Question

Calculate the mass of isoborneol in 2.5 mmol of isoborneol and the theoretical yield (in grams) of camphor from that amount of isoborneol

isoborneol = 154.25 g mol?1
Camphor, Molar mass = 152.23 g/mol

Answer 1

`m_(isoborneol )=0.39g\\\\m_(Camphor)=0.38g\\`

Step-by-step explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to infer that the reaction whereby isoborneol goes to camphor occurs in a 1:1 mole ratio, that is why the theoretical yield of the latter is also 2.5 mmol (0.0025 mol) but the masses can be calculated as follows:

`m_(isoborneol )=0.0025mol*(154.25g)/(1mol) =0.39g\\\\m_(Camphor)=0.0025mol*(152.23 g)/(1mol) =0.38g\\`

Because of the fact this is a rearrangement reaction whereas the number of atoms is not significantly modified.

Regards!