# Fertilisers (Sampling and Analysis) Regulations (Northern Ireland) 1996, 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:
Mo(%)=X×0.02305×———V×Da×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.

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Source: https://www.legislation.gov.uk/nisr/1996/513/schedule/2/part/I/chapter/26j./paragraph/8 (the official version)
Page: https://openlegislation.uk/fertilisers-sampling-and-analysis-regulations-northern-ireland-1996/schedule-2/part-I/chapter-26j./paragraph-8
Contains public sector information licensed under the Open Government Licence v3.0. Source: legislation.gov.uk, The National Archives.