3 replies to this topic

[EngRichard]

I'm focussing on the bottoms product of a continuous distillation column. In the column i'm trying to separate water and ethanol. I don't know W (the bottoms product flow rate), but I know that I need a bottoms product composition of water(99.5% by mass) and ethanol (0.5% by mass). Is it possible to covert these mass fractions to mol fractions, so I would know the composition of the bottoms products in mol fractions rather than mass fractions? If this can be done, can someone explain how?

Thank you!

[Zauberberg]

Wi = Mi*Xi / (M1*X1 + M2*X2 + ... + MiXi)

 

where:

Wi - mass fraction of component "i"

Xi - mole fraction of component "i"

Mi - molecular weight of component "i"

 

and the other way round:

 

Xi = (Wi/Mi) / (W1/M1 + W2/M2 + ... + Wi/Mi)

 

The details you can find in most of (if not all) Chemistry and Thermodynamics textbooks.

[breizh]

Hi ,

To support you work and understand :

Consider 100g of product >>>> 99.5 g EToh +0.5 g H2O >>>>

Mw EtOH =46 g/mol & MwH2O =18 g/mol

 

Nbre of Moll EtOH = 99.5/46

Nbre mol of H2O =0.5/18

 

%mol EtOH = (99.5/46)/(99.5/46+0.5/18) *100

%mol H2O =100 -%mol EtOH or (0.5/18)/(99.5/46+0.5/18)*100

 

Hope this helps

 

Breizh

Edited by breizh, 16 March 2016 - 04:43 PM.

[shantanuk100]

Hello Engrichard,

You have 99.5 % water by mass and 0.5 % water by mass.
Which means you have the ratio of their weight fractions.

 

Mass fraction of a component is the component weight's ratio with the total weight.

Similarly mole fraction is the ratio of the component moles with the total moles.

 

So, you have 99.5 wt% water and 0.5 wt% ethanol.

This means the weight ratio of water is 0.995 / (0.995+0.005) = 0.995

and weight ratio of ethanol is 0.005 / (0.995+0.005) = 0.005

 

So for mole ratio we need to consider the moles of substance, which is the weights divided by their respective molecular weights.

 

So your mole ratio for water is (0.995/18) / (0.995/18 + 0.005/46) = 0.05527 / (0.05527 + 0.000108) = 0.998 = 99.8 mole %

 

and mole ratio for ethanol is (0.005/46) / (0.995/18 + 0.005/46)  =  0.000108 / (0.05527+ 0.000108)  =  0.002 = 0.2 mole %

 

 

Regards,

Shantanu

Edited by shantanuk100, 17 March 2016 - 02:01 AM.