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When an exothermic reaction occurs, the energy in the chemical bonds of the reactants is partially transferred to the chemical bonds of the products. The remaining energy is released as heat.

For example:

$$\mathrm{N_2 + 3H_2 \to 2NH_3} \qquad \Delta G^\circ = -32.96 \,\rm kJ/mol$$

Therefore, when $1\,\rm mol$ of nitrogen reacts with $3\,\rm mol$ of hydrogen (under standard conditions), we get $32.96\,\rm kJ$ of heat.

Now, applying $E=mc^2$, this works out to be

$$m = 32.96 \times (3 \times 10^{-8})^2 = 2.96 \times 10^{-14} \,\rm kg \quad \text{or} \quad 29.6\, pg$$

Does this relationship hold? Do the products of an exothermic reaction really weigh ever so slightly less than the reactants?

In a more general sense, does removing energy from a system decrease its mass (or vice versa)?

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    $\begingroup$ You do realize that it is very difficult to measure a difference of 30 pg when the mass of the object is 34 g; a magnitude difference of 1,000,000,000,000 times. Actually, the error in Avogadro's Number only allows for accuracy of 1 in a 1,000,000. So, it will be some time before anyone can prove your statement. $\endgroup$ – LDC3 Jul 24 '15 at 4:08
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    $\begingroup$ Absolutely - this is (at least with current technology) impossible to measure. I'm more curious as to what our current understanding of physics has to say about this. $\endgroup$ – Chinmay Kanchi Jul 24 '15 at 4:13
  • $\begingroup$ @LDC3: The precision with which we know Avogadro's number is not relevant. There is no reason to invoke units of moles in any such experiment. $\endgroup$ – Ben Crowell Aug 25 '18 at 20:39
  • $\begingroup$ The question seems to be posed in terms of passive gravitational mass, as opposed to inertial mass or active gravitational mass. If we assume that the equivalence principle holds, then this can be tested instead using a test that probes inertial mass or active gravitational mass. If the hypothesis is, on the other hand, that mass-energy equivalence fails for chemical reactions for one type of mass but not others, then I think this is ruled out by existing high-precision tests of the equivalence principle. $\endgroup$ – Ben Crowell Aug 25 '18 at 21:06
  • $\begingroup$ An exothermic chemical reaction like this one turns the energy of electric fields into heat. If mass-energy equivalence were to fail for either of these two types of energy, then I think we would already have seen that in various experimental tests of GR, such as Kreuzer, Phys. Rev. 169 (1968) 1007. $\endgroup$ – Ben Crowell Aug 25 '18 at 21:12
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As far as the theory goes, you are absolutely correct, the (negative) binding energy between atoms in a molecule contributes to the total mass of that molecule, so a stable molecule is less massive than the sum of the masses of its constituent atoms.

However (as you yourself calculated), the mass difference is absolutely tiny, and as far as I know, it has never been measured. But the principle is no different from the mass deficit that occurs in nuclear reactions and that, in turn, is readily measurable. Consider the atomic mass of deuterium ($2.01410178\,\rm u$) vs. helium ($4.002602\,\rm u$), which is about $0.64\%$ less than the mass of two deuterium atoms. The difference is the energy that would be released in a fusion reaction.

So yes, in general, removing energy from a system decreases its mass, and conversely, adding energy to the system increases its mass. The most extreme example perhaps would be protons and neutrons: roughly $99\%$ of their masses come from the (positive) binding energy between their constituent quarks, and only about $1\%$ is attributed to the quark rest masses.

