REMDEMEN (CHEMISTRY)

In chemistry, yield, also referred to as reaction yield, is the amount of product obtained in a chemical reaction. The absolute yield can be given as the weight in grams or in moles (molar yield). The percentage yield (or fractional yield or relative yield), which serves to measure the effectiveness of a synthetic procedure, is calculated by dividing the amount of the obtained desired product by the theoretical yield (the unit of measure for both must be the same):

${\displaystyle {\mbox{percent yield}}={\frac {\mbox{actual yield}}{\mbox{theoretical yield}}}\times 100\%}$

The theoretical yield is the amount predicted by a stoichiometric calculation based on the number of moles of all reactants present. This calculation assumes that only one reaction occurs and that the limiting reactant reacts completely. However the actual yield is very often smaller (the percent yield is less than 100%) for several reasons:

• Many reactions are incomplete and the reactants are not completely converted to products. If a reverse reaction occurs, the final state contains both reactants and products in a state of chemical equilibrium.
• Two or more reactions may occur simultaneously, so that some reactant is converted to undesired by-products.
• Losses occur in the separation and purification of the desired product from the reaction mixture.
• Impurities are present which do not react

The ideal or theoretical yield of a chemical reaction would be 100%. According to Vogel's Textbook of Practical Organic Chemistry, yields around 100% are called quantitative, yields above 90% are called excellent, yields above 80% are very good, yields above 70% are good, yields above 50% are fair, and yields below 40% are called poor. These names are arbitrary and not universally accepted, and depending on the nature of the reaction in question, these expectations may be unrealistically high. Yields may appear to be above 100% when products are impure, as the measured weight of the product will include the weight of any impurities.

Purification steps always lower the yield, through losses incurred during the transfer of material between reaction vessels and purification apparatus or imperfect separation of the product from impurities, which may necessitate the discarding of fractions deemed insufficiently pure. The yield of the product measured after purification (typically to >95% spectroscopic purity, or to sufficient purity to pass combustion analysis) is called the isolated yield of the reaction. Yields can also be calculated by measuring the amount of product formed (typically in the crude, unpurified product) relative to a known amount of an added internal standard, using techniques like gas / liquid chromatography, or NMR spectroscopy. A yield determined using this approach is known as an internal standard yield. Yields are typically obtained in this manner to accurately determine the quantity of product produced by a reaction, irrespective of potential isolation problems. Additionally, they can be useful when isolation of the product is challenging or tedious, or when the rapid determination of an approximate yield is desired. Unless otherwise indicated, yields reported in the synthetic organic and inorganic chemistry literature refer to isolated yields, which better reflects the amount of pure product one is likely to obtain under the reported conditions, upon repeating the experimental procedure.

When more than one reactant participates in a reaction, the yield is usually calculated based on the amount of the limiting reactant, whose amount is less than stoichiometrically equivalent (or just equivalent) to the amounts of all other reactants present. Other reagents present in amounts greater than required to react with all the limiting reagent present are considered excess. As a result, the yield should not be automatically taken as a measure for reaction efficiency.

Limiting Reagent and Theoretical Yield

It’s a classic conundrum: We have five hot dogs and four hot dog buns. How many complete hot dogs can we make? 

Assuming the hot dogs and buns combine in a one-to-one ratio, we will be limited by the number of hot dog buns we have since we will run out of buns first. In this less than ideal situation, we would call the hot dog buns the limiting reagent or limiting reactant.

In a chemical reaction, the limiting reagent is the reactant that determines how much of the products are made. The other reactants are sometimes referred to as being in excess, since there will be some leftover after the limiting reagent is completely used up. The maximum amount of product that can be produced is called the theoretical yield. In the case of the hot dogs and hot dog buns, our theoretical yield is four complete hot dogs, since we have four hot dog buns. Enough about hot dogs, though! In the following example we will identify the limiting reagent and calculate the theoretical yield for an actual chemical reaction.

Problem solving tip: The first and most important step for any stoichiometric calculation—such as finding the limiting reagent or theoretical yield—is to start with a balanced reaction! Since our calculations use ratios based on the stoichiometric coefficients, our answers will be incorrect if the stoichiometric coefficients are not right. 

