# Molar mass of combustion gases

1. Apr 2, 2010

### treddie

I have a question about how to calculate the mass/mole of a combustion gas.

In my own research, I came across two methods. The first method is suspect in my opinion. The second I trust, although the discussion left some issues unexplained, so I attempted to fill in the details as best I could.

The following is a non-stoichiometric reaction in a rocket engine combustion chamber:

1C12H23.4 + 12.265O2 --> 11.7H2O + 6.415CO2 + 5.585C

It has an oxidizer to fuel mixture ratio (O/F) = 2.34.

As a check:

O/F =

12.265 * 31.998 g/mol (O2)
---------------------------------------------------------- = 2.34
(1 * 12 * 12.011 g/mol (C)) + (1 * 23.4 * 1.0079 g/mol (H))

Now we can find the total molar mass of the combustion products (NOT the
reactants) that actually create thrust. However, this method seems to miss an
important issue covered by the second method discussed later:

Moles of combustion products:
11.7 mol (H2O) + 6.415 mol (CO2) + 5.585 mol (C) = 23.7 mol

Total mass of combustion products:
(11.7*2*1.0079 g/mol(H)) + (11.7*1*15.999 g/mol(O)) +
(6.415*1*12.011 g/mol (C)) + (6.415*2*15.999 g/mol (O)) + (5.585*1*12.011 g/mol (C))
= 560.17233 g

Total molar mass of the combustion products as a gas mixture:
560.17233 g
------------ = 23.63596 g/mol
23.7 mol

But this seems incorrect, because it does not take into account the PROPORTION of the given mole
amounts with respect to their different molar masses. According to my references on Chemistry and
molar fractions, should not the following, second method be used?

Moles of combustion products:
11.7 mol (H2O)
6.415 mol (CO2)
5.585 mol (C)

Molar masses of the individual combustion products:
MolarMass (H2O) = (2 * 1.0079) + (1 * 15.999) = 18.0148 g/mol

MolarMass (CO2) = (1 * 12.011) + (2 * 15.999) = 44.009 g/mol

MolarMass (C) = (1 * 12.011) = 12.011 g/mol

Mass fractions of the combustion products:
f(H2O) =

11.7 [(2 * 1.0079) + (1 * 15.999)]
------------------------------------------------------------------------------------
11.7[(2*1.0079) + (1*15.999)] + 6.415[( 1*12.011) + (2*15.999)] + 5.585[1*12.011)]

= .3763

f(CO2) =

6.415 [( 1 * 12.011) + (2 * 15.999)]
------------------------------------------------------------------------------------
11.7[(2*1.0079) + (1*15.999)] + 6.415[( 1*12.011) + (2*15.999)] + 5.585[1*12.011)]

= .504

f(C) =

5.585( 1 * 12.011)
------------------------------------------------------------------------------------
11.7[(2*1.0079) + (1*15.999)] + 6.415[( 1*12.011) + (2*15.999)] + 5.585[1*12.011)]

= .1197

Total molar mass of the combustion products as a gas mixture:
MolarMass(mixture) = $$\Sigma$$fjMolarMassj = 30.3972 g/mol

Is the second method the correct approach, or am I totally missing the point?

2. Apr 2, 2010

### Staff: Mentor

I have troubles understanding what the second method is for. When you mix gases you can assume (with a good approximation) that volumes are additive and that final volume depends only on the number of moles. Masses of individual gases nor their mass fractions don't matter, so first method seems correct to me.

But could be I am missing something. What is it that you are trying to calculate? Observed "molar mass" of the exhaust gas?

--

Last edited: Apr 3, 2010
3. Apr 3, 2010

### treddie

I see where I screwed up. My logic was as follows:

I am assuming that if we have a gas mixture of more than one species (for instance, H2O and CO2 at a combustion temperature where H2O is a gas), then we use the following approach described in Rocket Propulsion Elements (3rd Edition):

From another source, a volumetric percentage (nj) is given as
Correct:
nj = (nj) / (n1 + n2 + n3 ... + nm), where (j) is the subscript for a given species, and (m) is the total amount of species present.

Where I screwed up was in calculating (nj). I had incorrectly multiplied all (nj) by (Mj), when in fact, as you say, mass has nothing to do with it at that point:
Incorrect:
nj = (njMj) / (n1M1 + n2M2 + n3M3 ... + nmMm)

So that was how I was arriving at my value for the total molar mass in the second method of my original post. As a check, I did a little bit of basic algebra, and once I made the fix, the two methods were found to be equivalent.

So thanks for jogging my brain, there.