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  • $\begingroup$ Another example where you can actually see this in experimentally measured values is the ionization of a hydrogen atom. Look at the mass of a proton, an electron, and an H atom, and you can (to about 2-3 sig figs if my memory holds) calculate the ionization energy. $\endgroup$ – Godric Seer Jul 24 '15 at 20:27
  • $\begingroup$ Is 2-3$\sigma$ possible? The ionization energy of H is $2.18\times 10^{-18}$ J, which is equivalent to $2.42\times 10^{-35}$ kg of mass. The proton mass is $1.67262178\times 10^{-27}$ kg; the electron, $9.10938291\times 10^{-31}$ kg. The total is $1.67353271\times 10^{-27}$ kg. The ionization energy is in the 8th significant digit. The best value I can find for the H atom mass is 1.007825 u; only 7 digits. Multiplying by 1 u = $1.660538921\times 10^{-27}$ kg and subtracting from the combined p+, e- mass gives $8\times 10^{-35}$ kg. Right ballpark, but beyond the last significant digit. $\endgroup$ – Viktor Toth Jul 24 '15 at 21:14
  • $\begingroup$ I remember doing it several months ago and got it within about 1 eV. I will hunt around and see if I can find what values I used. $\endgroup$ – Godric Seer Jul 24 '15 at 21:16
  • $\begingroup$ Can you please provide a reference for the last bit?... That most of the mass of protons and neutrons is in their binding energy. I've heard this a lot but am wondering about some solid sources. $\endgroup$ – N. Steinle Aug 25 '18 at 18:55
  • $\begingroup$ Look up the up and down quark masses, e.g., on Wikipedia (u: 2.3, d: 4.8 MeV). A proton is uud, so the quark rest masses amount to 9.1 MeV out of its mass of 938.3 MeV; a neutron is udd, so 12.1 out of 939.6 MeV. So 0.96% and 1.29%, respectively. The rest is binding energy. With error bars, its $0.96 \pm 0.19$% and $1.29 \pm 0.13$%, respectively, assuming I used my calculator correctly. $\endgroup$ – Viktor Toth Aug 25 '18 at 19:19
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Yes, bonds have mass, like every other kind of energy.

This can be significant; if you had a glueball (a hypothetical particle made of massless gluons), it would have mass, and all of the mass would be from the bond energy! Same would go if you somehow managed to bind photons together.

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  • $\begingroup$ As energy is given out when a bond is formed, wouldn't a glueball therefore have negative mass? That just seems weird... $\endgroup$ – Level River St Jul 24 '15 at 10:09
  • $\begingroup$ You can't assign mass to the "bond" unless you stipulate that every bond is in fact a particle. Which it isn't. $\endgroup$ – Carl Witthoft Jul 24 '15 at 13:13
  • $\begingroup$ @steveverrill Yes, it is negative. The thing you are really looking at is the mass of the whole system (for example, a water molecule), and subtracting the mass of the constituent atoms. If the atoms have less mass than the molecule, the "bond mass" is effectively positive, and breaking the molecule releases energy. If they have more mass than the molecule, the "bond mass" is negative, and breaking the molecule requires an energy input. The reason the mass (and energy) can be negative is that we're looking at mass relative to the system without the bond. $\endgroup$ – Luaan Jul 24 '15 at 14:16
  • $\begingroup$ @Luaan I wasn't talking about a molecule in which the component atoms have mass, I was talking about a "hypothetical particle made of massless gluons" as mentioned in this answer. $\endgroup$ – Level River St Jul 24 '15 at 16:56
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    $\begingroup$ You don't even need to look at something as exotic as a glueball to see a case where "bonding energy mass" is dominant. Take the simple neutron or proton. The cumulative rest mass of three u/d quarks is basically negligible compared with the overall mass of the composite object. Note, too, that in a related vein, when folks say that the higgs field is what "gives mass to" particles, they're talking about rest masses ONLY. So, in fact, most of the mass of a proton (and, by extension, most mass in the universe), is NOT a consequence of the Higgs field, but of binding energies. $\endgroup$ – Paco Jain Aug 21 '15 at 21:29
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I realize this isn't the main point of your question, but it seems worth mentioning that the Gibbs free energy change $\Delta G^\circ$ is not a measure of the amount of energy released in the reaction. It's an abstract quantity related to the entropy change of the universe. The relevant amount of energy released that you would use in the mass change calculation is the change in "internal energy", which for a reaction at constant volume and pressure is equivalent to the enthalpy change $\Delta H$. In your case where the number of moles of gas is decreasing, we would need more information about whether pressure or volume or neither is held constant.

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