Determining the water content of a hydrate crystal When you deal with a hydrated crystal, one common task is to determine the number of water molecules attached to each crystal. For example, copper II sulfate pentahydrate is CuSO4*5H2O. Therefore for every one molecule of copper II sulfate there are five molecules of water attached. In the lab, the normal way to find the water of hydration is to heat the solid crystal until all the water evaporates (think back to the reaction type lab when we heated the blue copper II sulfate pentahydrate until it turned to a white solid). You weigh the compound before you heat it and weigh it again after heating. All the mass that disappears should be water that boiled off; conversely, all the solid that remains should be the anhydrous crystal (the chemical you really want, like copper II sulfate). Percent to mass Mass to moles Divide by smallest Multiply ‘til whole The problem then becomes an empirical formula problem that jumps to the second line of the poem. The only difference between this problem and all the other empirical formula problems is that the pieces that you are finding the moles of are NOT ELEMENTS. The pieces are MOLECULES! One molecule is ALWAYS H2O. The other molecule is the crystal that you want. For example, you begin with 2.50 grams of a barium hydroxide hydrate crystal. After heating, only 1.36 g of the solid remains. This means that 1.14 g of water evaporated. Now that we have the mass of barium hydroxide that remains (1.36 g) and the mass of the water that evaporated (1.14 g) gsolid gH 2O molesBa(OH) 2 molesH 2O FM FM 1.36 1.14 7.94e 3molBa(OH) 2 .0633molH 2O 171.3 18 At this point, we have taken mass to moles. We divide by the smallest (which should be the .0633molH 2O 7.98H2O 8H2O crystal). So, the water of hydration number is .00794molBa(OH)2 1Ba(OH)2 1Ba(OH)2 8. The formula is Ba(OH)2*8H2O. Multiplying ‘til whole is not usually necessary with water of hydration.