Question

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

Help please!!

Answer 1

From the synthesis reaction we know the mole ration between ferrocene : monoacetylferrocene is 1 : 1

We use this mole ratio to get moles of monoacetylferrocene.

From the moles of it we can get its mass.

Step 1) Determination of limiting reactant.

Reactant which has less amount than required for stoichiometric limit is known as limiting reactant.

Mol ratio between ferrocene : acetyl chloride is 1 : 1

So we can now simply compare the moles of both reactant and will decide the one which has less moles.

            Calculation of moles:

            n ferrocene = mass in g / molar mass = 0.100 g / 186.04 g per mol = 5.38E-4 mol

            n acetyl chloride = (0.08 g x 1.1 g /mL) / 78.49 g per mol = 0.011212 mol

            From calculation we can say that ferrocene which has less moles and so the limiting reactant is ferrocene.

Step 2 :

Calculation of maximum amount of product formed :

The maximum moles of product can be predicted by using moles of limiting reactant.

Moles of product = 5.38 E-4 mol reactant x 1 mol product / 1 mol reactant

= 5.38 E-4 mol Product

Mass of product in g = 5.38E-4 mol x molar mass of product

= 5.38 E-4 mol x 228.07 g/mol

=1.23 E-1 g

Mass in mg

= 123.0 mg

Theoretical yield = 123.0 mg