Question

If you use 100 mg of ferrocene and 0.08 mL of acetyl chloride, what is your theoretical yield (in mg) of monoacetylferrocene?

Please help!

Answer 1

As per the reaction stoichiometry, one mole of ferrocene reacts with one mole of acetyl chloride to give one mole of monoacetyl ferrocene

Let us determine the limiting reagent:

Moles of ferrocene used = Mass of ferrocene / Molecular weight of ferrocene

Moles of ferrocene used = 0.100 grams / 186.04 g / mole = 5.38 X 10^-4 moles

Density of acetyl chloride is 1.1 g / mL

Mass of given volume of acetyl chloride = Density X volume = 1.1 g /mL X 0.08mL = 0.088 grams

Moles of acetyl chloride = Mass of acetyl chloride / Molecular weight = 0.088grams / 78.49g/mole = 11.21 X 10^-4 moles

So here ferrocene is the limiting reagent (less moles are present than required for complete reaction with given moles of acetyl chloride)

Moles of product formed = Moles of limiting reagent present (1:1 stoichiometry)

Moles of product formed = 5.38 X 10^-4 moles

Mass of monoacetyl ferrocene produced = Moles X molecular weight = 5.38 X 10^-4 moles X 228.07 g / mole

Mass of monoacetyl ferrocene produced = 0.123 grams = 123 mg (theoretical yield)