What do the numerals in the photosynthesis equation mean?

Answer 1

Assuming that this is the photosynthesis equation you mean:

6CO2 + 12H2O + light → C6H12O6 + 6O2 + 6H2O

(The numerals that are directly to the right of a letter should be subscripted -- written below the line, but I don't know how to do that with this keyboard. Also note that there are no zeros in this, all the O's are letter O's.)

The first 6 on the left indicates that six molecules of carbon dioxide (CO2) are participating in this chemical reaction.

The 12 after the first + sign shows that twelve water (H2O) molecules are involved.

Notice that the sugar molecule (C6H12O6) on the right of the arrow doesn't have a numeral in front of it. That means that there is only one of them. To be completely consistent it should say 1C6H12O6, but chemists don't write that way.

Then in a similar manner there are six oxygen (O2) molecules and six water (H2O) molecules.

(The arrow is showing that the reaction starts with the things on its left, and the reaction ends with the things to the right of the arrow.)

Within each of the molecular notations, the numeral after the letters show how many atoms of each element are contained in each molecule -- again with 1's omitted. So CO2 indicates one atom of Carbon and two atoms of Oxygen. H2O is two atoms of Hydrogen and one of Oxygen. Sugar (I'm being vague about what kind of sugar this is because I'm not certain, but probably sucrose.) is more complicated with six atoms of Carbon, twelve atoms of Hydrogen, and six atoms of Oxygen. Oxygen usually occurs as a molecule with two atoms of Oxygen in each molecule.

Overall, what this formula is telling us is that for each molecule of sugar produced, six molecules of carbon dioxide and twelve molecules of water are consumed, and six molecules of oxygen and six molecules of water are also produced.

This business of measuring how much is used and how much is produced by a chemical reaction is called "stoichiometry" and is fundamental to the study of chemistry as a whole.