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

##### 1 Answer

You can either use the density of

**USING THE DENSITY**

This is the more accurate way, since

#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

#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