# 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

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. 

Percent error = (theoretical – experimental) x 100%

theoretical

1
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 .

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