5 answers

Naturally occurring chlorine molecules, CF, spectrum above_ have They - occur masses 0f 70, 72 and 74 amy a5 seen

Question:

Naturally occurring chlorine molecules, CF, spectrum above_ have They - occur masses 0f 70, 72 and 74 amy a5 seen the average atomic percentages of 56.25%, 37.50% &nd 6.250%, in the mass mass of chlorine atoms and to find Use this data to calculate Click or tap here to enter text; the relative abundance of "scl , and "CI Isotopes:


Answers

Naturally occurring chlorine is composed of two isotopes: 75.76$\%$ Cl-35 (mass 34.9688 amu) and 24.24$\%$ Cl-37 (mass 36.9659 amu $)$ . Naturally occurring oxygen is composed of three isotopes: 99.757$\%$ O-16 (mass 15.9949 amu), 0.038$\%$ O-17 (mass 16.9991 amu), and 0.205$\%$ O-18 (mass 17.9991 amu). The compound dichlorine monoxide is composed of two chlorine atoms and one oxygen atom bonded together to form the Cl $_{2} \mathrm{O}$ molecule. How many $\mathrm{Cl}_{2} \mathrm{O}$ molecules of different masses naturally exist? Give the masses of the three most abundant $\mathrm{Cl}_{2} \mathrm{O}$
molecules.

.

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..

The Superb # 26. Uh And this Question says natural occurring chlorine is 75.53% chlorine. 35. So 75 points 53% Blowing 30 35 35 years. The mass of the urine which has an atom atomic masses of 30 point point 34.969 AM you.

And 24 point okay it has atomic must equal to 34 points 969 AM You. 36.969 M. U. This is one Second is 24.47% Chlorine. 37.

24 points for 7% Chlorine 37. And it has atomic mass equal to 39 36.966 a. m. U. 36.966.

I am you. So we need to conclude the visit during months of the chlorine. Okay we have formula for this. According to that formula. Every age atomic mars will be uh this is 75.53% of 13 34.969.

It was 34. Okay it's 75%, That is 34.969 in two 75 points 5 3 divided by 100 bless 24.43% of this. 36.966 into 24.47 by 100 M. U. Okay so we have to use calculators 34 points by nine 35 point So approximately it will be equal to 35 46.

I am you right. They should be the answer Thank you..

Let's first consider all the possible bro Mean chloride combination of these isotopes that we could have. The first one that we can have is 79 bro. Mean chlorine 35. The second one we can have is bro being 79. Chlorine 37.

Then we can also have roaming 81 Chlorine 35 and we can finally have bro Ming 81 quarry in 37. Now we need to find the percent of growing 79 chlorine 35. And based on this 1 to 1 molar ratio, we know based on the information, we're given its 50.69%. So we can find the mass of growing 79 chlorine 35 by taking its present percentage in our sample well supplied by the some of the atomic masses. So we'll have 0.5069 well supplied by 78.92 plus 34.969 and well funded.

The mass of this compound is 57.73 And now we can do this for the other isotope combinations in our list. So let's do bro Ming, 79 Chlorine 35 again. All of this information is given to you in this problem. We just have to figure out how to arrange it. Essentially.

So for this molecule have 0.243 multiplied by 78.92 plus 36.966 and we'll get that. The mass of this combination is 28.78 now for the Mass of bro Ming, 81 Chlorine 35. We'll have 0.4931 multiplied by 80.916 plus 34.969 and we'll find that our Mass is 57.14 And now for the mass of Bro Ming, 81 Chlorine 37 will have point to 4 to 3, multiplied by 80.9163 class 36.966 and we'll get the air masses 28.564 and now what we could do with all of these values of Mass. Based on these combination of isotopes, we can create what's called a mass spectrum, and this is what our graph would look like on our X axis. We have our molecular mass on our Y axis.

We have the relative number of molecules and obviously we have the percentage of the presence of these isotope combinations. So I hope that this problem helped you understand how we can find the mass of various combinations of isotopes and hopefully introduce you Teoh what's called the mass spectrum and further understand how understanding the mass spectrum relates to this chapter in atomic theory..

For part A. There are three peaks at 70 72 and 74. Atomic mass units do to the different combinations of eyes. It hopes that conform. The peak at 70 is a combination of, um, chlorine 35.

At 72 would be chlorine 30 five, paired with the core in 37 and the peak at 74 would be chlorine 37. So seo to with corn, 37 for be the peak at 70 a. M. U. Is is much more intense than the peak at 74 a.

M. U. Because Chlorine 35 Corinne CEO, 35 is much more abundant then Chlorine 37..

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