Hydrate lab
In this lab we calculated the percent,(by weighing the hydrates and heating them to eliminate the water) of water in two hydrated salts, and how to figure out the formula of two hydrated salts. We were given two unknown hydrates and the molar mass of the anhydrous salts. We then heated the anhyrdrous salt with the hydrate to burn off the water. After 5 minutes we took them of the flame and let them cool for another 5 minutes. After we re measured their weight, We repeated this process for the second hydrate.
This is what our station looked like before we started the heating process.
Weighing hydrate A before heating it
Weighing Hydrate A with the crucible after the first 5 minute heating process. It looked like a white powder.
Weighing Hydrate A after the second heating process. It looked the same but dryer and had brown spots on the powder.
FORMULA OF A HYDRATE
Data:
HYDRATE: A HYDRATE: D
Molar mass of anhydrous salt: 142.0 g/mol 161.5 g/mol
Mass of crucible & lid: 35.27 g 32.92 g
Mass of hydrate: 2.16 g 2.36g
Mass of crucible, lid
& contents (1st heating) 36.06 g 34.26 g
Mass of crucible, lid
& contents (2nd heating) 36.06 g 34.29 g
Calculations & Conclusion:
1. Use the given molar masses of the anhydrous salts to determine the moles of anhydrous salt present after heating each of the hydrates.
A- (35.27 + 2.16) – 36.06 = 1.37 / 142.0 = 0.00960 mol of salt
D- ( 32.92 + 2.36) – 34.29 = 0.99 / 161.5 = 0.00610 mol of salt
2. Determine the mass and moles of water lost for each hydrate.
A- ( 2.16 – 1.37) = 0.79g of H2O / 18g/mol = 0.0439mol of H2O
D- (2.36 – 0.99) = 1.37g of H2O / 18g/mol = 0.0761mol of H2O
3. Determine the formula of each hydrate in the form of X. __H2O (X = unknown salt) by comparing the ratio of moles of H2O: moles anhydrous salt.
A- 0.0439 / 0.0096 = 4.57 -> 5H2O per salt
D- 0.0761 / 0.00613 = 12.41 -> 12H2O per salt
4. Determine the % composition (mass) of water in each hydrated salt.
A- 0.79/ 2.16 = 36.6% of H2O
D- 0.0761/ 0.00613 = 58.1% of H2O