Example 1: Finding the limiting reagent

For the following reaction, what is the limiting reagent if we start with 2.80g of $\text {Al}$AlA, l and 4.25g of $\text {Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript?

$2 \text{Al}(s)+ 3\text{Cl}_2(g) \rightarrow 2 \text{AlCl}_3(s)$2Al(s)+3Cl​2​​(g)→2AlCl​3​​(s)2, A, l, left parenthesis, s, right parenthesis, plus, 3, C, l, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, right arrow, 2, A, l, C, l, start subscript, 3, end subscript, left parenthesis, s, right parenthesis

First, let’s check if our reaction is balanced: we have two $\text{Al}$AlA, l atoms and six $\text{Cl}$ClC, l atoms on both sides of the arrow, so we are good to go! In this problem, we know the mass of both reactants, and we would like to know which one will get used up first. In the first step, we will convert everything to moles, and then we will use the stoichiometric ratio from the balanced reaction to find the limiting reagent.

Step 1: Convert amounts to moles.

We can convert the masses of $\text {Al}$AlA, l and $\text {Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript to moles using molecular weights:

$$moles of Al=2.80g Al×1mol Al26.98g Al=1.04×10−1mol Al   (Convert g Al to mol Al)\text {Cl}_2

moles of Cl2=4.25g Cl2×1mol Cl270.90g Cl2=5.99×10−2mol Cl2  (Convert g Cl2 to mol Cl2)

Step 2: Find the limiting reagent using the stoichiometric ratio.

Now that our known quantities are in moles, there are multiple ways to find the limiting reagent. We will show three methods here. They all give the same answer, so you can choose your favorite. All three methods use the stoichiometric ratio in slightly different ways.

METHOD 1: The first method is to calculate the actual molar ratio of the reactants, and then compare the actual ratio to the stoichiometric ratio from the balanced reaction. 

Actual ratio=​moles of Cl​​moles of Al​​​2​​=​5.99×10​−2​​mol Cl​2​​​​1.04×10​−1​​mol Al​​=​1 mol Cl​2​​​​1.74mol Al​​

 The actual ratio tells us that we have 1.74 mol of $\text{Al}$AlA, l for every 1 mol of $\text{Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript. In comparison, the stoichiometric ratio from our balanced reaction is below:

$\text{Stoichiometric ratio}=\dfrac{2 \,\text{mol Al}}{3 \,\text {mol Cl}_2}=\dfrac{0.67 \,\text{mol Al}}{1\,\text{mol Cl}_2}$Stoichiometric ratio=​3mol Cl​2​​​​2 

mol Al​​=​1mol Cl​2​​​​0.67mol Al​​S, t, o, i, c, h, i, o, m, e, t, r, i, c, space, r, a, t, i, o, equals, start fraction, 2, space, m, o, l, space, A, l, divided by, 3, space, m, o, l, space, C, l, start subscript, 2, end subscript, end fraction, equals, start fraction, 0, point, 67, space, m, o, l, space, A, l, divided by, 1, space, m, o, l, space, C, l, start subscript, 2, end subscript, end fraction

This means we need at least 0.67 moles of $\text{Al}$AlA, l for every mole of $\text{Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript. Since our actual ratio is greater than our stoichiometric ratio, we have more $\text{Al}$AlA, l than we need to react with each mole of $\text{Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript. Therefore, $\text{Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript is our limiting reagent and $\text{Al}$AlA, l is in excess.

METHOD 2: A more guess-and-check way you can figure out the limiting reactant is by picking one of the reactants—it doesn’t matter which one—and pretending that it is the limiting reagent. We can then calculate the moles of the other reagent needed based on the moles of our pretend limiting reagent. For example, if we pretend that $\text {Al}$AlA, l is the limiting reagent, we would calculate the required amount of $\text{Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript as follows:

