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

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?

## marked as duplicate by jerepierre, M.A.R. ಠ_ಠ, ron, Klaus-Dieter Warzecha, Martin - マーチン♦Mar 27 '15 at 1:19

• Simply put: you can't. – Mithoron Mar 25 '15 at 0:00

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). 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 $7004\ \ce{mol/m^3}$

• Does it matter what is assumed for the mass of the wood? For example, to get the formula for beech wood I assumed 100 grams. However, that formula would be different if I assumed a different mass such as 25 grams. In the end, the mole concentration would be different too. – wigging Mar 24 '15 at 20:44
• No, the molar mass of a compound does not depend on how much of a sample you take. The formula for beech wood just gives the ratios of how many atoms are in a molecule, not their number in a sample. – ahemmetter Mar 24 '15 at 20:48
• @andynitrox But the formula I gave for beech wood $C_{4.084}\,H_{5.77}\,O_{2.81}\,N_{0.0086}$ was assumed to be from 100 g. If you assume it's from 25 g then the formula becomes $C_{1.02}\,H_{1.46}\,O_{0.7}\,N_{0.0021}$ which will give a molecular weight of $24.94\;g/mol$ which gives a mole concentration of $28,067\;mol/m^3$. – wigging Mar 24 '15 at 21:09
• But does the particle density of your sample change only because you select a different quantity? – ahemmetter Mar 24 '15 at 21:16
• Like @andynitrox said, the molecular formula is independent of quantity. Just because you divided the amount by 4, the formula shouldn't change. This is very tricky when it comes down to representative formulas without integer proportions, since we can't simply draw molecules. Honestly, what I'd recommend is to avoid usin moles for these kind of complex molecules. I'll update my answer with that. – Molx Mar 24 '15 at 21:32

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.

• As @Molx suggested in his comments, using moles may not be the best approach. Do you have any recommendations on other approaches for representing different types of biomass? – wigging Mar 25 '15 at 13:50
• I am not from this field, but I would suggest the density, humidity and fractional composition like in the question, that would probably specify the material sufficiently. – ahemmetter Mar 25 '15 at 18:33
• Please refer to my other question instead of this one, the new question will hopefully clarify what I'm trying to do, chemistry.stackexchange.com/questions/27895/… – wigging Mar 26 '15 at 16:17
• Also, delete your answer so I can delete the question. Refer to the new question if you would like to comment. Sorry for the confusion. – wigging Mar 26 '15 at 17:08
• You could ask a moderator to move the answers, otherwise Molx's is lost too – ahemmetter Mar 26 '15 at 17:38