Bromine has two naturally occurring isotopes (Br-79 and Br-81) and has an atomic mass of 79.904 amu. The mass of Br-81 is 80.9163 amu, and its natural abundance is 49.31%. Calculate the mass and natural abundance of Br-79.

Video transcript

Hi, everyone. Welcome back. Our next question says Bromine has two naturally occurring isotopes. BR 79 and BR 81 and has an atomic mass of 79.904 AM U. The mass of BR 81 is 80.9163 AM U and its natural abundance is 49.31%. Calculate the mass and natural abundance of BR 79. We have four answer choices for the natural abundance of BR 79 and the mass of BR 79 A, the abundance being 50.69% mass 79.98 AM UB abundance, 50.69% mass, 78.92 AM U C abundance. 50.87% mass, 74 4.74 AM U and D abundance. 50.87% mass, 76.67 AM U. So we have our element with just two isotopes, we have its atomic mass which would be that average mass of the mix of those isotopes found in nature and were given the mass and natural abundance of one of them. Can we calculate the mass and natural abundance of the other. So let's think about how we calculate that average atomic mass and that's the atomic mass is equal to. And now I'm going to jump straight to exactly how we'd solve it for this particular situation with two isotopes. And just to sort of keep it from getting really cluttered here, I'm going to label Bromine 79 as isotope one. Bromine 81 as isotope two. So keep that in mind as we go through. So the atomic mass that average overall mass is equal to I put in parentheses, the mass of isotope one. So subscript one multiplied by and I write fa of one fa stands for fractional abundance. So we got our abundance as a percent fractional abundance. We just change that into a decimal or fraction. So that's very simple. So I'm going to just put an arrow here to convert natural abundance to fractional abundance. I would divide by 100% to get that fractional abundance number. So back to our equation, we have mass subscript one multiplied by fa subscript one plus. And then in parentheses, the mass of isotope two. So mass subscript two multiplied by the fractional abundance of isotope two. Again, this makes sense because you're ta because you're taking into account the relative abundance of the two isotopes, which is not equal. Well, notice here, we've got some known factors. We know the atomic mass overall. And then if I stop one is VR 79 we need to calculate, so calculate these and then we know the mass and the fractional abundance of number two. So we have what we need, we do have two unknown factors. But since fractional abundance is just this percent of the whole, we can easily get the fractional abundance of bro means 79. So fractional abundance of isotope, one is equal to 100% minus the fractional abundance of isotope two because there's only two of them. So they must add up to 100%. So that would be 100% minus or excuse me, I said, fractional abundance here, I'm doing relative abundance. So let's erase fa have too many abundances here. So it's going to be the relative abundance or natural abundance. So they're calling it in this case was right. Abundance in a percent form abundance of number one is 100% minus the abundance of number two equals 100% minus 49.31% which is equal to our percent of 50.69%. So very simply, we have already calculated are natural abundance of BR 79. Since that's isotope one, we can go ahead and eliminate answer choices C and D since they have that value as 50.87%. So now let's look at, we've just got A and B left since they have the correct 50.69% figure, we just need to calculate the mass of the R 79 I'm going to go ahead and highlight in blue, our equation that we're going back to so that we can easily refer back to it. So we have our, we start with our atomic mass overall of bromine, which we know is 79.904 AM U. Then there 9:04 a.m. U equals and we have our mass of one multiplied by the fa of one. While the mass we're still trying to calculate 12 equals the mass of subscript one. Now, for fractional abundance, again, we're going to convert from a percent to a decimal. So our fractional abundance that we just calculated 50.69% we will write that as fractional abundance, zero point 5069. As we said to do to get that, we just divide our natural abundance by 100%. Then we're going to add that too. Our mass of isotope two bromine 81 which is 80.9163 AM U. And that's multiplied by its fractional abundance. We were given that its natural abundance is 49.31%. So its fractional abundance will be zero point 4931 squeeze that in at the edge of the screen there. So now let's just make those calculations there to get ourselves down to of equation, we can move around. So 79.904 AM U the U there is equal to the mass of isotope one multiplied by 0.5069. And now let's calculate that multiplication for isotope two. That term will be equal to 39.90 AM U. So now we just need to isolate our number for the mass of isotope one. So let's subtract that 39.90 AM U from both sides of the equation. So subtract 39.90 AM U from both sides that will end up giving us 40 0.00 AM U is equal to the mass of isotope one multiplied by 0.5069. And now we just divide both sides by that factor there by that 0.5069. And we're going to get that the mass of Bromine 79 is 78.92 AM U, the equal to the mass of isotope one. We'll check our significant figures. I actually dropped a number up here when I calculated 40.00 AM U because we're only going to have four significant figures when we look at the values that we've calculated. So this is with four sig bigs. Well, right here, since that's our smallest number of significant figures that we work with in our calculation. So we have our mask, go back to our answer choices, 78.92 AM U. That's choice B choice A is incorrect because that has 79.98 AM U. So we rule that out. So we've now successfully calculated our mass and natural abundance of br 79 natural abundance of 50.69% mass of 78.92 AM U. See you in the next video.