How can I calculate the expression of the abundance of a given element Xi in mass fraction from its abundance by number Ni/NH, where log(NH) = 12 is the abundance by number of hydrogen?

I've tried the following: suppose I have a star with, among others, an element Xi and hydrogen, then the mass fraction would be:

$X_{i} = \frac{m_{i}}{m_{tot}}$

and as the number of atoms is given by:

$N_{i} = \frac{m_{i}N_{A}}{M_{i}}$ ((($M_{i}$ is the molar mass)))

you can get:

$\frac{N_{i}}{N_{H}} = \frac{m_{i}M_{H}}{M_{i}m_{H}} = \frac{X_{i}M_{H}}{X_{H}M_{i}}$


$X_{i} = \frac{N_{i}M_{i}}{N_{H}M_{H}}X_{H}$

Now, let's take into account that $A = \log\left(\frac{N_{i}}{N_{H}}\right)+12$

so, finally:

$X_{i} = 10^{A-12}\frac{M_{i}}{M_{H}}X_{H}$

However, here: Difference in stellar abundance numbers they obtain this:

$X_{i} = \frac{N_{i}}{N_{H}}\frac{M_{i}}{M_{H}}\frac{1}{1-\frac{N_{i}}{N_{H}}}X_{H} = \frac{N_{i}}{N_{H}-N_{i}}\frac{M_{i}}{M_{H}}X_{H}$

the only difference is that I get $N_{H}$ and they get $N_{H}-N_{i}$.

Why do they get that and which is the correct way?

  • $\begingroup$ But, then there are more steps in that other calculation to compensate for Z. I think you have the simpler solution. $\endgroup$
    – eshaya
    Feb 18, 2023 at 18:37
  • $\begingroup$ @eshaya I don't understand... Suppose we have a start with many elements, Xi is one of them. My calculation is right? $\endgroup$
    – Elena
    Feb 18, 2023 at 19:33
  • $\begingroup$ Yes. The other is overly complicated and specific for Helium. $\endgroup$
    – eshaya
    Feb 18, 2023 at 19:59


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