[BADMIN]

My class has conducted an experiment in which we heat bluestone in a crucible to seperate the water from it.

Given the percent compositions of 68% CuSO4, and 32% H2O we need to find the formula of the hydrate. I'm not entirely sure how to approach this problem.

The textbook we're using is Nelson, Chemistry 11 if that helps.

I've also probably left out vital information as well in regards to sample masses or something but please bear with me and lend me a hand if you would.

[BADMIN]

not sure of its necessity but **Ignore me, I am impatient**

[allanf]

The only info you need is the mass percents (which you gave) and the molar mass of CuSO4 and H2O (which you can easily figure out from a periodic table, or look up in a table of physical data).

Consider what the mass percent of H2O means: 32% == 32g H2O per 100g CuSO4*nH2O

Similarly for CuSO4 (anhydrous), 68% == 68g CuSO4 per 100g CuSO4*nH2O

If you convert these masses (32g H2O, 68g CuSO4) to moles then you can figure out the number (n) of moles of H2O present per mole of the hydrate (since there is only one mole of CuSO4 in a mole of the hydrate)

[BADMIN]

Thanks, my porblems were solved long before that post by a friend from class but thank you very much in convincing me that this site and its members are valuable resources in times of need!