clarify definition of _pd_calib.std_internal_mass_percent

[rowlesmr]

The (relevant part of the) current definition is

100 times the ratio of the amount of standard added to the original sample mass.

which reads, to me, as: std wt% = 100 mass(std) / mass(original sample) -> 100 1 / 3 = 33.3 wt% internal std which I believe is wrong, as you've added 1 g std to 3 g of original sample, giving 4 g. The mixed sample now contains 25% std.

The new definition is

100 times the mass of standard added divided by the sum of the mass of standard added and the original sample mass.

which reads, to me, as: std wt% = 100 mass(std) / (mass(std) + mass(original sample)) -> 100 1 / (1+3) = 25 wt% internal std

[jamesrhester]

Given that the original definition seemed to quite clearly state something different, I would like confirmation from @briantoby that this was indeed an oversight. I assume that in QPA the corrected definition proposed here is normally used?

[briantoby]

This was an error on my part (though the wording does not seem mine, so perhaps someone "corrected" the style and introduced this for my benefit.) With the previous definition, if one adds 2 grams of standard to a 1 gram sample, the value would be 200% rather than the sensible 66%. I recommend the correction be accepted.

[jamesrhester]

Thanks Brian for the comment. The old definition text is found in IT Vol G 1st edition as well so I don't fancy our chances of finding when it was "adjusted".