How do you find the number of molecules in "2 L CO"_2?

1 Answer
May 8, 2017

You can either use the density of "CO"_2 at STP or assume that "CO"_2 is an ideal gas.


USING THE DENSITY

This is the more accurate way, since "CO"_2 is not quite ideal. Wikipedia gives the density as "1.977 g/L" at STP. We can use the molar mass and density to get the number of mols and hence the number of molecules.

2 cancel("L CO"_2) xx ("1.977 g CO"_2)/cancel("L CO"_2) = "3.954 g CO"_2

Therefore, the mols are:

3.954 cancel("g CO"_2) xx "1 mol CO"_2/(44.009 cancel("g CO"_2)) = "0.0898 mols CO"_2

(How can you get the molar mass of "CO"_2?)

Therefore, the number of molecules is:

0.0898 cancel("mols CO"_2) xx (6.0221413 xx 10^(23) "anything ever")/cancel("1 mol anything ever")

= color(blue)(5.41 xx 10^(22)) color(blue)("molecules CO"_2)

USING THE IDEAL GAS LAW

Of course, we could also use the ideal gas law by assuming "CO"_2 is ideal. In that case, we would assume that it has a molar volume of "22.414 L/mol" at 0^@ "C" and "1 atm". Therefore:

2 cancel("L CO"_2) xx cancel"1 mol ideal gas at STP"/(22.414 cancel"L")

= "0.0892 mols CO"_2

Therefore, the number of molecules is:

0.0892 cancel("mols CO"_2) xx (6.0221413 xx 10^(23) "anything ever")/cancel("1 mol anything ever")

= color(blue)(5.37 xx 10^(22)) color(blue)("molecules CO"_2)

There is a small error (about 0.7%), but not a bad one. It's a rather good assumption that "CO"_2 is ideal at STP. At higher pressures and lower temperatures though, the approximation fails (why?).