Example:

In the reaction below we calculated a theoretical yield of 1 mol, and obtained an actual yield of 0.55 mol. What is the percent yield?

Example:

First let’s consider a non-chemistry example. Suppose you work at McDonalds assembling Big Mac’s. You are told the formula for a Big Mac is: “two all beef patties on a sesame seed bun”. This is your balanced equation. Now suppose your manager gave you 12 patties and 10 buns. If you start making up Big Macs, what reactant (buns or patties) will you run out of first?

The answer is that patties are limiting. You have enough buns to make 10 Big Macs, but you can only make 6 Big Macs with the patties you were given (because you have to put two patties on every bun). In this example beef patties are the limiting reagent.

Example:

In the above example equation 1 is one way you might see this reaction written, but it is not balanced. Notice that it takes two moles of NaI to react with every mole of ClCH2CH2CH2Cl, giving one mole of the diiodide product and two moles of NaCl (equation 2 is balanced). Therefore, you need twice as much NaI as ClCH2CH2CH2Cl in this reaction. Also notice that acetone is a solvent in this reaction. It is important for the reaction, but isn’t a reagent so it doesn’t figure into the yield calculation. A balanced equation is needed to correctly determine percentage yield.

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A Step-by-step Guide to Calculating Limiting Reagent, Theoretical Yield, and Percent Yield

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Yield calculations are common in chemistry. I've helped many frustrated students with these calculations in the past, so I developed this guide to help. Calculating percent yield actually involves a series of short calculations. Follow this step-by-step guide and you will be able to calculate limiting reagent, theoretical yield, and percent yield.

1. Write a balanced equation for the reaction

2. Calculate the molecular weight of each reactant and product

3. Convert all amounts of reactants and products into moles

4. Figure out the limiting reagent

5. Calculate the theoretical yield

6. Calculate the percentage yield

1. Write a balanced equation for the reaction: To figure out percentage yield you need to know the correct ratio of each of the reactants and products of interest (this is called stoichiometry).

• Many times reactions are not written in balanced form. Make sure you are looking at a balanced equation before trying to do any yield calculations.

Example:

Note that in the examples above HCl functions in different ways. In equation 3, HCl is a catalyst, and does not get consumed in the reaction. However, in equation 4, HCl is consumed and is a reagent.

2. Calculate the molecular weight of each reactant and product: You will need to know these numbers to do yield calculations.

• To calculate the molecular weight of a molecule, simply add up the masses of the individual atoms.

3. Convert all amounts of reactants and products into moles: Usually reactants are measured out by volume or mass. You need to know these quantities in terms of moles to do yield calculations. The conversion of volume and mass into number of moles can be done using the density and molecular weight of the material

• Mass can be converted to moles using molecular weight. Be sure to include all units in your calculations. It will help you to avoid errors. By insuring that the mass units cancel in the calculation you can be sure you have the calculation setup properly.

Example:

This time let’s consider a chemistry reaction in the balanced equation 2 from above. If we started with 10 mol of ClCH2CH2CH2Cl and 12 mol of NaI, which reagent will get used up first?

The answer is that NaI is limiting. You have enough ClCH2CH2CH2Cl to make 10 mol of ICH2CH2CH2I, but you can only make 6 mol of this product with the NaI that you started with (because you use two NaI molecules on every ClCH2CH2CH2Cl). Therefore, NaI runs out first and it is the limiting reagent.

Example:

Let’s consider a simple example first, equation 3 from above. In this example, there is only one reactant (CH3)3COH, so this is the limiting reagent (remember HCl is a catalyst in this reaction). If we started with 1 mol of (CH3)3COH, how many moles of (CH3)2C=CH2 would we expect for a theoretical yield?

The answer is theoretical yield = 1 mol. The stoichiometry of this reaction is such that every molecule of the limiting reagent gives one molecule of (CH3)2C=CH2.

Example:

Now let’s consider an example from before. In part 4 above we determined that if we started with 10 mol of ClCH2CH2CH2Cl and 12 mol of NaI in the reaction below that NaI was the limiting reagent. Under these conditions, what is the theoretical yield of ICH2CH2CH2I?

The answer is theoretical yield = 6 mol. It takes two molecules of NaI to make one molecule of ICH2CH2CH2I.