<p>All you can do with an oxygen abundance is convert between various ways of expressing oxygen abundance.</p>
<p>You <em>could</em> assume some compositional mixture to estimate a metallicity. In your example, you <em>could</em> assume a compositional mix like the Sun and that the solar metallicity was <span class="math-container">$Z=0.02^{\dagger}$</span>. In which case, the formula would be
<span class="math-container">$$Z = 10^{(8.2-8.69)}\times 0.02\ ,$$</span>
since the oxygen abundances are expressed on a base 10 logarithmic scale.</p>
<p><span class="math-container">$\dagger$</span> Possibly too high - the Asplund et al. (2009) paper reporting the 8.69 (<span class="math-container">$\pm$</span> 0.05) figure for the oxygen abundance, has a photospheric <span class="math-container">$Z=0.0134$</span> and a bulk solar metallicity of <span class="math-container">$Z=0.0142$</span>.</p>

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2022-07-10T08:21:03.187Z
2022-07-10T08:37:21.513Z
2022-07-10T08:37:21.513Z
ProfRob
<p>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?</p>

<galaxy><metallicity><elemental-abundances>
Conversion of oxygen abundance to metallicity
SlightDecoy
# Conversion of oxygen abundance to metallicity

(Asked by: username_0 on 2022-07-09)

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?

**Tags:** galaxy, metallicity, elemental-abundances


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## 1 answers

### Answer by username_1 on 2022-07-10. Score: 2 | Accepted answer

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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