Stoichiometry

[stoi-kee-om-i-tree]

What is Stoichiometry?

Yes, I know what you're thinking. What the heck is that big word and why does it sound like it belongs in a math class?! Don't worry, it's much easier than it looks. (and sounds.)

Stoichiometry is the measurement and calculation of the amounts of reactants and products in chemical equations. Let me break that down for you. Stoichiometry is a part of chemistry that studies the substances in a balanced equation. See? It's just a gigantic word that describes a simple idea. Stoichiometry is all about the numbers. What falls into Stoichiometry is molar masses, mole to mole conversions, mass to mass conversions, the limiting and excess reactants, theoretical and percent yield, and much more. Stoichiometry is the art of counting without counting and weighing without weighing.

How to solve a Stoichiometry problem!

Lets use Barium hydroxide and lead (ii) nitrate as an example!

  • Ba(OH) + Pb(NO₃)₂ → Ba(NO₃)₂ + Pb(OH)₂

Hmmm.. this looks balanced. Let's get started!

  1. Write a balanced chemical reaction. (That was easy.)
  2. Write the molar masses for each reactant and product under their respective formulas. The molar mass is the mass of a given chemical element or chemical compound divided by the amount of the substance. This is when we need to grab our periodic table and find the masses. We find the molar mass by multiplying the element to the following subscript. The product should be your molar mass! Our results for this reaction is (by the order of the reaction above) 171.34 g/mole, 331.2 g/mole → 225.32 g/mole, 224.2 g/mole.
  3. Mole to mole ratio- the conversion factor for any two reactants or products in a chemical equation. The first chart below shows mole to mole conversion. Mole A (the ?) should be filled in as the given, mole B should be the coefficient of the last product and last but not least mole A should be the coefficient of the reactant that was given to you. In this case we used my birthday as a given. 8.30Ba(OH)₂ + Pb(NO₃)₂ → Ba(NO₃)₂ + Pb(OH)₂ I followed and substituted what the chart tells me and my results were 8.30.
  4. Mass to mass conversion. Basically the same thing as mole to mole but more complex. The second chart below shows mass to mass conversion. Mass A (?) is the given, mole A is always one, mole B is the going to be the last product, molar mass B is the molar mass of the last product, molar mass A is the molar mass of the first reactant, mole A is the first reactant and finally, mole B is always one. In this case, 12.1 is our given mass. By following the chart, the results were 15.57 g.
  5. Limiting and excess reactants to find Theocratical Yield. The limiting reactant is the chemical that is used up first in a chemical reaction. It limits the amount of product that can be actually made. The excess reactant is the other reactant. Theoretical Yield is the amount of product produced by the limiting reactant. Our mass to mass conversion shows us the excess reactant which is Ba(OH)₂. This leaves Pb(NO₃)₂ as the excess reactant by using our given as 20.15. This leaves us our theoretical yield which is 13.64. (the product of our second mass to mass conversion using 20.15)
  6. Finally, percent yield. It's simple. Our given is 6.5 and we divide it by our theoretical yield. We multiply by 100 to create the percent and tada! 46.9%

Some diagrams and tips to help you solve!

Easy, isn't it? If not, you'll get the hang of it!

Jamie Lam 5th Period Chemistry PAP Mr. Mikesell