What is percent yield?

What is percent yield?

When you pop a bag of popcorn you expect the bag to fill up as the corn expands, but

you know that there will be un-popped kernels. In other words, we know that poten- tially all of the kernels could pop or open up, but we know from experience that this is not what actually happens. Although the popping of popcorn is not a chemical reaction it is a good analogy for the percentage yield of a chemical reaction.

Often with chemical reactions you do not obtain the exact amount of product you predicted with your stoichiometric calculation. This is not because you did the calculation wrong! It is simply because in real life things are often not perfect, for a variety of reasons. In the laboratory when we perform an experiment, it is rare that we obtain 100% of the desired product(s).

percent yield - the ratio of the amount of product actually obtained by experiment (actual yield) as compared to the amount of product calculated theoretically

(theoretical yield) multiplied by 100.

actual yield - the amount obtained in lab by actual experiment.

theoretical yield - the calculated amount produced if everything reacts perfectly.

Section 11.2 Percent Yield and Concentration

Percent yield

In the Lab As an example of how percent yield can be obtained in the laboratory lets investigate a common reaction. When baking soda is heated it decomposes according to the following equation: 2NaHCO

(s) → Na

CO

(s) + H

O(l) + CO

(g). We can weigh the baking soda before heating and weigh the solid product Na

CO

after heating. Stoichiometry allows us to calculate the amount of product we should obtain from our initial mass of baking soda, NaHCO

. We can then compare our actual mass of the solid product Na

CO

with the calculated mass and see how successful we were with our experiment.

Question: “Can you think of reasons why the final mass of the remaining solid, Na

CO

may not be accurate?

Reaction yields are often not 100%

When we mass the solid at the end of the experiment we generally obtain less than the full amount of product. There are several reasons for this: 1) generally there is some human error involved in experiments–such as not measuring exact amounts carefully; 2) not heating long enough for the reaction to be fully complete, so not all of the NaHCO

decomposes; 3) drying the product completely so that the water liquid does not increase the mass of Na

CO

and make it too heavy; 4) weighing the product before it is cool.

Chemicals do not always fully react to form product due to impurities in the samples, small surface area, or insufficient reaction time. Chemical reactions can also produce side reactions that can cause the desired product to turn into something else, less desirable.

Chemists are often faced with the challenge of “stopping” a chemical reaction so that it ends with the desired product. Lastly, some product is often lost during the isolation and purification steps.

TABLE 11.2. Data 2NaHCO

(s) → Na

CO

(s) + H

O(l) + CO

(g)

Initial Mass of NaHCO3

Final Mass of Na2CO3

10.0 g 4.87 g

Applying the formula for percent yield

Lets use the data obtained from heating 10.0 grams of baking soda (on the previous page) to calculate the percentage yield.

First we need to calculate the theoretical yield, using our Flow Chart According the balanced equation:

2NaHCO

(s) → Na

CO

(s) + H

O(l) + CO

(g)

Step 1: Convert 10.0 g of NaHCO

to moles.

Molar mass of NaHCO

= 22.99 + 1.0079 + 12.011 + (15.999 x 3) = 84.01g/mole

10.0 g NaHCO

x = 0.119 moles NaHCO

Step 2: Convert moles NaHCO

to moles of Na

CO

0.119 moles NaHCO

x = 0.0595 moles Theoretical

yield is the

calculated yield Step 3: Convert moles Na

CO

to grams.

Molar mass of Na

CO

= (22.99 x 2) + 12.01 + (15.999 x 3) = 105.99 g/mole

0.0595 moles Na

CO

x = 6.31 g Na

CO

We just determined our theoretical yield to be 6.31 g Na

CO

. From the experiment a mass of 4.87 g was obtained. Now we can calculate the percentage yield.

Actual yield is obtained by experiment

Percent Yield = x 100 = 77.2 %

1 mole NaHCO 3 84.01 g NaHCO

3 ---

Na 2 CO 3 2 mole NaHCO 3 ---

105.99 g Na 2 CO

3 1 mole Na 2 CO 3 ---

4.87 g (actual yield)

6.31 g (theoretical yield)

---

Section 11.2 Percent Yield and Concentration

Practice calculating percent yield

Another example

To calculate the percent yield you need to know the theoretical yield and the actual yield.

Recall that the theoretical yield is calculated based on the amounts of reactants used and the mole relationship obtained from the balanced equation. The actual yield can only be obtained by experiment, so it will either be given to you in the problem or you determine it in the lab.

Now we use the theoretical yield to calculate the percentage yield.

1.Calcium carbonate, which is often found in seashells, decomposes when heated. CaCO

(s) → CaO(s) + CO

(g)

If 30.5 g of CaCO

is heated, calculate the theoretical yield of CaO produced.

