# What is the area that 1 litre of gasoline would spread above the sea?

I am studying for a general chemistry test with Atkins and Jones "Principles of Chemistry" 5th edition.

In the Chapter 18, page 743, section 18.3, there is this funny question:

How would you estimate the area that 1 litre of gasoline would spread above the water surface?

Observation: I translated it from the Brazilian Portuguese edition. The English version may have different units, but the idea should be the same.

Let's say that the water body is the sea. But it could be a big lagoon, too. It doesn't really matter.

It got me thinking and now I'm stuck. How can we do that?

• Estimate the thickness of the layer, take 1 litre and divide by the thickness.. otherwise you probably need more data – orthocresol Apr 6 '17 at 19:25
• If we had more data, what is the best way to do this? I thought about using surface tension properties and equations, but I am not sure if it is the easiest way... – Guilherme Cruz Apr 6 '17 at 19:52
• You can calculate the surface area per molecule, using surface tension measurements – orthocresol Apr 6 '17 at 20:10
• For an estimation I might start with this table regarding the dispersal of oil spills on water. I would assume that gasoline would spread at the very thin end of that table, and guess something on the order of 50 nm thickness and say that's the amount you can see or some such thing. There is no reason it couldn't spread much thinner, too thin to see even. – airhuff Apr 6 '17 at 20:18
• This site offers a hand rail: mathcentral.uregina.ca/QQ/database/QQ.09.08/h/phillip4.html Be careful, as it mixes SI and non-SI units; but the principle is legible. – Buttonwood Apr 6 '17 at 20:20

$$1 \text{ L} = 1 \text{ dm}^3 = 10^{24} \text{ nm}^3$$
Deviding by 168 nm provides with googles built-in calculator an area of about $$5.95 \times 10^{21} \text{ nm}^2 = 5952 \text{ m}^2$$
To set this into a context, UEFA and FIFA agreed on the size of European Football / Association (soccer) field ($105 \times 68 \text{ m}$) which equals $7140 \text{ m}^2$.
Addendum: According to this, an American Football field measures "$120.0 \text{ yd } (109.73 \text{ m}) \times 53.3 \text{ yd } (48.74 \text{ m})$ or $6400 \text{ yd}^2 (5351.2 \text{ m}^2$)".