Formula of a Hydrate Lab
Formula of a Hydrate Lab
Wednesday October 29, 2014
Chemistry Honors
Purpose
This lab was done to determine the percentage of water in a hydrate, which was CuSO4 ?H20. Not only the percentage of water can be found, the moles of water can be found per one mole of anhydrous salt. An anhydrous salt is a hydrate that lost its water. Using various lab equipment such as burners, crucible, and balance, and techniques such as the mass-to-mole ratio and mass to percentage, the percentage of water in a hydrate was determined. Salts appear to be dry, yet when heated, surprisingly large quantities of water are driven off, because water is loosely bonded to the hydrate.
Procedure
1. Set the necessary lab equipment properly.
2. Weigh dry and clean crucible and its cover on the balance. Record.
3. Measure about 1 to 2 grams of the hydrate and record the weight using the weighing boat.
4. In moderate flame, heat the empty crucible with tilted lid for 3 to 5 minutes. Record what happens during the heating process. After recording, wait for crucible and lid to cool, and weigh, and record.
5. Repeat step 4 until the mass of the hydrate doesn’t change after each heating. If the mass changed 0.55 grams or less, stop heating the hydrate. This means all of the water has been driven off.
6. After recording all the measurements and the mass of the hydrate, now anhydrous salt, add 1 or 2 drops of water.
7. Record what happens to the anhydrous salt.
8. Pour and scrape the anhydrous salt into trash, and clean up. Make sure to wash crucible with soap.
Results & Calculations
Data:
Mass of empty crucible: 32.67 grams
Mass of crucible and hydrate: 34.1 grams
Mass of crucible and anhydrous salt: trial 1 (33.93g); trial 2 (33.51); trial 3 (33.51)
Mass of water: 0.59g
Mass of anhydrous salt: 0.84g
Calculations:
Water: 0.59g/18=0.033; 0.033/0.00526= 6.27
Anhydrous Salt: 0.84/159.61= 0.00526; 0.00526/0.00526=1
Coefficient of H20: 6
Percent Error
According to actual percentage
[Experimental
Wednesday October 29, 2014
Chemistry Honors
Purpose
This lab was done to determine the percentage of water in a hydrate, which was CuSO4 ?H20. Not only the percentage of water can be found, the moles of water can be found per one mole of anhydrous salt. An anhydrous salt is a hydrate that lost its water. Using various lab equipment such as burners, crucible, and balance, and techniques such as the mass-to-mole ratio and mass to percentage, the percentage of water in a hydrate was determined. Salts appear to be dry, yet when heated, surprisingly large quantities of water are driven off, because water is loosely bonded to the hydrate.
Procedure
1. Set the necessary lab equipment properly.
2. Weigh dry and clean crucible and its cover on the balance. Record.
3. Measure about 1 to 2 grams of the hydrate and record the weight using the weighing boat.
4. In moderate flame, heat the empty crucible with tilted lid for 3 to 5 minutes. Record what happens during the heating process. After recording, wait for crucible and lid to cool, and weigh, and record.
5. Repeat step 4 until the mass of the hydrate doesn’t change after each heating. If the mass changed 0.55 grams or less, stop heating the hydrate. This means all of the water has been driven off.
6. After recording all the measurements and the mass of the hydrate, now anhydrous salt, add 1 or 2 drops of water.
7. Record what happens to the anhydrous salt.
8. Pour and scrape the anhydrous salt into trash, and clean up. Make sure to wash crucible with soap.
Results & Calculations
Data:
Mass of empty crucible: 32.67 grams
Mass of crucible and hydrate: 34.1 grams
Mass of crucible and anhydrous salt: trial 1 (33.93g); trial 2 (33.51); trial 3 (33.51)
Mass of water: 0.59g
Mass of anhydrous salt: 0.84g
Calculations:
Water: 0.59g/18=0.033; 0.033/0.00526= 6.27
Anhydrous Salt: 0.84/159.61= 0.00526; 0.00526/0.00526=1
Coefficient of H20: 6
Percent Error
According to actual percentage
[Experimental