Asked: Calculate the amount of CaO produced. (theoretical yield) Given: 30.5 g of CaCO

reacts

Relationships: mole ratio = 1 mole CaCO

= 1 mole CaO

molar mass of CaCO

= 40.078 + 12.011+ (15.999 × 3)

=100.09 g/mole. Molar mass of CaO=40.078+15.999=56.08 g/mole

Solve:

Answer: The theoretical yield is 17.10 g of CaO 30.5 g CaCO

3

1 mole CaCO 3 100. 09 g CaCO 3 ---

× 0.305 mole CaCO

= 3

0.305 mole CaCO 3

1 mole CaO 1 mole CaCO 3 ---

× = 0.305 mole CaO

0.305 mole CaO 56.08 g CaO 1 mole CaO ---

× = 17.10 g CaO

2. Calculate the percentage yield if 12.5 g of CaO is actually obtained.

Asked: Calculate the percentage yield of CaO.

Given: The actual yield of 12.5 g CaO and the answer in part 1.

Relationships: Formula for percent yield

Solve:

Answer: The percent yield is 73.1%

Discussion: The percent yield tells the experimentalist how good their isolation techniques were.

Percent yield =

12.5 g (actual) 17.10 g (theoretical)

--- × 100 73.1% =

Calculating amounts from reactions in solution

Chemical reactions often take place between solids and liquids and between two liquids.

Here we will look at a solid zinc metal reacting with hydrochloric acid.

Zn(s) + 2HCl(aq) → H

(g) + ZnCl

(aq) To be able to calculate the amount of a substance that reacts in an aqueous solution, it is helpful to review our units of concentration.

Recall that our concentration unit is molarity, M, which tells us the number of moles in one liter of solution, M = mole/L.

As you can see above you must always find moles of the reactant. If the substance is a solid you calculate moles by dividing by the molar mass. If the substance is dissolved in an aqueous solution, you must find moles using your molarity equation. In this case molarity (M) x volume (L) will give you the amount of moles. So when determining moles you need to be aware of the physical states of the chemicals you are using.

A sample of Zinc metal (Zn(s)) reacts with 50.0 mL of a 3.0 M solution of hydrochloric acid (HCl). How many grams of hydrogen gas (H

(g)) will be produced? Assume zinc metal is in excess.

Zn(s) + 2HCl(aq) → ZnCl

(s) + H

(g) Asked: How many grams of H

will be produced?

Given: 50.0 mL of 3.0 M of HCl reacting with excess zinc

Relationships: M = , so mole HCl = 3.0 M × 0.050 L = 0.150 moles HCl Excess Zn means more than enough, so the exact amount is not important.

Mole Ratio: 2 moles HCl = 1 mole H

Molar mass of H

= 1.0079 × 2= 2.02

Solve:

Answer: 0.15 grams of H

are produced.

mole ---L

g mole---

0.150 mole HCl

1 mole H 2 2 mole HCl ---

× = 0.075 mole H 2

0.075 mole H 2

2.02 g H 2 1 mole H 2 ---

× = 0.15 g H 2

Section 11.2 Percent Yield and Concentration

Stoichiometry with solutions

Mass percentage

One of the most common ways to express the concentration of a solution is to use percent by mass of the compound in solution. For instance vinegar contains about 5% by mass acetic acid. Acetic acid comes from apples and is responsible for the sour taste of vinegar. So if vinegar is 5% acetic acid by mass that means it contains about 5 grams of acetic acid for every 100 g of solution. You can use the formula:

Example where you determine the amount of solute.

A solution is made by dissolving 15.6 g of glucose (C

H

O

) in 0.100 kg of water. Calculate the mass percent of glucose in the solution.

Asked: Determine the mass percent of the solute glucose, C

H

O

. Given: 15.6 g of C

H

O

in 0.100 kg of water

Relationships: Mass% = (mass of glucose / mass of solution) x 100

Solve: Mass % C

H

O

=

Answer: 13.5 % of glucose, C

H

O

15.6 g C 6 H 12 O 6

(15.6 g + 100 g)

--- × 100 13.5 = %

Commercial vinegar is reported to be 5% acetic acid by mass. How many grams of acetic acid are in 120 mL of commercial vinegar? (assume the density of vinegar is the same as pure water = 1.0 g/mL.

Asked: How many grams of acetic acid, C

H

O

are in 120 mL of vinegar?

Given: 120 ml of vinegar and 5% acetic acid by mass Relationships: 120 ml = 120 g given a density of 1.0 g/mL

Mass% = (mass of glucose / mass of solution) x 100 Solve: 5 % C

H

O

/ 100 = 0.05

0.05 × 120 g of vinegar = 6.0 g of acetic acid, C

H

O

are in 120 mL of vinegar

Answer: 6.0 g of acetic acid, C H O