9
$\begingroup$

This question already has an answer here:

I was thinking about some physics (relativity in particular), when it suddenly occurred to me that all my life I had been balancing chemical equations assuming conservation of mass, but I was disregarding energy!

For example, consider combustion: $$\rm CH_4 + 2O_2 \to 2H_2O + CO_2 + {Energy}$$

However, since energy was released, some mass should have been converted to energy right? Why is the equation reflecting a balance in mass?

$\endgroup$

marked as duplicate by BowlOfRed, user191954, Martin, Kyle Kanos, ZeroTheHero Mar 13 at 2:12

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I don't see any mass in this equation. $\endgroup$ – immibis Mar 11 at 23:11
8
$\begingroup$

Adding to TechDroid's answer, energy is also present in chemical bonds. When some higher energy (less stable) bonds are broken to form lower energy (more stable) ones (i.e. exothermic reactions), that energy difference can be released as energy.

So, almost all of that "+ energy" is due to the energy being released from the bonds themselves, and not the matter.

$\endgroup$
  • 1
    $\begingroup$ You can say so just because the changes are negligibles. It doesn't really add. The same could be true at nuclear level replacing chemical bonds with nucleons ones. Just in that case the involved masses undergo much change. $\endgroup$ – Alchimista Mar 11 at 9:11
  • 2
    $\begingroup$ It is funny that you take a clear distinction between mass and energy for granted while the exchangeability of both terms appears to be at the core of the question. $\endgroup$ – Peter A. Schneider Mar 11 at 10:21
  • 3
    $\begingroup$ -1 This answer seems to imply that the change in energy from chemical reactions does not correspond to a change in mass. It is precisely this misconception that needs to be dispelled. (Yes, the change in mass is hardly detectable, but it's there.) $\endgroup$ – jkej Mar 11 at 11:03
  • 1
    $\begingroup$ For someone as uninitiated as I, why does a change in energy correspond to a change in mass? Do you mean that a bond itself has a mass? How is that mass distributed in that case (i.e. I guess it would have it's own center of mass, inertia tensor, etc.)? $\endgroup$ – HelloGoodbye Mar 11 at 16:00
  • 2
    $\begingroup$ @F16Falcon There still seems to be a misconception here. Your phrasing seems to suggest that some of the energy comes from the bond and some from the matter. This distinction is meaningless. All the energy comes from a reconfiguration of chemical bonds, but this is equivalent to a change in mass, so it is also true that all the energy comes from the matter. $\endgroup$ – jkej Mar 12 at 9:49
23
$\begingroup$

Let's do an analysis and see how much of a difference this makes. The relevant enthalpies of formation are

  • Methane: −74.87 kJ/mol
  • Oxygen: 0
  • Water(vapor): −241.818 kJ/mol
  • Carbon dioxide: −393.509 kJ/mol

Therefore: $$\rm CH_4 + 2O_2 \to 2H_2O + CO_2 + 802.3 \text{kJ}$$ The mass of the products and reactants not worrying about the energy would be: $$12.01 + 4(1.01) + 4(16.00) = 80.04\text{g/mol}$$ Now checking the energy released: $$m/\text{mol} = \frac{E/\text{mol}}{c^2} = \frac{802.3\text{kJ/mol}}{(3.0\times 10^8 \text{m/s})^2} = 8.9 \times 10^{-9}\text{g/mol}$$ or just a bit more than 1 part in $10^{10}$.

So the amount of mass that is missed by not considering the energy is well below the level of precision that is normally used. The amount of mass lost in reactions can generally be ignored until you reach nuclear energies.

