a. Suggest possible molecular formulas for a compound that has a ...

Question

a. Suggest possible molecular formulas for a compound that has a molecular ion with an m/z value of 86.

Accepted Answer

All right. Hi, everyone. So for this question, let's determine the possible numbers of carbon in a compound that contains carbon, hydrogen and oxygen with a molecular ion peak equal to 98. Option A says four carbons. Option B says five carbons. Option C says six carbons. Option D says both A and B option E says both B and C and option F says both A and C. So before we begin, let's recall that the number of carbons found in a compound as well as the molecular structure in general can be determined by what we refer to as the rule of 13. Right. Now recall that um using the rule of 13, you can figure out the number of carbon and hydrogen in a compound by dividing your molecular ion by 13, right? And so that's going to give you a quotient which we represent as the letter N and then your remainder. I'm going to put in parentheses represented as the letter R. So using these two numbers, right, your N and your R, you can figure out the number of carbon and hydrogen like so, right. So in your molecular formula, your number of carbons is going to be equal to your quotient or your N and then your hydrogens are going to be equal to the sum of your quotient plus your remainder. So N plus R and the art looks a little weird. So let me just fix that. There you go. And plus R. So before we can apply our rule of 13, it's important to understand that we have to subtract the mass of any other atoms that we might have in this molecule, including in this case oxygen. But the problem is that we don't know the exact number of oxygens that we have in this molecule. So what we're going to do is we're going to assume that we have different numbers of oxygens and we're going to factor out the mass of them before calculating our hypothetical molecular formula, right. And from there, what we do is we calculate the I H D just to ensure that the molecular formula that we derive can correspond to an existing molecule. So let's begin by assuming that we have just one oxygen in our molecule here. So before we can apply a rule of 13, we have to subtract the mass of our oxygen here from our molecular ion of 98. So that's going to be 98 minus Or subtracted by 16. And that gives me 82. So now 82 divided by 13, gives me a quotient of six with a remainder equal to four. So now let's apply our molecular formula, right, according to the number of our quotient, we're going to have six carbons And the number of hydrogens is going to be equal to 6-plus 4, which is 10. And for the sake of this hypothetical here, we assumed one oxygen. So here's our one oxygen leading to our first molecular formula of C six H 10. Oh. So when we calculate the I H D R I H C, if you recall Is equal to two, multiplied by the number of carbons plus two plus any nitrogens, you have minus any hydrogens minus any halogens And all of that, you're going to divide by two. So for this molecular formula here, for our C six H 10 0, our I H D is going to be equal to two multiplied by six plus two, we have no nitrogen. So I'm not going to write that, But we do have 10 hydrogen. So that's -10. And all of this is going to be divided by two. So that is going to give us an IHD of two, which is valid, right? Because recall that an I H D value must be a positive whole number. So this does check out, right. So this can be a possibility. Now, from here, let's go ahead and assume if maybe there are two oxygens in this molecule, Right? So now when we subtract the mass of oxygen from our molecular ion of 98, it's going to be 16, multiplied by two, Right? So that's 98 minus 16, multiplied by two, which gives you 66. So when I apply my rule of 13, I'm going to do 66, divided by 13 Is going to equal five with a remainder of one. So here, my molecular formula is going to be C5 H five plus one. So that's six and we assumed that there would be two oxygens. So that's C five H 602. And so my I H D is going to be equal to two multiplied by five plus two. There are no nitrogens once again, but there are six hydrogens. So it's two multiplied by five plus two minus six. and all of that will be divided by two, which is going to give you an I H D equal to three. Now, this is also a valid possibility because once again, we have a positive whole number. So lastly, right, what if we had three oxygens in this molecule? So now when we're calculating our, Our mass, after accounting for oxygen, right, that's going to be 98 -16 times three. So that gives me 50. And when I apply my rule of 13, 50, divided by 13 Is equal to three with a remainder of 11. So now my molecular formula is C H three plus 11. So that's 14 and we assume that there will be three oxygens, right? So it's C three H 14 oh three. So when we calculate our I H D here We do two times 3 plus two minus 14, All of which gets divided by two, Which gives me -3, right? So this molecular formula cannot be valid because my I H D value can never be negative, right? So we know for a fact that this possibility here does not make sense, which means that the only ones that do make sense are when I have either six carbons or five carbons. So that means that options B and C are both correct. Meaning that my final answer is going to be options E option E which is both B and C. So if you stuck around to the end, thank you so much for watching and I hope you found this helpful.

a. Suggest possible molecular formulas for a compound that has a molecular ion with an m/z value of 86.

Key Concepts

Molecular Ion

Empirical Formula

Isomerism

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