11th gr. chem (composition of a hydrate)

[monica_e826]

i might get confusing cause i don't understand this very much
but here is the copy of my lab i need help with #1 & 6 of the POST-ALAYSIS
here is the word doc. of my lab
http://courses.digitaldapp.org/documents/Inquiry Lab - Experiment 03 - Composition of a Hydrate.doc


btw.
5 grams of copper II sulfate was used
when it became an anyhride it had a mass of 0.15g


i know i made a mistake somewhere in my lab and i was wondering if you tell where my mistake is and what i should do. i would highly appreciate it.

so i know...
CuSO₄ . 5H₂O
molar weight is 249.69g


by itself is
CuSO₄ = 159.61g
5H₂O= 90.81

so i took the
mass of the water/total mass of hydrate ^x100= 36% of H₂O removed in anhydrous copper (II) sulfate.

am i correct?
this is where i got confused...
-determine the # of moles of water present in each mole of hydrous copper (II) sulfate.
so what i did was:
5H₂O

H₁₀= 11%=11/1.0079=10.91=2
O₅=88%=88/15.9994=5.5=1
is this the correct Empirical formula?
2H₁₀O₅ & can be simplified to 2H₂O

so now what do i use to find the amount of moles of water present?

 

[chemisttree]

i might get confusing cause i don't understand this very much
but here is the copy of my lab i need help with #1 & 6 of the POST-ALAYSIS
here is the word doc. of my lab
http://courses.digitaldapp.org/documents/Inquiry Lab - Experiment 03 - Composition of a Hydrate.doc


btw.
5 grams of copper II sulfate was used
when it became an anyhride it had a mass of 0.15g

Are you sure you got that right? I assume that "it had a mass of..." means that the anhydride had a mass of... Is that right?

i know i made a mistake somewhere in my lab and i was wondering if you tell where my mistake is and what i should do. i would highly appreciate it.

so i know...
CuSO₄ . 5H₂O
molar weight is 249.69g


by itself is
CuSO₄ = 159.61g
5H₂O= 90.81

Good so far...

so i took the
mass of the water/total mass of hydrate ^x100= 36% of H₂O removed in anhydrous copper (II) sulfate.

am i correct?
this is where i got confused...
-determine the # of moles of water present in each mole of hydrous copper (II) sulfate.

Here is where you strayed off course. The formula for your unknown hydrate is CuSO₄ . XH₂O. You need to solve for 'X', not assume that it is 5 and do some math without using any of the data you collected.

Show us the raw data you collected.

 

[Blueshift5]

Dear student,

First of all, it is great that you are seeking help and clarification on your lab report. it is important to always question and seek understanding in our experiments.

To answer your question, let me first clarify the concept of a hydrate. A hydrate is a compound that contains water molecules that are chemically bound to the other molecules in the compound. In your case, copper II sulfate pentahydrate (CuSO4·5H2O) means that for every one mole of copper II sulfate, there are five moles of water molecules attached. The water molecules are not simply mixed with the copper II sulfate, but they are actually part of the compound.

Now, let's address your mistake in the calculation. You have correctly calculated the percentage of water removed in the anhydrous copper II sulfate. However, you did not take into account the fact that you started with 5 grams of copper II sulfate PENTAHYDRATE, not anhydrous copper II sulfate. This means that your calculation should have been:

Mass of water removed/total mass of copper II sulfate pentahydrate x 100 = 0.15g/5g x 100 = 3%

Next, to determine the number of moles of water present in each mole of copper II sulfate pentahydrate, we need to use the molar ratios from the compound's formula. In this case, for every one mole of copper II sulfate, there are five moles of water. So, we can say that one mole of copper II sulfate pentahydrate contains 5 moles of water. This means that in your experiment, you should have removed 5 moles of water for every mole of copper II sulfate pentahydrate used.

To find the amount of moles of water present, you can use the following formula:

Number of moles of water = mass of water removed/molar mass of water

In your case, the mass of water removed is 0.15g, and the molar mass of water is 18.015 g/mol. Plugging in these values, we get:

Number of moles of water = 0.15g/18.015 g/mol = 0.00832 mol

This means that for every mole of copper II sulfate pentahydrate used, there are 0.00832 moles of