Analysis of Alum Al K SO4 2 12H2O PDF

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Analysis of Alum, AlK(SO4)2 12H2O

Table of Contents: Purpose of Experiment Background Pre-Lab Data Calculations Post-Lab

Purpose of Experiment: To analyze alum, AlK(SO4)2⋅12H2O by three techniques in order to verify the identity of alum. The melting point, mole ratio of hydrated water to anhydrous aluminum potassium sulfate, and percent of sulfate ions contained in the compound will be determined. These three properties will also be verified against other known calculated values.

Background: In order to identify a compound’s properties with certainty, a certain number of its properties must be verified experimentally. In this experiment, three properties of Alum, AlK(SO4)2⋅H2O, will be identified -- alum’s melting point, mole ratio of hydrated water to anhydrous aluminum potassium sulfate, and percent of sulfate ions contained in the compound. The data gathered from the experiment will be compared with other reported and known values.

Pre-Lab: 1. When measuring a melting point, why is it necessary to raise the tem perature very slowly when approaching the melting temperature? a. It is necessary to raise the temperature very slowly when approaching the melting temperature because the solid should have the same temperature throughout. Raising the temperature too fast would melt the portions of the solid closest to the heating device, which would result in incorrect data. 2. Washing soda is a hydrated compound whose formula can be written Na2CO3•xH2O, where x is the number of moles of H2O per mole of Na2CO3. When a 2.123 g sample of washing soda was heated at 130  C, all of the water of hydration was lost, leaving 0.787 g of anhydrous sodium carbonate. Calculate the value of x. a. Moles Na2CO3 = 0.787 g Na2CO3 * (1 mol Na2CO3 / 106g Na2CO3) = 0.007425 mol b. Moles H2O = (2.123g - 0.787g) H2O * (1mol H2O / 18.02g H2O) = 0.0741398 mol c. Mole to mole ratio of Na2CO3 to H2O = 0.0741398/0.007425 = 10 d. x = 10, Na2CO3•10H2O 3. The formula for Epsom salts is MgSO4•7H2O. If 1.250 g of the compound are dissolved in water, calculate the number of milliliters of 0.200 M Ba(NO3)2 that would be required to precipitate all of the sulfate ions as barium sulfate. Repeat the calculation for 1.000 g of alum. a. Ba(NO3)2 + MgSO4 • 7H2O ---> BaSO4 + Mg(NO3)2 + 7H2O b. 1.250 MgSO4•7H2O * (1 mol MgSO4•7H2O / 246.47g MgSO2•7H2O) * (1 mol Ba(NO3)2 / 1 mol MgSO4•7H2O) x*(1L / 0.200 mol Ba(NO3)2) * (1000 mL / 1L) = 25.36 mL c. KAl(SO4)2•12H2O(s) + 2Ba(NO3)2(aq) ---> 2BaSO4(s) + KNO3(aq) + Al(NO3)3 + 12H2O(l)

d. 1.000 KAl(SO4)2•12H2O x (1 mol KAl(SO4)2•12H2O / 474.39g KAl(SO4)2•12H2O) * (2 mol Ba(NO3)2 / 1 mol KAl(SO4)2•12H2O) * (1L / 0.200 mol Ba(NO3)2) * (1000 mL / 1L) = 21.08 mL

Data: 1. Melting Point Trial #1

Trial #2

Measured Melting Point

91.6° C

92.0° C

Literature Value

92.5 °C

92.5 °C

Source of Literature Values: https://pubchem.ncbi.nlm.nih.gov/compound/Potassium_alum#section=Melting-Point - PubChem

2. Determination of the Number of Waters of Hydration

Mass of crucible and cover

34.8130 g

Mass of crucible, cover, and alum crystals

36.8192 g

Mass of alum crystals

2.0062 g

Mass of crucible, cover, and alum after heating #1

36.2925 g

Mass of crucible, cover, and alum after heating #2

36.2841 g

Mass of water driven off

1.4711 g

Mass of anhydrous alum, AlK(SO4)2

0.5351 g

Moles H2O

0.0814

Moles AlK(SO2)2

0.007202

Mole ratio; moles H2O/moles AlK(SO4)2

11.3

3. Determination of the Percent Sulfate in alum

Mass of alum

1.0200 g

Mass of filter paper

0.9993 g

Mass of filter paper + BaSO4

2.1374 g

Mass of BaSO4

1.1381 g

Mass of sulfate in the precipitate

0.4689 g

Experimental percent sulfate in alum

41.20 %

Theoretical percent sulfate in alum

40.51 %

Percent error

0.02 %

Calculations: Mass water driven off: 36.8192-36.2841 = 1.4711 g Mass Anhydrous Alum: 36.2841-34.8130 = 0.5351 g Moles H2O: 2 - 0.5351 = 1.4649 hydrate h2o 1.4649 hydrate H2O * 1mol H2O = 0.0814 moles hydrate H2O Moles Anhydrous Alum: 0.5351 anhydrous alum * 1 mol anhydrous alum = 0.007202 moles anhydrous alum Mole Ratio: 0.0814/0.007202 = 11.3 Molar mass BaSO4 = 137+32+64 = 233 96/233 * 100 = 41.20% 1.1381 / 233.4 = 0.0048 mol BaSO4

0.0048 * 96.07 = 0.461 g 0.461/1.1381 = 40.5% Theoretical percent sulfate in alum = (192.2/474.4) * 100 = 40.51% Percent error: 0.02%

Post-Lab: 1. Why must objects be cooled before their mass is determined on a sensitive balance? a. The objects could damage the surface of the balance. b. Hot objects also warm the air next to it. The warm air next to hot objects would cause the object to seem to have a lighter weight because it expands. 2. Comment on the results of the different tests used to verify that the sample tested was alum. a. The melting point was a test that we used to verify that the substance tested was alum, since we knew that the crystal alum would melt at 92.5 degrees Celsius. Our average showed that we were off by 0.7 degrees Celsius, yet close enough. 3. What other tests could be used to verify the composition of alum? a. Some other tests that could be conducted to verify the composition of alum could include the determination of the freezing point of alum and boiling point of alum, and finding the mole ratio using the percentage of sulfate in alum to determine the amount of H2O in the alum....