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Question about CaCl2 and CaCl2


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#1 Doraid

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Posted 05 June 2017 - 02:57 PM

Hello,

A recipe for calcium phosphate transfection calls for 2M CaCl2. I have CaCl2.H2O. My basic understanding is CaCl2 is higher in salt concentration compared to CaCl2.H2O. 

 

Can I still use CaCl2.H2O to make the 2M solution? I did the numbers and I arrived at 2.58 g of CaCl2.H2O in 10mL (that's how much I want to make). Is that fine, or did I not account for the H2O?

Thanks.



#2 bob1

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Posted 06 June 2017 - 07:01 AM

You need to account for the water. 



#3 Doraid

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Posted 06 June 2017 - 10:03 AM

Okay, how do I do that?

I know CaCl2.H2O has 24.5% water, and CaCl2 is 111.1g. 


Edited by Doraid, 06 June 2017 - 10:11 AM.


#4 bob1

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Posted 06 June 2017 - 10:41 AM

Look up the molar mass of CaCl2.H2O and work from there.



#5 Doraid

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Posted 06 June 2017 - 11:06 AM

Okay.

Molar mass of CaCl2.2H2O is 147.0146 g/mol. Molar mass of CaCl2 is 110.9840 g/mol.

H2O makes 24.5% of CaCl2.2H2O. 147.01-36.0 = 111.01 g/mol of CaCl2.

I need 2 M, and I'm making 10 mL. 2 M=x/(0.010L) -> x= 0.02 mol.

(0.02 mol)(111.01 g/mol) = 2.22g of CaCl2.2H2O in 10 mL. 

Yes?



#6 bob1

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Posted 06 June 2017 - 12:49 PM

No, you need to use the mass with the water in it -the water is part of the crystal, without it, you will get the wrong amount. CaCl2 can exist in 5 different hydration states from anhydrous to .6H2O (hexahydrate) in multiples of 2. Make sure that the hydration state is the same as the chemical in your lab, and that you use the mass associated with this hydration state. 

 

It also helps if you write it out like they taught you in school, that way you are less likely to make a mistake.

 

n=c*v

n= 2 mol/L * 0.01 l

n= 0.02 mol

 

m=n*mr

m=0.02* ...






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