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# 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. [2]

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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