Question

Given the molality of isoborneol is 0.701 mol/kg, what is the percent by mass of camphor in the product?

Answer 1

Isoborneol can be oxidised to camphor via the following reaction scheme, where the oxidation can be carried out using a suitable oxidizing agent.

It is easy to realise that 1 mol of isoborneol will give 1 mol of camphor when we have 100% yield in a perfectly controlled reaction.

Given that the concentration of isoborneol taken = 0.701 mol/kg.

The number of moles of camphor expected from 1 kg of reaction mixture is 0.701 mol.

Molecular formula of camphor =

Molar mass of camphor =

Hence, mass of 0.701 mol of camphor =

Hence, assuming 100% yield in the conversion, the mass of camphor in 1 kg of reaction mixture is 106.6 g approximately.

Hence, the mass percentage of camphor can be calculated as

Since the concentration of isoborneol is given with 3 significant figures, we express our answer with three significant figures.

Hence, the mass percent of camphor in the product is 10.7 % approximately.

Note: if this question pertains to a particular case where the percent yield is not 100%,  use the percentage yield data to recalculate the mass percent.