Preparation of Alum from Aluminum Metal
Preparation of Alum from Aluminum Metal
By: Rachel Buckley, Alexa Kirchharr
Abstract:
The purpose of this experiment, Preparation of Alum from Aluminum Metal, was to extract alum from aluminum metal, and to find the percent yield of the alum from the aluminum metal. The experiment was carried out through a series of processes using aluminum foil, potassium hydroxide, and sulfuric acid.
Materials and Methods:
The first part of this experiment was to dissolve the aluminum foil in 1.5 M of potassium hydroxide by placing both in a beaker and placing it on a hot plate heated to a temperature of 250° C. Once the solution started to bubble vigorously, it was removed from the hot plate and continually stirred to dissolve all of the aluminum foil. The solution turned a grey/black color. During this process, hydrogen gas is released.
Once the aluminum was fully dissolved, the solution was filtered through a Buchner funnel lined with filter paper, which removed the undissolved plastic lining and paint that was present from the aluminum foil. The solution that was drained was then transferred into a beaker and 20 mL of 9 M sulfuric acid was added to it. This solution was then placed in an ice bath for about 15 minutes, to allow the solution to produce white crystals, which are the alum.
While the crystals were forming, the filter paper from the draining was rinsed through the Buchner funnel using water, to collect the paint and undissolved plastic in a pile to weigh using an analytical balance. The mass was then recorded and used to calculate the percent yield of alum.
Once the crystals were formed, the solution was drained again into the Buchner funnel, except with a different piece of filter paper. Ethanol was used to dry up the crystals and remove all the water. Once the crystals were drained, they were weighed along with the filter paper and the mass was recorded.
Observations and Discussions:
| Weight of Aluminum Metal (g) | 1.9648 |
| Weight of Filter Paper (g) | 0.3596 |
| Weight of Filter Paper and Product (g) | 0.94904 |
| Weight of Plastic and Paint Removed (g) | 0.58944 |
| Weight of Product (g) | 27.0104 |
| Theoretical Yield of Alum (g) | 34.547 |
| Percent Yield of Alum | 78.18% |
The amount of aluminum foil used to perform this experiment was 1.9648, and this mass was used to calculate the theoretical yield using the equation: (grams of product)=(grams of reactant)x(1 mole of reactant/molar mass of reactant)x(mole ratio of product/reactant)x(molar mass of product/1 mole). The theoretical yield turned out to be 34.547, because the mole to mole ratio of reactant to product was one to one, and the molar mass of the product was 474.39 g. The percent yield was calculated using the equation (mass of actual yield/mass of theoretical yield)x100. The mass of actual yield was the weight of the product, and the mass of theoretical yield was the number calculated using the first equation.
The percent yield found was about what was expected, for there was some error due to water and ethanol not being fully removed, because the crystals were not given enough time for water and ethanol to evaporate. There was also some missing due to the fact that alum is very soluble in water, and the addition of water caused some of the alum to dissolve. The alum could be recovered by repeating the crystallization process again, using the solution that was drained from the crystals in the last part of the experiment.
The reason the percent yield was under 100% could be due to the fact that some of the reactant did not react, instead it just stayed in its original form. Also, some of the product may have been lost when transferred from one container to another.
Conclusion:
The experiment performed, Preparation of Alum from Aluminum Metal, was used to separate and extract alum from the paint and plastic materials present in aluminum foil. The experiment used potassium hydroxide and sulfuric acid to extract the alum, and in turn, was part of the formation of the alum. The experiment was overall successful, and it is also a successful way to reduce waste and recycle aluminum of all kinds to produce new materials.
Editor’s note (October 28, 2026): The student’s text and table are left as written. Four points need clarifying.
- Alum is made, not extracted. The foil dissolves in KOH as soluble K[Al(OH)4], filtering removes only undissolved plastic and paint, and H2SO4 plus cooling crystallizes alum, a new compound. So “extract” in the Abstract and Conclusion should read “prepare” (LibreTexts). Balanced equations: 2Al + 2KOH + 6H2O → 2K[Al(OH)4] + 3H2, and K[Al(OH)4] + 2H2SO4 + 8H2O → KAl(SO4)2·12H2O (Truman State). Each Al gives one K[Al(OH)4] and then one alum, so 1 mol Al gives 1 mol alum.
- Theoretical yield. 1.9648 g ÷ 26.98 g/mol = 0.07282 mol Al, × 474.39 g/mol = 34.55 g, matching the recorded 34.547 g (PubChem lists 474.4 g/mol). This counts the whole foil mass as aluminum, although plastic and paint were removed, so the true theoretical yield is somewhat lower.
- Table inconsistency, flagged only. Filter paper and product (0.94904 g) minus filter paper (0.3596 g) is 0.58944 g, the value listed as plastic and paint removed, yet the product is listed as 27.0104 g. The 78.18% yield (27.0104/34.547) cannot be checked against the recorded weights. If that 0.589 g were all non-aluminum, at most 1.3754 g of Al would remain, giving at most about 24.2 g of alum, less than 27.0104 g. No corrected masses are supplied here.
- Safety. KOH and sulfuric acid are corrosive, and the hydrogen produced is highly flammable. Wear goggles, gloves and a lab coat, work in a fume hood away from flames, and add the acid slowly to control heat and splattering (LibreTexts).
For other alum syntheses from foil, see The Synthesis and Gravimetric Analysis of Alum and The Formula, Synthesis, and Analysis of Alum.