Paragraph 7 EXPRESSION OF RESULTS

One mg of barium sulfate corresponds to 0.137 mg of S or to 0.343 mg of SO₃.The percentage S content of the fertiliser is obtained as follows:

$$\mathrm{S}\left(\mathrm{\%}\right)=\mathrm{w}\times 0.0137\times \genfrac{}{}{0.1ex}{}{\mathrm{v}1}{\mathrm{v}2\times \mathrm{m}}$$ $$\mathrm{SO}3\left(\mathrm{\%}\right)=\mathrm{S}\left(\mathrm{\%}\right)\times 2.5$$

where:

• w = the mass of the barium sulfate precipitate in milligrams,

• v₁ = the volume of the extraction solution in millilitres,

• v₂ = the aliquot volume in millilitres,

• m = the mass of the test sample in grams.