Csdrhrt

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?

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.

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 kilograms 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.

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:
$$12.01 + 4(1.01) + 4(16.00) = 80.04text{g/mol}$$
Now checking the energy released:
$$m/text{mol} = frac{E/text{mol}}{c^2} = frac{802.3text{kJ/mol}}{3.0times 10^8 text{m/s}} = 8.9 times 10^{-9}text{g/mol}$$ 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.

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.