# Fertilisers (Sampling and Analysis) Regulations (Northern Ireland) 1996, Paragraph 8: EXPRESSION OF RESULTS


The percentage of trace element (E) in the fertiliser is equal to:
E(%)=[(xs-xb)×V×D]/(M×104). If method (25c) has been used:
E(%)=[(xs-xb)×V×2D]/(M×104), where:
E = the amount of the trace element determined, expressed as a percentage of the fertiliser;
xs = the concentration of the test solution (6.2), in μg/ml;
xb = the concentration of the blank solution (7.1), in μg/ml;
V = the volume of the extract obtained by Method 25a or 25b, in ml;
D = the factor corresponding to the dilution carried out in (6.2);
M = the mass of the test sample taken in accordance with Method 25a or 25b in grams.
Calculation of dilution factor D:
If (a1), (a2), (a3), . . . , (ai) and (a) are the aliquot portions and (v1), (v2), (v3), . . . , (vi) and (100) are the volumes in ml corresponding to their respective dilutions, the dilution factor D will be equal to:
D=(v1/a1)×(v2/a2)×(v3/a3)x . . . x(vi/ai)×(100/a).

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Source: https://www.legislation.gov.uk/nisr/1996/513/schedule/2/part/I/chapter/25d./paragraph/8 (the official version)
Page: https://openlegislation.uk/fertilisers-sampling-and-analysis-regulations-northern-ireland-1996/schedule-2/part-I/chapter-25d./paragraph-8
Contains public sector information licensed under the Open Government Licence v3.0. Source: legislation.gov.uk, The National Archives.