C5H8 combustion

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Viertelstein
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C5H8 combustion

Post by Viertelstein »

Hi , I can easily derive on formula basis what happens if polyisoprene is burned: C5H8+7O2→5CO2+4H2O.
I also can use this tool on webqc()org to get the atomic weight.
But how to get a conclusion for "non-chemists" like one litre Diesel emit 2640 grammes of CO2/l.

If I have 1 kg of rubber (1,4 cis-polyisoprene) and burn it (in an abundant supply of oxygen), how much CO2 it will emit? Answers in the web are from 3,8 to more then 20 kg.
Thanks,
Viertelstein
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ChenBeier
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Re: C5H8 combustion

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For a combustion you have to know the molar mass of the compound.
1 Isopren has 68 g/ mol. But a Polyisopren can have x molecules of it.
If you have 68 g/ mol then you can obtain 5 * 44g => 220 g CO2 and 4 * 18 = 72 g H2O .1 kg contain 14,7 mol Isopren this will create 3,234 g CO2 and 1,058 g H2O
A Polymer has a different molar mass depending on the grade of polymerisation. At least its x number of 68 g/mol. If you have a molar mass of 10000 g /mol the molecule contain 147 Single single isopren molecules. These multiplied with 220 g/mol will give 32,3 kg CO2 and 10,6 kg H2O

For Diesel you need the mixture composition of the fuel. If you know each compound and the amount the CO2 and H2O can be calculated.
Viertelstein
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Re: C5H8 combustion

Post by Viertelstein »

Hi Chen, Thanks a lot for this introduction.
I found this on a website of a natural rubber supplier:

"When considering the macrostructure of natural rubber, one must note the very high molecular weight of natural rubber. With a molecular weight range of from 3 x 104 to 1 x 107, it is several times more, at the high end, than most synthetic rubbers.5 This high molecular weight results in less chain ends and more entanglement than an equal weight of synthetic rubber."

How to find out the "molar mass" for natural rubber? Searching in the web, I found this which make its not easier I guess:

"The molecular weight of natural rubber varies greatly, from only a few hundred thousand to more than one million. Liquid natural rubber (LNR) products generally use natural rubber in the form of latex with a relatively low molecular weight, which ranges from a molecular weight of 600,000 g / mol. This study aims to determine the molecular weight of LNR products made from the chemical depolymerization process of natural rubber latex with a molecular weight of more than one million. The natural rubber used latex with 20% Dry Rubber Content (KKK). The natural rubber depolymerization process used NaNO2 and H2O2 degrading agents with a CoCl2 catalyst at 80 °C. H2O2 levels were kept constant at 1 parts per hundred rubber (phr), and NaNO2 levels vary at 0; 2; 3; and 4 phr. After the reactor temperature reaches 80°C, the CoCl2 catalyst was added, which level varied at 0; 1; 3; and 5 phr. The reaction time was varied, which 7, 8, and 9 hours. The results of LNR product was purified to form a semi-gel, and molecular weight was analyzed using gel permeable chromatography (GPC). The results showed that the lowest molecular weight of the resulting LNR product was 108,335 g / mol from the initial molecular weight of 1,384,638 g / mol."

Kind regards,
Viertelstein
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ChenBeier
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Re: C5H8 combustion

Post by ChenBeier »

Nature Rubber has a molar mass between 500.000 to 2.000.000 g/ mol.

The molar mass can be measured by combustion analysis. A known mass will be burnt and the obtained mass of CO2 and water will be measured.
From CO2 => C and H2O => 2H can be found, what give the empiric formula.
From the known mass the molar mass can be calculated. There several other methodes to measured it.
Viertelstein
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Re: C5H8 combustion

Post by Viertelstein »

Thanks again !!
Viertelstein
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Re: C5H8 combustion

Post by Viertelstein »

I'm sorry, I have to come back to this calculation again.
I read somewhere that the factor between C5H8 and CO2 after burning is 3,66. I think I know why:

Before we found that "(C5H8 with 68.1170 g/ mol ) leads to 5 * 44g => 220 g CO2 and 4 * 18 = 72 g H2O .1 kg contain 14,7 mol Isoprene this will create 3,234 g CO2 and 1,058 g H2O"

If we calculate C5H8 together, i.e. with a weight of 68 then it leads to 220/68=3,2.
Is it correct to take the 8 H toms into that calculation? Because we only want to know how much CO2 is inside the rubber! Only taking into account 5 Carbon atoms with atomic weight of 12 (x5) we have a weight of 60. Dividing the weight of 220 g CO2 by 60 it becomes 3,66.

Thus, which statement is correct? Combusting 1 part rubber creates 3,2 parts CO2 or 3,66?
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Re: C5H8 combustion

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In one case it is calculated as the Rubber. 220g/68 = 3,23 is the right answer for this.
But if only 5 carbon is considered then 220g/60 = 3,66. But this is not Rubber only 5 C out of the molecule. Make no sense in my opinion.
Maybe you tell us the paper where you have read it.
Viertelstein
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Re: C5H8 combustion

Post by Viertelstein »

I can't find it quickly any more. However.

We cannot (from a chemical point of view) reduce the rubber to its carbon content, then it would be pure carbon and then we can/have to treat him as such. So the calculation using stoichiometry is certainly correct, even if the following statement is perhaps clearer for laypeople:

One gram (or whatever) of pure rubber (assuming pure C5H8) contains as much carbon as there is in 3.6 grams of carbon dioxide. hhhmmmm.. :?:

....next step would be to say (if you don't care about units) one part of rubber "contains" 5 parts of CO2 because the 5 C atoms becomes 5 CO2 molecules.... okay....now we are in Vegas... :lol:
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