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How can you convert oxygen abundance values (12+log(O/H)) to metallicity values z. Like oxygen abundance of 8.69 is a metallicity of about 0.02 (solar metallicity). Thus, given a random abundance value, say 8.2, will a ratio of (8.2/8.69)*0.02 give the metallicity value?

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All you can do with an oxygen abundance is convert between various ways of expressing oxygen abundance.

You could assume some compositional mixture to estimate a metallicity. In your example, you could assume a compositional mix like the Sun and that the solar metallicity was $Z=0.02^{\dagger}$. In which case, the formula would be $$Z = 10^{(8.2-8.69)}\times 0.02\ ,$$ since the oxygen abundances are expressed on a base 10 logarithmic scale.

$\dagger$ Possibly too high - the Asplund et al. (2009) paper reporting the 8.69 ($\pm$ 0.05) figure for the oxygen abundance, has a photospheric $Z=0.0134$ and a bulk solar metallicity of $Z=0.0142$.

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