If you added 0.832 g of Mg to a solution containing 1.794 g of CuSO_4, to reduce

Question

If you added 0.832 g of Mg to a solution containing 1.794 g of CuSO_4, to reduce any copper ion present, what is the least volume of 6.0 M HC1 you would need to dissolve the excess Mg? Explain why you chose this volume. Calculate the formula weight of an unknown copper compound from this data that a student collected(Assume 1 Cu atom per formula unit.) mass of the test tube and unknown sample 14 603 g mass of test tube without the unknown sample 13.437 g mass of filter paper 0.633 g mans of filter paper and copper 1 096 g

Explanation / Answer

3. moles of Mg = 0.832/24.305 g/mol = 0.034 mols

moles of CuSO4 = 1.794 g/159.609 g/mol = 0.011 mols

excess Mg remain = 0.034 - 0.011 = 0.023 mols

Volume of HCl required = 0.023 mol x 2/6 M = 7.66 ml

4. mass of Cu in sample = (1.096 - 0.633) = 0.463 g

mass of Cu compound = (14.603 - 13.437) = 1.166 g

Percent Cu in sample compound = 0.463 x 100/1.166 = 39.71 %

Percent Cu in CuSO4 is close to the percent we have calculate above.

Thus the unknwon compound here is CuSO4

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