The term, yield, carries the same meaning in chemistry as it does in any field where an amount is produced. Farmers look at yield of a crop per acre. Bankers look at the yield (interest earned) on savings, investments, etc. Chemists look at the yield of a chemical reaction. Yield consistently means the amount produced to all these individuals.
The theoretical yield of a reaction is determined from the balanced equation. The balance tells us that for some input (reactants for chemists; seed, water, and fertilizer for farmers) we should get a certain output. Recall that all chemical reactions express amounts in terms of atoms, molecules, or moles. Chemical reaction equations cannot express masses directly because chemicals interact part to part and masses are not parts. Think about nuts bolts, and washers. In a normal application, one nut and one washer are attached to a single bolt. The bolt, the washer, and the nut each have their own mass but the interactions are part to part. The masses have no real bearing on the interaction. We can, however, measure matching numbers of bolts, washers, and nuts by weighing some (figuring out the individual masses just like we figure out atomic or molecular masses) and then weighing large batches of each just as a chemist weighs chemicals. Just as bolts, washers, and nuts are counted individually, moles, as should be understood, is a count of parts. A single mole is always 6.022 x 1023 parts, whether the parts are atoms, molecules, or what have you.
Here's an example:
Iron (II) oxide is heated in a furnace with carbon. This produces Fe metal and carbon dioxide. The basic translation of these English statements into "chemistry" is:
The theoretical yield of Fe product from FeO is 2 moles for every two moles reacted. This ratio can be reduced to 1:1 because 2:2 is equivalent to 1:1. BUT we can't measure a mole directly. We can only measure mole equivalents through the use of mass. Chemists use atomic masses for elements and molar masses for compounds. Thus one needs to calculate the molar mass of FeO which is 55.847 g + 15.999 g = 71.846 g. The atomic mass for Fe is read directly from the Periodic Table and is 55.847 g. So the reaction equation above indicates that reacting 143.692 g (2 moles) of FeO with at least one mole of C will produce 111.654 g (2 moles) of Fe. I say at least one mole of C because the reaction equation shows that the minimum ratio of FeO:C is 2:1. Any more than 1 mole C will simply leave some unreacted C.
Exact moles are not always used in an experiment. Some chemicals are expensive; some can be quite toxic. Various constraints may cause the chemist to work with parts of a mole BUT the balance above tells us we are fine as long as we maintain the ratios of reactants. This means that for any mole amount of FeO, we need half that amount (2:1) of C for a balanced reaction.
As an example, lets assume someone brought in an FeO sample that weighed 215.0 g. If the sample is pure, we can calculate the theoretical yield of pure Fe using the reaction equation shown above. The first step is to figure the number of moles of FeO to which 215.0 g is equivalent.
1 mole x moleSolving for x we find x= 2.99 moles
-------- = --------
71.846 g 215.0 g
From the equation above, this means that we should produce 2.99 moles of Fe since the reaction is 1 mole:1 mole. See that 2.99:2.99 is equivalent to 1:1. Knowing that the theoretical yield is 2.99 moles, we can calculate the theoretical yield as a mass instead of as moles, since we can't weigh moles without "Dr. Boyle's Mole Balance(tm)". The calculation is found in the following manner.
x g 55.847 gSolving for x, we find that the original 215.0 g of FeO should (theoretically) produce (yield) 166.98 g of Fe.
----------- = ----------
2.99 moles 1 mole
This brings us from the world of theory to the world of reality. Of course, nothing we do is 100%. Our bodies are not 100% efficient at using the food we eat. Our autos are not 100% efficient at burning the fuel we pour into them.
In refining the ore (FeO) above, reality sets in. In all lab procedures, there is a dose of reality. The amount of material that is produced when the experiment is run is known as the actual yield. Just as a farmer has an expected yield per acre (theoretical yield), the same farmer has an actual yield (harvest) per acre. Chemists are no different. The farmer might figure the percentage yield and so might chemist. In both cases, the calculation is simple. In both cases, the actual yield is measured at the final harvest stage. The chemist removes the Fe from the reaction container. The farmer hauls the produce to the barn, silo, or what have you and the amount that was really produced is measured. Actual yields only come from doing the experiment. They cannot be obtained any other way. The reaction equation tells us the theory. Running the experiment gives us the real results.
And so we come to calculating percent yield.
The percent yield for a chemical reaction is calculated identically to percent yield of farm produce or a savings bond. We take the actual yield divided by the theoretical yield ad multiply by 100 to normalize the data. Percents are numbers normalized to the range of 0-100. Normalizing means adjusted from what the range is actually to a desired range.
I haven't run the experiment but I can give you an actual yield off the top of my head just for the sake of calculating % yield.
Assume, after the reaction was carried out, the Fe was weighed and found to be 125 g. The % yield is found thus:
125 g FeSo here you have the various parts of theoretical yield, actual yield, and percent yield. Remember to determine the theoretical yield, one MUST have a balanced reaction. Then IF the amount of reactants is given in any unit other than moles, one must convert to moles so that the amount of product can be determined. Lastly, one must convert back to the starting units, say grams, to express the yield in terms that can be directly measured.
% yield = ----------- x 100 = 74.86%
166.98 g Fe
Go try some problems!