# How to perform stoichiometric calculations

Stoichiometry is the calculation of reactants and products in a chemical reaction.

**What’s a mole?**

Before we dive in, let me introduce you the concept of *mole*: a mole (symbol *mol*) is the quantity of matter, equal to the mass of a substance divided by its molar mass. But wat is this?

**Molar mass**

The molar mass of a substance is defined as the mass of a mole of that substance and is therefore measured in *grams per mole* (g/mol).

These are the things you need to know in order to proceed.

Let’s take as an example the reaction between hydrogen and oxygen, already balanced:

Two molecules of hydrogen react with one of oxygen to give two molecules of water.The molar mass of biatomic hydrogen is 2 g/mol (two times that of H), that of biatomic oxygen is 32 g/mol (two times that of O) and that of water is 18 g/mol (molar mass of hydrogen*2+molar mass of oxygen). You can check for the molar masses on Google. So we will say that 4 g (2 g/mol * 2 mol) of hydrogen react with 32 g (32 g/mol * 1) of oxygen to give 36 g (18 g/mol * 2) of water. To check that everything is correct, add up the quantities on the left side and see if the result is the same for the right side. 4 g + 32 g = 36 g. Everything is right!

**Formulas**

where *M* is the molar mass, *m* the mass and *n* the number of moles.

Suppose you want 50 g of water; you first calculate how many moles that corresponds to: 50 g/(18 g/mol)=2.80 mol. Now, in our reaction we have two molecules of water, so we have to divide our result by two (1.40 mol). Now we are ready to find out how much hydrogen and oxygen we need. Multiply 1.40 mol by the coefficient of H2, which is 2. The result, 2.80 mol, is the moles of biatomic hydrogen needed, but we want to know the mass, so we are going to use the formula m= M*n: m=(2 g/mol)*2.80 mol = 5.60 g. We need 5.60 g of hydrogen. The same thing is for oxygen: divide 1.40 mol by the coefficient of O2, which is 1. You obviously get 1.40 mol. m= M*n –> m= (32 g/mol * 1.40 mol)=44.80 g. In order for everything to be correct we must get 50 g (the mass of water) from the sum of the mass of the reactants: 5.60 g + 44.80 g = 50.4 g. I know, this is slightly more, but this is due to theapproximation of 2.77777… to 2.80 mol. My advice is to approximate only at the end when you get the grams, especially when you want to perform a reaction, in order to be as accurate as possible.

Hope this explanation was clear; if you have any doubt don’t hesitate to ask in the comments, I’ll be happy to help!

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