Determining stoichiometry in ICP-OES data analysis
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.