Preparation of Alum from Aluminum Metal

Categories: Full Metal Jacket

The objective of the laboratory is to synthesize alum (KAl(SO4)2.xH2O) from aluminum powder and to determine the proportion of water in the alum crystals. Alum is a product from the reaction between potassium hydroxide and sulfuric acid. The reaction include several steps, as followed:

Aluminum powder reacts with potassium hydroxide to generate Al(OH)4- ions and release hydrogen. 2 Al(s) + 2 KOH(aq) + 6 H2O 2 K[Al(OH)4](aq) + 3 H2 (g)

A gelatinous precipitate of aluminum hydroxide was created when sulfuric acid was added to the aqueous solution of Al(OH)4- ions.

2 K[Al(OH)4](aq) + H2SO4 (aq) 2 Al(OH)3 (s) + K2SO4 (aq) + 2 H2O

Later, excessive addition of the acid causes the precipitate to dissolve in the solution. 2Al(OH)3 (s) + H2SO4 (aq) Al2(SO4)3 (aq) + 6 H2O

Precipitation of alum was resulted from cooling in ice water bath. K2SO4 + Al2(SO4)3 + 2x H2O 2 KAl(SO4)2.xH2O

It is noticeable that alum is a hydrate (a hydrate consists of water molecules in its ionic structure), which leads to its solubility in water.

Get quality help now
Writer Lyla
Writer Lyla
checked Verified writer

Proficient in: Chemistry

star star star star 5 (876)

“ Have been using her for a while and please believe when I tell you, she never fail. Thanks Writer Lyla you are indeed awesome ”

avatar avatar avatar
+84 relevant experts are online
Hire writer

However, a minimum amount of cold water will cause the alum to crystallize. The amount of water incorporated in the alum structure should be clearly defined to derive the full formula of alum, which makes it possible for calculations of theoretical, actual and percent yield of alum.

Experimental Methods

The experiment was constructed based on the guidelines from Franklin and Marshall Lab Manual1. In a 400 mL-beaker, 0.5 g of aluminum and 2.

Get to Know The Price Estimate For Your Paper
Number of pages
Email Invalid email

By clicking “Check Writers’ Offers”, you agree to our terms of service and privacy policy. We’ll occasionally send you promo and account related email

"You must agree to out terms of services and privacy policy"
Write my paper

You won’t be charged yet!

01g of potassium hydroxide was prepared and mixed together. An amount of 25 mL of distilled water was poured into the beaker in the hood. The mixture was then continuously stirred to help disperse the heat generated from the exothermic reaction. As observed, hydrogen was liberated from the solution, along with aluminum powder gradually darkening and disintegrating into insoluble flakes. It took the solutions 15 minutes to complete when there were no signs of hydrogen released. The solution was then filtered into a new 250 mL beaker. The residue left on the filter paper was carefully washed into the filtrate.

A portion of 10 mL of 9M sulfuric acid was added slowly and attentively to the filtrate, with gentle stirring. The presence of acid will neutralize the solution, generating a gelatinous precipitate known as Al(OH)3. The precipitate was later dissolved when excessive addition of acid was poured into the solution, combined with gentle heating on hot plate. The acidity of the solution was confirmed when tested with litmus paper: the paper turned into red. The solution was filtered for the second time to eliminate any undissolved residues remaining.

The solution was set aside to cool at room temperature. The crystallization process was conducted by placing the solution beaker into an ice water bath for 20 minutes. After crystallization, white, soft crystals were formed. The mixture was filtered through a Buchner funnel.

A wash solution was prepared by combining 5 mL of ethyl alcohol and 5 mL of distilled water. The crystals were washed twice with proper wash solution. Then, the solution was put through suction again to dry out completely. The crystals were spread in a recrystallization disk. Large crystals were broken into small ones with a stapula. The crytals were allowed to air dry in one week.

The weight of the air-dried crystals was then recorded.  Two porcelain crucibles were supported on ceramic triangles and heated to red heat with a Bunsen burner for 10 minutes each. The crucibles were set
aside cool, then was placed into the desiccator to cool to room temperature. Their weighs were recorded. An amount of 0.5 g of the crystallized alum was placed into each of the crucibles. The crucibles (with alum inside) were carefully heated on ceramic triangles to red heat. The alum inside the crucibles appeared to melt, transforming into a kind of liquid solution. After 5 to 10 minutes of continuous and gentle heating, the content inside the crucibles started to solidify again, yielding white, soft crystals. The crystals were heated at maximum heat for 5 minutes. The crucibles were placed back to the desiccator. After cooling to room temperature, the masses of the contents inside the crucibles were carefully weighed.


