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:
Then:
Next:
Last:
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 gIf 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!