# Hydrated copper (II) sulfate mass calculation

calculate the precent by mass of water in a sample of hydrated copper (II) sulfate.

mass of beaker-165.3g
mass of beaker with bluestone-173.9g
mass of dehydrated bluestone-170.7g

Would this be a ballanced equation for this reaction?
CuSO4•5H2O+ heat= CuSO4+5H2O

so would you do this

%by mass=mass of part/mass of whole
=90.1g/249.7g/molx100
= 36%

is this correct?

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Borek
Mentor
No. But it is so wrong I have no idea how to help you :(

Start calculating mass of the copper sulfate before roasting and after roasting.

No. But it is so wrong I have no idea how to help you :(
lol, its ok i figured that part out

i have different questions now

Suppose you heated a sample of hydrated ionic compound in a test tube. What might you expect to see inside the test tube, near the mount of the test tube? Explain

You would see water droplets and steam. This is because a hydrate indicates the presents of water, when you heat a hydrated ionic compound the water will separate from the compound, and you will see some of the water collecting near the mount of the test tube. The steam could be a result of the water, or the compound evaporating

would that be right?

****
Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the molecular formula you determined?

you would get more molecules of water???

or the compound evaporating
Ionic compounds generally are nonvolatile solids and have high boiling points. Stick to the water answer.

Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the molecular formula you determined?

you would get more molecules of water???
No, think about what the question is saying here. Some water molecules are leaving the compound, just not all of them. When all the water has gone from the compound, then you will have the anhydrous product.

^ok thanks

7. Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the calculated percent by mass of water in the compound?

If the hydrate was not completely converted to the anhydrous compound, the mass of anhydrous Cu(II)SO4 would increase because it would not be pure anhydrous Cu(II)SO4, instead it would be CuSO4•5H2O, in other words you would be weighing the compound plus water. In turn, this would change the mass of H2O, making to higher (for example instead of 3.2g it might be 4.2g.) the results of the percent composition calculation would also change to a higher number, for example %= (4.2/8.6)100 would equal 48.8% of water, compared to 37.2% of water, which is what we got by completely convert the hydrate to the anhydrous compound.