CO2 and CO are soluble in hydrocarbons, so they partition between the light vapor (like ethane and methane) and the liquids near the top of the column. They phase behavior of the pure components does not provide a good indication of where this happens. For example, CO2 dissolves in liquid hydrocarbons under conditions where there would be no liquid CO2 if it were a pure component.
CO2 is heavier than methane but is azeotropic with ethane (and ethylene for that matter). It is lighter than propane. What spec are you imposing on the LPG? If your spec is 5% ethane, you should expect some CO2 there are well. If your spec is 1% Propane, I'd actually be surprised to see CO2 there.
What makes CO2 hard to predict (at least at a first glance) is the fact that it does not have a normal boiling point. At atmospheric pressure, it can only exist as a vapor or a solid.
CO is a much much lighter component. I am surprised to read that you have it in LPG. Are you talking about PPM levels or several percentage points?
Larry, Farid and Hazem, thank you a lot for your answers. It was very helpful to me.
I want to share some extra information about the process we spoke about. After obtaining the LPG, we separate C3 from C4 fraction. C3 contains propane and propylene (FCC unit of oil refinery) and problem with CO2/CO was occurred in propylene. It is quite sure that some quantity of C2 fraction transfers to LPG, but we don't do such analysis to know how much. Forming the azeotropic between C2 and CO2 is reasonable explanation for CO2 appearance in LPG. The spec amount of CO2/CO (shown together) is 5 ppm in propylene from LPG, but we have 7.5 ppm.
OK, I think we now understand better what is going on. Basically, you are depropanizing. The CO2 and CO have no place to go but up the column. If this quantity (7.5 ppm CO & CO2 in the distillate) is excessive, then the problem is not the depropanizer itself, but too much CO2 & CO in the feed.