$\endgroup$
  • 2
    $\begingroup$ This is the answer I was coming here to write. Another way to see it is that ionization and excitation energies for an atom or molecule are typically measured in eV, while the mass of an atom or molecule is typically measured in $\rm dalton = amu \approx GeV/\mathit c^2$. A change of a few eV in a system like $\rm CH_4 + 2O_2$ with total mass $\sim 48\,\mathrm{GeV}/c^2$ is a correction to the mass starting in the tenth decimal place. $\endgroup$ – rob Mar 11 at 5:12
  • 1
    $\begingroup$ In order for mass to be converted to energy a sub atomic particle must be annihilated which is not the case in a chemical reaction. While it is interesting you can use E=mc2 to convert the energy into an apparent mass, it is off base. When an 100g apple falls from a 1m tree do we say 0.98J was converted to mass and the apple is lighter? I think not but if it's true tell me which particle in the apple was annihilated? $\endgroup$ – PhysicsDave Mar 11 at 15:42
  • 3
    $\begingroup$ @PhysicsDave no, no particle has to be removed for the mass to be different. See also the duplicate that was found which more explicitly goes into the equivalence (which I only assumed in my answer). physics.stackexchange.com/questions/11449/… $\endgroup$ – BowlOfRed Mar 11 at 16:07
  • 2
    $\begingroup$ @PhysicsDave We tend to regard the elements (like the apples) as identical, and the mass is lost from the system as a whole. This treats the potential energy(mass) as a property of the system. So the rest mass of the apple on the ground is the same as on the tree, but the gravitational system has less energy. You could apply the same to any other potential energy system (chemical, nuclear, spring, etc.) $\endgroup$ – BowlOfRed Mar 11 at 16:48
  • 1
    $\begingroup$ @PhysicsDave You're taking an assumption "mass -> energy means particles were annihilated" and using it to prove your conclusion that "mass -> energy means particles were annihilated". Mass is a property of a system - you're focusing on the system "all of the particles that form the apple", and completely ignoring the system "the apple and the rest of the planet". Mind you, as already noted, the difference is tiny (for both gravitational potential energy and chemical reactions), but it's very much there. $\endgroup$ – Luaan Mar 12 at 9:30
7
$\begingroup$

Bowl's answer is spot on, but I want to correct one thing that probably lead to your confusion - chemical equations do not balance mass. They balance moles (or individual atoms, if you prefer to think of it that way) - two carbon in, two carbon out, regardless of their configuration. This doesn't change, unlike the mass - carbon dioxide has a (very slightly) lower mass than one carbon plus two oxygen, but it still has one carbon and two oxygen.

$\endgroup$
  • 2
    $\begingroup$ Up voted but it is already taking into account the physics as it is. Practically in chemistry the balance involves both the moles and the masses coming along. Simply because there is no practical difference between the two. $\endgroup$ – Alchimista Mar 11 at 9:18
  • 4
    $\begingroup$ @PeterA.Schneider Yes, certainly. But that's also experiments from times where we didn't even realize that water has two hydrogen atoms, much less the absurdly tiny mass changes from absorbed/released energy. Chemistry is an old science. Modern stochiometric equations still represent the same idea, but aren't really tied to it. Two hydrogen atoms and an oxygen still form one water molecule, even if those hydrogen atoms have an extra neutron and thus mass twice as much (though sometimes that uses D instead of H, given how obvious the difference is). $\endgroup$ – Luaan Mar 11 at 10:31
  • 2
    $\begingroup$ It is not a funny remarks as for chemistry is certainly non separate from physics. Chemistry nowadays balances amount of substances rather than mass. Not to say that fixed proportions were as much as important as weight, in chemistry. $\endgroup$ – Alchimista Mar 11 at 11:52
  • 4
    $\begingroup$ @Luaan Ah, I see your distinction now -- yes, the equations are only counting atoms; any mass equivalent is implied only if we imply constant mass for the involved atoms. $\endgroup$ – Peter A. Schneider Mar 11 at 11:55
  • 1
    $\begingroup$ @PeterA.Schneider Yes, as soon as you work with any practical experiment, you use the masses and the assumption is "the mass of a molecule is equal to mass of the individual atoms making up the molecule". We just always used the molar mass as a simple conversion factor between the abstraction of the equation and the reality of our measurements (even though the rules for chemical equations originally came from the experiments, of course). I agree the distinction is very tiny, and entirely negligible for any practical purpose, at least in a relatively small-scale setting. $\endgroup$ – Luaan Mar 12 at 11:52
3
$\begingroup$

It actually does, but the amount converted is so small it's considered insignificant in the real world context. Based on the Einstein's famous equation ($E=mc^2$), a lot of energy can be extracted from a really small mass, and the reaction of methane and oxygen produces relatively small amount of energy which equates to a lot more smaller merely insignificant mass. The atomic bomb testiments to the amount of energy just some few grams of mass can decay into.

