May 6-10

This week was anther continuation of stoichiometry and how to apply it. Last week, I said that stoichiometry can be used to calculate the number of moles of the products will be created based on the ratios and number of moles of the reactants. But, what if there were a limited or excess reactant?

In this case, the same steps can be applied. The only difference is that too much of a product will be left over while not enough or just enough of another product can be used to make the products.

The first step in finding the excess and limited reactants is to know the ratios of the compounds, which can be calculated by balancing the chemical equation.

The second step is to do some mental math. If the ratio of the reactants was 1:3 and the ratio of the actual amount of elements is different from that, then this will affect the yield that will be produced but keep the ratios the same or a different compound will form if the ratios aren't proportionate. For example, if the actual ratio were 1:2, then the larger number of reactants would be used up to maintain the ratio of reactants and to leave the smaller number of reactants left. Since all of 2 would be used up (33% less than 3), 33% of 1 would be used up (leaving 2/3 of the reactant left). The 2/3:2 is equivalent to a 1:3 ratio since 2/3 is 1/3 of 2. By analyzing the changing ratios from the original ratio of compounds to the actual compounds, you can find the excess ingredient as well as the limited reactant.

The above steps are how you solve simple stoichiometry problems. But, complicated ones may involve converting x g of this element to moles. In order to do so, you need to know the molar mass of the element, which can be found on the Periodic Table of Elements, hence making the Periodic Table of Elements quite useful. If you don't see molar mass in the squares, look for atomic mass because the two mean the same thing. The only difference is that atomic mass isn't necessarily attached to moles. It just determines how much mass an element contains. But, the molar mass is associated with the atomic mass per mole (6.02 * 10^23 particles).

Once you have found the molar mass, you can set up a proportion for the molar mass and the given mass of the element. Or, you could divide or multiply the given mass by the molar mass to find the number of moles depending on if the given mass greater than or less than the molar mass. If less than the molar mass, then it must be less than one mole, so you would simply divide the given mass by the molar mass. But, if the given mass is greater than the molar mass, then the molar mass must be divided by the given mass to find the number of moles.

Another complicated process of stoichiometry is if the compound contained more than one element. In this case, add up all the atomic masses together to find the molar mass. Another point to bring up is that molar mass can represent the total atomic mass of the compound whereas the atomic mass represents only one element.

One thing that wasn't mentioned was that stoichiometry only calculates the yield that could be formed. Theoretically speaking, for example, if an experiment involved combining a certain amount of sodium chloride with a certain amount of aluminum sulfate but you have a limited or excess amount of a compound, this will affect the yield you will produce. For example, if you could theoretically produce thrice the amount of sodium sulfate to twice the amount of aluminum chloride using 6NaCl + Al^2(SO4)3 ------- 3Na2(SO4) + 2Al(Cl3) but produced only 200g of sodium sulfate instead of the required 300g, then the yield is about 67%, or a ratio of 2:3. The yield percentage between moles and grams is proportional.

The yield percentage between moles and grams is proportional.

In order to calculate the yield produced, you must go through all the steps of stoichiometry. At the end, once you know the moles of a product produced compared to what could have been produced, you can divide the actual mass or moles of a compound by the theoretical mass or moles. Think of taking a test, and the test consists of 100 questions, but you only get 80% correct. The percentage of questions you got right are 80%. This concept can be applied to find the yield percentage of the products produced.