moles of Cl2=1.04×10−1mol Al×3mol Cl22mol Al=1.56×10−1mol Cl2

Based on this calculation, we would need $1.56\times 10^{-1}\,\text{mol Cl}_2$1.56×10​−1​​mol Cl​2​​1, point, 56, times, 10, start superscript, minus, 1, end superscript, space, m, o, l, space, C, l, start subscript, 2, end subscript if $\text{Al}$AlA, l is actually the limiting reagent. Since we have $5.99 \times 10^{-2} \,\text {mol Cl}_2$5.99×10​−2​​mol Cl​2​​5, point, 99, times, 10, start superscript, minus, 2, end superscript, space, m, o, l, space, C, l, start subscript, 2, end subscript, which is less than $1.56\times 10^{-1}\,\text{mol Cl}_2$1.56×10​−1​​mol Cl​2​​1, point, 56, times, 10, start superscript, minus, 1, end superscript, space, m, o, l, space, C, l, start subscript, 2, end subscript, our calculation tells us that we would run out of $\text {Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript before we fully reacted all of the $\text{Al}$AlA, l. Therefore, $\text{Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript is our limiting reagent.

METHOD 3: The third method uses the concept of a mole of reaction, which is abbreviated as mol-rxn. One mole of reaction is defined as occurring when the number of moles given by the coefficients in your balanced equation react. That definition can sound rather confusing, but the idea is hopefully more clear in the context of our example. In the current reaction, we would say that 1 mole of reaction is when 2 moles of $\text{Al}$AlA, l react with 3 moles of $\text{Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript to produce 2 moles of $\text{AlCl}_3$AlCl​3​​A, l, C, l, start subscript, 3, end subscript, which we can also write as

$1\text{mol-rxn}=2\text{mol Al}=3{\text{mol Cl}}_2=2{\text{mol AlCl}}_3$​​1, space, m, o, l, negative, r, x, n, equals, 2, space, m, o, l, space, A, l, equals, 3, space, m, o, l, space, C, l, start subscript, 2, end subscript, equals, 2, space, m, o, l, space, A, l, C, l, start subscript, 3, end subscript

We can use the above relationship to set up ratios to convert the moles of each reactant to moles of reaction:

 1.04×10−1mol Al×1mol-rxn2mol Al=5.20×10−2mol-rxn   (Convert mol Al to mol-rxn)

5.99×10−2mol Cl2×1mol-rxn3mol Cl2=2.00×10−2mol-rxn   (Convert mol Cl2 to mol-rxn)

The more moles of reaction you have, the more times the reaction can occur. Therefore, the reactant with fewer moles of reaction is the limiting reagent since the reaction can be carried out fewer times with that reactant. We see that this method also verifies that $\text{Cl}_2$Cl​2​​C, l, start subscript, 2, end subscript is our limiting reagent because it makes $2.00 \times 10^{-2} \,\text {mol-rxn}$2.00×10​−2​​mol-rxn2, point, 00, times, 10, start superscript, minus, 2, end superscript, space, m, o, l, negative, r, x, n, which is less than $5.20 \times 10^{-2}\,\text {mol-rxn}$5.20×10​−2​​mol-rxn5, point, 20, times, 10, start superscript, minus, 2, end superscript, space, m, o, l, negative, r, x, n from $\text{Al}$AlA, l.

Percent Yield

Amounts of products calculated from the complete reaction of the limiting reagent is called theoretical yields, whereas the amount actually produced of a product is the actual yield. The ratio of actual yield to theoretical yield expressed in percentage is called the percentage yield.

Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. Thus, the theoretical yield can be calculated from reaction stoichiometry. For many chemical reactions, the actual yield is usually less than the theoretical yield, understandably due to loss in the process or inefficiency of the chemical reaction.

The theoretical yield is the maximum amount of product you would expect from a reaction based on the amount of limiting reagent. In practice, however, chemists don’t always obtain the maximum yield for many reasons. When running a reaction in the lab, loss of product often occurs during purification or isolation steps. You might even decide it is worth losing 10% of your product during an extra purification step because it is more important to have extremely pure product—as opposed to having a larger amount of less pure product. 

Despite how nice and tidy a balanced reaction appears, reactants can also react in unexpected and undesirable ways such as doing an entirely different reaction—sometimes called a side reaction—to give products that we don't want. Your actual yield may change based on factors such as the relative stability of reactants and products, the purity of the chemicals used, or the humidity on a given day. In some cases, you might be left with all starting materials and no products after your reaction. The possibilities are endless!

