## % Yield from Experiments

(C) - Copyright, 2000 F.W. Boyle, Jr. Ph.D.

Chemical reactions seldom if ever produce 100% of the expected (theoretical) amounts. Usually the mass of material produced is less than expected. The fact that not all reactants are transformed perfectly and completely into products is a major factor in determining the cost of products. Manufacturers often search for the best reaction to produce a substance since the ability to make money from the reaction is dependent upon the amount of products made.

Recently in Chem 111 lab, the experiment involved producing alum from potassium hydroxide, sulfuric acid, and metallic aluminum. The following set of reactions were postulated:

2Al(s) + 2KOH + 2H2O --> 2KAlO2 + 3H2(g)

Then:

2H2O + 2KAlO2 + H2SO4 --> 2Al(OH)3(s) + K2SO4

Next:

2Al(OH)3(s) + 3H2SO4 --> Al2(SO4)3 + 6H2O

Last:

Al2(SO4)3 + K2SO4 + 24H2O --> 2KAl(SO4)2.12H2O

All the separate reactions shown above are a bit confusing. These can be combined into a single reaction which will give the necessary information. The reactions are separated in order to show the student that the production of alum is not a simple, single step but a series of steps which depend upon each other. Substances appearing on both sides of the --> are struck out just like one would cancel variable that appear on both sides of an algebraic expression.

```
2Al(s) + 2KOH + 2H2O --> 2KAlO2 + 3H2(g)

2H2O + 2KAlO2 + H2SO4 --> 2Al(OH)3(s) + K2SO4

2Al(OH)3(s) + 3H2SO4 --> Al2(SO4)3 + 6H2O

Al2(SO4)3 + K2SO4 + 2422H2O --> 2KAl(SO4)2.12H2O
___________________________________________________

2Al(s) + 2KOH + 4H2SO4 + 22H2O --> 2KAl(SO4)2.12H2O + 3H2(g)

```
This last equation shows the reaction in which Al(m) was transformed into alum, a salt. The student should note that these reactions release a lot of heat. This is known as being exothermic. The heat allows for the alum to be more soluble than at room temperature. As the mixture cools a supersaturated state is reached. A seed crystal was added which gives a growth site for the supersaturated (too much alum in solution for the current temperature) alum to precipitate out.

In order to determine % yield in an experiment, one must first determine the theoretical yield. Theoretical yields are based upon the stoichiometry of the reaction. In the above reaction, we see the following mole ratios for the chemical substances involved.

```
2 moles  :  2 moles   : 2 moles  : 22 moles  : 2 moles : 3 moles
aluminum   potassium    sulfuric     Water      alum     hydrogen
hydroxide      acid                              gas
```
The ratio of Al as a reactant to alum as a product is:

```                    2 moles Al : 2 moles alum
```
OR

```                    1 mole of Al : 1 mole alum
```
This means that for any number of moles of Al reacted, the same number of moles of alum will be produced. For example:

```                  23 moles Al produces 23 moles alum

OR

0.01 moles Al produces 0.01 moles alum
```
This is an important concept to learn. Moles produce moles but we cannot weigh moles. We weigh chemicals in g (grams). So gram masses must be converted to moles using molar masses in order to evaluate how well the reaction proceeded. If we wanted to produce 474 g of alum (1 mole) then we would need to start with 1 mole (27 g) of Al. These numbers are calculated in the following manner.

Alum has the formula: KAl(SO4)2.12H2O

The molar mass of alum is: (I round the masses to whole numbers for demonstration purposes. In research exact values are used.)

```                K     39.1
Al    27.0
S     32.0*2
O     16.0*8
H      1.0*24
O     16.0*12
---------
alum   474.1 g/mole

Al is read from the Periodic Table.

```
Thus our theoretical yield in mass would be 474.1/27 which equals a 17.56 times increase in mass. Note that the mole ratio was 1:1 while the mass ratio is 1:17.56. This is a clear example of why one cannot use chemical equations to do direct mass calculations.

In the experiment, the mass of Al was around 0.05 g. Using the 17.56 times increase, we should have produced 0.878 g of alum.

```(1)               0.05 g x 17.56 = 0.878 g

```
If for example, a particular reaction produced 0.37 g of alum, we would calculated the % yield as follows:

```                Mass of alum produced
------------------------ X 100
theoretical mass of alum
```
Substituting of mass values:

```                    0.37 g alum
(2)               -------------- x 100 = 42.1 % yield
0.878 g alum
```
You can calculate your % yield by using formula (1) and your mass of Al. Replace the denominator in formula (2) with the mass of alum your quantity of Al should have produced. Then use the mass of alum you found by weighing at the end of your experiment in the numerator and finish the calculation using formula (2).

I hope this helps!