# 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 ${\text{CO}}_{2}$ at STP or assume that ${\text{CO}}_{2}$ is an ideal gas.

USING THE DENSITY

This is the more accurate way, since ${\text{CO}}_{2}$ is not quite ideal. Wikipedia gives the density as $\text{1.977 g/L}$ 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 ${\text{CO}}_{2}$?)

Therefore, the number of molecules is:

$0.0898 \cancel{\text{mols CO"_2) xx (6.0221413 xx 10^(23) "anything ever")/cancel("1 mol anything ever}}$

$= \textcolor{b l u e}{5.41 \times {10}^{22}}$ $\textcolor{b l u e}{{\text{molecules CO}}_{2}}$

USING THE IDEAL GAS LAW

Of course, we could also use the ideal gas law by assuming ${\text{CO}}_{2}$ is ideal. In that case, we would assume that it has a molar volume of $\text{22.414 L/mol}$ at ${0}^{\circ} \text{C}$ and $\text{1 atm}$. Therefore:

$2 \cancel{\text{L CO"_2) xx cancel"1 mol ideal gas at STP"/(22.414 cancel"L}}$

$= {\text{0.0892 mols CO}}_{2}$

Therefore, the number of molecules is:

$0.0892 \cancel{\text{mols CO"_2) xx (6.0221413 xx 10^(23) "anything ever")/cancel("1 mol anything ever}}$

$= \textcolor{b l u e}{5.37 \times {10}^{22}}$ $\textcolor{b l u e}{{\text{molecules CO}}_{2}}$

There is a small error (about 0.7%), but not a bad one. It's a rather good assumption that ${\text{CO}}_{2}$ is ideal at STP. At higher pressures and lower temperatures though, the approximation fails (why?).