Question for the day – Einstein, special relativity and atomic weights.

Sometimes a (scientific) thought just pops into one’s mind. Most are probably best not shared with anyone, but since its the summer silly season, I thought I might with this one.

Famously, according to Einstein, m  = E/c^^2, the equivalence of energy to mass. Consider a typical exoenergic chemical reaction:

 A → B, ΔG -100 kJ/mol.  

According to the above, the molecule looses 100 kJ ≡ 1.112650056053618e-18 g after transformation from A to  B. Not much, but possibly measurable using today’s very best technology.

Now for the questions that might arise.

  1. What sort of energy applies above?  If its a free energy, then thermal (zero point and entropic vibrational) energy must clearly contribute. Or is it total energy without thermal and entropic contributions? 
  2. Is the mass loss distributed equally amongst all the atoms. In other words, how much mass does any particular atom lose after reaction or is this question meaningless?
  3. Since clearly the atoms must each lose some mass, that must mean that their atomic weight is a function of the energy content of the molecule they are part of.  A molecule with a lot of internal energy (lets say octanitrocubane, which decomposes to carbon dioxide and nitrogen) must have heavier atoms in the form of cubane than as nitrogen gas.
  4. And to recapitulate the question above, how many orders of magnitude away (if any) might we be from being able to measure this? Or, one can repose this question by asking whether one can measure the mass lost by a battery after discharging?

As with most spontaneous questions, the answers are probably all out there somewhere. Just a matter of finding them!

Here is a real-world example. At the large hadron collider at CERN, about 1011 protons are accelerated to almost the speed of light. During this process, they acquire a mass approaching kgs (I do not recollect the exact value). It certainly is a surprisingly large mass! And it is a surprisingly large amount of energy that has to be injected to achieve this. And when the beam is quenched, that mass is very quickly lost (and a lot of heat is generated in the quenching tunnel).

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7 Responses to “
Question for the day – Einstein, special relativity and atomic weights.

  1. Cina Foroutan-Nejad says:

    Very interesting questions; I would like to try to answer question 2. To understand how much energy each atom has lost, we first need an energy decomposition method, which is able to break molecular energy into atomic ones. Let’s pick up QTAIM atoms for example. Now, having an energy decomposition approach like IQA, we can define how much the energy of each part of the molecule has changed. It includes both nuclear and electronic parts. Employing the famous mass-energy relationship, E = mc^2, we can define how much mass is lost/gained by each part of the molecule.
    Now, I have another question!
    We can easily define nuclear share in mass change. However, changes in molecular energy are all a result of electrostatic interactions, of course by Born-Oppenheimer approximation. Then at Born-Oppenheimer approximation neutrons do not change their masses upon chemical reactions. Right?

    Have a wonderful weekend

    • Henry Rzepa says:

      Re Energy decomposition: Like charges, that is of course very dependent on the method used. I was thinking that eg thermal energies are associated with vibrations which have normal modes. These are a function of more than one atom. So how does one partition these thermal energies into atoms and hence atomic weights?

      Re Neutrons. Well, how would one even begin to design an experiment to measure that? And if the mass of eg the neutron changes relative to the proton, would that affect nuclear properties?

      • Cina says:

        For a small molecule I can imagine that one can measure variation of the atomic energy as the molecule vibrates by following each normal mode and recomputing atomic energy as the interatomic distances vary. I agree that this will largely depend on the type of EDA used, but recently I submitted a manuscript in which discuss what types of EDA should not be used. I do not say what types are useful (that directly) but only a few, maybe just one EDA survives if we want to follow the conditions necessary to select a secure EDA. I don’t know if you are interested to know more about it, or not, but in short, we must make sure that our EDA energy components are state function, not path function as the interaction/bond dissociation energies are state function. This seems trivial but almost all EDA approaches suffer from this issue.

        About neutrons; it sounds pretty strange but perhaps worth trying to see if nuclear properties change in different molecules. Can we make a nucleus more/less stable by forming chemical bonds? If yes, to what extent?

  2. Raphael says:

    Well, one can weigh A and B in a rovibrational eigenstate, the difference should correspond to a total (inner) energy difference. Another experiment is to weigh an ensemble, then you’d obtain the corresponding thermodynamic quantity.

    For the rest: I don’t believe in atoms in molecules as well defined physical entities. It’s something from the world of models one can use to explain phenomena.

    But its a very nice thought experiment, that gives an alternative definition to chemical reaction energies!

    • Cina says:

      IMHO the concept of atom is not really important here. You can divide a molecule into subgroups (based on any criteria) then measure the changes in energy in each section, whether we call them atoms or not. To do so, you can use already known atomic partitioning approaches, or invent something from the very beginning.

  3. Peter W says:

    I hadn’t really thought about this, but I’m not seeing the same mass difference. This is what I am seeing. The protons in the LHC contain a great amount of energy and are traveling at a great speed. If we were to compare another group of protons with the same energy, but at a much reduced speed, they would have to have a much larger mass. In my way of thinking, it isn’t the mass that has changed, but the speed.

    If I apply this thinking to the exergonic reaction, then the effect must be either a change in mass as being asked or a change in velocity as E=mc^2 equates. With my thinking, you can not detect a change in mass independent of the velocity.

  4. Silly answer:

    Doesn’t it only apply to kinetic energy? that m in Einstein’s equation relates to rest masses. From those 100 kJ/mol lost only a fraction of the corresponding partition function would correspond to the translation component (kinetic energy) and that fraction of Energy is the one that corresponds to the change in mass and not the entire 100 kJ/mol.

    Thanks for the post; a fun exercise indeed!

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