Bromine has two naturally occurring isotopes (Br-79 and Br-81)
an...

Question

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.

Accepted Answer

hey everyone in this example, we need to calculate mass and natural abundance of the given isotope magnesium were given to other isotopes. The first being magnesium 24 with its associated atomic mass and percent abundance were also given our second isotope, magnesium 25 with the associated atomic mass and percent abundance. So we should first recall our formula for the atomic mass of a natural element. And in this case because our isotopes are magnesium, we're focusing on natural magnesium. So we should recall that this is going to be equal to the products where we have our first isotope, which in this case is magnesium 24 multiplied by its natural abundance or percent abundance Added to the second isotope being magnesium 25 times its percent abundance. And then lastly added to our third isotope, magnesium 26 multiplied by its percent abundance. Now so far we can go ahead and reformulate this so that we have that our percent abundance because that's what we're trying to solve for in this question Is equal to 100% for our natural magnesium minus Our first isotope, which is given as a abundance of 78. And then minus our third isotope, which abundance is given as 10% for Magnesium 25. And so this is going to allow us to calculate our percent abundance of our Unknown abundance of Magnesium 26, our third isotope. So when we take the difference of these percentages, we're going to get a value of 11 point oh 1%. And so that means that this is the percent abundance of our magnesium 26. So so far we've at least answered that portion of our question and now we need to get that mass of our third isotope. So we should go ahead and convert the abundances from percentages to decimals. So for our first isotope magnesium 24 we're going to convert from 78.99% to a decimal as .7899. For our magnesium 25 we're going to follow the same step given its percent abundance as 10%. We're going to convert that to a decimal as .1000. And then lastly we have magnesium 26 which we just found above its percent abundance from taking the differences between our natural magnesium abundance from our Other two isotope abundances. We got the difference of .1101 As a decimal. And so now we can go ahead and take note of magnesium atomic mass on the periodic table. So we would refer to the periodic table and we would find magnesium in group two a Which is associated with the given atomic mass of 24.305 am use. And so now we can go ahead and plug this into our equation above. So for our natural magnesium on the left hand side we're going to plug in that atomic mass 24.305 A. M. U. S. And we're setting that equal to the known atomic masses for our other isotopes. So for our first isotope magnesium 24 we have 23.9850. And Muse which is given to us times its percent abundance which we converted to a decimal as .7899. This is then added to the um the atomic mass of our second isotope Magnesium 25 which is given as 24.9858 am use and then multiplied by its percent abundance which we converted to a decimal 2.1000. And lastly we're going to do the same thing for our Third Isotope Magnesium 26. I'm just going to scoot this all over so we have enough room. So we're going to add The atomic mass of our Magnesium 26 which in this case we're just going to say is X. So we can just say magnesium 26. So we can just say X here And then we're going to multiply this by its percent abundance as a decimal which we found above as .1101. So this completes our expression and we're going to go ahead and solve for X. So we can simplify this so that we have 24. A M U. S. Is equal to four. First product 18.9458. And this should be an eight. And we have the units. And you added to our second product which gives us 2.49859 A. M. U. And that nine should actually be an eight there. Sorry about that. And then lastly we're going to add our last product which should be .1101 times x. Where again exes are unknown atomic mass of our third isotope magnesium 26. And so now we can Further simplify this so that we have 24.305 am use equal to 21.4444 plus .1101 x. Now we're going to go ahead and subtract 21.4444. And this should be an A. M. U. S. From both sides. And so this allows us to now get 2.8606 A. M. U. S. Equal to 0.1101 X. And actually the Amy's will cancel out there. So we would just have 22.8606 equal 2. 01X. So now we're just going to divide by .1101 on both sides to isolate for X. And so what we would get is X. Is equal to 25.9828 am us. Because we should recall that our X. Variable is representing the atomic mass of our third isotope magnesium 26. So this is our atomic mass for magnesium 26. And then above we calculated that its percent abundance Is going to be equal to .11. Or rather we can say 11.01%. And so to complete this example, this is our final answer. So I hope that everything we reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

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.

Watch next