Water Lost: 24.45-23.71 = .74 g
Molar Mass of Hydrate used: CuSO4●5H20 = 63.55+32.07+64+10.10+80 = 249.72 g/mole
Molar Mass of Anhydrous Salt: CuSO4 = 63.55+32.07+64 = 159.62 g/mole
Molar Mass of all the Waters in Compound: 5H2O = 80+10.10 = 90.10 g/mole
Theoretical Percentage of Water in Hydrate: 90.10 * 100 = 36.98%
249.72
Experimental Percentage of Water in Hydrate: .74 * 100 = 30.58%
2.42
Percent Error in the Lab: 36.08 - 30.58 = 15.24%
36.08
Questions:
1) Regardless of the mass of hydrate, the percentage of water lost stays the same. But, the value is not always the theoretical value, often close; it may vary, depending on the time you have kept the hydrate over the fire.
2) If I used 5 grams of hydrate, I would expect to lose around 36.98% of water. That’s about 1.849 grams.
Conclusion:
The purpose of this lab was to determine the percentage of water in a hydrate and to calculate the coefficient of the water of hydration of a hydrate. To make it all happen, first, we obtained a clean crucible and cover. Thereafter, the mass of crucible was obtained. Then, the group determined the mass of hydrate it will use (below 3). The crucible with hydrate was heated for about 15 minutes. Then, the crucible was left to cool for a few minutes and the mass of it was obtained. Lastly, our group dumped the contents of the crucible in the garbage and cleaned the lab. This report has shown that regardless of the mass of hydrate, the percentage of water loss stays the same. But, the value is not always the theoretical value; it may vary, depending on the time you have kept the hydrate over the fire. Sometimes you can overheat it and sometimes you can under-heat it. In our case, we under-heated the hydrate. We were aiming to lose 36.98% of water but only 30.58% was evaporated, that is, .74 grams out of 2.42 grams. Throughout the lab, we have faced only a single problem, we have not distributed the hydrate evenly through the crucible, one hydrate was over another hydrate.