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Thread: Why use a spectrometre?

  1. #1 Why use a spectrometre? 
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    Can someone please explain why you would use a spectrometer rather than a simple flame test to identify the atoms in an unknown mixture? I'm guessing that the spectrometre would divide the spectra into a much more specific pattern, so that you could identify the individual elements in the unknown mixure, wheras the flame would only give you a mix of colours, thus giving the flame a single coulour consisting of all of these different colours. So the flame test would make guessing the elements of the unknown mixture impossible. But I'm not sure if this is the best answer I can give to this question? Maybe they are looking for something more specific, or am I just overcomplicating things?


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  3. #2  
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    A flame test observation takes place in a lab. With light, I can see the evolution of the Universe.


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  4. #3  
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    Oh come on guys, that's not really even an answer. Can someone please confirm if I am on the right lines or not? If I gave an answer like that in a test my teachers would laugh at me.
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  5. #4  
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    I don't remember much but.... a flame test would yield a similar result to a spectrometer IF-and-only-IF the gasses were in very very low density state (such as in outer-space, like in nebulae). The reason is because: atoms in outer-space is spaced very far apart, but when the atom is close together: the atom cannot emit on similar frequency because it violated 1 principle of quantum mechanic: "2 object cannot have the same quantum state" (in this case: densely populated atoms has a same quantum-state at a very similar location thus it conflicted with the quantum law, thus each atoms must go to different state and emit different frequency). As a result: atom from a dense gasses will tend to emit a "broadband" of light frequency (like a smear...) which is inaccurate for material identification, but an atom from a dispersed gas shows a uniform/similar frequency at all time (a precise & distinct spectral line, like Hydrogen gas in space).

    The problem was with the quantum mechanic's principle/law.
    Last edited by msafwan; August 14th, 2011 at 04:03 PM.
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  6. #5  
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    And if you don't mind... I think this principle ("2 quantum state can't be same") can also complete the picture of why electron rather jump (in atomic orbit) rather than gradually decaying (like a space orbit) to a lower state in an atoms. Because: quantum scale has a specific "pixel size" to it: called the "planck scale" which denoted the smallest scale for any quantum changes, where any smaller changed can't exist, and we also know that electron orbited an atom and by classical-principle-of-electromagnetism we know that this orbit will cause an oscillation of electric field which would induce light-wave.... but a real electron can't behave like classical electromagnetism (and thus can't emit continuous frequency, but discrete one) because it must at least accumulate 1 "planck size" of photon energy before it can emit it, and this caused the electron to "jump" state rather than decaying.... As a result: they behave like "quantum pixel", and that is IMHO fascinating.
    Last edited by msafwan; August 14th, 2011 at 03:55 PM.
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  7. #6  
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    Quote Originally Posted by msafwan View Post
    I don't remember much but.... a flame test would yield a similar result to a spectrometer IF-and-only-IF the gasses were in very very low density state (such as in outer-space, like in nebulae). The reason is because: atoms in outer-space is spaced very far apart, but when the atom is close together: the atom cannot emit on similar frequency because it violated 1 principle of quantum mechanic: "2 object cannot have the same quantum state" (in this case: densely populated atoms has a same quantum-state at a very similar location thus it conflicted with the quantum law, thus each atoms must go to different state and emit different frequency). As a result: atom from a dense gasses will tend to emit a "broadband" of light frequency (like a smear...) which is inaccurate for material identification, but an atom from a dispersed gas shows a uniform/similar frequency at all time (a precise & distinct spectral line, like Hydrogen gas in space).

    The problem was with the quantum mechanic's principle/law.
    Dude I dig where you are coming from, but I suspect you are waaay overcomplicating things for the question I asked. At a guess I would say that it is just because a spectrometer would give a higher degree of specificity regarding the constituent elements of the mixture than a flame test would, particularly as a flame test may show the same result for different elements. (Boron copper and barium for example all give the same green flame). But I just want someone to confirm that this is right.
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  8. #7  
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    LOL. k, I think you are right.

    P/S: but that is based on what I've learnt. I've (sadly) never done any of those test myself! I think maybe you wanted someone to point out an anecdote for a similar test that confirmed your guess (which I have none).
    Last edited by msafwan; August 14th, 2011 at 05:32 PM.
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    raid517, I'm guessing that you're talking about a UV/Visible spectrometer, is that correct? A flame test is typically used to test for specific atoms (in fact there are only a few atoms that give specific flame colors Flame test - Wikipedia, the free encyclopedia). The flame test destroys any molecules in the solution. If you wanted to study molecules you would have to perform a different test, and there are a few different spectrometers that you could use. Conjugated molecules (such as benzene and napthalene) absorb UV/Visible photons, and that can be observed by a UV/Visible absorption spectrometer. You could also use a mass spectrometer to measure the mass of almost any molecule (not just ones that absorb particular wavelengths of light).

    Also, take a look at the guy doing that flame test in the wiki page above. A kung fu flame test?
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  10. #9  
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    I just thought of another thing, if you're looking at metals a flame test would tell you which metals are present, but you would most likely destroy the ligands. So if you wanted information about the ligands you could try to look at them by UV/Vis spectroscopy or mass spectrometry (with a UV/Vis spectrometer or a mass spectrometer). In this example the flame test and the spectrometer measurement are complimentary.
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  11. #10  
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    Hello,

    you'd use a flame test to tell you what is there and a spectrometer can tell you not only what is there but are A & B identical. In other words the flame test is qualitative (a color observed by the human eye only so subject to error) and the spectrometer is quantitative (provides actual information about the molecular structure by quantifying things like stretching vibrations, bending, twisting, etc). This allows you to compare the IR spectrum of one compound and tell if that compound is in fact identical to another compound. I will briefly explain this below - but it is not all inclusive..
    You can find a lot of info on this online though, because there are entire courses dedicated to molecular spectroscopy.

    Take for example IR, you can learn a lot about the environment of each atom in a molecule. Plenty of molecules have the same chemical formula - but different connectivity - therefore, you can basically learn a lot about the identity of a compound from getting an IR, because in no case of two different molecules will their absorption patterns be the same. Moreover, the IR can tell you the type of bonds that are present in your molecule.. say for example the difference between a single, double, and triply bound carbon-carbon bond.

    If you get something like an IR/UV-vis/MS this gives you structural information about your compound that you can then use to verify it's identity.

    Hope this was helpful,
    Cheers
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