# Using gas laws to find density

An unknown gas at $\mathrm{63.1\ ^\circ C}$ and $\mathrm{1.05\ atm}$ has a molar mass of $\mathrm{16.04\ g/mol}$.
What is the density of the gas?

I know I need to use the $pV=nRT$ equation but it gives me g/mol and not moles. I'm not really sure what to do here.

$$1.05\ V = n\ 0.08206 \cdot (63.1+273.15)$$

Ok that's as far as I got. I have no idea how would I even go about converting g/mol back to moles so I can solve for V, and after that I would also need to figure out a way to find grams so I can divide g/v to get the density.

• Put 1 L for V so you can see how many moles there’d be in a liter. Convert between mass and moles by the equation mass = moles * molar mass. Divide your mass by the volume (1 L) and you’ll get the density in grams per liter. – lightweaver Oct 3 '15 at 1:39
• @lightweaver Thank you very much, so if a variable in any equation isn't specified is it automatically assumed to be 1? – user3882522 Oct 3 '15 at 1:52
• In this case you can set one of the variables to 1 as density is known as an intensive variable, meaning that its value is independent of how much material there is. So it could be any number, but the value will remain the same. We just set it to 1 as it simplifies the calculations. – Nanoputian Oct 3 '15 at 1:56
• In my answer I have done the same thing, but I made the number of moles 1 and then see how many litres there are in 1 mole. Either way, you will get the same answer – Nanoputian Oct 3 '15 at 1:58
• The molar density is n/V, so the mass density is Mn/V, where M is the molar mass, n is the number of moles, and V is the volume. So Mn/V= (pM)/RT. – Chet Miller Oct 29 '15 at 2:55

Okay, to find the density of the gas, to need to know its mass and its volume. So to do this, lets say we have one mole of this substance. So its mass will be $16.04~\mathrm g$. Now all we need to do is find its volume.
This can be done by rearranging the ideal gas equation: $$V = \frac{nRT}{p}$$ Now all we have to do is plug in the values. However you have be careful that you use the write units. This is a common mistake, especially when using the ideal gas law however it is good to see that you have used the correct units: \begin{align}V &= \mathrm{\frac{1~mol\times0.08206~L~ atm~ mol^{-1}~K^{-1}\times336.25~K}{1.05~atm} }\\ &=\mathrm{26.28~L}\end{align}
Now, density can be calculated: $$\rho =\mathrm{\frac{16.04~g}{26.28~L} = 0.61~g~L^{-1}}$$