Ap Chemistry Crucible lab

Question

A student heats up a crucible and in the end after doing the math, the hydrate is 1.51 g, the dehydrate is 0.98 g and the water lost is 0.53g.

This needs to be converted into moles of Copper (II) Sulfate and moles of water lost.

The original values were (crucible, 20.15 g, crucible and hydrate 21.66, crucible and dehydrate 21.13) I don't know if the orginal values are needed. But I need answers to structure the table for the problem properly.

Answer 1

mass of CuSO₄•XH₂O (hydrate) = 21.66 g - 20.15 g = 1.51 g

mass of H₂O lost/present = 21.66 g - 21.13 g = 0.53 g

mass of CuSO₄ (dehydrate; anhydrous) = 1.51g - 0.53 g = 0.98 g

molar mass CuSO₄ = 159.60 g/mole

molar mass H₂O = 18.02 g/mole

moles CuSO₄ (dehydrate; anhydrous) present = 0.98 g x 1 mol / 159.60 g = 6.140x10^-3 moles

moles H₂O lost/present = 0.53 g x 1 mol / 18.02 g = 0.0294 moles

How many moles of H₂O are present per mole of CuSO₄?

0.0294 moles H₂O / 6.140x10^-3 moles CuSO₄ = 4.8 ~ 5 moles H₂O per mole of CuSO₄

Formula of the original hydrate = CuSO₄•5H₂O