The Ideal Gas Law: Molar Mass - Video Tutorials & Practice Problems
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The Ideal Gas Law can be further extended to find the molar mass of a gas.
The Ideal Gas Law (Molar Mass)
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concept
The Ideal Gas Law: Molar Mass
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Up to this point, we'd use 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 molar mass. Now recall, molar 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 lowercase m here equals the mass of the gas in grams, and n we know from the ideal gas law equals the amount of the gas in moles. So molar mass equals grams per mole. Now that we remember what molar mass is, we're gonna take a look at how we can use the ideal gas law to find the molar mass of a gas. So for now, click on the next video and let's take a look at an example question.
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example
The Ideal Gas Law: Molar Mass Example 1
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Here we're told, calculate the molar mass of a gas if 2.50 grams, occupies 0.995 liters at 715 Tor and 40 degrees Celsius. Alright. We need to calculate molar mass. K. Now molar 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 2.50 grams, but we don't have the moles. However, in the question we see that we have volume, we have pressure, and we have temperature. V, p, and t we know are part of the ideal gas law. So we're gonna say p v equals n r t, With this information we can isolate the moles of our gas. So divide both sides here by r t, and we're gonna say here that moles equals pressure times volume over r t. We need to convert our pressure into atmospheres, so remember we have 715 tours, and then we're gonna say here for every one atmosphere that we have that's equal to 760 torres. Tors here would cancel out and we get as our pressure point 9408 atmospheres. Take those atmospheres and plug them in. So we have 0.9408 atmospheres. Our volume is already in liters so we're good there. Remember r is our gas constant which is 0.08 206 liters times atmospheres over moles times k. And temperature, temperature has to be in Kelvin. So add 273.15 to this. So when we do that that's gonna give us 313 point 15 Kelvin. Plug that in. What units cancel out? Atmospheres cancel out, liters cancel out, kelvins 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 0.0364 moles, take those moles and plug them into the molar mass formula, and doing that gives us a value of 68.681 grams per mole. Now here 2.50 has 3 sig figs, 3 sig figs, 3 sig figs. 40 has only 1 sig fig. So if we wanted to go by that 1 sig fig that will come out to 70 grams 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 grams per mole would be the molar mass of our unknown gas.
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concept
The Ideal Gas Law: Molar Mass
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1m
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Besides finding the moles of a gas sample, the ideal gas law can be extended further to find the molar mass of a 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 only look at that video if your professor cares on showing how you derive this new version of the ideal gas law. If your professor doesn't care, then just ignore that video and just go on to the next question where we have to solve for a problem. Alright. So here to help us remember this one we're gonna say more mass really tests our valuable patients. So here molar and then mass really tests r t. They are over valuable patients, which is volume and pressure. Here I just kept them in the order of p v. So this is the formula we can utilize to find the molar mass of a gas immediately, instead of doing it piece by piece like we did in previous example questions. So just remember, molar mass really tests our valuable patients. Capital m equals lowercase m r t over p v. 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.
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concept
The Ideal Gas Law: Molar Mass
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1m
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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 molar mass formula. Remember molar mass equals mass m divided by moles n. Here we're gonna algebraically rearrange this formula to isolate n. So first you're gonna multiply both sides by n, so now it becomes m times n equals lowercase m. To isolate our n moles, we divide out molar mass. So we're gonna say here moles equals mass divided by molar mass. Going to the ideal gas law formula, we can now substitute in for n this value here. So p v equals n r t now becomes p v equals mass over molar mass. We need to isolate our molar mass, so what I'm gonna do here is I'm gonna multiply both sides by molar mass. Here it's a denominator so I can multiply to get rid of it, and that equals mass times r t. Finally to isolate our molar mass we divide both sides by p v. And that's how we got that more mass equals mrt overpv. 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 molar mass really tests our valuable patience. Remembering that will help you remember this version of the ideal gas law.
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example
The Ideal Gas Law: Molar Mass Example 2
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2m
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Here it states in this example question, an unknown gas with a mass of 0.1727 grams is placed into a 100 and 20 5 milliliter flask. If its pressure is 0.833 atmospheres at 20 degrees Celsius, what is the identity of the gas? Here we're given different gases 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 we'll be able to find the molar mass of our unknown gas, and then all we have to do is see which one has that molar mass, as their actual weight. So we're gonna say here molar mass equals rt overpv. We're gonna plug in the mass of our unknown. We're gonna plug in r which is 0.08206 liters times atmospheres over moles times k. Remember temperature has to be in Kelvin, so add 273.15 to this number to give us 293.15 k. Divided by our pressure which is 0.833 atmospheres. And then our volume we convert those milliliters into liters so that gives us 0.125 liters. Here we look and see what units cancel out, and we see all that's left is grams over moles. So we have molar mass. When you plug that in you get approximately 39.90 grams per mole. All we have to do now is see which one of these gases has a molar mass closest to our answer. So here n 2, the 2 nitrogens comes out to 28.02 grams per mole. R gone from the periodic table is approximately 39.95 grams per mole. Oxygen is 32 grams per mole, grams per mole, grams per mole, grams per mole. Neon, 1 neon is approximately 20 or so grams per mole according to the periodic table. And then c h 4 is roughly around 16 grams per mole. Our closest answer would have to be argon. It's the one that's closest to this 39.90 grams per mole. So here we've just utilized the molar mass version of ideal gas law to find the molar mass of our unknown gas.
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Problem
Problem
To identify a homonuclear diatomic gas, a chemist weighted an evacuated flask with a volume of 3.9 L then filled it with the gas at a pressure of 2.00 atm and 29.0 ºC. The chemist then re-weighted the flask and recorded the difference in mass as 8.81 g. Identify the gas.
a) H2
b) N2
c) Cl2
d) F2
e) O2
A
H2
B
N2
C
Cl2
D
F2
E
O2
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Problem
Problem
What is the molecular formula of a compound that contains 39.0% carbon, 16.0% hydrogen, and 45.0% nitrogen, if 0.1576 g of the compound occupies 125 mL with a pressure of 0.9820 atm at 295.15 K?
A
CH3N
B
CH5N
C
C2H3N2
D
C2H5N2
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