Question
A 6.25 gram sample of copper(II) chloride hydrate was heated until all the water was liberated. The residue massed 4.45 grams. How many water molecules are present for every formula unit of copper(II) chloride?
Answers
First, we get the molar masses to the CuCl2 and H2O. Just get a periodic table and add the individual masses of elements in the chemical formula:
CuCl2: 63.55 + 2*35.45 = 134.45
H2O: 2*1 + 1*16 = 18
Then we get their moles. We divide the mass by the molar mass to get the moles. Note that the 4.45 grams is for anhydrous CuCl2 (or the one without water):
4.45 g / 134.45 g/mol = 0.033 mol
For water, we first subtract 6.25 by 4.45 grams, because the difference is the mass of the released H2O. Then divide by the molar mass:
(6.25 - 4.45) / 18 = 0.1 mol
Finally, we get the ratio of moles:
CuCl2 : H2O
0.033:0.1 = 1:n
n = 3.03 or approximately equal to 3
Thus, the formula of the hydrate is CuCl2 * 3H2O
Hope this helps~ :3
CuCl2: 63.55 + 2*35.45 = 134.45
H2O: 2*1 + 1*16 = 18
Then we get their moles. We divide the mass by the molar mass to get the moles. Note that the 4.45 grams is for anhydrous CuCl2 (or the one without water):
4.45 g / 134.45 g/mol = 0.033 mol
For water, we first subtract 6.25 by 4.45 grams, because the difference is the mass of the released H2O. Then divide by the molar mass:
(6.25 - 4.45) / 18 = 0.1 mol
Finally, we get the ratio of moles:
CuCl2 : H2O
0.033:0.1 = 1:n
n = 3.03 or approximately equal to 3
Thus, the formula of the hydrate is CuCl2 * 3H2O
Hope this helps~ :3
Thank you! I really love how you took the time to help me out!
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