Paragraph 8 EXPRESSION OF RESULTS

1 mg of molybdenyl oxinate, MoO₂(C₉H₆ON)₂, corresponds to 0.02305 mg Mo.

The percentage of molybdenum in the fertiliser is given by:

$$\mathrm{Mo}\left(\mathrm{\%}\right)=\mathrm{X}\times 0.02305\times \mathrm{———}\genfrac{}{}{0.1ex}{}{\mathrm{V}\times \mathrm{D}}{\mathrm{a}\times \mathrm{M}}$$

where:

• X is the mass in mg of the molybdenyl oxinate precipitate;

• V is the volume in ml of the extract solution in accordance with Methods 26a or 26b;

• a is the volume in ml of the aliquot portion taken from the last dilution;

• D is the dilution factor of the aliquot portion;

• M is the mass in g of the test sample.