Answer to Question #155526 in Organic Chemistry for Raneem

Question #155526



Moles of water H2O 11.38- 7.28 = 4.1g 

Moles of anyhydrate ( CuSO4)= 7.28/159.62 = 0.0456 moles of copper sulfate



n = m/M

m = 4.1g, M = 18.02g

n = 4.1/ 18.02 = 0.227 moles of water


 

molecules of water 0.227(6.02 x 10^23)

                                          = 1.37 x 10^23


Answer this:

  1. Your value probably does not match the actual value for the hydrate. Use the following formula to calculate the percent error for the experiment. Remember to report an absolute value. [2]

Percent error = (theoretical – experimental) x 100%

    theoretical



1
Expert's answer
2021-01-15T06:38:30-0500

Moles of water H2O 11.38- 7.28 = 4.1g 

Moles of anyhydrate ( CuSO4)= 7.28/159.62 = 0.0456 moles of copper sulfate



n = m/M

m = 4.1g, M = 18.02g

n = 4.1/ 18.02 = 0.227 moles of water


 

molecules of water 0.227(6.02 x 10^23)

                                          = 1.37 x 10^23



% error = ( 0.227 - 0.0456 )× 100 / 0.227

= 79.9 % .


For given moles of water and Anhydrous cupric Sulphate .


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be first!

Leave a comment

Ask Your question

LATEST TUTORIALS
New on Blog
APPROVED BY CLIENTS