Question

1. If 147 mg of cyclohexanol and 2.2 mL of Clorox are mixed in excess acetic acid and 130 mg of cyclohexanone are produced, the percent yield is ___ %

2. If 152 mg of cyclohexanone and 0.96 mL of concentrated nitric acid are mixed and allowed to react and 199 mg of adipic acid are formed, the percent yield is ___ %

Answer 1

1. If 147 mg of cyclohexanol and 2.2 mL of Clorox are mixed in excess acetic acid and 130 mg of cyclohexanone are produced, the percent yield is ___ %

Moles of cyclohexanol = 147 mg = 0.147 g / 100.158 g/mol

= 1.47*10^-3 moles

One cyclohexanone is produced from one cyclohexanol.

Now calculate the moles of cyclohexanone :

1.47*10^-3 moles cyclohexanol =1.47*10^-3 moles cyclohexanone

Amount of cyclohexanone = 98.15 g/mol *1.47*10^-3 moles

= 0.144 g

Or 144 g

the percent yield = actual yield / theoretical yield *100

= 130 mg /144 mg*100

= 90.3%

2. If 152 mg of cyclohexanone and 0.96 mL of concentrated nitric acid are mixed and allowed to react and 199 mg of adipic acid are formed, the percent yield is ___ %

Moles of cyclohexanone = 152 mg = 0.152 g / 98.15 g/mol

= 1.55*10^-3 moles

One cyclohexanone is produced one mole adipic acid.

Moles of adipic acid = 1.55*10^-3 moles

Amount of adipic acid =1.55*10^-3 moles *146.1412 g/mol

= 0.226 g

= 226 mg

the percent yield = actual yield / theoretical yield *100

= 199 mg /226 mg*100

= 88.1%