<p>Meteorite abundances are referenced to their silicon abundance and then that is bootstrapped onto the hydrogen scale by assuming that the silicon abundance of meteorites is the same as that in the Sun. See for example <a href="http://arxiv.org/abs/0909.0948" rel="nofollow">Asplund et al. (2009)</a>. 
This means there is an additional uncertainty of a couple of hundredths of a dex when comparing meteoritic and solar photospheric abundances.</p>

<p>So the answer to your second question is work out the abundances relative to Si and then use the solar Si/H ratio to put them on the hydrogen scale.</p>

CC BY-SA 3.0
2016-02-25T23:39:43.217Z
Hi thanks for your answer! It certainly helped and I managed to find a Journal paper describing the process that you just wrote. They add a **1.540** to the log(El/Si) value which turns out is a constant. Although, this they can only do for a certain no. of elements whose photospheric uncertainties are less than 0.1 dex.

The Journal paper: Solar System Abundances
of the Elements (Palme & Jones). 

Thanks again!
Gourav Mahapatra
CC BY-SA 3.0
2016-02-25T19:16:53.660Z
2016-02-25T23:46:28.337Z
2016-02-25T23:46:28.337Z
ProfRob
<p>In astronomy, solar abundances are often calculated or tabulated per $10^{12}$ H atoms. I understand that in case of the Sun this can be done because H atoms are in majority or are present for a scale comparision. </p>

<p>How can the same element abundance be calculated for an example meteorite which does not have any volatile gas such as H? In particular, I am referring to the table 1 of <a href="http://astro.uni-tuebingen.de/~rauch/TMAP/grevessesauval2001.pdf" rel="nofollow">http://astro.uni-tuebingen.de/~rauch/TMAP/grevessesauval2001.pdf</a>. It lists Solar abundance and lists meteorites abundance as a comparision. How have they calculated the element abundance for a meteorite in the $log_{10} \epsilon_{H} = 12$ scale?</p>

<p>Finally, I have a sample rock which has its's composition given in Weight $\%$ of various oxides such as $SiO_{2}, MgO, FeO$ etc. I have to convert these into elemental abundances in $log_{10} \epsilon_{H} = 12$  scale. How should I do this?</p>

<p>Thanks in advance!</p>

<the-sun><cosmology><meteorite><elemental-abundances>
How are the element abundances calculated for a meteorite in the Hydrogen log10 scale?
Gourav Mahapatra
# How are the element abundances calculated for a meteorite in the Hydrogen log10 scale?

(Asked by: username_0 on 2016-02-25)

In astronomy, solar abundances are often calculated or tabulated per $10^{12}$ H atoms. I understand that in case of the Sun this can be done because H atoms are in majority or are present for a scale comparision.

How can the same element abundance be calculated for an example meteorite which does not have any volatile gas such as H? In particular, I am referring to the table 1 of <http://astro.uni-tuebingen.de/~rauch/TMAP/grevessesauval2001.pdf>. It lists Solar abundance and lists meteorites abundance as a comparision. How have they calculated the element abundance for a meteorite in the $log\_{10} \epsilon\_{H} = 12$ scale?

Finally, I have a sample rock which has its's composition given in Weight $\%$ of various oxides such as $SiO\_{2}, MgO, FeO$ etc. I have to convert these into elemental abundances in $log\_{10} \epsilon\_{H} = 12$ scale. How should I do this?

Thanks in advance!

**Tags:** the-sun, cosmology, meteorite, elemental-abundances


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## 1 answers

### Answer by username_1 on 2016-02-25. Score: 4 | Accepted answer

Meteorite abundances are referenced to their silicon abundance and then that is bootstrapped onto the hydrogen scale by assuming that the silicon abundance of meteorites is the same as that in the Sun. See for example [Asplund et al. (2009)](http://arxiv.org/abs/0909.0948).
This means there is an additional uncertainty of a couple of hundredths of a dex when comparing meteoritic and solar photospheric abundances.

So the answer to your second question is work out the abundances relative to Si and then use the solar Si/H ratio to put them on the hydrogen scale.

> **Comments:**
> **username_0**: Hi thanks for your answer! It certainly helped and I managed to find a Journal paper describing the process that you just wrote. They add a **1.540** to the log(El/Si) value which turns out is a constant. Although, this they can only do for a certain no. of elements whose photospheric uncertainties are less than 0.1 dex.  The Journal paper: Solar System Abundances of the Elements (Palme & Jones).   Thanks again!  

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