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In: Chemistry

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

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

Expert Solution

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.

queen_honey_blossom answered 1 year ago

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