December 12, 2019, 09:12:52 PM
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### Topic: Can someone help me out with % yield of this multistep lab?  (Read 193 times)

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#### yourdeath01

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• Mole Snacks: +1/-1 ##### Can someone help me out with % yield of this multistep lab?
« on: December 04, 2019, 04:03:37 AM »
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: https://imgur.com/xQqgecC

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!

#### AWK

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« Reply #1 on: December 04, 2019, 04:40:16 AM »
A month ago I gave you an example of how to calculate the theoretical yields of all stages of the synthesis.
If you know the masses of intermediate products, then calculation the performance of each stage should be trivial.

https://www.chemicalforums.com/index.php?topic=101929.msg357472#msg357472
AWK

#### yourdeath01

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•   • Posts: 30
• Mole Snacks: +1/-1 ##### Re: Can someone help me out with % yield of this multistep lab?
« Reply #2 on: December 04, 2019, 07:48:40 PM »
Well the theoretical yields are calculated and all good and I am probably overthinking the % yield part but here is what I got:

- For part A

Benzaldehyde used: 15.28 mL (so that would be 15.28 g)

Actual mass of benzoin collected: 5.681 g

Calculuate % yield of benzoin, guessing this is 37.18 % yield? Cos I know that 15.92 is my theoretical yield so 5.681/15.92 = 37.18% yield?

- For part b:

benzoin used 5.681 g

Crude benzil: 5.118

Pure benzil: 3.84 g

For % yield since I know benzil theoretical yield is 15.77 and I collected 3.84 pure that means 24.35% yield and % recovery is 75.13% (3.84/5.118)

- for part c:

More or less the same

Pure benzil mass: 3.84 g

Crude benzilic mass: 9.299 g

Pure benzilic acid mass: 0.451 g

For % yield here since I know 17.12 is my theoretical yield that means 9.299/17.12 = 54% yield and % recovery is 0.451/9.299 = 4.85%

Is this all good? Just wanna be sure im not skipping anything, again prob overthinking but really I calculated theoretical yield before lab based off the 150 mmol we used and then once I got actual ammounts I just divided those by the theoretical yield to get me % yields and % recoveries.

Only question I have left is that I am being asked to calculuate % yield of overall benzylic acid and I am not sure why because I already calculauted its % yield, not sure what they want here...

#### AWK

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« Reply #3 on: December 05, 2019, 01:42:59 AM »
Quote
- For part A

Benzaldehyde used: 15.28 mL (so that would be 15.28 g)
[/quote]
15.28 g is not true - density of benzaldehyde missing
AWK