Introduction
This series of posts seeks to present the material covered in the first semester of a college level general chemistry course, in an easily digestible steemit blog post format. The series is intended to be read, and experienced in sequential order starting with Post 1. The material will build upon itself, and potential exercises included (problem sets), will pertain to the post they are contained in, or any previous post. Each post will pick up immediately where the previous in the series left off. Please check out the #chemistry-lesson tag for all posts in this series. I hope you find this series to be informative and beneficial toward your understanding of chemistry and science in general.
Immediate Preceding Post
Part 4: Calculating Atomic Mass and Writing Chemical Reactions
Legend for This Section
As subscripts do not currently work on steemit, to symbolize when a number should be sub-scripted I will be writing it in the text as follows: /x/.
Using a Chemical Reaction to Calculate Theoretical Yield of a Product
So imagine we are in the lab, we have some materials sitting around to do a reaction, and we want to figure out what is the maximal amount of the product that could form from those materials. Well we can do this using balanced chemical reactions, just like the ones we learned to create in Part 4. In fact, let’s do just that using a reaction from the previous blog:
What is the maximum theoretical yield of H/2/O in grams if we were to combust 10 g of C/4/H/10/O?
Well the first thing we must know is the molar mass of C/4/H/10/O (butane), because whenever we use a reaction to convert between molecules or atoms in the reaction we must always be doing the conversions in moles. So the molar mass of C/4/H/10/O is equal to the sum of the atomic masses of each atom multiplied by the number of that atom present in the compound so C (4 * 12.01 g) + H ( 10 * 1.008 g) + O (16 g) = 209.3 g/mol C/4/H/10/O. We must also know the molar mass of the H/2/O product we are interested in, H (2* 1.008g) + O (16 g) = 18.02 g/mol H/2/O. Now that we know these we can use the coefficients in the reaction to convert:
Here what we did was first convert to moles using the molar mass we calculated, we then used the ratio of the stoichiometries from the reaction to convert between butane and water. Looking at the reaction we see that for every 2 butanes that are combusted we generate 10 H/2/O’s. So we set up this conversion. Now we must convert back from moles to grams using the molar mass of H/2/O, and generate our answer 4.3 g H/2/O. As a fun side note water weighs 1 g / mL so this generates 4.2 mL of H/2/O or 0.71 % of a 20 oz soda bottle (that is...if it were a liquid, however it’s not, it’s a gas as indicated by the (g) in the reaction).
Chemical Reactions with a Limiting Reagent
So when we set up a reaction, we often times have more of one component then the other, and the reaction is only able to proceed until its components run out. Since we have more of one than the other, the component that runs out first is known as the Limiting Reagent. Let’s go through an example dealing with limiting reagents. For this example let us use the reaction so many of us have done when making silly volcanoes for science fair projects, baking soda (sodium bicarbonate, NaHCO/3/) + vinegar (acetic acid HC/2/H/3/O/2/):
We react 20g of NaHCO/3/ together with 14 g NC/2/H/3/O/2/, what is the limiting reagent? How many grams of CO/2/ would be generated?
Here what we did first was to calculate the number of moles of both of our reactants, we can see that despite having a greater mass of material we actually had less of the Sodium Bicarbonate (it has the smaller number of moles) so it is our limiting reagent (and I put it in a red box so it would stand out! Hooray for emphasis!). Having identified the limiting reagent, we know that the reaction can only proceed until that reactant is consumed, so we use it to determine the theoretical yield of CO/2/.
Calculating Percent Yield
One thing that we can count on as chemists is that our reactions do not go to 100% completion, so our true yields are never what the theoretical maximum predicts they should be. However we can get a pretty good gauge of how everything went by calculating our percent yield. This value is just the ratio of the actual yield (in grams) over the theoretical yield (in grams). Let us go through an example of how to calculate percent yield. We will use the following reaction known as the Haber process:
Let’s say we reacted together 0.8 mol of N/2/ with 2.5 mols of H/2/ and got back 21 g of NH/3/. What is our percent yield? Let’s get to the calculation!
First we started by determining what the limiting reagent was. Since we were given moles this was easy, we just needed to use the coefficients of the reaction to determine which of the two starting materials makes more products. Here I say products because I am not using an actual product, there is no need since we just want to see how many moles the reaction is driven forward by. So, as we can see from our calculation, we make 0.8 moles with the N/2/ and 0.833 moles with the H/2/, since the reaction stops when the first one runs out, we can only go forward until N/2/ is gone, thus it is our limiting reagent. Next we use the limiting reagent to determine the amount of NH/3/ formed, and convert from moles of NH/3/ to grams with its molecular mass. This gets us our theoretical yield of 27.25 grams. Finally we determine the percent yield by dividing our experimental yield (21 g) by the theoretical one (27.25 g) and multiply by 100%. This shows that we had 77% of the maximal yield for the reaction, and as far as many reactions go, this is a fairly good yield!
End of Chemical Reactions continued Problem Set
We’ve reached the end of this section, and as with the end for previous sections, I will provide you with a problem set so you could test yourself to see if you are fully understanding the material.
Problem Set 4
Problem Set 4 Answer Key
Future Posts
Subsequent posts will cover: Acids Bases and other Aqueous Reactions, Electronic Configuration of Atoms, Chemical Bonding, and Molecular Geometry, and more.
Reference Figure: Periodic Table
Other References
Constants and Conversions List
Source for Additional Constants
Some Common Ions
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