The masses of alum, KAl(SO4)2 and water recorded were given in Table I.
Table I. Masses of Alum, KAl(SO4)2 and water in two different crucibles.

| Crucible 1| Crucible 2|
Alum| 0.5000 g| 0.5000 g|
KAl(SO4)2| 0.2721 g| 0.2696 g|
H2O| 0.2279 g| 0.2304 g|
x= nwaterndry product| 12.00| 12.24|

According to the values of x obtained from the table above, the average result of x is 12.12. We can define the formula of alum as KAl(SO4)2.12,12H2O (Molar Mass M = 476.16 gmol-1). Finding the formula of alum makes it possible to calculate the theoretical yield and the percent yield of alum. After calculations from the equations demonstrated in the introduction, the theoretical number of moles of alum would be 0.019 moles. The theoretical yield, as a result, would be mtheoretical = 9.69 g. The actual yield recorded after the laboratory was 4.77 g. Combining all the yields gives us the final result of the percent yield: 52,71%.


Several steps of heating the alum crystals and calculations took place to find out the formula of alum. Concerning the first crucible, an amount of 0.5 g of alum was added to the crucible. After heating, there was 0.2521 g of contents (KAl(SO4)2) left in the crucible. That means there was 0.2479 g of H2O fully evaporating. In this case, x= nH2Ondry product= 0.2279180.2721258= 12.00. Concerning the second crucible, an amount of 0.5 g of alum was added to the crucible. After heating, there was 0.2496 g of contents (KAl(SO4)2) left in the crucible. That means there was 0.2504 g of H2O fully evaporating. In this case, x= nH2Ondry product= 0.2304180.2696258= 12.24.

The average result of x: x= 12.00+ 12.242= 12.12. With calculations concerning the masses of contents in the crucibles before and after heating, it is observed that 12.12 molecules of water in a mole of alum. The general formula of alum, therefore, is KAl(SO4)2.12.12H2O. The literature value of portions of water molecules in alum is 12, which makes the formula of alum KAl(SO4)2.12H2O. The proximity of the calculated result and the literature result reflected to efficiency and accuracy of the laboratory.

Through a series of chemical reactions, alum (the double salt with incorporated water molecules, with the calculated formula of KAl(SO4)2.12H2O) was formed from aluminum powder, potassium hydroxide and sulfuric acid. The reactions lead to the formation of alum are summarised as followed: (I) 2 Al(s) + 2 KOH(aq) + 6 H2O 2 K[Al(OH)4](aq) + 3 H2 (g) (II) 2 K[Al(OH)4](aq) + H2SO4 (aq) 2 Al(OH)3 (s) + K2SO4 (aq) + 2 H2O (III)2Al(OH)3 (s) + H2SO4 (aq) Al2(SO4)3 (aq) + 6 H2O

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

The theoretical yield was accumulated over a few steps:

There are 0.019 moles in 0.5 g of Aluminum. Similarly, there are 0.036 moles in 2.01 g of potassium hydroxide. We used a portion of 10 mL of 9M sulfuric acid, meaning that we use 0.09 moles of sulfuric acid.

In reaction (I) that potassium reacted with aluminum powder with the presence of water, the aluminum played the role of the limiting reagent. In reaction (II) that sulfuric acid was added into the solution of Al(OH)4- ions, the ions were the limiting reagents. The gelatinous precipitate formed in reaction (II) by pouring in acid was soon dissolved in the solution in the reaction (III) by the addition of excessive sulfuric acid. The alum crystals were formed in the reaction (IV) by cooling. From the four reactions, we can easily see that the number of moles of alum formed is equal to the number of moles of aluminum in the aluminum powder. nalum = naluminum = 0.019 moles.

The theoretical yield is the product of the number of moles and alum’s molar mass: malum= n × M= 0.019 × 476.16= 9.05 (g). The actual yield is 4.77 g (as stated in the results). The percent yield is calculated by dividing the actual yield by the theoretical yield: %Yield= Actual YieldTheoretical Yield = 4.77 g9.05 g = 52.71%.

About 47% of alum was lost during the crystallization. From 0.5 g of aluminum, 2.01 g of potassium hydroxide and 10 mL of 9M sulfuric acid at the beginning, the product obtained after crystallization was only 4.77 g of alum, compared to the theoretical value of 9.05 g. A significant amount of alum was lost during filtration, suction and crystallization, because of the fact that the filter paper was not wet enough and the crucibles were not dry enough due to short maximum heating time.


1. Franklin and Marshall College Chemistry 111/112 Laboratory Manual, Fall 2012/Spring 2013, p. 39-41.

Updated: Mar 15, 2022
Cite this page

Preparation of Alum from Aluminum Metal. (2016, Dec 27). Retrieved from

Preparation of Alum from Aluminum Metal essay
Live chat  with support 24/7

👋 Hi! I’m your smart assistant Amy!

Don’t know where to start? Type your requirements and I’ll connect you to an academic expert within 3 minutes.

get help with your assignment