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Topic: Copper (II) Sulfate * xH2O Lab  (Read 9049 times)

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Offline RockThis52

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Copper (II) Sulfate * xH2O Lab
« on: November 01, 2010, 08:45:24 PM »
So our teacher gave us CuSO4 * xH20 and we must find out how many hydrates are present in the sample. She gave each group a different mass sample. I was given 4.25 g. This was the mass of CuSO4* xH20 before I heated it up. When I heated it up the mass was 2.97g. The difference is 1.28g.

The first question asks to find the percent by mass of water in the sample.

I did.... 4.25g-2.97g = 1.28 g

1.28g/4.25g X100% = 30.12 % By Mass of Water (No idea if that's a right number but I'm sure I did the process right)

Next question asks to determine the number of hydrates in the sample.

I did.... Moles of CuSO4 = Cu-63.55g/mol + S-32.07g/mol + O-16.00(4) = 159.62g/mol.

I then divided 2.97 (Mass of anhydrous compound) by 159.62 (molar mass of CuSO4) to get...

2.97/159.62 = 0.019 mol of CuSO4

Now I did the same for water

Moles of H2O = H-1.01(2) + 16.00 = 18.02 g/mol

Divide 1.28 (Difference of hydrous sample and anhydrous sample) by 18.02 (molar mass of H2O)

1.28/18.02 = 0.071 mol of H20

Now I mist divide 0.071 mol of H2O by 0.019 mol of CuSO4

0.071/0.019 = 3.737

I know that the correct answer should be 5, but I have no idea why I'm getting these answers. Please help I will surely give out the best answer fairly.

Offline Borek

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Re: Copper (II) Sulfate * xH2O Lab
« Reply #1 on: November 02, 2010, 04:18:14 AM »
we must find out how many hydrates are present in the sample

There is one hydrate present - copper sulfate.

Apart from atrocities you do to the naming of substances and things you calculate, your approach is correct. I haven't checked the numbers, but the logic behind is correct. You should not round down intermediate results when using them for calculations (reporting them as rounded is correct though).

Quote
Next question asks to determine the number of hydrates in the sample.

Number of water molecules per salt molecule.

Quote
Moles of CuSO4 = Cu-63.55g/mol + S-32.07g/mol + O-16.00(4) = 159.62g/mol.
Quote
Moles of H2O = H-1.01(2) + 16.00 = 18.02 g/mol

It is not "Moles", it is molar mass.

You are right final answer should be around 5. It rarely is exactly 5, although below 4 seems to me a little bit too low.

Think about possible sources of experimental errors. What will happen if the sulfate after roasting is not completely anhydrous?
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