English Resume: Predict Rendement of Product a Reaction

Yield

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):

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:

Ø  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.

Ø  Two or more reactions may occur simultaneously, so that some reactant is converted to undesired by-products.

Ø  Losses occur in the separation and purification of the desired product from the reaction mixture.

Ø  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.

Example:

This is an example of an esterification reaction where one molecule acetic acid reacts with one molecule ethanol, yielding one molecule ethyl acetate (a bimolecular second-order reaction of the type A + B → C):

120 g acetic acid (60 g/mol, 2.0 mol) was reacted with 230 g ethanol (46 g/mol, 5.0 mol), yielding 132 g ethyl acetate (88 g/mol, 1.5 mol). The yield was 75%.

Ø  The molar amount of the reactants is calculated from the weights (acetic acid: 120 g ÷ 60 g/mol = 2.0 mol; ethanol: 230 g ÷ 46 g/mol = 5.0 mol).

Ø  Ethanol is used in a 2.5-fold excess (5.0 mol ÷ 2.0 mol).

Ø  The theoretical molar yield is 2.0 mol (the molar amount of the limiting compound, acetic acid).

Ø  The molar yield of the product is calculated from its weight (132 g ÷ 88 g/mol = 1.5 mol).

Ø  The % yield is calculated from the actual molar yield and the theoretical molar yield (1.5 mol ÷ 2.0 mol × 100% = 75%).

How to Predict Products in Chemical Reactions

Chemistry students typically experience difficulty in predicting the products of chemical reactions. With practice, however, the process becomes progressively easier.

The first step identifying the type of reaction involved is usually the most difficult. The primary reaction types students encounter are displacement, acid-base and combustion. They are easily identified if the tell-tale signs are known. Displacement reactions involve two ionic compounds with cations and anions, such as sodium sulfate, in which sodium (Na⁺) is the cation and sulfate (SO₄²^─) is the anion. Ionic compounds always consist of a metal and a nonmetal or polyatomic (multiple-atom) anion. Decomposition reactions involve a single compound breaking into two or more compounds. Acid-base reactions must involve an acid (identified by its chemical formula that begins with “H,” such as HCl). Combustion reactions involve hydrogen or a hydrocarbon (such as CH₄) reacting with oxygen (O₂).

DISPLACEMENT REACTIONS   

Identify the cation and anion of the compounds involved in the reaction, as well as their charges. If necessary, refer to tables of cations and anions, such as the one available at Penn State University’s website (See Resources). Sodium chloride (NaCl), for example, consists of a sodium ion (Na⁺) and a chloride ion (Cl^─).

Exchange the anions of the two reactants to determine the products of the reaction. Displacement reactions take this general form:

AB + CD → AD + CB

Thus, for a reaction between sodium chloride (NaCl) and silver nitrate (AgNO₃):

NaCl + AgNO₃ → NaNO₃ + AgCl

Determine whether the products are soluble. This may require referring to a list of “solubility rules,” such as that at Southern Methodist University (see Resources). In the example from Step 2, NaNO₃ is soluble and thus remains in solution, but AgCl is insoluble and will form a precipitate.

Verify that the reaction is balanced by adding coefficients in front of the reactants and products as necessary to ensure that each type of atom is present on each side of the reaction arrow in equal numbers. In the example from Step 2, the left side of the equation contains 1 Na, 1 Cl, 1 Ag, 1 N, and 3 O; the right side contains 1 Na, 1 Cl, 1 Ag, 1 N, and 3 O. Thus, the reaction is balanced.

ACID-BASE REACTIONS

Identify the acidic compound (containing H⁺ in its formula) and the basic compound (usually a hydroxide, OH^─).

Determine the products according to the general reaction:

acid + base → salt + water

For example, the reaction of hydrochloric acid (HCl) with sodium hydroxide (NaOH) produces sodium chloride and water:

HCl + NaOH → NaCl + H₂O

Determine if the salt is soluble or insoluble by referring to the solubility rules.

Balance the reaction. In this case, the reaction from Step 2 is already balanced.

COMBUSTION REACTIONS    

Determine the fuel (the source of carbon and/or hydrogen) and the oxidant (the source of oxygen) (see Resources). If the combustion is carried out in air, the oxidant is assumed to be molecular oxygen (O₂). Other oxidants, such as nitrous oxide (N₂O), are possible, but this would require special reaction conditions.

Predict the products by assuming this general reaction:

Fuel + oxidant → CO₂ + H₂O

For example, propane (C3H8) combines with O₂ during combustion as:

C₃H₈ + O₂ → CO₂ + H₂O

Balance the reaction. For the example in Step 2:

C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O

References

https://en.wikipedia.org

http://sciencing.com/