Finding the Value of 'n' in CuSO4*nH20 Lab

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The lab involved determining the value of 'n' in the hydrate CuSO4*nH2O using given masses of the hydrate and anhydrate. The masses recorded were 0.1989 g for the hydrate and 0.1271 g for the anhydrate, with 0.0718 g of water evolved. By converting the masses to moles and dividing by the smallest mole value, 'n' was calculated to be 5. This confirms the compound as Copper(II) Sulfate Pentahydrate, assuming a 100% yield. The calculations were validated by peers in the discussion.
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Doing a lab on the computer we received the data of a hydrate to be .1989 g and the dehydrate .1271 g last mass of water evolved from reaction .0718g. We heated CuSO4*nH20 and we needed to find n. I changed the anhydrate and water to moles then dividing the number by the smallest amount of moles getting n to be 5 by dividing .000796 mol./.00398 mol. The name of the chemical formula ends up being Copper(II) Sulfate Pentahydrate. Overall we assume 100% yield. Did I do this problem right?
 
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Looks good.
 
Score, I wasn't to sure, but thanks for the confirmation.
 

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