A student synthesized a cobalt amine complex with 11.0323 g of CoCl2·6H2O (limiting reactant, MW = 237.83 g/mol) and collected 9.4705 g of the cobalt amine complex (MW = 250.45 g/mol). What was the percent yield?

Question

Kathy Wolfe

Chemistry

20 October, 17:15

A student synthesized a cobalt amine complex with 11.0323 g of CoCl2·6H2O (limiting reactant, MW = 237.83 g/mol) and collected 9.4705 g of the cobalt amine complex (MW = 250.45 g/mol). What was the percent yield?

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Answers (1)

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Sammy Bridges · Answer 1

The percent yield would be 81.51%

Explanation:

Number of moles of CoCl2·6H2O would be calculated by using the formula

number of moles = given weight/molecular weight.

Thus, by putting values,

= 11.0323/237.83

= 0.046387

Number of moles of CO+2 = Number of moles of CoCl2·6H2O = 0.046387

As we know that CoCl2·6H2O is a limiting reactant, therefore, the number of moles of complex = Number of moles of CO+2 = 0.046387.

Hence, we can calculate the mass of the complex as follows:

Molecular weigth x number of moles = 250.45 x 0.046387

Thus, the theoratical yield would be 11.6176.

We know that actual yield as given in statement is 9.4705 g. Therefore, percent yield would be,

= (9.4705 / 11.6176) x 100

= 81.51%