General Chemistry Concepts that Show up Again and Again – Stoichiometry, Limiting Reagent, and %Yield.

Dr. Grice again! My last blog post was about solvent properties (see here) and I wanted to make another post.

Disclaimer: I recognize it is April Fool’s day, but unfortunately this post does not contain any April Fool’s jokes or pranks. There are some good chemistry-related ones online, like this one.

Ok, on to the topic for today: General Chemistry I, II, and III introduce many key foundational topics that appear again and again in other chemistry courses, in research, and beyond. In our General Chemistry I Lecture, we cover the concepts of Stoichiometry, Limiting Reagent, and Percent Yield. I wanted to highlight these topics because they are used in all areas of chemistry, including in organic chemistry laboratory, other classes, and our research labs.

These topics are particularly important whenever you make (or “synthesize”) a chemical, which is why they show up so much in organic chemistry lab, inorganic chemistry, other labs involving synthesis, and in research.

Stoichiometry is the concept that we can use the “coefficients”, or numbers in front of chemical species in a balanced chemical reaction, to find how many moles of reactants are needed, how many moles of reactants will be consumed, or how many moles of product will be formed in a reaction, given some other information. A balanced chemical equation can be interpreted in terms of molecules reacting with molecules, or moles reacting with moles. So, in this reaction below, you can think of it as 1 molecule of N₂ reacting with 3 molecules of H₂, or you can think of it as 1 mole of N₂ reacting with 3 moles of H₂. A mole is a huge number of molecules (6.022 × 10²³ things per mole!), and brings the teeny tiny atoms up to the scale of things we can weigh.

We often start with masses of species (in grams or milligrams, g or mg) and obtain products as solids that we can weigh, so we can use molar masses to go from mass to moles of reactant or product, to moles or another reactant or product, to mass of that other reactant or product. You can calculate molar masses by hand from the periodic table, look them up for common chemicals, or use a handy calculator like this one if you know your formula. This is the general idea for using molar masses in stoichiometry:

For example, here’s a simple balanced chemical reaction, an acid-base reaction of triethylamine (sometimes abbreviated TEA) and hydrochloric acid to form triethylammonium chloride. Notice there are no numbers, or coefficients, in the chemical equation… that’s because they are all 1 (we usually just leave off the “1”). It’s 1 mole of triethylamine reacting with 1 mole of HCl to make 1 mole of triethylammonium chloride.

If you want to react 25.0 g of triethylamine, you can use stoichiometry to figure out how many g of HCl you will need. You use the molar masses and the coefficients as conversions in your calculation. Here’s how I would set it up to find how much HCl I need to react with 25.0 g of triethylamine. Start with what you know, the mass of triethylamine. Students often ask “do I mutiply or divide the values?”, and my answer is: “I don’t know, let’s check the units!”. When you arrange a series of conversions, make it so the units cancel out from what units you start with to get to the units you want. Notice how the units cancel on top and bottom (usually diagonally) and you are left in the unit you are trying to find. We call that approach of using units as our guide Dimensional Analysis.

So, to react with 25.0 g of TEA, we would need at least 9.01 g of HCl. Also, you can calculate the mass of product you will obtain from 25.0 g of TEA using stoichiometry in a similar fashion (we would obtain 34.0 g of TEAHCl):

Now, triethylamine is a liquid, so you can use it’s density to figure out the volume you will need to use. Many small organic compounds are liquids, and weighing out a liquid can be annoying, so this approach of using volume and density is common. Densities are often on the bottle of liquid chemicals, or can be found online in places like Wikipedia. Let’s say I wanted to make 75.0 g of triethylammonium chloride, how many milliliters of triethylamine would I need? We can calculate that, again starting with the known info:

Also, we often use reactants or reagents as solutions. For example, pure HCl is a gas, but is most commonly encountered as an aqueous solution, hydrochloric acid (HCl(aq)). It’s often sold as concentrated as possible, which is about 37% by mass (37 g of HCl for every 100 g of solution), which is about 12 M (M is molarity, which is moles of solute per liter of solution).

We often make more diluted solutions to use (using the dilution equation C₁V₁=C₂V₂). The HCl could be provided as a 1.00 M solution of HCl. You can always use volume multiplied by concentration to find the amount of solute in the solution (in this case, moles of HCl), and so those can also be used to find out how much 1.00 M HCl solution to use to react with 25.0 g of TEA. Here’s how many mL of 1.00 M HCl you would need to react with 25.0 g TEA. Again, start with your known value, the 25.0 g TEA you have that you want to completely react with the HCl:

So, you can use densities to convert between mass and volume of liquids, and molarity to convert between moles of solute and volume of solution. We can revise the stoichiometry scheme to include those ideas like this, and there are many different paths you could take, depending on if you are working with a solid, a pure liquid, or a solution.

