A few days ago, I learned the lengthy process of salt analysis during my practicals. The process is based on deductive reasoning but is too lengthy.

What I am thinking is, can't it be done using the fact that all molecules have unique emission spectra? I searched for it and found that it's the same thing as the flame test which was covered in my class.

But a flame test is only used for metal ions. So, my question is: Is there a way in which I can determine the metal ion as well as the radical part(anion part) of the salt using spectroscopic methods? What equipment would I need for it?

  • $\begingroup$ Not sure what you mean by "the radical part of the salt" ... I am assuming you mean the anions? Beyond that, it's a good enough idea, but not one that you could implement very easily. To cover all kinds of salts, you'd need a wide range of spectrometers, since different ions might only give detectable signals in specific spectral regions. $\endgroup$ Jan 9, 2015 at 19:57
  • $\begingroup$ Yes. I mean the anions. I don't know why don't they show up during flame test. $\endgroup$
    – Yashbhatt
    Jan 10, 2015 at 13:27
  • $\begingroup$ Well, a full answer to that is pretty complicated, but it's easier if you start by thinking about why the metal ions do show up. They show up because the high-energy environment of the flame excites electronic transitions of the metal ions that relax by emitting light that shows up in the visible region of the spectrum. So, by inference then, the anions either don't have electronic transitions in this region of the spectrum, or in cases where they do have such transitions, the emission isn't intense enough to be detected by your eye. $\endgroup$ Jan 10, 2015 at 16:57
  • $\begingroup$ Well, if I pass the light through a prism and observe the emission spectrum. Would that work? $\endgroup$
    – Yashbhatt
    Jan 10, 2015 at 17:47
  • 2
    $\begingroup$ You are assuming that the atoms will stay bonded in the flame, which will probably happen for some strong bonds, while other weaker bonds will break. The chemistry of flames is quite complicated. That is why this technique of atomic emission spectroscopy/flame analysis is usually restricted to elemental analysis. You put the sample in an extremely high energy environment that is guaranteed to break almost every bond, so that only atomic ions are present. See inductively coupled plasma AES for example. $\endgroup$ Jan 10, 2015 at 17:58

1 Answer 1


Two of the methods are:

Atomic Absorption Spectroscopy - similar to the flame test, but is usually quantitative. The sample is heated (in a flame or furnace) and the absorbance at a defined wavelength is measured against standardized solutions. It will work for any atom if you use a tunable laser.

ICP/MS (Inductively Coupled Plasma/Mass Spectrometry) - a solution of the compound is ionized and passed into a mass spectrometer. Some atoms (especially halogens) cannot be properly ionized and passed into the mass spectrometer. Oxygen and nitrogen (and a few others) would be difficult to quantify since it may come from impurities in the sample.

  • $\begingroup$ X-ray fluorescence is another possibility. Best for Na, and elements above Na in atomic number. (Z>= 11) $\endgroup$
    – MaxW
    Mar 16, 2017 at 20:28

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