Percent Yield

Why there is a percent yield...

Reactions rarely produce the predicted amount of product from the masses of reactants in the reaction .An example of this is the reaction of carbon with oxygen. Normally we expect a 1 mol yield of carbon dioxide for every mol of carbon burned. This does not always happen. 

C(s) + O₂(g) --- > CO₂(g)

If you burn 12 grams of carbon to make CO₂, then amount of carbon dioxide expected is one mol of CO₂ or 44 grams of CO₂.

Sadly the amount you will get will probably be less than 44 grams and more like 34 grams of CO₂. The problem is a competing reaction that happens. Some carbon reacts to make CO.

2 C(s) + O₂(g) --- > 2 CO(g)

The carbon participating in this "side" reaction will not be able to make CO₂. The reaction will not yield 100% of the expected CO₂.

The amount of carbon dioxide produced, 34 grams of CO₂ is only 77% and not 100 % of the expected 44 grams.

Percent Yield = 100 x ( 34 grams CO₂ actual / 44 grams CO₂ predicted ) = 77 %

The percent yield is defined as

The predicted yield is determined by the masses used in a reaction and the mole ratios in the balanced equation. This predicted yield is the "ideal". It is not always possible to get this amount of product. Reactions are not always simple. There often are competing reactions. For example, if you burn carbon in air you can get carbon dioxide and carbon monoxide formed. The two reactions occur simultaneously.  Some carbon atoms end up in CO and others end up in CO2. The typical calculation in a starting class assumes that there is only one path for the reactants. This is an over simplification.You know for example from real life that food is not always converted to energy. If you eat a cookie, some of it could end up stored as "fat" Ugh!

Example:

What is the percent yield for a reaction if you predicted the formation of 21. grams of C6H12 and actually recovered only 3.8 grams?

1. Recall definition of percent yield.

2. Substitute the actual and predicted  yields.

3. Answer: The percent yield is 18 %.

Example 2:

A reaction between solid sulfur and oxygen produces sulfur dioxide.

The reaction started with 384 grams of S6 (s). Assume an unlimited supply of oxygen. What is the predicted yield and the percent yield if only 680 grams of sulfur dioxide are produced?

1 S6 (s)	6 O2 (g)		6 SO2 (g)

Step 1 : Calculate the molar masses for S6 (s) and SO2(g). The oxygen has no effect on the answer because there is more than you need. 1 mole S6 (s)= 193 grams S6 (s); 1 mole SO2(g) = 64 grams SO2(g)

Step 2 : Mole ration method Determine the mole ratio for 1 mole S6 (s) to mole SO2(g) The balanced equation indicates 1 mole S6 (s) to 6 mole SO2(g)

Step 3 : Calculate the number of moles of S6 (s) moles S6 (s) = [384 g S6 (s)][ 1 mole S6 (s)/ 192 g S6 (s)] = 2 moles S6 (s)

Step 4 : Calculate the moles of SO2(g) expected using the mole ratio 6 moles SO2(g) / 1 mole S6 (s) moles SO2(g) = 2 moles S6 (s)[6 SO2(g) / 1 S6 (s)] = 12 moles SO2(g)

Step 5 :  Calculate the grams of SO2(g) predicted using 1 mole SO2(g) = 64 grams SO2(g) grams SO2(g) = 12 moles SO2(g)[64 grams SO2(g)/1 mole SO2(g)] = 768 g SO2(g)

Step 6 : Calculate the percent yield using the definition

Percent yield = 100[actual yield/ predicted yield] = 100[680 grams SO₂(g)/ 768 g SO₂(g)]= 89%