Preparation of an Ionic Compound from Aluminum Metal Essay Example

ten minutes, the blue litmus paper used to test the acidity of the solution turned red, thus showing that the solution was acidic. 1 mL more acid was added to the solution to dissolve the remaining solid particles. After about two minutes, most of the solid had dissolved; any solid particle still present was filtered out by gravity. The solution was set aside to cool to room temperature. It was then placed in an ice-water bath for 30 minutes.

The product had crystallized out of the solution. The bottom of the beaker was scratched using a glass stirring rod to induce crystallization. The mixture was thoroughly filtered through a Buchner funnel. The crystals left in the funnel were washed using a solution of 5 mL of 95% ethyl alcohol and 5mL of distilled water. The suction was continued for about five more minutes to ensure that the product was dry. Once the suction was complete, the crystals were spread out into a dry, 106.53 g recrystallizing dish and left to air dry for one week. One week later, the dried crystals were weighed at 111.4 g. Bunsen burners heated two crucibles supported on clay triangles to red heat for no more than five minutes. After allowing the crucibles to cool, they were placed in a desiccator to cool to room temperature. The crucibles were weighed to 0.1 mg once they had reached room temperature. About 0.5 g of the air-dried product was cautiously weighed into each crucible. The crucibles were once again placed on clay triangles and heated gently. After the product began to solidify, the Bunsen burners were adjusted so that maximum heat was provided

to the crucibles for about five minutes.

The crucibles were set aside to cool and then placed in a desiccator to fully cool to room temperature. Once room temperature was reached, the crucibles were weighed to 0.1 mg. The mass of water of crystallization in the sample is represented by the loss in mass of the sample.

Discussion Number of Waters of Crystallization

Based on the results obtained from the experiment, the average number of waters of crystallization of the two samples was 12 molecules H2O. Therefore, it can be concluded that the proper formula for the product Alum is KAl(SO4)212H2O for any given sample size. This was determined by calculating the number of waters of crystallization for each sample, then averaging those results.

To calculate the waters of crystallization for each sample, the mass ratio of KAl(SO4)2 to H2O for the sample was divided by the molecular mass of each species. Then, this ratio is converted to find the molecular value of H2O for 1 molecule of KAl(SO4)

Theoretical Yield

The overall reaction for the synthesis can be found by adding reactions 1-4 (below) and cancelling like species. 2Al(s)+2KOH+6H2O > 2K[Al(OH)4]+3H2(1) 2 K[Al(OH)4] + H2SO4 > 2 Al(OH)3(s) + 2 H2O + K2SO4(2) 2 Al(OH)3 + 3 H2SO4 > Al2(SO4)3 + 6 H2O(3) Al2(SO4)3 + K2SO4 + 24 H2O > 2 KAl(SO4)2•12H2O(4) 2Al(s)+2KOH+4H2SO4+22H2O > 2 KAl(SO4)2•12H2O + 3 H2(5)

After determining that there are 2 moles of Aluminum for every 2 moles of KAl(SO4)2•12H2O produced, it was then necessary to calculate the quantity, in moles, of alum that should theoretically be produced from the quantity (in moles) of aluminum, used 0,5 grams. Knowing the number of moles of alum that are

theoretically obtainable from 0.5 g Aluminum, the mass of alum expected could be found, thus providing the theoretical yield. 0.5 g Al x (1 mol Al/26.98g Al) = . 0185 mol Al 2 mol KAl(SO4 ) 2 • 12H2O x . 0185 mol Al =. 037 mol KAl(SO4)212H2O 2 mol Al .037 mol KAl(SO4)212H2O x (474.39 g KAl(SO4)212H2O/mol) = 17.5 g KAl(SO4)212H2O theoretical yield = 17.55 g KAl(SO4)212H2O

Percent yield of the 17.55 g KAl(SO4)212H2O that could theoretically be produced from the reaction of aluminum with potassium hydroxide and sulfuric acid in the water, only 27.98% (4.910 g) of this amount was actually yielded. The actual yield was previously calculated from first weighing an empty crystallization dish, then weighing the air-dried crystals with the dish one week later. The sum was subtracted by the weight of the crystallization dish to produce the mass of the crystals, or the actual yield of the reaction: 4.91 g KAl(SO4)212H2O.

To find the percent yield, the actual yield was divided by the theoretical yield, 17.55 g, and then multiplied by 100%. The discrepancy between the theoretical yield and the actual yield obtained, as seen by the percent yield, could have been caused due to the Bunsen burner not consistently heating the 2nd crucible containing the product.

• Actual yield x 100 = % yield
• Theoretical yield (13.214 gc KAl(SO4)212H2O /17.55 g KAl(SO4)212H2O ) x 100 = 75.29%
• Actual yield x 100 = % yield
• Theoretical yield (13.214 gc KAl(SO4)212H2O /17.55 g KAl(SO4)212H2O ) x 100 = 75.29%
• The percent yield is 75.29%