# How do chemists measure purity in high purity metals and other reagents?

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?

• I think XRF is the "gold" standard for this analysis, and can measure to better than 4-nines fine (999.9 = "4-nines fine"). – airhuff Oct 2 '17 at 21:35
• – Mithoron Oct 2 '17 at 22:18
• How about ICP-MS or ICP-AES? – logical x 2 Oct 3 '17 at 9:48
• @deusexmachina Why use destructive methods which require probe preparation when there is XRF? :) – andselisk Oct 3 '17 at 17:45
• I did XRF for years. For such a high purity (99.99%) you'd have to analyze for the impurities not the gold itself. – MaxW Oct 11 '17 at 19:19

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.