We did a multi step 3 part lab where we went from benzaldehyde all the way to benzilic acid.Here is the overall chemical equation for all 3 parts
We are given 150 mmol of benzaldeyhde and looking at the equations above we see that 2 molecules benzaldehyde make 1 molecule of benzoin so 150/2 = 75 mmol of benzoin
Part b: 75 mmol of benzoin = 75 mmol of benzil
part c: 75 mmol of benzil = 75 mmol benzilic acid, sound good so far?I went ahead and calculated the theoretical yield using the given mw of each of these parts and this is the theoretical yield I got for each:
Part a: 150 mmol benzaldehyde makes 75 mmol benzoin and I got 15.92 theoretical yield for benzoin and here is my math here: https://imgur.com/L27k2HV
Part b: 75 mmol benzoin makes 75 mmol benzil and so we just multiply 0.075 mols by mw of benzil and get 15.77 g theoretical yield for benzil
Part c: same thing, 75 mmol of benzil makes 75 mmol of benzilic acid, so 0.075 mmol x mw of benzilic acid and get 17.12 g theoretical yield benzilic acid.
My professor checked these theoretical yields and said I am all good!
So now that you guys have some background on the numbers and the equations, I am being asked to calculuate % yield of benzil, benzoin, % recovery of benzoin, % yield of benzilic acid, % recovery of benzilic acid and then finally the overall % yield of benzilic acid.
Before I do all these calculations I want to start off with the % yield of benzil first and have you guys check that part and then I will move forward.
SO, when I did the lab, I started off with 15.28 mL of benzaldehyde as required and reacted with thiamine and NaOH and made benzoin. The mass of my actual benzoin was 5.681 g
So, since I knew for a fact my theoretical yield for benzoin is 15.92 g , and my actual yield was 5.681 g, therefore, % yield of benzoin is 35.68%? If this is correct math let me know and I will do the next 2 parts!