What is the mass of 7.00 liters of Neon if the volume is measured at 0 degrees C and at a pressure 101.3 kPa?

Expert Answers
txmedteach eNotes educator| Certified Educator

This combines two main portions of chemistry: the Ideal Gas Law and Stoichiometry. The basic steps are as follows:

1) Find how many moles of gas you have.

2) Use the number of moles and the periodic table to calculate the mass.

Step 1:

We're going to use the Ideal Gas Law:


P = Pressure (kPa)

V = Volume (L)

n = moles of gas (mol)

R = Ideal Gas Constant (8.314 L*kPa/K/mol)

T = Temperature (KELVIN ONLY)

All we need to do to continue is to start filling in the variables, but before we do...

ALWAYS CONVERT TEMPERATURE TO KELVIN FIRST! I swear, the most-often missed parts of these questions involve someone forgetting to convert to Kelvin, and it's easy when you get temperature in Celcius:

T (in Kelvin) = T (in Celcius) + 273.15

For example, in this problem, T is given as 0 degrees Celcius, so to get the right temperature in Kelvin we just add 273.15

0 degrees Celcius + 273.15 = 273.15 degrees Kelvin

Now that we have the right temperature, we can start filling in the Ideal Gas Law equation

P = 101.3 kPa (given in problem)

V = 7.00 L (given in problem)

n = ? (Since we have no idea what this is yet, it looks like the value we'll be solving for)

R = 8.314 kPa*L/mol/K (Make sure the units here line up with the units for the other variables...there are a few "R"s)

T = 273.15 K (What we just calculated)

Here's what we have now for the Ideal Gas Law:

101.3*7.00 = n*(8.314)*(273.15)

We will now solve for the number of moles of gas to finish step 1 of the problem:

101.3*7.00 = n*(8.314)*(273.15)    Combine terms here

709.1 = 2271*n     Now, solve for "n" by dividing by 905.2

n = 0.3122 mol  Yay! Finished with Step 1

Step 2:

Now that we have the number of moles, we simply look at the periodic table to calculate our mass. Because we have Neon according to the problem, we just go look at Neon and find the standard atomic mass: 20.18 g/mol

Now, we finish off by multiplying our number of moles with the mass number/atomic mass/whatever you want to call it:

0.3122 mol * 20.18 g/mol = 6.301 g

I know that sounded long-winded, but really we're just using two relations that will bombard you all the time in chemistry: the ideal gas law and converting from moles to grams.

txmedteach eNotes educator| Certified Educator

Here's another way to do it. There's a concept called Standard Temperature and Pressure (STP). It used to be defined as follows:

T = 0 degrees Celcius or 273.15 Kelvin

P = 1 atm or 101.3 kPa

At these conditions, the volume of 1 mole of an ideal gas is 22.4 L.

We have 7.00 L, but the same temperature and pressure. Therefore, we will have a proportional difference in the number of moles, which you can see in the following equation:

n/1 mol = V/22.4 L

This makes it easy for us when we simplify:

n = 7.00/22.4 = 0.313

Notice the difference between this number and the one we got from the ideal gas law. It simply has to do with the number of significant figures and shouldn't affect your answer. Don't worry about it!

Now that you have the number of moles, just go and solve like we did above by using the periodic table to convert moles of neon to grams of neon, and you're done!

Side note: Apparently IUPAC, the body that decides on things like what "STP" means, decided that the pressure term of STP is 100 kPa instead of 101.3 kPa (which is interesting because it looked like they made this call in 1997, way before I was taught that standard pressure is 101.3 kPa).

This does not affect our answer because now the volume that one mole of an ideal gas is 22.1 L instead of 22.4 L. That's a little more than 1%, and if new textbooks have this number, I'm sorry for telling you that 22.4 L is the volume at STP when it is apparently supposed to be 22.1 L. However, in the engineering world, for the most part, a difference of 1% doesn't really matter. Chances are, you'll get pretty much the right answer if you use 7.00 L/22.1 L instead of 7.00 L / 22.4 L.

Just use whichever number is in your textbook, I suppose!