Determing The Chemical Formula of a Hydrate

AI Thread Summary
Copper (II) sulfate hydrate loses water upon heating, transforming from a blue crystalline solid to a white anhydrous form. The mass of water driven off during the process is calculated as 0.0554g, and the moles of water present are approximately 3.07 x 10^-3. The moles of CuSO4 are calculated to be about 6.13 x 10^-4. To determine the ratio of moles of water to moles of CuSO4, the previously calculated values should be divided. This ratio is essential for understanding the chemical formula of the hydrate.
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Homework Statement


When copper (II) sulfate hydrate, a blue crystalline solid containing embedded water molecules (called a hydrate), is heated in air, it loses the water molecules and the blue solid is transformed to a white anyhydrous (no water) crystal known as copper (II) sulfate.

Homework Equations


CuSO4 .x H2O +heat --> CuSO4 (s) + H2O (g)Observations: Mass of copper (II) sulfate hydrate = 0.1534g
Mass of anyhydrous copper (II) sulfate = 0.0980g

1a) Calculate mass of H2O that was driven off.
b) Calculate the number of moles of water present.
c) Calculate the number of moles of CuSO4 present.

My attempt to answer these questions:
1a) Since anhydrous copper sulfate has no water the mass of H2O driven off is 0.1534g - 0.0980g =0.0554g
b) n = m/M n= 0.0554/18.02 therefore, n= 3.07 x10 to the power of -3.
c) Mcuso4 = 159.62g/mol so, n=m/M n= 0.0980g/159.62g/mol which = 6.13 x10 to the power of -4
 
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So far so good (that is, I have not checked the numbers, just the general idea).

What is the ratio?
 
The numbers just don't look right for some reason...
No ratio given, that's a whole new question "Determine the ratio of moles of water to moles of CuSO4" that i have yet to attempt. If my answers are right then i can move on.
 
Come on, "whole new question" means just divide two numbers you got so far.
 
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