# Fertilisers (Sampling and Analysis) Regulations (Northern Ireland) 1996, 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:
S(%)=w×0.0137×v1v2×m SO3(%)=S(%)×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.

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Source: https://www.legislation.gov.uk/nisr/1996/513/schedule/2/part/I/chapter/23./paragraph/7 (the official version)
Page: https://openlegislation.uk/fertilisers-sampling-and-analysis-regulations-northern-ireland-1996/schedule-2/part-I/chapter-23./paragraph-7
Contains public sector information licensed under the Open Government Licence v3.0. Source: legislation.gov.uk, The National Archives.