Ethanol - Correction factor ? 2

[AlexEss]

Hi all

Hopefully a simple question for you -
I am measuring the ethanol content of a gas stream, using a hand-held analyzer.

The result is displayed in mg/Nm3 of Ethanol.. say, 400mg/Nm3 for example.

How do I determine the Carbon content ( mgC/Nm3 ) from my result ?
Is it a simple correction factor ?

I'm not a Chemist, so please excuse me if the question is dumb.

Thanks in advance for any replies,


Alexess

 

[pierreick]

Hi,
Considering ethanol C2H5OH, 1 mole ethanol, 2 moles de C
Molecular weight Ethanol: 46 g/mol
Molecular weight C: 12g/mol

Calculation : base 400 mg ethanol/Nm3 >>> 400/ 46 E3 = 0.008696 mole C2H5OH /Nm3 >>>> 0.017391 mole C /Nm3 >>>> 0.017391 *12 E3

= 208.69 mg C/Nm3

Pierre

 

[AlexEss]

Pierre Hi

Thank you for your super-fast response, and also for the clear and concise explanation.
That is really helpful !

Many thanks, Merci Bien !

Alex
[/indent]

 

[shvet]

Considering ethanol C2H5OH, 1 mole ethanol, 2 moles de C
Molecular weight Ethanol: 12*2+1*6+32*1=76 g/mol
Molecular weight C: 12g/mol
Mass fraction of C is 12*2/76=0.316

Calculation : base 400 mg ethanol/Nm3 >>>> 400*0.316 >>> 126.3 mg C/Nm3

 

[pierreick]

Wow Shvet!
Mw O is 16 g/mol. Even your sum is wrong.
Too much vodka?
Pierre

 

[shvet]

@pierreick
Seems yes )))

@Alex
Ignore my post

 

[Latexman]

shvet,

Cheers!

Good Luck,
Latexman

 

[georgeverghese]

Analysers are notorious for wrong readings - always check that there is a recent calibration record.
Where possible, get a sample of vapor and send it to your lab for manual analysis.

 

[AlexEss]

Thanks a lot for your input Guys, really interesting.

Just FYI, my analyzer is a RAE Systems MiniRae3000.
The Calibration Gas is Isobutylene.
One can call up an extensive list of measured gases, the analyzer then applies a correction factor specific to the gas of interest.

In my case, Ethanol, this results in a factor of 10.0 and the result is displayed in mg Ethanol/m3. One then normalizes the result to obtain mg/Nm3

Going back to the Mass Fraction calculation @Pierreick,
<Calculation : base 400 mg ethanol/Nm3 >>> 400/ 46 E3 = 0.008696 mole C2H5OH /Nm3 >>>> 0.017391 mole C /Nm3 >>>> 0.017391 *12 E3
= 208.69 mg C/Nm3>

Is it possible to use the percentage: 208.69mgC/400mgE = 52.0725% - and apply this to other readings? For example: 950mg/Nm3 x 52.0725% = 494.6mgC ???
or would this be considered sloppy practice ?

This is new territory for me.

Thanks for your help,

Alex

 

[pierreick]

Hi,
Yes, you are right the ratio 208.69/400 is a Constante using Ethanol as a reference.
Pierre

 

[AlexEss]

Pierre, thank you for confirming this - it is extremely helpful.

Regards

Alex