[jrpestana]

Hello:

I just finished an exp. where I used 5g isoborneol in 15 acetic acid to form CAMPHOR. I did the following to obtain the theoretical yield but am uncertain about how accurate this is:

5.00 g Isoborneol x 1 mole / 154.25 g (mol.wt)Isoborneol = 0.0324 moles Isoborneol

0.0324 moles camphor x 152.24 g/ 1 mol = 4.93 g of camphor

It is safe for me to assume 1 mol Isoborneol = 1 mol camphor... should I be doing a calculation for this? Can someone give me a hand... I don't have my product weight which makes things even worst (I just know its a small number) Thanks in advance

[arnyk]

Could you write out the rx equation?

[jrpestana]

Isoborneol over NaOCl and CH3C02H to give camphor

I believe isoborneol is C10H18O and camphor is C10H16O

[movies]

It is safe for me to assume 1 mol Isoborneol = 1 mol camphor... should I be doing a calculation for this?

Where do all of the atoms in the camphor come from?  You can't create or destroy matter, right?  So everything is in there somewhere.  In a theoretical calculation you assume that the reaction goes to completion and you don't lose any material from transfer or isolation.

[alphahydroxy]

It is safe for me to assume 1 mol Isoborneol = 1 mol camphor... should I be doing a calculation for this?

[/b]

Yes you can assume this because the reaction equation shows that 1 mole of isoborneol goes to 1 mole of camphor.

A theoretical yield assumes that the reaction goes to completion and you get 100% yield of your product.

Thus, using the reaction equation, the theoretical yield when using 0.0324 mol of starting material is 0.0324 mol of product.

To calculate the actual yield you have obtained, you need to make a note of the mass of product you have isolated and calculate the number of moles from this.

Once you have the number of moles, you can divide it by the theoretical number of moles (i.e. 0.0324) and multiply by 100 to give the percentage yield

[jrpestana]

So in this case it is correct. All I need now is my weight so that I can calculate the % percent yield... thanks for the reply.