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

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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.

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3. Answer: The percent yield is 18 %.

Example:

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% return to top of page