The Diluted Heroin Exhibit
A multi-bag seizure yields 40 individual glassine bags. Ten bags are randomly selected and composited. The composite tests positive for heroin by GC-MS, with a measured concentration of 8.2% heroin base. The remaining 30 bags are not analyzed. The total gross weight of all 40 bags is 12.6 g; the average empty bag weighs 0.10 g. The defendant is charged under a statute that specifies weight of the mixture, not the pure drug. The defense moves to suppress the unanalyzed bags as unsupported.
1. Calculate the total net mixture weight and the estimated heroin content, and state what can and cannot be asserted.
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Empty bags: 40 × 0.10 g = 4.0 g. Net mixture = 12.6 − 4.0 = 8.6 g. Estimated heroin = 8.6 × 0.082 = 0.71 g. What can be asserted: the 8.6 g mixture weight is a direct measurement; the 0.71 g is an estimate from a 10-bag composite applied to 40 bags. What cannot be asserted: that every bag contains heroin or that the 8.2% applies uniformly. The composite result establishes presence in the sampled set and an approximate concentration for the lot.
2. The statute charges on mixture weight. Does the sampling gap matter to the charge, and how should the report be framed?
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If the statute is mixture weight, the charge rests on the 8.6 g measured weight, which does not depend on the 10-bag composite; the composite only establishes that the mixture contains a controlled substance. The sampling gap matters to the purity estimate, not the mixture-weight charge. Frame the report as: (1) total net mixture weight measured; (2) controlled substance identified in a random sample of 10 of 40 bags; (3) estimated purity with explicit caveat that unsampled bags were not individually tested. Do not overstate the estimate as a measured value.