Since chemists know that the actual yield might be less than the theoretical yield, we report the actual yield using percent yield, which tells us what percentage of the theoretical yield we obtained. This ratio can be very valuable to other people who might try your reaction. The percent yield is determined using the following equation: 

p, e, r, c, e, n, t, space, y, i, e, l, d, equals, start fraction, a, c, t, u, a, l, space, y, i, e, l, d, divided by, t, h, e, o, r, e, t, i, c, a, l, space, y, i, e, l, d, end fraction, times, 100, percent

Since percent yield is a percentage, you would normally expect to have a percent yield between zero and 100. If your percent yield is greater than 100, that probably means you calculated or measured something incorrectly.

Example:

The following reaction is performed with 1.56g of $\text {BaCl}_2$BaCl​2​​B, a, C, l, start subscript, 2, end subscript, which is the limiting reagent. We isolate 1.82g of our desired product, $\text {AgCl}$AgClA, g, C, l.

${\text{BaCl}}_2\left(aq\right)+2{\text{AgNO}}_3\left(aq\right)\to 2\text{AgCl}\left(s\right)+{\text{Ba(NO}}_3{)}_{\mathrm{}}$B, a, C, l, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis, plus, 2, A, g, N, O, start subscript, 3, end subscript, left parenthesis, a, q, right parenthesis, right arrow, 2, A, g, C, l, left parenthesis, s, right parenthesis, plus, B, a, left parenthesis, N, O, start subscript, 3, end subscript, right parenthesis, start subscript, 2, end subscript, left parenthesis, a, q, right parenthesis

What is the percent yield of the reaction?

First, we check to see if the reaction is balanced. It looks like we have equal numbers of all atoms on both sides, so now we can move on to calculating the theoretical yield.

Step 1. Find moles of limiting reagent.

We can calculate the moles of limiting reagent $\text {BaCl}_2$BaCl​2​​B, a, C, l, start subscript, 2, end subscript using the molecular weight:

1.56g BaCl2×1mol BaCl2208.23g BaCl2=7.49×10−3mol BaCl2

Step 2. Calculate moles of product.

We can calculate how many moles of $\text {AgCl}$AgClA, g, C, l we would expect to make using the stoichiometric factor from the balanced equation. The balanced equation tells us that we expect 2 moles of $\text{AgCl}$AgClA, g, C, l for every 1 mole of $\text{BaCl}_2$BaCl​2​​B, a, C, l, start subscript, 2, end subscript:

7.49×10−3mol BaCl2×2mol AgCl1mol BaCl2=1.50×10−2mol AgCl

Step 3. Convert moles of product to grams.

We can convert moles of $\text{AgCl}$AgClA, g, C, l to the mass, in grams, using the molecular weight, which will give us the theoretical yield in grams:

1.50×10−2mol AgCl×143.32g AgCl1mol AgCl=2.15g AgCl

We can use the theoretical yield and actual yield to calculate the percent yield using the following equation:

$\begin{aligned}\text{percent yield}& = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\%\\ \\ &=\dfrac{1.82 \,\text{g AgCl}}{2.15 \,\text{g AgCl}} \times 100\%\\ \\ &= 84.6\% \,\text{yield}\end{aligned}$​percent yield​​​​​​​=​theoretical yield​​actual yield​​×100%​=​2.15g AgCl​​1.82g AgCl​​×100%​=84.6%yield​​

Summary

• The theoretical yield for a reaction is calculated based on the limiting reagent. This allows researchers to determine how much product can actually be formed based on the reagents present at the beginning of the reaction.
• The actual yield will never be 100 percent due to limitations.
• Percent yield measures how efficient the reaction is under certain conditions.
• The limiting reagent is the reactant that gets used up first during the reaction and also determines how much product can be made. We can find the limiting reagent using the stoichiometric ratios from the balanced chemical reaction along with one of the many nifty methods.
• Once we know the limiting reagent, we can calculate the maximum amount of product possible, which is called the theoretical yield. Since the actual amount of product is often less than the theoretical yield, chemists also calculate the percent yield using the ratio between the experimental and theoretical yield.