# Chemistry 11

So I did a lab determining the chemical formula of a hydrate. We had to find the molecular formula of a hydrate of copper (2) sulphate, CuSO4 .xH2O. My observation table looks like this:

Mass of clean, dry test tube: 21.6g
Mass of test tube + hydrated copper (2) sulphate: 24.1 g
Mass of test tube + anhydrous copper (2) sulphate

The work I did:
Mass of total: 24.1g - 21.6g= 2.5g
Mass of CuSO4: 23.3g- 21.6g= 1.7g

Determine the percent by mass of water in your sample of hydrated copper (2) sulphate. This is what I did:

Percent by mass of water in CuSO4 .xH2O=
2.5g - 1.7g/ 2.5g x100= 32%

Do you expect the mass percent of water that you determined to be similar to the mass percent that other groups determined? Explain.

Not exactly. Other groups could have added more CuSO4 than us or added less. Other groups could have insufficiently heated the test tube and the weight could differ.

a) Calculate the number of moles of H2O:
Molar mass of H2O: 18.02 g/mol
Mole of H2O: 2.5g - 1.7g divided by 18.02g/mol = 0.044395 mol of H2O

b) Calculate the number of moles of CuSO4:
1.7g divided by 159.62 g/mol (the molar mass of CuSO4) = 0.01065 mol of CuSO4

0.044395 mol of H2O divided by 0.01065 mol of CuSO4 = 4.16, so approximately 4.

0.01065 mol of CuSO4 divided by 0.01065 mol of CuSO4= 1 mol

So the mole ratio would be 4:1 ? I heard 3's and 5's rolling around the classroom and I'm really not sure if I'm doing this right.

I really appreciate the help!
If more or less CuSO4 was added into the test tube, would this affect my answer?

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1. Bob Pursley answered this for you just a few minutes ago and you've hardly let the ink dry before re-posting as if you had heard nothing. No, the amount of material you took makes no difference in the formula nor the percent.
If I have 5g out of 10 g sample, that's 50%.
Suppose you take 20 g sample, you will get 10 g of whatever which is still 50%.

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2. Actually, I made a mistake in my previous question & fixed it, additionally I looked at the link he gave me and didn't find it useful. I appreciate Bob Pursley taking the time out of his day to help me with this problem but I didn't quite grasp what he was getting at. I thought my question got old and people weren't seeing it. I'm new here and I don't know exactly how this website works. I never meant to offend anybody and I'm grateful for those who answer my questions.

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3. I've looked at your data again. %H2O appears to be done correctly. I disagree with the answer to the question about getting results similar to your neighbors results. If everyone is on the same page and everyone has good lab technique then the results should be comparable. As I said in my first response taking different amounts has no effect on the results.
b is calculated right.
c is calculated right. As Bob P pointed out the correct value is CuSO4.5H2O and it would have helped if you had used a balance with better accuracy (been able to read it to at least to one more place). One cause for low results is inadequate heating; i.e., not all of the water was driven off. In analytical chemistry we guard against that by heating, cooling, weighing once as you did, then heat again, cool, weigh again, and continue that until we get no change upon heating. That way we know all of the water is gone (at least at that tempeature).
Welcome to Jiskha and thank you for showing your work. That's the way to get answers quicker. By the way you omitted the mass of the CuSO4 and dish (from both post) but I assume that 23.3 you used in your calculation is that mass.

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