calcuations-hydrated copper (II) sulfate

[Aka]

calculate the precent by mass of water in a sample of hydrated copper (II) sulfate.

 

mass of beaker-165.3g

mass of beaker with bluestone-173.9g

mass of dehydrated bluestone-170.7g

 

Would this be a ballanced equation for this reaction?

CuSO4•5H2O+ heat= CuSO4+5H2O

 

 

so would you do this

 

%by mass=mass of part/mass of whole

=90.1g/249.7g/molx100

= 36%

 

is this correct?

[encipher]

I beleive the reaction would be as follows:

 

 

CuSO4 • 5H2O <--> CuSO4 + 5H2O

 

(<---> is a double headed arrow (equilibrium sign) and i guess to represent the heat a triangle on top)

 

for percent mass you subtract the weight of the beaker from the weight of the beaker with salt.

 

173.9g - 165.3g = 8.6 grams of hydrated Cu(II)SO4

 

then get the mass of the dehydrated Cu(II)SO4

 

170.7g - 165.3g = 5.4 grams.

 

8.6g - 5.4g = 3.2 grams of H2O

 

for percent by mass:

 

(3.2 / 8.6) x 100 = 37.2 % water

[Aka]

^thanks

 

But why do you include the dehydrated Cu(II)SO4 and use that instead of using the molar mass of CuSO4•5H2O?

 

I also need help on this question

 

based on your obeservations. determine the molecular formula of CUSO4 • xH2O

 

I don't understand this question, would'nt it be CuSO4 • 5H2O, how are you supposed to do this based on observations?

[encipher]

to get the number of water molecules to copper (II) sulfate you do this:

 

 

Molar mass of CuSO4 is 159.62 g/mol

Molar mass of H2O is 18 g/mol

 

 

now there is 5.4 grams of dehydrated CuSO4

 

(5.4 grams) / (159.62 g/mol) = .034 moles of CuSO4

 

then for water

 

(3.2 grams) / (18 g/mol) = .17 moles of H2O

 

Next to get the number of water molecules to copper (II) Sulfate you divide the moles of water by the moles of copper (II) sulfate:

 

.17 / .034 = 5 Therefore, there are 5 moles of Water for every one mole of Copper(II) Sulfate.

[Aka]

Suppose you heated a sample of hydrated ionic compound in a test tube. What might you expect to see inside the test tube, near the mount of the test tube? Explain

 

You would see water droplets and steam. This is because a hydrate indicates the presents of water, when you heat a hydrated ionic compound the water will separate from the compound, and you will see some of the water collecting near the mount of the test tube. The steam could be a result of the water, or the compound evaporating

 

would that be right?

 

****

Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the molecular formula you determined?

 

you would get more molecules of water???

[encipher]

Sounds like a chemistry lab at school =D

 

Your answer is somewhat right. I remember when I took AP chemistry, they stressed on clarifying my statements. First of all, the compound does not evaporate. Only the water does. A better word to use instead of compound would be salt, becuase you might forget to say 'ionic' before it and you would lose points. Again, the salt itself does NOT evaporate, only the water.

 

If you did not heat the hydrated salt enough you would end up with excess water, giving you a larger mass at the end which means a smaller difference (betwen hydrated and dehydrated) in turn, giving you LESS grams of water, which turns out to be less moles. End result: Less moles of water.

 

(Personally, I hated questions like these, always got me confused :S )

[Aka]

^ok thanks but i have one more question (sorry..)

 

Suppose that you did not completely convert the hydrate to the anhydrous compound. Explain how this would affect the calculated percent by mass of water in the compound?

 

If the hydrate was not completely converted to the anhydrous compound, the mass of anhydrous Cu(II)SO4 would increase because it would not be pure anhydrous Cu(II)SO4, instead it would be CuSO4•5H2O, in other words you would be weighing the compound plus water. In turn, this would change the mass of H2O, making to higher (for example instead of 3.2g it might be 4.2g.) the results of the percent composition calculation would also change to a higher number, for example %= (4.2/8.6)100 would equal 48.8% of water, compared to 37.2% of water, which is what we got by completely convert the hydrate to the anhydrous compound.

 

 

i dont know if it makes sence at all

[encipher]

I apologize if I'm being picky, but you CANNOT say stuff like:

 

"the mass of anhydrous Cu(II)SO4 would increase"

 

"change the mass of H2O, making to higher"

 

etc..

 

You don't make the mass increase, not does the mass of the H2O increase. A good way to state questions like these is this way:

 

If the Copper(II) Sulfate was not completely dehydrated of H2O, The measured mass of the apparent anhydrous Copper (II) Sulfate would be slightly higher than It's actual value. Therefore causing ......

 

This is just a quick example.. think up of something to write. Hopefully you got the idea.. People who grade exams and tests are very picky when it comes to wording. Hope this helped.

[Aka]

^ Ok thank you for all your help I really appreciate it

 

people who grade exams and tests are very picky when it comes to wording

 

I'll keep that in mind...