The Ideal Gas Law: Molar Mass

up to this point. We've used the ideal gas law to help us isolate pressure volume, moles or temperature. But now we're gonna be faced with a new but old topic. The idea of Moeller Mass. Now recall Moeller Mass represents the mass of a substance divided by the amount of that substance. If we take a look here, we're going to say that this capital M represents the molar mass of the gas and lower case M here equals the mass of the gas Ingram's and And we know from the ideal gas law equals the amount off the gas in moles. So more mass equals grams promote. Now that we remember what more Masses, we're gonna take a look at how we can use the ideal gas law to find the more mass of a gas. So for now, click on the next video and let's take a look. An example question

here, we're told. Calculate the molar mass of a gas if 2.50 g occupies 0.995 leaders at 715 Tour and degrees Celsius. All right, we need to calculate Mueller Mass. Okay, Now more mass here equals grams over moles. So let's just say that grams over moles within the question they give us grams right off the bat. We have 250 g, but we don't have the moles. However, in the question, we see that we have volume. We have pressure and we have temperature. The P and T, we know are part of the ideal gas law. So we're gonna say PV equals NRT. With this information, we can isolate the moles of our gas so that by both sides here by rt and we're gonna say here that moles equals pressure. Times volume over rt. We need to convert our pressure into atmospheres. So remember, we have 715 tours and then we're going to stay here for everyone atmosphere that we have that's equal to 760 tours. Tours here would cancel out and we get as our pressure 0.9408 atmospheres, take those atmospheres and plug them in. So we have 9408 atmospheres. Our volume is already in leader, so we're good there. Remember, our is our gas constant, which is 0.8 to 06 leaders times atmospheres over moles, times K and temperature temperature has to be in Kelvin, so add to 73 15 to this. So when we do that, that's gonna give us 313. 15. Kelvin plug that in What units? Cancel out atmospheres. Cancel out leaders, Cancel out Kelvin's cancel out. And that's how we see. We'll have moles at the end. So here, when we do that, we get our moles as 0.364 moles take those moles and plug them into the Mueller mass formula. And doing that gives us ah value of 68.681 g per mole. Now here, 2.50 has three sig figs, three sig figs. Three sig figs, 40 has only one sigfig. So if we wanted to go by that one sigfig that will come out to 70 g per mole. It's not as accurate, but it's not a big deviation from our initial answer. So if we're going by significant figures, we can say that 70 g promote would be the more mass of our unknown gas.

Besides finding the moles of the gas sample, the ideal gas law could be extended further defined the molar mass of the gas. Here, we're gonna look at the easy way on how we were able to derive the ideal gas law to this new version. Besides this easy method, we have a harder method. You can click on the next video to see how I derived this formula. But again on Lee. Look at that video. If your professor cares on showing how you derive this new version of the ideal gas law, If you're professor doesn't care, then just ignore that video and just go on to the next question where we have to solve for problem. Alright, so here to help us remember this one, we're gonna same or mass. Really? Tests are valuable patients. So here Moeller and then mass really tests rt They are over valuable patients, which is volume and pressure here. I just kept them in the order of PV. So this is the formula we can utilize to find the more massive A gas immediately, instead of doing it piece by piece, like we did in previous example questions. So just remember more mass, really. Tests are valuable patients. Capital M equals lower case m r t over PV. So now that we've seen this easy method, you can click on to the next video to look at how I derived this formula. Or you can skip that video and just head straight to questions or utilize this particular newly derived ideal gas law formula.

So for those of you who clicked on this video, let's see how we derived this new version of the ideal gas law. So we start out with the mower mass formula. Remember more mass equals mass m divided by moles. End here we're gonna algebraic rearrange this formula toe isolate, end So first you're gonna multiply both sides by end So now it becomes m Times n equals lower case m toe Isolate our and moles We divide out Moeller Mass So we're gonna say here moles equals mass divided by Mueller Mass going to the ideal gas law formula We can now substitute in for n this value here. So PV equals NRT now becomes PV equals mass over Moeller mass. We need toe isolate our Moeller mass. So what I'm gonna do here is I'm gonna multiply both sides by Mueller Mass Here is the denominator, so I can multiply to get rid of it. And that equals maths times rt finally toe isolate our Mueller mass. We divide both sides by PV and that's how we got that more mass equals MRT over PV again. This is the way we derived it. If your professor wants to show wants you to show the work. But again, it's always easier to remember that Moeller Mass really tests are valuable patients, remembering that will help you remember this version of the ideal gas law.

here, it states in this example question An unknown gas with a mass of 0.1727 g is placed into millimeter flask. If it's pressure is 0.833 atmospheres at 20 degrees Celsius. What is the identity of the gas here? Were given different gasses. And we need to realize here that they're giving us our grams, which is mass. They're giving us volume. They're giving us pressure and they're giving us temperature with this Will be able to find the Moller Mass off are unknown. Gas. Then all we have to do is see which one has that Mueller Mass, um, as their actual weight. So we're gonna say here Mueller Mass equals rt over PV. We're gonna plug in the mass of our unknown. We're gonna plug in our, which is 0.8 to 06 leaders times atmospheres over moles. Times K. Remember, temperature has to be in Kelvin. So I had to 73.15 to this number to give us to 93.15 k divided by our pressure, which is 0.833 atmospheres and then our volume we convert those milliliters into leaders, so that gives us 0.1 to 5 leaders. Here we look and see what units cancel out and we see all that's left is grams over moles. So we have Moeller mass. When you plug that in, you get approximately 39 90 g promote. Yeah, All we have to do now is see which one of these gasses has a more mass closest to our answer. So here and to the two nitrogen is comes out to 28.2 g per mole are gone from the periodic table is approximately 39.95 g per mole. Oxygen is 32 g per mole grams per mole grams per mole grams per mole. Neon one neon is approximately 20 or so grams promote according to the periodic table, and then ch four is roughly around 16 g per mole. Our closest answer would have to be are gone. It's the one that's closest to this 39.90 g per mole. So here we've just utilized the molar mass version of the ideal gas law to find the molar mass of our unknown gas