How to do Stoichiometry Woo~

By Kenny Vu

What is this "stoichiometry" you want to learn about?

Stoichiometry is defined as the relationship between reactants and products in a chemical reaction to find quantitative data.

Where would you start?

We will use the reaction between Sodium Sulfate and Strontium Nitrate as our example reaction. The reaction that'll occur will be a double replacement where Sodium Sulfate and Strontium Nitrate will yield to make Strontium Sulfate and 2 Sodium Nitrate.

Balanced Equation: Na2SO4(aq)+Sr(NO3)2(aq)->SrSO4(s)+2NaNO3(aq)

Molar mass for each reactant and product

142.01(sodium sulfate)+211.58(strontium nitrate)->183.65(Strontium sulfate)+169.94(sodium nitrate)

Now for the great math part!

mole to mole conversions

First we have to have a given so we'll use my birthday, February 2nd, which would be 2.02 and we'll use the moles of sodium sulfate to find the moles of Sodium Nitrate. You'd have to make a small table of 4 squares. The bottom left will be crossed and the top left will have the given of 2.02 moles of sodium sulfate. The top right square will have the coefficient for the moles of sodium nitrate and the bottom right will have the coefficient for the moles of sodium sulfate. The amount of moles you can make should be 4.04 g/mol.

mass to mass conversions

Now to do mass to mass we'll just use a given mass of sodium sulfate of 12.1g and find the mass of our second reactant, strontium nitrate. To do this, you'll need to use a longer 2x8 rectangle equation where on top the squares will be given mass of A, mole which is always 1, coefficient for mole B, and molar mass of B. The bottom squares will first box is crossed out, molar mass of A, coefficient for mole A, and then mole for B which will always 1. In the end, 12.1g of sodium sulfate could make 18.03g of strontium nitrate.

Limiting and excess reactant

This part is very important in an actual experimental test because the limiting reactant will show how much you should be able to make and excess is just the reactant that has more than enough for the reaction. To find this you just have to find mass to mass between the same given for both reactants and seeing how much they'll make of the same product. In our example, we'll use 12.3g of each reactant to find how much strontium sulfate they can produce. Sodium Sulfate will make 15.9g and strontium nitrate will make 10.68g, so since strontium nitrate made less, it is our limiting reactant and sodium sulfate will be our excess reactant.

Theoretical yield, actual yield, and percent yield

The theoretical yield is basically how much product can be made in theory because of our limiting reactant, so for this experiment, it would be 10.68g of Strontium Sulfate.

Actual yield is how much product is actually made from the experiment, for us it's given as 11.3g of strontium sulfate.

Percent yield is the percentage of how much product is made in comparison to how much should've been made and you find that by dividing actual yield by theoretical yield and then multiplying by 100. The percent yield for the example reaction should then be 105%.