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Chemistry · Ch 12 — Organic Chemistry – Some Basic Principles and Techniques

Sulphur

12.10.4

Sulphur

The Principle: Oxidising Sulphur to a Precipitate

Sulphur in an organic compound is not directly measurable. The method used here is to destroy the organic framework and convert all the sulphur into a single, weighable inorganic compound — barium sulphate (BaSO4\text{BaSO}_4). The key is that every atom of sulphur in the original sample ends up in one molecule of barium sulphate, so the mass of the precipitate tells you exactly how much sulphur was present.

A known mass of the organic compound is heated strongly in a sealed Carius tube with either sodium peroxide (Na2O2\text{Na}_2\text{O}_2) or fuming nitric acid (HNO3\text{HNO}_3). Both are powerful oxidising agents. Under these harsh conditions, the carbon and hydrogen are burned off, and any sulphur present is oxidised to sulphuric acid (H2SO4\text{H}_2\text{SO}_4). The reaction can be represented simply as:

S (in compound)→heatoxidising agentH2SO4\text{S (in compound)} \xrightarrow[\text{heat}]{\text{oxidising agent}} \text{H}_2\text{SO}_4

Once the digestion is complete and the tube is opened, the contents are treated with water. To this solution, an excess of barium chloride (BaCl2\text{BaCl}_2) solution is added. The barium ions (Ba2+\text{Ba}^{2+}) react with the sulphate ions (SO42−\text{SO}_4^{2-}) from the sulphuric acid to form a dense, white precipitate of barium sulphate:

BaCl2+H2SO4→BaSO4↓+2HCl\text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4 \downarrow + 2\text{HCl}

The precipitate is filtered, washed thoroughly to remove any soluble impurities, dried to constant weight, and then weighed. From this single mass measurement, the percentage of sulphur in the original compound is calculated.

Watch out

A common mistake is to forget that the barium sulphate precipitate must be washed free of chloride ions (tested with silver nitrate solution) before drying. Any trapped barium chloride would add extra mass and give an inflated sulphur percentage.

The Calculation: From Mass of Precipitate to Percentage of Sulphur

The entire calculation rests on the stoichiometric relationship between barium sulphate and sulphur. One mole of BaSO4\text{BaSO}_4 contains exactly one mole of sulphur atoms.

Let:

  • mm = mass of the organic compound taken (in grams)
  • m1m_1 = mass of barium sulphate precipitate obtained (in grams)

The molar mass of BaSO4\text{BaSO}_4 is:

137.3 (Ba)+32.1 (S)+4×16.0 (O)=233.4 g mol−1137.3 \, (\text{Ba}) + 32.1 \, (\text{S}) + 4 \times 16.0 \, (\text{O}) = 233.4 \text{ g mol}^{-1}

For practical purposes, the textbook uses the rounded value of 233 g mol−1^{-1}. The atomic mass of sulphur is taken as 32 g mol−1^{-1}.

Therefore:

  • 233 g of BaSO4\text{BaSO}_4 contains 32 g of sulphur.
  • 1 g of BaSO4\text{BaSO}_4 contains 32233\frac{32}{233} g of sulphur.
  • m1m_1 g of BaSO4\text{BaSO}_4 contains m1×32233m_1 \times \frac{32}{233} g of sulphur.

This mass of sulphur came entirely from the original mm g of the organic compound. The percentage of sulphur is therefore:

Percentage of sulphur=mass of sulphurmass of compound×100\text{Percentage of sulphur} = \frac{\text{mass of sulphur}}{\text{mass of compound}} \times 100

Substituting the expression for the mass of sulphur:

% S=32×m1×100233×m\boxed{\% \, \text{S} = \frac{32 \times m_1 \times 100}{233 \times m}}

% S=32×m1×100233×m\% \, \text{S} = \frac{32 \times m_1 \times 100}{233 \times m} …

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