Question

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

Answer 1

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?).