By Dude156

2019-03-11 02:57:34 8 Comments

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?


@Dude156 2019-03-11 16:36:35

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.

@TechDroid 2019-03-11 03:31:37

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.

@safesphere 2019-03-11 03:46:10

@TechDroid 2019-03-11 04:24:18

Well, Einstein has taken it all. But thanks for the link, it added something.

@Alchimista 2019-03-11 09:15:54

I think the mass involved in atomic bombs is the order of grams.

@UKMonkey 2019-03-11 12:29:08

-1 for misquoting Einsteins equation (light has energy too)

@Luaan 2019-03-11 08:28:05

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.

@Alchimista 2019-03-11 09:18:42

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.

@Peter A. Schneider 2019-03-11 10:25:27

That is a funny remark. Very obviously classical chemistry is based on the notion of the preservation of mass; it's all about transformation, not creation or destruction. One of the earliest school experiments captures the gas resulting from a combustion and shows that actually no masses disappear. That's the foundation of all that follows.

@Luaan 2019-03-11 10:31:53

@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).

@Peter A. Schneider 2019-03-11 11:05:37

@Luaan I don't see how what you say questions the fundamental idea of the preservation of mass in chemical reactions. Wouldn't we be amazed if after an oxyhydrogen explosion we would have heavy water, with neutorns added from nowhere in particular?

@Luaan 2019-03-11 11:42:53

@PeterA.Schneider Of course. I've never talked about chemical reactions, only about stochiometric equations - which are an inherently imprecise description of potential reactions.

@Alchimista 2019-03-11 11:52:01

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.

@Peter A. Schneider 2019-03-11 11:55:19

@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.

@Peter A. Schneider 2019-03-12 10:22:19

@Luaan But then: How do you "count" atoms when you prepare a chemical reaction? Consider: You want to produce 1 mol of water. You know from the reaction balance that you'll need 1 mol of molecular hydrogen and 1/2 mol of molecular oxygen; but you obtain the mol of hydrogen by measuring 2g, and the half mol of oxygen by weighing 16g. The underlying assumption (pre-Einstein) is that the reaction does not change the atoms' or molecules' masses. A stochiometric equation is at the same time a mass equation, and is in practice always used that way.

@Luaan 2019-03-12 11:52:12

@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.

@BowlOfRed 2019-03-11 04:43:55

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.

@rob 2019-03-11 05:12:29

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.

@PhysicsDave 2019-03-11 15:42:21

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?

@BowlOfRed 2019-03-11 16:07:41

@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).…

@PhysicsDave 2019-03-11 16:28:35

Thx BORed, I see that excellent answer from David Z. Relativity is neat in all its implications, but there is a lot of other more practical physics going on in this reaction that is being ignored ? such as entropy and other thermodynamics ... It's really annoying that identical apples, one on the tree and one on the ground have different masses.

@BowlOfRed 2019-03-11 16:48:30

@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.)

@PhysicsDave 2019-03-12 00:32:00

If we carry the 80 grams of material up a 110m hill it looks to me then that the mass is conserved again!.

@rob 2019-03-12 05:05:40

@PhysicsDave In the example of a falling apple, we say that the energy stored in the gravitational field between the two objects has changed. In the example of a chemical reaction, we say that the energy stored in the electromagnetic field around the atoms has changed. In the chemical case, for simple systems, it's sometimes possible to compute the energy of the photons that will be emitted, but lots of chemistry is done using calorimeters rather than spectrometers.

@Luaan 2019-03-12 09:30:21

@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.

@PhysicsDave 2019-03-11 04:17:25

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.

@Peter A. Schneider 2019-03-11 10:29:19

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?

@my2cts 2019-03-11 14:05:14

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.

@PhysicsDave 2019-03-11 16:30:06

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.

@Dude156 2019-03-11 16:42:38

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.

@Luaan 2019-03-12 12:08:49

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.

@F16Falcon 2019-03-11 03:38:43

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.

@Alchimista 2019-03-11 09:11:31

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.

