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Chinese Amidol Analysis

Photo Engineer

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Chemical analyses of this sort start by giving the theoretical % of C, H, O, N, S etc, and then the actual values present.

Thus: Calc for C10H14N2O: C, 67.38; H, 7.92. Found C, 66.62; H, 7.66. And etc.... The error in ratios of C, H, N and O in this type of representation will tell you how much amidol there is and how much is organic and metallic junk. (this example is not Amidol, it is hypothetical taken from an off the shelf text)

The analysis as given in the reference is not revealing in any way.

PE