I have questions regarding data analysis for an inductively coupled plasma (ICP) experiment from a paper by Wang et al. [1].

The generic formula for the chemical compound is $\ce{Na_{2−x}Fe[Fe(CN)6]}$.

In the paper, it was mentioned that chemical formula and ICP data are

\begin{array}{lrr} \text{Formula} & \ce{Na} / \mathrm{ppm} & \ce{Fe} / \mathrm{ppm} \\\hline \ce{Na_{1.53}Fe[Fe(CN)6].{4.2}H2O} & 112300 & 357200 \\ \ce{Na_{1.67}Fe[Fe(CN)6].{3.9}H2O} & 116100 & 339400 \\ \ce{Na_{1.73}Fe[Fe(CN)6].{3.8}H2O} & 118400 & 332400 \\ \ce{Na_{1.68}Fe[Fe(CN)6].{3.9}H2O} & 112000 & 332700 \\ \end{array}

My question is how they determined the $\ce{Na}$ number ($1.53$, $1.67$, and so on) on each cases based from the $\mathrm{ppm}$? I tried to make a $\mathrm{ppm}$ ratio of $1:2$ for $\ce{Na}$ and $\ce{Fe}$ however the number does not match with the paper.


  1. Wang, W.; Gang, Y.; Hu, Z.; Yan, Z.; Li, W.; Li, Y.; Gu, Q.-F.; Wang, Z.; Chou, S.-L.; Liu, H.-K.; Dou, S.-X. Reversible Structural Evolution of Sodium-Rich Rhombohedral Prussian Blue for Sodium-Ion Batteries. Nature Communications 2020, 11 (1), 980. DOI: 10.1038/s41467-020-14444-4

1 Answer 1


Let R be the ratio of number of Na atoms to number of Fe atoms. Then, for the first compound, we have

$$\pu{R = }\frac{\pu{112300 ppm } /\ \pu{22.98977 g mol^{-1} } }{\pu{357200 ppm } / \ \pu{55.845 g mol^{-1} } } = \pu{0.7637 }$$

Since there are two Fe atoms in the compound, the Na number for the first compound is simply 1.527, i.e., twice the value of R calculated above.

Following the same procedure for the remaining three compounds, I get Na numbers of 1.662, 1.731 and 1.635. The last one is furthest from the value of 1.68 the authors published, but maybe 1.63 was read as 1.68.

  • $\begingroup$ Thank you very much for your concise explanation. It is really helpful. I also fell a bit strange about the last number but I guess it might be a mistake. Once again, thank you. $\endgroup$
    – Alvero
    Commented Jun 3, 2020 at 1:54
  • $\begingroup$ Thank you as well and glad to be of assistance! $\endgroup$
    – Ed V
    Commented Jun 3, 2020 at 1:55

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