Conservation of Mass and Energy 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?
 A: 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.
A: 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.
A: 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.

A: 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.
A: 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.
A: 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. 
