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When calculating the molar mass, is there any measurable difference in total mass of the molecule due to the loss of binding energy?

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    $\begingroup$ The difference is real, so it is measurable. What kind of balance do you have? $\endgroup$ – Karsten Theis Apr 30 '19 at 15:41
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    $\begingroup$ Einstein showed that mass and energy are interchangeable through the equation $E=mc^2$. So an exothermic reaction loses mass, an endothermic reaction gains mass. However the mass change in chemical reactions is vanishing small and is ignored. For example in chemistry the molar mass of HCl is the atomic mass of H plus the atomic mass of Cl. $\endgroup$ – MaxW May 1 '19 at 5:37
  • $\begingroup$ @karsten what do you mean when you say "what kind of balance do you have"? $\endgroup$ – suse May 1 '19 at 14:26
  • $\begingroup$ @MaxW Thanks for the explanation- I was trying to reconcile the concept of molar mass and binding energy. $\endgroup$ – suse May 1 '19 at 14:51
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    $\begingroup$ @suse - She means what kind of balance do you use to weigh chemicals. There isn't a balance that can weigh with enough precision so as to measure the mass change of an explosion. $\endgroup$ – MaxW May 1 '19 at 15:30
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The difference is the best observed for products of nuclear reactions, related to nucleon binding energy due the strong force.

By comparing the mass of $\pu{2 mol}$ of $\ce{^1H2}$ and $\pu{1 mol}$ of $\ce{^4He}$, the difference is about $0.7\%$.

The chemical bonds have difference more then million times smaller.

E.g. if there is an exothermic reaction producing $\pu{500 kJ/mol}$, the mass difference per mol is $$\Delta m = \Delta E/c^2 = \pu{-5.56 ng }$$

That means, we have hardly ever scales with the needed accuracy.

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  • $\begingroup$ Mass energy equivalence give the relevant answer and it is very small. A hydrogen molecule weighs about 4ng per mol (2g) of hydrogen less that two hydrogen atoms (using a little mental arithmetic so don't hold me to the exact number). $\endgroup$ – matt_black May 1 '19 at 13:46
  • $\begingroup$ Sure, I do not say otherwise. $\endgroup$ – Poutnik May 1 '19 at 13:59
  • $\begingroup$ I didn't mean to imply you didn't: I was just using your calculation on a specific chemical reaction to put it in context. $\endgroup$ – matt_black May 1 '19 at 14:00
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Yes the binding energy of the electrons results in a loss of mass.

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2} \begin{array}{lrl} \hline \text{Particle} & \text{Mass in AMU} \\ \hline \text{Rest mass electron} & 0.&\hspace{-1em}000548579909070(16) \\ \text{Rest mass proton} & 1.&\hspace{-1em}007276466879(91)\\ \text{Sum rest mases} & 1.&\hspace{-1em}007825046788\\ \hline \text{Hydrogen atom} & 1.&\hspace{-1em}00782503224(9)\\ \hline \text{Hydrogen atom - Sum Rest masses} & -0.&\hspace{-1em}000000014548\\ \hline \end{array} $$

Difference is $1.4548\cdot10^{-8}$ amu.

1 amu is equivalent to an energy of 931.5 MeV.

$1.4548\cdot10^{-8}\times 931.5\cdot10^6 = -13.55$ ev

Guess what the binding energy of the hydrogen electron is?

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  • $\begingroup$ A slightly better example would be to use the binding energy of two hydrogen atoms forming dihydrogen since that is a stable chemical molecule with a bond in it. A hydrogen atom forming from an ion plus an electron isn't really forming a chemical bond. $\endgroup$ – matt_black May 2 '19 at 0:25

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