In addition the notion of the energy gained to achieve freedom for each atom reacting has to be given up to form a stable bond (that which sounds logical but I'm not entirely certain since I've not explored that domain very much) is also a solid argument to consider.

$\endgroup$
  • $\begingroup$ scientificamerican.com/article/… $\endgroup$ – safesphere Mar 11 at 3:46
  • $\begingroup$ Well, Einstein has taken it all. But thanks for the link, it added something. $\endgroup$ – TechDroid Mar 11 at 4:24
  • $\begingroup$ I think the mass involved in atomic bombs is the order of grams. $\endgroup$ – Alchimista Mar 11 at 9:15
  • $\begingroup$ -1 for misquoting Einsteins equation (light has energy too) $\endgroup$ – UKMonkey Mar 11 at 12:29
0
$\begingroup$

I think I figured out the answer (however, it may be incorrect, please do let me know). Also, this answer is very conceptual, rather than theoretical. We assume the following:

All protons experience a force of repulsion and electrons do the same. To counter this force of repulsion, intermolecular forces brought upon by hydrogen bonding, dipole dipole bonding, etc. keep the atom together.

For example, let's say that the repulsion force between two O2 atoms is 5 newtons (obviously not to scale). Thus, to maintain the particles together there has to be a counter force of 5 Newtons. When we break these intermolecular bonds (easily achievable by exciting the atoms through adding heat, causing the atoms to shake vigorously and loosening the intermolecular forces) the intermolecular force falls and the result is that the repulsive force applies a force of around 5 Newtons for some distance d. This Force*Distance is the very definition of Energy.

The excited particles now find other atoms to bond to. Since they have broken free, they can now bond to atoms that will allow a smaller force of repulsion. The favorable compounds in combustion are water and carbon-dioxide. No matter was every converted to energy.

$\endgroup$
-2
$\begingroup$

All the energy released is in the form of potential energy (of the electrons) falling to a lower (in general closer average positions) to the positive nuclei. This is similar to an apple falling off a tree. When this happens photons are released (no mass), molecules/atoms speed up and vibrations within the molecules and atoms increase (kinetic energy). All your chemical equations will have an energy balance but in addition you will need to take into account hidden thermodynamics, such as increased pressure and expansion of gases for example. This stuff is first year university, you will also learn about entropy ( why does salt melt ice?) which is another thermodynamic related energy concept required to balance.

In these reactions NO mass is converted to energy, mass is always conserved. In a nuclear reaction you again get photons, increased atomic/molecular motion but in addition you get high velocity sub-atomic particles like neutrons. Most (like >99% if I recall from wiki) of the mass is again conserved! You just get new types of atoms formed and isotopes (atoms that have absorbed a neutron). A few photons are indeed a result of a complex nuclear reaction where E=mc2 applies. But these are not of the same nature of the photons produced in a chemical reaction.

$\endgroup$
  • 1
    $\begingroup$ So tell me: What is the difference "in nature" of photons produced by chemical reactions as opposed to those produced by nuclear reactions? The latter ones usually have a higher frequency. Does that qualify as different nature? $\endgroup$ – Peter A. Schneider Mar 11 at 10:29
  • $\begingroup$ The mass of an atom or molecule includes the potential and kinetic energy of its constituents. It is useless to separate thus into mass and bond energy. $\endgroup$ – my2cts Mar 11 at 14:05
  • $\begingroup$ OK if we are talking relativity than it's really annoying that identical apples, one on the tree and one on the ground have different masses. $\endgroup$ – PhysicsDave Mar 11 at 16:30
  • $\begingroup$ I believe that this is the correct answer. Lol, I didn't even read your answer, but we both thought of the problem in the same way. $\endgroup$ – Dude156 Mar 11 at 16:42
  • $\begingroup$ Mass is only conserved if you only consider all of the products of the reaction as one system - that is, if you add the energy released (either the kinetic energy of the particles or the energy of the photons) to the mass of the products themselves. Of course that's true - that's what mass is in the first place. But that doesn't change the fact that if you precisely measure the mass of the reactants, and then do the same with the products, you'll get two very slightly different numbers; and that's what we're talking about when considering if mass is conserved in an interaction. $\endgroup$ – Luaan Mar 12 at 12:08

Not the answer you're looking for? Browse other questions tagged or ask your own question.