Okay, I’m reading a website for an electronics manufacturer that makes a device intended to help you save gas.
Cool.
I notice, however, that they make a statement I feel contradicts some basic physics rules.
Here’s the statement:
*Carbon Footprint:
1 gallon = 19.4 pounds CO2
Every gallon of gas you use produces 19.4 pounds of CO2. An average driver in a 4 door sedan produces about 8 tons (16,000 pounds) of CO2 per year. Putting things into perspective, an adult african elephant weighs 6 tons. * [1]
How does one gallon of gasoline, having a weight of 5.8 to 6.5 lbs, turn into 19.4 pounds of anything?
combustion of a fuel typically involves the addition of oxident from another source. In an automobile’s case, the oxident comes from air. I’ll leave the bookkeeping to someone else.
What YamotoTwinkie said. Each mole of carbon, weighing 12.017 grams, combines with a mole of oxygen, weighing 31.9998 grams, to make one mole of CO2 weighing 44.0158 grams.
This is one reason why I’d rather not use moles in the first place. If you’re counting atoms, then say “atoms”, and if you’re counting molecules, say “molecules”. What purpose is served by throwing the amu-to-grams conversion factor around willy-nilly?
I’ll make a ruling. If atoms were meant than atoms would have been said. It would be unreasonable to talk about a mole of monotomic oxygen under standard conditions. Given that Bill Door explicitly stated 32 Da (grams per mole otherwise known as Daltons), it seems the correction was unecessary. In general, I would assume that a mole of oxygen meant a mole of oxygen gas unless otherwise specified.
This chemistry court hereby rules in favor of the plaintiff Bill Door.
The OP did ask about the mass of CO2 being produced, and there is no way to get there from atoms without going through moles.
>This is one reason why I’d rather not use moles in the first place.
Yay, Chronos. This has always been a dumb irritant. Besides, it helps keep this “molecular weight” thing alive in its non-SI form, which people usually don’t even notice.
Those of us who learned chemistry back when it was easier and had just those four elements on the periodic table find it difficult to let go of the old. I don’t even know how to get from a weight of gasoline to a weight of carbon dioxide without passing through moles. Plus, if I don’t use moles, what am I to do with Avogadro’s number banging around up in my head?
You have no choice but to go through moles. (I will grant you, however, that most people, myself included, are not perfect in differentiating between moles of atoms and moles of molecules. Usually, though, moles of molecules is implied.) And the conversion is not amu to grams but moles to grams.
For the complete combustion of one pound of octane (which is not 100% of gas, but is the easiest component to deal with), the stoichiometric calculations look like this:
2 C8H18 + 25 O2 --> 16 CO2 + 18 H2O
g 454.5 1597 1404 646.4
g / mol 114 32.0 44.0 18.0
mol 3.99 49.9 31.9 35.9
mol ratio 2 25 16 18
for the equation: 2 C[sub]8[/sub]H[sub]18[/sub] + 25 O[sub]2[/sub] → 16 CO[sub]2[/sub] + 18 H[sub]2[/sub]O, and where the top row of numbers is mass, the middle is molar mass (or formula weight), the thrid row is moles, and the bottom is molar ratio.
Thus 454.5g of octane (one pound) uses 3.51lbs of O[sub]2[/sub], and produces 3.09lbs of CO[sub]2[/sub] and 1.42 lbs of H[sub]2[/sub]O.
Assuming an average density of gasoline at 6.60lbs/gal, 20.4 lbs of CO[sub]2[/sub] are produced for every gallon of pure octane completely combusted. (Except that gasoline is not pure octane, which has an average density of about 6.54lbs/gal.)
Sure there is. On the molecular level, the reaction is two molecules of C[sub]8[/sub]H[sub]18[/sub] (approximating gasoline as pure octane, here) reacting with 25 molecules of O[sub]2[/sub] to form 16 molecules of CO[sub]2[/sub] and 18 molecules of H[sub]2[/sub]O. No moles needed for that. The two molecules of gasoline each have a molecular mass of 114 amu/molecule, and the 16 molecules of carbon dioxide each have a molecular mass of 44 amu/molecule. Note that I am using molecular masses here (measured in amu per molecule), not molar masses (measured in grams per mole): I nowhere need to know Avogadro’s Number; I’m just counting up protons and neutrons. So in each molecular reaction, I start off with 228 amu of gasoline, and end up with 704 amu of carbon dioxide. Taking the ratio of these, I find that the mass of carbon dioxide is 3.09 times as great as the mass of the gasoline. This is a dimensionless ratio, so it works just as well on the macroscopic scale as on the molecular, so I can then say that for every kilogram of gasoline that I start with, I end up with 3.09 kilograms of carbon dioxide.
The conversion factor I was referring to was Avogadro’s number, 1 gram = 6.02e23 amu. Sure, you can multiply and then divide by that number if you like, but you don’t need to. In fact, you don’t need to use any particular unit system: I could have gone directly to using slugs or tons or solar masses in my final answer, without ever needing to use grams as an intermediate.