Lithium has only two naturally occurring isotopes. The mass of lithium-6 is 6.01512 amu and the mass of lithium-7 is 7.01601 amu. Calculate the relative abundances of the two isotopes.

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Welcome back. Everybody. Here is the next problem. Lithium has only two naturally occurring isotopes. The mass of lithium six is 6.0152 AM U and the mass of lithium seven is 7.01601 AM U calculate the relative abundances of the two isotopes. And we have choice a percent abundance of lithium 6 92.588% percent lithium 77.412%. Choice B percent abundance. Lithium 67.412% of lithium 7 92.588%. Choice C percent abundance of lithium 6 92.506% of lithium 77.494%. And choice D percent abundance of lithium 67.494% lithium 7 92.506%. So we have masses given to us, we need to calculate relative abundances. So the formula we want to look at here, it is the one that says that the atomic mass of our element, which is sort of an average atomic mass of all the lithium that exists in both forms. And So, atomic mass is equal to, we have mass of our first isotope. So I'll put mass subscript one. Uh In this case, I'm going to call lithium six isotope one and lithium seven isotope two just to cut down on how much clutter I have here. So mass of one multiplied by fa subscript one where fa is fractional abundance and that will be added to. And then in parentheses, the mass of iso tripp two multiplied by B fa of isotope two in closed parentheses. And this makes sense because again, that atomic mass is sort of an average figure. So we add together the mass of a particular isotope multiplied by that fractional abundance. So what percent of all the lithium does it make up? Well, our atomic mass, we can look up in the periodic table and then our masses of the two isotopes are known. Now, our fractional abundances are what we're looking for. But we should note that these two are related to each other because an equation with two unknowns is going to be difficult if not impossible. But since one is related to the other, we essentially just have one unknown. So let's plug in our figures. So we'll go to our periodic table and look up that atomic mass of lithium, which is going to be 6.941 AM U that will equal. So then we have the mass of lithium six, which is our isotope. One is 6.01512 AM U. So I'm putting that in parenthesis. Now, let's take a break here. We're looking for how are the fractional abundance of one and the fractional abundance of two related to each other? Well, we have our equation has fractional abundances. We're looking for relative abundances. Relative abundance is going to be the fractional abundance multiplied by 100%. Note that our total relative abundance must equal 100% the total amount of it we have. And so in that case, our total fractional abundance which is relative abundance divided by 100% must equal one. Well, what does that mean for us? That means that if we call the fractional abundance of mass one X, then the fraction abundance of element two must be one minus X. Since the two together must equal added up must equal one. So that's the way we relate them to each other. So as we've writing our equation, we have the mass of isotope one, we're going to call the fractional abundance of isotope one X. So we multiply that by X and then add the mass of isotope two, which is 7.01601 A U put that in parentheses that will be multiplied by one minus X. So now let's expand this out a bit 6.941 A U equals 6.01512 AM U multiplied by X. We'll expand out the next equation. So we'll have plus 7.01601 AM U multiplied by. Well, it's multiplied by one. So that can just stay. And then we have minus 7.01601 A U multiplied by X. So now we combine our factors that have X in them and our factors that are just numbers. So we have, we'll subtract 7.01601 A U from both sides and then combine our two uh expressions that have X as a coefficient. So we have 6.9 for 1 a.m. U minus 7.01601 A U. Yeah, and that's going to equal, we have negative 0.07501 AM U equals. And then our factor that has the coefficient X in it is negative 1.00089 AM U multiplied by X. So now we just divide both sides by that negative 1.00089 AM U. And that's going to give us zero 0.07494. It's not AM U our AM us have canceled out when we divided. So 0.07494 equals X. We've gotten ourselves to X. Let's recall what X is, which is that X is the fractional abundance of isotope one. And that means that one minus X equals one minus 0.07494 which equals 0.92 506 equals the fractional abundance of isotope two. So we've calculated them both. Now, our question did ask us for the percent abundance. So we just need to multiply both of these by 100%. So that's pretty simple. So do that and not that, that will give us. So we end up with 7.494% 92.506%. So when we look at our answer choices, we see that choice A is incorrect as both percent abundances don't match choice. B has 7.412 92.588. So neither of those match choice C has 92.506% and 7.494%. So choice C looks like it's our answer. Just double check. Choice D has 7.494%. So that's correct. But then it has 92.506%. So that's the correct number. But do we have the right isotopes? So we've got the correct number, which do they belong to? Well, we said that the 7.494% is X which is our fractional abundance of isotope one, which is lithium six. So that means we've got to erase choice C and choose choice D because our 7.494% belongs to lithium six and the 92.506% belongs to lithium seven. So just something to be aware of it. Often we're looking at these multiple choice we're just quickly grabbing. Oh which matches. But in this case, that would, could make us lead us to make a mistake like that. So choice B is incorrect or choice C is incorrect because these two are reversed. Now note we can have a quick gut check here. Is this correct? Because back when we looked up the average atomic mass of lithium as a whole, we get 6.941 AM U which is almost seven. So it makes sense that lithium seven is the far more abundant isotope. So just a way to double check that answer. So finally, we had our two isotopes of lithium lithium six and lithium seven. We were given the masses how to calculate the relative abundances. We end up with choice D the percent abundance of lithium six is 7.494%. The percent of abundance of lithium seven is 92.506%. See you in the next video.

Lithium has only two naturally occurring isotopes. The mass of lithium-6 is 6.01512 amu and the mass of lithium-7 is 7.01601 amu. Calculate the relative abundances of the two isotopes.

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Key Concepts

Isotopes

Relative Abundance

Weighted Average