@Peter A. Schneider 2019-03-11 10:21:05

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.

@jkej 2019-03-11 11:03:40

-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.)

@Alchimista 2019-03-11 11:46:43

@Peter A. Schneider. While you comment is in line with mine and the other one, one has to carefull with "exchangeability". Big heads got confused about that and it is still matter of dispute. But in this context I agred and I can also be more critic to the answer than in the first comment. The answer is misleading.

@HelloGoodbye 2019-03-11 16:00:12

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.)?

@jkej 2019-03-11 17:00:06

@HelloGoodbye The short answer to your first question is Mass–energy equivalence. You could view it as the bonds having negative mass, but I don't know if that is very helpful.

@F16Falcon 2019-03-12 01:06:46

@jkej Hi, I meant to add on to another answer that already explained that there will be a small (basically negligible) change in mass, hence why I wrote "most of" rather than "all of" energy. Sorry if it was unclear.

@HelloGoodbye 2019-03-12 08:09:56

@jkej I understand, but where is that mass located? Energy itself usually has no position, as far as I know, but is more a property of a whole system. Mass on the other hand, does have a position. If mass has a position but energy doesn't, they can't truly be equivalent, can they?

@Luaan 2019-03-12 09:40:02

@HelloGoodbye Mass is exactly the same - it's a property of a system (with the exception of a few elementary particles that have intrinsic mass, at least to the best of our knowledge). The mass of two atoms of hydrogen and one atom of oxygen is higher than the mass of a molecule of water. The difference is released as (mostly) heat when you react hydrogen and oxygen to form water. The same way, the mass of the hydrogen atom itself is smaller than the mass of an isolated electron and an isolated proton, and it works the same way for the electron and the proton.

@jkej 2019-03-12 09:49:26

@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.

@jkej 2019-03-12 09:50:53

@HelloGoodbye This is an interesting question. Maybe you should post it as a new question?

@Ed999 2019-03-12 11:21:23

Until I noticed the 11 comments, I thought it the most straightforward explanation possible for the question asked. It is the answer I wanted to give: that it is not the energy bound up in the particles themselves which is being released, but merely a part of the binding energy. There is no conversion of mass into energy occuring: the mass (i.e. the particles) survive unchanged, but they are now bound together in a way that requires less binding energy, and the released heat of combustion represents the surplus binding energy.

@HelloGoodbye 2019-03-12 15:14:59

@Ed999 According to what is stated here in the comment section, the mass is not only constituted of the particles, but also of the bindings (which add a negative mass). You seem to assume that bindings are mass-less. That is, you seem to disagree on weather bindings are massive.

@Ed999 2019-03-12 15:40:36

Smile! We aren't allowed to wrangle about mere comments, or we'll get moved to chat! Because energy has a mass equivalence, if the binding energy in the gluon field is reduced this will manifest as a reduction in the mass of the particle. There is no such thing as negative mass, but if there is less binding energy present (after the chemical reaction) it will show as a reduction in mass. Both the bond between the atoms in a molecule, and that between the particles and the spacetime field, seem to lose energy in the reaction: in both cases, this energy loss reduces the mass equivalence.

@Ed999 2019-03-12 15:47:24

... All those bonds are composed of energy, and all energy present has a mass equivalence. But the quantity of energy in the bonds is very little (compared to the energy locked-up inside the quarks), so small changes in the binding energy comprising those bonds will make only tiny changes in the mass equivalence of the atom/molecule in question. No energy is mass-less, but to contribute significant mass requires huge amounts of energy.

Related Questions

Sponsored Content

1 Answered Questions

5 Answered Questions

3 Answered Questions

1 Answered Questions

0 Answered Questions

Would impact angle matter on relativistic impactor?

2 Answered Questions

[SOLVED] Energy & Mass of a Photon

1 Answered Questions

[SOLVED] Why does mass change in to energy during a nuclear change?

1 Answered Questions

[SOLVED] Can non-free forces change the rest mass?

Sponsored Content