Conservation of Mass and Energy

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.

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.

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.