[narrator] Let’s talk about elemental symbols in three scenarios.
One where we’re reading the periodic table, another where we’re looking at isotopes
of the same element, and then finally when we’re talking about the most common or any
ion of an element.
In the beginning here, we have a block that I’ve extracted from the periodic table,
for the third element lithium, and we see the elemental symbol, or abbreviation, the
atomic number, often abbreviated as capital Z , and here we have the average atomic mass
of all isotopes of a particular element.
That’s reading the periodic table.
This mass is not the mass number because we’re not talking about a particular
isotope we’re talking about the average of all the abundances of the element that’s
given.
When we’re talking about isotopes, however, we will write the elemental symbol, lithium
here, and we’ll put the atomic number in the bottom left.
We’ll also put the mass number, say we’re talking about lithium-7 versus lithium-6,
and this mass number is abbreviated as A, and it tells us something.
It tells us how many protons plus neutrons there are in a particular isotope of an element.
If we calculate this for lithium we don’t even need the three here.
Often times you won’t see the three.
We can leave it on there for now but when we’re calculating the number of protons,
well that’s given to us, and the number of electrons for a neutral element is going
to be equal to the number of protons.
The number of neutrons is going to be equal to the mass number minus the atomic number,
or A - Z.
Here, we have 7 - 3 is 4 for the number of neutrons.
4 + 3 is 7 is another way to look at it.
What about if we have lithium-6, instead?
Protons doesn’t change.
We still stay with three, electrons we’re going to stay with three, because it’s a
neutral species, and then for the number of neutrons it’s going to be the mass number
minus the atomic number, 6 - 3 equals 3.
No tricks there, except recognizing that the atomic number and the elemental symbol really
tell us the same information.
The number of protons is what gives us the identity of an element and they will never
change from one isotope to another.
The number of neutrons can change, which is the violation of Dalton’s atomic theory.
One of the violations where he says that all atoms of a particular element are the same,
well isotopes violates that particular rule of Dalton’s atomic theory.
Then most common ion, if we were to represent this as a symbol, we could say for lithium-7
the most common ion since it’s a group 1 element is +1.
We might represent this as lithium-7+.
It’s number of protons is going to be the same, always three for lithium, but the number
of electrons, because we’ve got a positive charge--we’re missing an electron and that’s
why three protons minus two electrons gives us a +1 charge.
The number of neutrons will not change from the other lithium-7 that we looked at.
But, with lithium-7 plus one ion the number of electrons changes.
We can do the same with lithium-6 and you’ll see that the number of protons never changes,
that’s what makes it lithium.
The +1 charge will take away an electron and the neutron number from lithium-6 stays the
same, its atomic--excuse me, mass number minus atomic number.
We can do this for any element.
Suppose we’re dealing with isotopic abundances and we say lithium, this is an approximation,
is composed entirely of lithium-6 and lithium-7, so we’re representing our two symbols that
we did above, in words--the nomenclature here.
We’re just representing them in words.
That second--or that number, that mass number comes after the name of the element.
What is the natural abundance of each?
What percentage of each would make up all of lithium if we’re assuming that the exact
masses of the elements are the same as the mass number and they’re whole numbers?
It’s a little bit different than a whole number for reasons we don’t want to get
into, at this time.
For more advanced courses.
However, we can do this using the elemental symbol that I’ve drawn above.
What this is is a weighted average.
The average is going to be equal to 100%, or excuse me ‘x’ out of 100.
Some percentage of lithium-6, and we’ll say 6.0--or we could say even, let’s keep
it--keep it generic.
‘x’ over 100 times mass A plus 100 minus ‘x’, whatever’s left over.
Whatever percentage is leftover.
Out of 100 means percent.
Whatever’s left over times mass B.
We could keep going if we had knowledge of other isotropic abundances, but given the
two abundances we have above we’ll use this formula.
Let’s look at our elemental symbol. 6.94 is our average.
If we plug in 6.94 atomic mass units, or Daltons, is equal to ‘x’ over 100 times the mass
A.
We’ll call that 6.00.
Which mass you define as the first it doesn’t matter, you can use either one.
We’ll choose 6.00 and then the rest is going to be mass B, which is 7.00.
We’re carrying it out to three significant figures because that’s what our periodic
table gave us.
We’re using whole numbers because the problem said to assume that the mass is a whole number,
just like the mass number.
When we do this we can multiply both sides by 100 and we’re going to get 6.9--694 equals
‘x’ times 6.00 plus 100 minus ‘x’ times 7.00.
We can distribute our values and when we solve for ‘x’, at the end.
I’m just shortening, working this out from the algebra, but when we finish working this
out we’re going to get ‘x’ is equal to 6.00%.
That means that if we have 6.00% then 100 minus ‘x’, whatever’s left over--lithium-6,
whatever’s left over, 100 minus ‘x’ is going to be equal to 94.0%.
To the proper number of significant figures or we could say .00 since we are subtracting
an exact number, equals the abundance of lithium-7.