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I have noticed that gold bars are often marked "999.9" fine:

Whenever I have metallurgical assays done, they always seem to do them with ISP or DSP or whatever, some spectrometer-based system. These systems have errors of 1% to 2%. When I complain about the high error, they say "it's good enough" and "nobody does the old fashioned stuff anymore". So, if ± 2% is the new standard for assayists, how are these banks determining that the gold is pure to a 9999 out of 10,000 standard?

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    $\begingroup$ I think XRF is the "gold" standard for this analysis, and can measure to better than 4-nines fine (999.9 = "4-nines fine"). $\endgroup$ – airhuff Oct 2 '17 at 21:35
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    $\begingroup$ related chemistry.stackexchange.com/questions/65180/… $\endgroup$ – Mithoron Oct 2 '17 at 22:18
  • $\begingroup$ How about ICP-MS or ICP-AES? $\endgroup$ – logical x 2 Oct 3 '17 at 9:48
  • $\begingroup$ @deusexmachina Why use destructive methods which require probe preparation when there is XRF? :) $\endgroup$ – andselisk Oct 3 '17 at 17:45
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    $\begingroup$ I did XRF for years. For such a high purity (99.99%) you'd have to analyze for the impurities not the gold itself. $\endgroup$ – MaxW Oct 11 '17 at 19:19
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As the comments alluded, to analyze purity at high purity, you analyze the impurities. I'll use XRF for this example, but ICP, MS, HPLC, or other methods could work in the same way.

With XRF's $1-2\%$ precision directly measuring the gold would give you a maximum supportable purity of $98-99 \%$. To verify a metal as $99.99\%$ the impurities are measured which are usually in the parts per million range, added up and subtracted from $\pu{100\%}$.

For example, let's say you have a gold bar with $\pu{35ppm}$ silver, $\pu{20ppm}$ copper, $\pu{10ppm}$ lead, and $\pu{5ppm}$ mercury with all other elements below a , $\pu{0.5ppm}$ detection limit (also called a threshold limit). The total impurity would be the total of the measured impurities and the detection limit of the unmeasured impurities. Note: most analysis is by metal basis which excludes the other elements that may be present.

$$\mathrm{Impurity = 35 + 20. + 10. + 5 + [0.5 \times (59-5)] = \pu{97 ppm} }$$

The $59$ is for the $59$ naturally occurring metal elements and the $5$ is for the five elements we measured ($\ce{Au, Ag, Cu, Pb, Hg}$). $\pu 2 \pu\%$ of $\pu{97ppm}$ is $\pu{1.94ppm}$ and the measurement plus the maximum error is $\pu{99\!.ppm}$. Taking $\pu{99\!.ppm} = \pu{0.0099\%}$ and subtracting from $\pu{100.00\%}$, we get an inferred purity of $\pu{99.991\%}$ Gold ($\mathrm{100\% - 0.01\% = 99.991\%}$). The gold would still be sold at $\pu{99.99\%}$ regardless of it being purer than needed as selling the gold by lots would be unnecessarily complicated and time intensive.

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