Paragraph 8 EXPRESSION OF RESULTS

The percentage elemental sulfur (S) content of the fertiliser is as follows:

$$\text{Impure S (\%) of the fertiliser}=\genfrac{}{}{0.1ex}{}{\mathrm{P}1-\mathrm{P}0}{\mathrm{m}}\times 100$$ $$\text{Purity of sulfur extracted (\%)}=\genfrac{}{}{0.1ex}{}{\mathrm{P}2-\mathrm{P}3}{\mathrm{n}}\times 100$$ $$\text{Pure S (\%) of the fertiliser}=\genfrac{}{}{0.1ex}{}{\left(\mathrm{P}1-\mathrm{P}0\right)\left(\mathrm{P}2-\mathrm{P}3\right)}{\mathrm{m}\times \mathrm{n}}\times 100$$

where:

• m = the mass of the test sample of fertiliser in grams,

• P₀ = the mass of the Soxhlet flask in grams,

• P₁ = the mass of the Soxhlet flask and the impure sulfur after drying,

• n = the mass of impure sulfur to be purified in grams,

• P₂ = the mass of the Petri dish and the sample,

• P₃ = the mass of the Petri dish after sublimation of the sulfur.