The idea of Limiting Reagent is that in general, when we react things, we don’t measure out things to be perfectly the right values needed for the exact ratio in the balanced chemical reaction. In reality, we usually have one thing that will get used up first (the limiting reagent), and the other reactants or reagents will still be there, so they were “in excess” to begin with. You had more than you needed of those. Using the limiting reagent, you can figure out how much product you expect to form.

My favorite way to deal with limiting reagent calculations is to just take all reactants on to the product you want to form and see how much mass of product is formed. Whichever reactant forms the least product will get used up first and therefore is the limiting reagent. You now also know how much product you expect to form! There’s also a way where you calculate the moles of each reagent and compare ratios of moles with the stoichiometric coefficients, but I like the way I do it, because you get both which species is limiting and also how much product you will form.

Let’s say we react 45.5 mL of triethylamine with 105 mL of 2.00 M HCl, how much NEt₃HCl should we get? I’d do both calculations to product:

So, my limiting reactant is HCl the most NEt₃HCl I could make would be 28.9 g!

Now, we can talk about Percent Yield. This is the percentage of the expected mass or moles of product that you actually obtain from your reaction. Percent yield is just your actual obtained value (in mass or moles) divided by expected value from your stoichiometry calculations (in the same unit) multiplied by 100. You can never get above 100% yield. If your product weighs more than it should, then it’s impure! For example, if you isolate a product from water by filtration, and get a mass that is 115% yield, maybe there is still water in there, and you need to dry your solid by rinsing the water off with a solvent that has a high vapor pressure, or removing the water with heat or vacuum. This is why it’s so important to use spectroscopy such as NMR spectroscopy to check that your product is what you think it is and is pure. It can happen that you actually form something totally different than you expect, so your %yield calculations would be wrong if your actual product had a different molar mass or different coefficients in the balanced chemical reaction.

Looking back up at our last limiting reagent calculations, let’s say we react 45.5 mL of triethylamine with 105 mL of 2.00 M HCl (which are the values in the calculations above, that showed us we coulld only make 28.9 g TEAHCl), and then isolate our product and when it’s pure and dry, it weighs 17.7 g. We can calculate the percent yield like this: 

Ok, so how does this work in research? We do stoichiometry calculations every time we do a reaction! First we do it in planning to see how much we can make, and then use the actual numbers we measured out and obtained to get the percent yield. I often work with metal complexes and we generally make 50-200 mg of a complex to work with. So, I often use mmol in my calculations (you can use the molar mass number as g/mol or as mg/mmol!).

Let’s say I want to do this reaction below to form this cool copper cluster that emits different colors under blacklight when it’s room temperature or liquid nitrogen temperature (we make this cluster and similar ones in Inorganic Chemistry Laboratory!). Note that the stoichiometry requires 4 moles of CuI per 4 moles of pyridine to make 1 mole of cluster. When trying to figure out how much reagent to use, I often pick a round number like 100 mg, and calculate how much of the reactants I would need to make 100 mg of product.

Here’s the calculation for how much CuI I need for 100 mg of product:

Here’s the calculation for how much pyridine I need for 100 mg of product (in microliters, because it’s a liquid! Density in g/mL is also the same value in mg/µL!)

Note that KI and ascorbic acid are additives in the reaction, but not in the balanced chemical equation (they help increase the yield and make sure we get the product we want and not other stuff). If I want to make more than 100 mg, I can just “scale up” the reaction by multiplying everything by the same number. If I want 300 mg, I just triple the amount of each reagent. Also, even if I try to measure the values I calculated, I will add a little more of one reagent so it’s in excess.

Now, let’s say I reacted 625 mg of CuI with 275 µL pyridine and I isolated 584 mg of product, what’s my percent yield? Here’s what I would do: first do the limiting reactant calculation and at the same time find out what my 100% yield would be:

So, CuI was limiting, and I would make 885 mg of the cluster for 100% yield. Since I only obtained 584 mg, here’s my %yield calculation:

Some folks really care about high yields, and try to “optimize” a reaction to give the highest yield possible. For my research, as long as I have enough of the product for what I want to do, and it’s pure, I’m happy. A low yield is just fine. A higher yield is better, but not critical for a project where the main goal is some application of the metal complex.

Whenever you publish a paper about making something, you need to report how you did it in the “Experimental Section” and you need to include masses and moles (or mmoles) of reagents, volumes of solvents, time and temperature, and mass of product, as well as a percent yield! That way, other scientists can reproduce your work and should expect to get a similar percent yield if everything is done right. Maybe writing a good experimental section will be a topic of a future blog post!