Quantitative Analysis - Sulphur with Example

Description:

A known mass of organic compound is heated in a Carius tube with Na₂O₂ (sodium peroxide) or fuming nitric acid.

Sulphur present in the compound is oxidized to sulphuric acid.

By adding excess BaCl₂, BaSO₄ is precipitated.

The precipitate is filtered, washed, dried and weighed.

The percentage of Sulphur can be calculated from the mass of barium sulphate.

Let the mass of organic compound = m g

• Mass of barium sulphate formed = m₁ g

• 1 mol of BaSO₄ = 233 g BaSO₄ = 32 g sulphur

• m₁ g BaSO₄ contains

  32 × m₁

  /

  233

  g Sulphur

Percentage of sulphur =

32 × m₁ × 100

/

233 × m

Problem

In Sulphur estimation, 0.2 g of organic compound gave 0.48 g of BaSO₄. Determine the percentage of Sulphur in the given organic compound.

Solution

Given, mass of organic compound = 0.20 g

molar mass of BaSO₄ = 137 + 32 + 64 = 233 g

233 g of BaSO₄ contains − 32 g sulphur

Thus, 0.48 g of BaSO₄ contains =

0.48 × 32

/

233

= 0.066 g bromine

Percentage of Sulphur =

0.066 × 100

/

0.20

= 32.96%