Methanol (CH₄O) is converted to bromomethane (CH₃Br) as follows: CH₄O + HBr → CH₃Br + H₂O
If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?

Question

Quizplus · Accepted Answer

To find the theoretical yield of bromomethane, we need to calculate the moles of methanol and use stoichiometry to convert to moles of bromomethane. 5.00 g CH₄O x 1 mol CH₄O/32.04 g CH₄O x 1 mol CH₃Br/1 mol CH₄O = 0.1551 mol CH₃Br (theoretical yield) To find the percentage yield, we divide the actual yield (12.23 g) by the theoretical yield (calculated above) and multiply by 100%. % yield = (12.23 g / (0.1551 mol x 94.94 g/mol)) x 100% = 82.6%

Methanol (CH4O) Is Converted to Bromomethane (CH3Br) as Follows: CH4O

Methanol (CH₄O) Is Converted to Bromomethane (CH₃Br) as Follows: CH₄O