Boron has only two naturally occurring isotopes. The mass of boron-10 is 10.01294 amu and the mass of boron-11 is 11.00931 amu. Calculate the relative abundances of the two isotopes.

Video transcript

Hello, everyone. Today, we have a following problem. Nitrogen has only two naturally occurring isotopes. We have the mass of nitrogen 14 and the mass of hydrogen 15 calculate the relative abundances of the two isotopes. So to do this, we can use the Fong equation that states that the atomic mass is equal to, we have the mass of our nitrogen. 14 isotope multiplied by its fractional abundance. And then we add that to the mass of our nitrogen 15. And we multiply that by its fractional abundance. So the atomic mass of nitrogen per the periodic table is 14 AM U or atomic mass units. So we equal that to the mass of nitrogen 14, which we were given is 14.00307. And then we multiply that by its fractional abundance, we don't have that. So we can actually just denote it as why we add that to the mass of our nitrogen 15 isotope. And that was stated to be 15.00011 atomic mass units. Now we will take that and since we don't have the fractional abundance of that, either, we can simply just say that this will be one minus Y. So when we have the value of Y, we can determine the fractional abundance of that. So since we have 14, we can just distribute such that we have 14. Is it good to 00307? Why we add that to our 15.00011? And then lastly since we distributed that 15 to the one and the Y you have minus 15.00011 Y. And combining like terms on the left hand side of the equation, we have 0.99 337 is equal to negative 0.99 704. Y dividing both sides by the value that contains the Y, we can isolate that Y such that it is equal to 0.99 632. We also said Y was equal to the fractional abundance of our nitrogen 14 isotope. So that means that to find the factional abundance of our nitrogen 15 isotope, we simply do one minus Y or one subtract by 0.99 632 for a value of 0.00368. Now all that's left is just to multiply both of these fractional abundances by 100%. Such that for our fractional bonus of nitrogen 14, we get 99.632%. And for our fraction abundance of nitrogen, 15 isotope, we get 0.36809%. Looking at our answer choice says we see the answer choice B best reflects these numbers overall. I hope this helped until next time.