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

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The lab discussion focused on determining the value of 'n' in the hydrate formula CuSO4*nH2O, where the mass of the hydrate was 0.1989 g, the mass of the anhydrate was 0.1271 g, and the mass of water evolved was 0.0718 g. The calculation involved converting the masses to moles and dividing by the smallest mole value, resulting in n being determined as 5, confirming the compound as Copper(II) Sulfate Pentahydrate. The assumption of 100% yield was also noted as part of the analysis.

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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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