte) / (Mass of precipitate)

Hence,

Mass of analyte = GF × Mass of precipitate

Mass of analyte = Mass of S = (0.1374) × (5.46 mg) = 0.75 mg

Mass of S = 0.75 mg

%analyte = (Mass of analyte / Mass of sample) × 100%

Hence,

%S = (Mass of S / Mass of sample) × 100%

%S = (0.75 mg / 8.19 mg) × 100% = 9.157% = 9.16%

%S = 9.16%

%S = (1 × Molar mass of S / Molar mass of the compound) × 100%

Hence,

9.16% = (32.065 g/mol / Molar mass of the compound) × 100%

0.0916 = (32.065 g/mol / Molar mass of the compound)

Molar mass of the compound = (32.065 g/mol) / (0.0916) = 350.05 g/mol = 350 g/mol

Molar mass of the compound = 350 g/mol

Answers:

The percentage of sulfur (S) in the compound is 9.16%

The molar mass of the compound is 350 g/mol