If 8.89 g of 2-methylcyclohexanone (112.17 g/mol) was reduced to 5.14 g of 2-methylcyclohexanol (114.19 g/mol), what is the percentage yield of the product?

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Answer:

56.8%

Explanation:

The reaction of the problem is 1:1. That means 1 mole of 2-methylcyclohexanone produce 1 mole of 2-methylcyclohexanol.

Percentage yield is defined as 100 times the ratio between actual yield of the reaction (5.14g) and the theoretical yield.

The theoretical yield (All reactant produce products) is obtained from the mass of the reactant as follows:

Theoretical Yield:

8.89g 2-methylcyclohexanone * (1mol/112.17g) = 0.07925 moles 2-methylcyclohexanone

Assuming all reactant produce the product in a 100% of yield, the moles of 2-methylcyclohexanol are 0.07925 moles and the mass (Theoretical yield) is:

0.07925 moles 2-methylcyclohexanol * (114.19g/mol) = 9.05g

Percentage yield:

5.14g / 9.05g * 100 = 56.8%

The percentage of the mass successfully converted into a new product is 57.2%.

Mass of the reactant

The mass of the reactant (2-methylcyclohexanone) before the reduction is given as 8.89 g.

Mass of the product yielded

The mass of the product ( 2-methylcyclohexanol) produced is given as 5.14 g.

Percentage yield of the product

The percentage of the mass successfully converted into a new product is calculated as follows;

[tex]= \frac{5.14}{8.99} \times 100\% \\\\= 57.2 \ \%[/tex]

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