Today, we messed with Copper (II) sulfate pentahydrate, CuSO4 * 5H2O, a crystalline compound with a blueish color when hydrated.

We gathered some of the hydrate in a dish, boiled away the water, massed it, boiled it again to make sure we got all the water out of the crystal structure, and massed it a final time. Out of 2.29 grams of Hydrate, my partner and I calculated the mass of the water to be .85 grams, giving a percentage of 100(.85)/2.29= 37%, or that there were .37 grams of water for every gram of the hydrate.

My picture

I can also say that my results were reliable to one-hundredth of a gram, the finest mass that the scales we used could measure. They were also reliable because I followed the rules of using significant digits taught to us by our Chemistry teacher and because we didn’t lose any great quantities of the hydrate, which would’ve messed up our future calculations. For example, if we had shaken the dish we were boiling the hydrate in, then some of the hydrate or the anhydrous solid would’ve fallen out, and even if we tried to put it back into the dish, the amount would be lower than what it should’ve been and the percentage of water would appear much larger that it really was.

My picture

The 37% should be a constant that can be used to show how much of any mass of the hydrate is water alone. Using units, the percentage would be .37gWater / 1gHydrate, and 6.0 grams of hydrate would cancel out and leave you with about 2.2 grams of water.

Update:

Today, we calculated the molecular mass of copper sulfate pentahydrate. The formula weight of the compound is derived from the atomic weights of the elements in it; copper gives 63.5; sulfur, 32.1; oxygen, 4*16.0 (because there are four oxygen atoms in the compound); hydrogen, 5*1.0; and the second amount of oxygen gives 80 amu. the total mass is 249.6 amu’s, and using the mass of water, 90.0 amu’s, the percentage of water in any amount of the compound should be (90*100)/249.6= 36.05% by mass. This means that from the 2.29 grams my group used, we should’ve calculated (2.29*90)/249.6= .826 grams of water.

However, my group got .85 grams of water, so there must be some percent error. That error is 100(.85- .826)/.826= 2.9% more mass due to water than there should’ve been. To remedy this, we must realize that in calculating the mass of water lost from the difference between the hydrate before and the anhydrous after boiling, any bit of anhydrous solid lost during the process, such as being left as residue on our stirring rods or having been splattered out of the dish, would be calculated as the mass of the lost water. however, the lost mass equates to about two-hundredths of a gram, so I think I can say my group’s results are still pretty reliable.