Chemical Reactions and Moles

When you’re working with chemical reactions, moles can help you figure out how much of a product you can expect to get based on how much of the reactants you have. Take a look at my favorite reaction, the Haber process, which is a method of preparing ammonia (NH3) by reacting nitrogen gas with hydrogen gas:

N2(g) + 3 H2(g) ↔ 2 NH3(g)

we use this reaction over and over again for various examples (like We said, it’s my favorite reaction) and explain that you can read the reaction like this: one molecule of nitrogen gas reacts with three molecules of hydrogen gas to yield two molecules of ammonia:

N₂(g) + 3 H₂(g) ↔ 2 NH₃(g)

1 molecule + 3 molecules ↔ 2 molecules

If you like, you can scale everything up by a factor of 12:

N₂(g) +3 H₂(g) ↔ 2 NH₃(g)

12 molecules + 3(12 molecules) ↔ 2(12 molecules) 1 dozen molecules + 3 dozen molecules ↔ 2 dozen

molecules Or how about a factor of 6.022 × 1023? But wait a minute!  Is ’t 6.022 × 10²³ a mole? So you can write the equation like this:

N₂(g) + 3 H₂(g) ↔ 2 NH₃(g)

6.022 × 10²³ molecules + 3(6.022 × 10²³) molecules ↔ 2(6.022 × 10²³ molecules) 1 mole + 3 moles ↔ 2 moles That’s right — not only can those coefficients in the balanced chemical equation represent atoms and molecules, but they can also represent the number of moles.

If you know the formula weight of the reactants and product, you can calculate how much you need and how much you’ll get. All you need to do is figure the molecular weights of each reactant and product and then incorporate the weights into the equation. Use the periodic table to find the weights of the atoms and the compound (see the earlier section “Counting by Weighing” for directions) and multiply those numbers by the number of moles, like this: 1 mol(28.014 g/mol) + 3 mol(2.016 g/mol) = 2 mol(17.031g/mol)  28.014 g of N₂ + 6.048 g of H₂ = 34.062 g of NH₃