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Why molecular weight of cetearyl alcohol is 512.94 instead of ~250
Molecular weight of stearyl alcohol is 270.49
Molecular weight of cetyl alcohol is 242.44
So how according to this website and all all websites in google search molecular weight of cetearyl alcohol is 512.94?
https://www.upichem.com/products/ceto-stearyl-alcohol-3070-flke/
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This discussion was modified 1 year, 2 months ago by
Perry44.
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This discussion was modified 1 year, 2 months ago by
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12 Replies
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I think they just added two masses like it is one molecule. But as I understand cetearyl alcohol is not a molecule but a mix of ingredients? I am not a chemist and I am not sure if molecular weight can be applied to a mix of ingredients?
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This reply was modified 1 year, 2 months ago by fareloz.
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This reply was modified 1 year, 2 months ago by
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Yeah, I think this is just an error. These brochures are usually put together by people who don’t have a chemistry background.
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I just googled “what molecural weight has cetearyl alcohol” and every website claims it has 512+…
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This reply was modified 1 year, 2 months ago by fareloz.
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I asked ChatGPT and it said…
“The molecular weight of cetearyl alcohol is 392.62 g/mol.“
I don’t think that is correct either.
More from ChatGPT -
“The sum of the molecular weights of cetyl alcohol (C16H32O) and stearyl alcohol (C18H38O) is approximately 410 g/mol.
To calculate this, we first need to determine the atomic weight of each element in the molecule and then multiply it by the number of atoms of that element in the molecule. The atomic weights of carbon (C), hydrogen (H), and oxygen (O) can be taken from the periodic table.
For cetyl alcohol, the molecular weight is:
16 (C) * 12.01 (atomic weight of C) + 32 (H) * 1.01 (atomic weight of H) + 1 (O) * 16 (atomic weight of O) = 192.48 g/mol
For stearyl alcohol, the molecular weight is:
18 (C) * 12.01 (atomic weight of C) + 38 (H) * 1.01 (atomic weight of H) + 1 (O) * 16 (atomic weight of O) = 218.36 g/mol
Finally, the sum of the molecular weights of cetyl alcohol and stearyl alcohol is:
192.48 + 218.36 = 410.84 g/mol
Note: This is an approximate value, and the actual molecular weight of cetearyl alcohol may slightly vary depending on its specific composition.“
So, I think the average molecular weight is (x)192.48 + (y)(218.36)
In this case x = fraction of cetyl alcohol and y = fraction of stearyl alcohol-
I see your point, but could you explain how it actually should be calculated? I tried to search what is molar mass (I still suspect molecular weight is wrong term since as I understand cetearyl alcohol is a mix of two substances in some ratio) of mixes but can’t find easy answer.
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See my edited previous comment. The calculation is in the last paragraph.
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This reply was modified 1 year, 2 months ago by
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P.S. my avatar is generated by this NN.
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@Perry Sorry but when i calculated the molecular weight of cetyl and stearyl alcohol from their atomic weight, it was 242.44116 & 270.49432 . Are you sure there is no mistake in your calculation?
Then some new questions.
1. Is cetearyl alcohol a new molecule by itself or just a mixture of two other molecules?
2. Why did you sum the molecular weight of cetyl and stearyl to get the molecular weight of cetearyl alcohol?
3. If in a formula we add 1% ceyl alcohol & 1% stearyl alcohol, would it be the same as 2% cetearyl alcohol 50:50?
In this lecture the instructor says molecular weight of cetearyl alcohol is 250, and it is the weight of cetearyl alcohol 70:30 if we calculate it from atomic weight. What is your comment on this one?
https://youtu.be/TYFDRkRclK0-
@Abdullah Your molar weights for both Cetyl and Stearyl alcohol are correct. Perry used the values ChatGPT gave him, but what he wanted to show is how to calculate the average molecular weight for Cetearyl alcohol using mole fractions of Cetyl and Stearyl alcohol.
You do not sum the molecular weight of Cetyl and Stearyl alcohol, since they are independent molecules present in the mixture (they are not a dimer, in which case the addition would be valid).
The instructor in your video (Dr. Ricardo Diez) uses a value of 250 g/mol for Cetearyl alcohol since he’s considering a 50/50 mixture (it should be around 256 g/mol, but he doesn’t need to be precise to show how to make those calculations, nor that the difference would change his results much).
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Got it
Thanks a lot
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It becomes a big problem when making LGN and calculating moles of fatty alcohol.
There is a bit difference. Almond two time.
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