Lithium has only two naturally occurring isotopes. The mass
of li...

Question

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.

Accepted Answer

Hey everyone in this example, we need to calculate the relative abundance of our isotopes Braun 10 and born were also given their corresponding atomic masses. What we should first recall is our formula for Atomic mass for any natural element and that's going to be equal to our first isotope. Which in this problem is given as Boron 10 multiplied by its fractional abundance or relative abundance. And then this is going to be added to our second isotope given as born multiplied by its fractional abundance. We should recognize that this portion of our example means that This quantity is going to give us 100% of our natural element or we can say equal to one. And so that means we can go ahead and represent our fractional abundances for each of our isotopes by the following variables for our first isotope we're going to represent it as X. And then for our second isotope we're going to represent the fractional abundance by the variable -1. So we can go ahead and calculate our atomic mass for natural boron from our periodic tables. And we're going to see that it's going to equal a value of 10.811 am use when we find boron In Group three a. of our periodic tables. So we can go ahead and plug this into our formula so that we have 10.811 am use for our natural boron equal to the molar mass or atomic mass given for our first isotope born 10 as 10.1 to 9 a.m. use multiplied by the variable X. That we assigned for the fractional abundance of our first isotope. And then this quantity is added to our given atomic mass for our second isotope born 11 given as 11.93 AM use. And then multiplied by the variable for the second fractional abundance Of our second isotope represented as one -X. So we're going to go ahead and simplify this by substituting into our parentheses our integers here. So what we would have in our next line is 10.811 am us on the left still. And then on the right hand side we should now have 10.0129 x plus 11. -11.0093 x. And so our next step is to combine like terms which I'm going to underline here. So we're going to take 10.129 x. And Subtract that from our negative or sorry, from our 11.093 X term. And we can also go ahead and subtract from both sides 11.093 to get rid of that. And so this is going to simplify to the difference where we have 0 .1983 on the left hand side equal to The difference between our like X terms is going to be negative 0.9964 X. And we can actually go ahead and cancel out the units am us because our 11.093 when we multiplied by one still carries its units of am us. But it cancels out on the right hand side ultimately getting rid of that term. So now that we have simplified up to this point, we can now focus on finally isolating for X completely by dividing both sides by negative point 9964. And so what we're going to get is that X is equal to 0.199. And so now we can go ahead and now that we know this decimal value for our X value, we can go ahead and write out our relative a K, a fractional abundances. And so we would say that the percent abundance for our Brunton is going to be equal to .199 which we can just multiply by 100 to get as a percent. And that's going to give us 19.9%. And again, because we know that our isotopes should make up 100% of our natural boron. That means that we can just take -19.9%. Or we can actually just go ahead and plug in to our variable for our second isotope representing its fractional abundance as again, 1 -1. We're going to actually plug in our X value. So we can say one minus .199. So this is the same thing here, and these both would equal a value of 80.1%,, Which is our fractional abundance for our born 11. And so to complete this example, this would be our final answer As well as 19.9% for our born isotopes. So these are the calculated relative abundances for our given isotopes. If you have any questions, please leave them down below. Otherwise, I will see everyone in the next practice 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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