Fossil fuels are stored hydrocarbons. They were removed from the atmosphere long before human history. When fossil fuels are burned, they release carbon dioxide.

Gasoline is mostly octane with some other hydrocarbons or added alcohols. “mostly” means about 90%, and those other hydrocarbons and alcohol burn almost the same way. When octane is burned, this happens:

Eq. 1: <math> \fbox{ C_{8}H_{18} + 17O_{2} \rightarrow 8CO_{2} + 9H_{2}O } </math>

This means that 1 molecule octane produces 8 molecules carbon dioxide and 9 molecules water.

How many molecules in a litre of octane?

• Density of Octane = 0.703 g/mL
• Molecular Mass of Octane = 114.22852 g/mol

<math> L = \frac{0.703kg}{0.1142kg/mol} </math>

Eq. 2: <math> \fbox{ = 6.154 mol/L } </math>

Note that “mol” is a molar, or <math>6.02*10^{23}</math> molecules.

So how many molars of <math>CO_2</math> are produced from one litre of octane?

Per eq. 1, <math> 1 mol \; C_8 H_{18} \rightarrow 8 mol \; CO_2 </math>

Per eq. 2, <math> 1L \; C_8 H_{18} = 8(6.154) </math>

Eq. 3 <math> \fbox{ 1L \; C_8 H_{18} = 49.232 mol \; CO_2 } </math>

49.232 molars. But this is a gas, so it takes up a lot more volume than a liquid.

How much volume does 49.232 mol of <math>CO_2</math> occupy at ground level?

<math> N_A = 24.7 L/mol \; \text{ at STP} </math>

Per eq. 3, <math> 1L \; C_8 H_{18} = N_A \times 49.232 mol \; CO_2 </math> <math> \fbox { = 1215.03 L \; CO_2 } </math>

So 1L liquid octane burns into 1216.03L pure gaseous <math>CO_2</math> at standard temperature and pressure.

The atmosphere is currently at 387ppm <math>CO_2</math>. 407ppm is considered unprecedented (+20ppm). How much atmosphere is required to dilute this <math>CO_2</math>?

<math> 1216.03 L \; CO_2 \times \frac{1,000,000L}{20L} </math>

Eq. 5 <math> \fbox { = 60,801,500L } </math>

This means that it takes almost 61 million litres of atmosphere to dilute the <math>CO_2</math> produced by 1L of octane.

1L is 10x10x10cm. 1 cubic meter is 1000L. A cube 10m x 10m x 10m is 1 million litres. This means that if you burned a litre of gasoline in your neighbourhood, it would bring up the CO2 levels of the neighbourhood lawns to high levels. Then it would float away and dissipate. Eventually the trees would reclaim it and everything is good right?

If I use 1 tank of gas every 2 weeks in my car, my car has a 35L tank, and I've been driving for 10 years, how much atmosphere is required to absorb the <math>CO_2</math> produced by my driving?

Per eq. 5, <math> 26 tanks \: \times 10 years \: \times \: 36L \: \times \: 60,801,500L </math>

Eq. 6 <math> \fbox { = 5.53 \times 10^{11} L } </math>

How can we visualize this number?

The atmosphere is 120km high, but realistically, the density drops off to zero out there, so we can half that to get a reasonable number for a “dense” atmosphere. Assuming an atmosphere of uniform density and an altitude of 60km, what percentage of a square km is this?

Volume of 1 square km of atmosphere:

<math> 60km \: \times \: 1km\times \: 1km </math> <math> = 60\times 10^4dm \times 1*10^4 \times 1*10^4 </math> <math> 6.0\times 10^{13} L </math>

Divide the polluted atmosphere (eq. 6) by the total atmosphere in 1 square km (above) and we have:

<math> \fbox { \frac{ 5.53 * 10^{11} L \: polluted }{6.0 \times 10^{13} L\: volume} = 0.922% } </math>

So my one car for 10 years has raised the CO2 levels to unprecedented amounts for almost 1% of a square km of atmosphere.