Miss Science, you have all the facts but you missed out the most important one, which is that the percentage by mass of water in CuSO4.5H2O is always the same.
the Mr of CuSO4 is 159.6 and that of 5 water molecules is 90.0. no matter what mass of hydrated copper(ii)sulphate, the masses of CuSO4 and H2O will always be in the ratio of 159.6:90.0
this means that the percentage mass of water would be 90/(90+159.6)x100% = 36.06%
2006-11-04 22:22:44
·
answer #1
·
answered by Anonymous
·
0⤊
0⤋
This problem is a mass percentage of a hydrate. Hydrate means a salt that has water molecules attached to the ions in its crystal network. The salt without the water molecules attached is called an anhydrous salt. To solve the problem, you first find the masses of the water and the anhydrous salt by going to the periodic table, and multiplying their masses by the number of atoms in the compound.
Ex: CuSO4 means 1 Copper atom times its mass from the periodic table, plus 1 sulfur atom x its mass, plus four times the mass of 1 oxygen atom.
The 5 in the formula means that for every mole of CuSO4, there are 5 moles of water attached. This means when you find the mass of the water, you have to multiply by 5
Once you have the mass of the CuSO4 and the mass of the water, add them together, and that is the mass of the entire hydrate.
Then, divide the mass of the water by that total mass, and multiply by 100...there you go...the percentage of water
2006-11-05 04:51:43
·
answer #2
·
answered by Miss Science 2
·
0⤊
0⤋