Analysis of Alum KAl (SO4) 2 . 12H2O

Analysis of Alum KAl (SO4)2 . 12H2O

Laayla Muhammad
10/31/06 -11/01/06


Purpose: To do several tests to determine if the resulting crystals are really alum, to use a Thiele melting point tube to determine the melting point of synthesized sample of alum and to determine the amount of water in a synthesized sample of alum and also the percent sulfate in a synthesized sample of alum.

Procedure/Method: First I have to find the melting point of alum and to do that, I have to pulverize the small amount (0.5g) of dry alum. Then I’ll pack the alum in a capillary tube (1 cm) and cut a 1mm length of rubber tubing and fasten the capillary tube to a thermometer. Then I’ll fasten that to a ring stand. Next step is to immerse bottom of capillary and thermometer in a beaker of H2O and heat. I will have to remember to heat slowly as closer to the melting point to get an accurate value. Then I’ll record the temperature at which my alum crystals melt. I will compare the experimental & published values. Next I have to determine the amount of water of hydration in alum crystals. Then I’ll find the mass of crucible with the cover on a sensitive balance. I will add two grams of alum crystals to the crucible and then find the mass of crucible, the cover, and the crystal alums. I will heat and the alum crystals will melt and water of hydration will vaporize. After heating it for 5 minutes, I will cool and find the mass of crucible, cover, and anhydrous alum. I will calculate the mass driven off. Next step is to find the moles of anhydrous KAl (SO4)2 and the moles of H2O. Then I will calculate the ratio of moles H2O: moles KAl (SO4)2 and give the calculate formula of alum, KAl (SO4)2 * X H2O, where X = the ratio of moles H2O: moles KAl (SO4)2. I will compare the values with that of alum. Then I will determine the percent sulfate in alum by doing the following. I will use suction to pull distilled water through filter pad and dry it in oven. I will determine mass by a sensitive balance and measure the mass of filter paper. I will weigh 1 gram of alum into a 250 mL beaker and dissolve 50 mL of distilled water. Calculate volume of 0.2M Ba(NO3)2 and add twice this volume of Ba(NO3)2, stirring. I will heat and cool it over night. I will filter precipitate and use rubber policeman so that every particle is transferred from beaker into crucible. I will wash beaker and precipitate several times with small quantities of water and transfer filter crucible to beaker and dry in oven (at 500C so it doesn’t char). I will let it cool and mass it. That is how I will calculate the percent sulfate in alum and compare it to the value of its formula.


Data:

Melting Point (temperature in degrees Celsius) of alum crystals:

Trial 1 Trial 2 Trial 3 Average


90.0



90.1



90.3



90.13

The published data value for alum crystals melting point is 92.5 meaning our melting point was about 2.37 degrees Celsius off.

Massed Objects Mass (g)
Massed out crystals 2 grams
Crucible + Cover 30.1105 g
Crucible + Cover + Alum Crystals 32.1105 g
Crucible + Cover + Anhydrous Alum 31.1591 g
Calculated Anhydrous Alum .9514 g
Calculated Water driven off 1.0486 g
Gooch Crucible + Filter Paper 15.8050 g
Gooch Crucible + Filter Paper + Dried Precipitate 16.0230 g
Calculated Dried Precipitate .218 g



Calculations:

Calculations made to find out the amount of Ba(NO3)2 needed to totally precipitate all of the sulfate ion present in the solution plus twice this volume:

1 gram alum X 1 mol alum X 1 mol SO42- X 1 mol Ba2+ X 1 mol Ba(NO3)2
474.4 g alum 1 mol alum 2 mol SO42- 1 mol Ba2+

X 5 L Ba(NO3)2 = About 21 mL ( multiplied by 2) = 42 mL needed
1 mol Ba2+





(Other Calculations)
32.1105-30.1105 = 2 grams of crystals
32.1105-31.1591 = .9514 anhydrous alum
2.000-.9514=1.0486 grams of hydrated H2O


.9514g anhydrous alum X 1 mol anhydrous alum = .003628 mol anhydrous alum
262.22 g anhydrous alum

1.0486g hydrated H2O X 1 mol H2O = .05825 mol H2O
18g H2O

.05828 mol H2O = 16.056 mol (About 16) = X
.003628 mol anhydrous alum

KAl(SO4)2 * 16 H2O




1.0145g BaSO4 X 1 mol of BaSO4 X 2 mol SO4 X 96.066 g SO4 = .4176 g SO4
233.393g BaSO4 1 mol BaSO4 2 mol SO4

100 X .4176g SO4 = 41.2% SO4 present in the alum
1.0145g alum



Questions/Answers:
1. Objects must be cooled before their mass is found on a sensitive balance because their accurate mass when still warm could alter the result in finding out the actual mass of the object. This could be due to the fact that when heated, objects have more energy so they weigh more and have a higher mass than the cold objects.

2. The different tests used to verify that the substance tested was alum was the melting point at which we knew that the crystal alum would melt at 92.5 degrees Celsius.

3. Other tests could be made to verify the compositions of alum. For example, we can use the percentage of the sulfate in the alum and use it to find its mole ratio and figure out how much water is in the synthesized alum.

Conclusions: To conclude, the precipitate that came out in as a result were alum crystals according to many tests we took. For example, its melting point was average 90.13 and we know that alum crystals published melting point is 92.5. We figured out by calculation that there needs to be total of 42 mL of Ba(NO3)2 to filter the precipitate completely. There is about 16 mol of H2O, as calculated, in the synthesized alum. After calculating, we also figured out that there was about 41.2 percent of sulfate in the synthesized alum.

Experimental Sources of Error: There could have been several errors made while performing this experiment. While figuring out the melting point for the alum crystals, I could have not analyzed when the alum crystals started melting exactly, leading me to maybe assume it was the first temperature that my eyes noticed. Another error could have been made while massing out the crucible, the cover, or the anhydrous alum, which would throw off my calculations when figuring out the mol to mol ratio of the given formulas. I could have also made an error filtering the precipitate Ba(NO3)2, in a way where not all the precipitate was filtered thoroughly from the beaker to the gooch crucible. Such imprecise and maybe even inaccurate measurements could be responsible for altering the following results for figuring out the exact percentage of sulfate in the alum.