What's the mole ratio of xCuSO4* xH20? (See The Whole Question)?

Question

From CuSO4*H20, The first weight was 1.57g, when heated, .57g of Suso4 was Left, How many Moles of Water(H20) was Lost?

1g/18g of H2O=.0556g/mol
.57g/(63.55*32.07*16*4) -->> .57/ 130435.104= 4.3699x10^-6

So now we divide both of them by teh lowest mole which for this equation is .0556g, right? But is the CuSO4 right? (The Mole)

Can someone check and answer this? Thank You!

Answer 1

"1g/18g of H2O=.0556g/mol
.57g/(63.55*32.07*16*4) -->> .57/ 130435.104= 4.3699x10^-6"

emm, I dont really understand what u'r trying to do here, molar mass of CuSO4 should be calculate this way:

M[CuSO4] = M[Cu] + M [S] + 4*M[O], not M[Cu]* M [S]*4*M[O].

ANY molar mass has the unit of g mol^-1, not g. If you divide 1g by 18g, u will have a UNITLESS number.

"So now we divide both of them by teh lowest mole which for this equation is .0556g, right? But is the CuSO4 right? (The Mole)"

it should be 0.0556 mol, not 0556g. No, it's the mole of water lost.

This is what u should do.

find out the MOLAR MASS [unit g/mol] of CuSO4 and H2O [CuSO4 should be around 160 g/mol, H2O should be around 18 g/mol]

then, find out the MOLE of CuSO4 and H2O in the compound by dividing the mass of CuSO4/ H2O by their molar mass.

Then divide the mole of H2O in the compound by the mole of CuSO4 in the compound, find out ur ratio.

the ratio turns out to be 15.6, a very strange number in my point of view [not an integer, and it's a little bit large for this compound]

Answer 2

I looked at your math, and it looks like you are multiplying instead adding up the individual elements in CUSO4 in order to get the formula weight. CuSO4 formula weight would be 63.55 + 32.07 + 4*16.0 = 159.62 . Thus if you have 0.57 g. of the residue left, you would have 0.00357 moles of CuSO4. The ratio of water to CuSO4 would come out as 15.57, which I think is too high and is not an integer. There may be experimental error to consider, though. I hope that helped!