Determine molecular weight of biomass based on ultimate analysis [duplicate]








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[Representing woody biomass concentration for kinetic reactions](/questions/27895/representing-woody-biomass-concentration-for-kinetic-reactions)

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The ultimate analysis (by weight percent) for beech wood is:
$$\ce{C}=49.05\,\%,\; \ce{H}=5.83\,\%,\; \ce{O}=45\,\%,\; \ce{N}=0.12\,\%$$
Using the ultimate analysis values, the molar mass of each element, and assuming 100 grams of wood (therefore C = 49.05 g, H = 5.83 g, etc.), a representative formula for the beech wood is:
$$
\ce{C} = 49.05/12.01 = 4.0841\; \ce{mol} \\
\ce{H} = 5.83/1.01 = 5.77\; \ce{mol} \\
\ce{O} = 45/16 = 2.81\; \ce{mol} \\
\ce{N} = 0.12/14.01 = 0.0086\; \ce{mol} \\
\ce{C}\_{\,4.0841\,}\ce{H}\_{\,5.77\,}\ce{O}\_{\,2.81\,}\ce{N}\_{\,0.0086\,}
$$


What is the molecular weight of the beech wood based on the representative formula?




As a general rule, it is not a good idea to treat complex materials like wood as something you can calculate a mole of. You can of course analyze the material and get elemental rates, but that does not mean they are bonded in such way that this rate forms a minimum fraction which represents the whole thing.


When you calculate the number of moles of each atom in 100g, what you have is an estimation of the expected quantities in that sample. Undoubtedly, if you took different samples of the wood you would find different values, specially if sampling from different parts of the material.


As an exaggerated example, think about the elemental constitution of the human body, which is mainly 65% Oxygen, 18% Carbon, 10% Hydrogen and 3% Nitrogen in mass ([source](http://en.wikipedia.org/wiki/Composition_of_the_human_body)). With that in mind you can calculate a mole ratio of these elements and create a "representative formula". Obviously, though, there's no such thing as the *human molecule*, and you can't say that someone's body is X or Y moles of *human*.


Yet, like I originally answered, if you define what the representative formula is, you can calculate it's molar mass.


$$\ce {M}=4.0841\*12.0107 + 5.77\*1.00794+2.81\*15.9994+0.0086\*14.0067 = 99.947 \ \ce{g/mol}$$


Knowing ${\ce M}$, the number of moles is ${\ce 700,000/99.947 \approx 7004}$, and the concentration of biomass is $7004\ \ce{mol/m^3}$


 2

Using your mass percentages and the molar masses of the elements in your sample, you have correctly found out how many mol of each element are in 100g of your sample.


However, you need to keep in mind that wood is more than one simple molecule but instead rather a collection of quite a number of molecules. The masses of the elements that show up in all of those molecules add up to the percentages you found by this analysis. You see now that trying to formulate this distribution as a molecular formula has its limitations. Depending on how you select your sample, you will once include maybe $\ce{1 mol}$ of molecule A and $\ce{18 mol}$ of molecule B, whereas if you take twice the amount, you will count a total of 38 molecules as one.


As @Molx stated in his answer, if you always select the same mass for comparision, this might be a reasonably suitable unit.


To rephrase this: if you only had one molecule or a specific ratio of known molecules in your sample, it would be clear how the different elements distribute themselves to the different molecules. Since you do not know what molecules beech wood is made of, this molecular formula is not very useful for such a case.


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