Stoichiometry- Chemistry Simplified!

If you're a high school student in Chemistry, you're diving into stoichiometry, and you either love it or hate it. Some students are all about these "mark booster" questions, while others would rather endure a root canal than tackle stoichiometry!

So what exactly is "stoichiometry"? Trust me, it is not as hard as you think! Let's take a simple chemical equation as an example:

1 N2 (g) + 3 H2 (g) -> 2 NH3 (g)

Interactions of elements in a molecular compound

Take good notice of the "coefficients" (or numbers) that surround the chemical compounds here. Those are the underlined numbers, NOT the numbers that are actually in the compound! These "coefficients" inform chemists of the mole ratios that each compound needs to be added in order to create the products. In other words, if you want to create 2 moles of ammonia (NH3) gas, you need to react 1 mole of nitrogen (N2) gas with 3 moles of hydrogen (H2). Simple, right?

The key to getting these coefficients is to balance the chemical equation. Balancing an equation means making the number of individual atoms equal on both sides of the reaction. Let's go back to our example:

1 N2 (g) + 3 H2 (g) -> 2 NH3 (g)

There are 2 Nitrogen atoms and 6 Hydrogen atoms on the reactants (left) side, and 2 Nitrogen atoms and 6 Hydrogen atoms on the products (right) side. We have to add the "coefficients" in order to balance these atoms!

These coefficients play a role in the mole ratios, which help us calculate any product produced, or the yields of any part of the chemical reaction. Let's say that we have reacted 10 g of N2 with excess H2 (we will go over limiting and excess reactants in the next post), and we need to find the mass of product produced. Let us first rewrite the reactant:product molar ratios based on the BALANCED chemical equation.

There is 1 N2: 2 NH3

There is 3 H2: 2 NH3

Since the ratios are given in moles, we must first convert our mass to moles. The equation to calculate moles is mass (in g)/molar mass (in g/mol), and the molar mass of N2 is 28 g/mol. So the moles of N2 (g) is 10 g/28 g/mol= 0.357 mol. So far, so good!

In order to get to the moles of the product, we need to use our calculated ratios. Since N2 is our limiting reactant (and we will go over this in the next post), it is the only reactant that matters for molar ratios, and the ratio of N2 to NH3 is:

1 N2:2 NH3

So for every mole of N2 reacted, there is 2 moles of NH3 reacted! So, in order to get to the moles of NH3, you need to double the moles of N2= 0.357 mol N2*2 mol NH3/1 mol N2= 0.714 mol of NH3.

We now have the moles of NH3, but not the mass of NH3. The equation to convert moles to mass is moles*molar mass of the compound, and the molar mass of NH3 is around 17 g/mol. If we have 0.714 mol of NH3, then multiply 0.714 mol*17 g/mol= 12.14 g of NH3. We finally got it!

It's as simple as that! Of course, in most stoichiometry questions, we will have to balance the equation first, as well as figure out the limiting and excess reactants. Look out for the post discussing limiting and excess reactants, and I will see you next time! If you liked this blog post, consider booking a session with me today.