In this problem, we will be discussing the concept of isotopes. So we see in the problem that we want to see how many different combinations of Sealed 2 0 molecules with different masses naturally exist. So looking first at um chlorine, there's two different isotopes um cl 35 cl 37. So the various combinations of cl two, we can make out of this RCL 35 And see a 37, Or we could do two of the CL 35. Yeah, Are we due to the CY 37s? So we have three different combinations.
And with oxygen, there's just one oxygen in the seal to a molecule And the different sizes up his oxygen. RO 16 I was 17 And 18. So this is three combination as well. So if we multiply these together three times three equals nine. So we have nine combinations.
And now we can figure out which are the three most abundant cl 20 molecules. And looking at the problem, we see that the percentage of 016 I so 16 Is extremely high at 99.757%. So are three of us abundant Cl 20 molecules will be molecule with 0 16 isotope. And then the three different combinations of these cl isotopes. So now all we do is plug in the numbers from our equate or from the problem To figure out the masses of the L 2 0 molecules.
So if we start off with the 30 cl 35 and cl 35 0 16, we have to times 34.9688 because that's the mass of the cl 35 isotope Plus the 15.9949. Because that's a massive the Um oxygen 16 I said hope Um that gives us 85.93 to 5 atomic mass units. And then we have The 35. c. l 37 cl I'm 16 And that will be a 34.9688.
Which is the mass of the c 35 Isotope plus a 36.9659 because that's also the cl 37 isotope. And then plus AR 15.9949. And that gives us 87.9296 atomic mass units. And then lastly, we're going to do or to see how 37's In our 16 for the oxygen. And that's two times 36.9659 Plus 15.9949.
And that's going to equal 89.9267, topic last units. So these are um the masses of the famous abundant cl 20 molecules..