The atomic mass of an element includes the masses of all 3 subatomic particles of neutrons, protons and electrons.
Determining Atomic Mass
1
concept
Atomic Mass
Video duration:
2m
Play a video:
so atomic mass is the mass of an element that includes the masses of all three subatomic particles. So we're looking at all the protons, neutrons and electrons forgiven element. We can also say that atomic masses of elements can be found by simply looking at the periodic table. So if we take a look here, the first element of the periodic table is H H stands for hydrogen. Now, later on, we'll go into how elements symbols are related to names, but for now realize that is hydrogen, this number below it. This 1.8 here that we have that is its atomic mass. And if we look, we can see that a majority of the elements have atomic masses that are not whole numbers. Well, that's because the atomic mass oven element is an average of all of its isotopes. So hydrogen has several different isotopes, and from those isotopes were taking the average mass. That's why we do not have whole numbers. Now. We can say here that these have this atomic mass can come in different units. Now we're used to one of them, and that would be grams per mole. That's our normal units for atomic mass, but can also be expressed as atomic mass units or Dalton's now recall that an atomic mass units itself is equal to 1.66 times 10 to the negative kg. So just realize when it comes to the periodic table, we have the elements symbol for each of these elements, and this number on the bottom, which is not a whole number, is the atomic mass, the number on the top, which will always be a whole number that is our atomic number. And remember, your Tomic number uses the variable Z. Okay, so keep this in mind when looking at any periodic table. Now that we've talked about the basic parts of a typical periodic table, let's continue on with additional videos.
2
example
Atomic Mass Example 1
Video duration:
2m
Play a video:
so here for this example question. It says Which of the following represents an element from the first column with the greatest atomic mass? All right, so our first column. If we look at this periodic table, our first column is this with all of these different elements. And remember the number of red, which is not a whole number. Normally, that represents the atomic mass of any of these given elements. Now here, if we take a look, we have barium. Be a again. Later, we'll learn about how the names are attached to the elements. Symbol B. A is not in the first column here. It's in the second call, so this cannot be a choice. Then we're gonna say next that we have a l A. L stands for aluminum. Aluminum is over here in the third column. Well, all the way over here in this 13th column, actually, So this is out. Next, we have C s just session. Here it is, right here. It's in the first column. It's pretty low down there. It's 1 32.91 for its atomic mass. Remember, that could be in grams per mole. Atomic mass units or Dalton's. So far, it looks like it's the highest one. The only one higher than that would be F R. Notice that in the bottom rose here. Most of them are whole numbers. These are super large mass elements that are pretty unstable. They typically don't have numerous isotopes. As a result, they have no decimal places. So next, so so far. See, looks like it's our best choice. If we look at D, we have alli, which is up here not hiring mass, not greater atomic mass. And then we have a which is right here. So it looks like, see is our best choice. It has the greatest mass atomic mass from column one from the choices provided. So just remember, we have our element symbols. We have our atomic masses, which normally are not whole numbers, and then we actually have whole numbers. Those represent our atomic numbers
On the Periodic Table, the atomic mass is represented by the number with decimal places.
3
Problem
Problem
Which of the following choices has the greatest atomic mass?
A
Element A (0.283 kg)
B
Element B (3.20 x 1024 amu)
C
Element C (0.350 kg)
D
Element D (4.14 x 1026 Da)
4
concept
Atomic Mass
Video duration:
2m
Play a video:
now we can look up. The atomic mass is off a known element by simply looking at the periodic table. But there's going to be instances where you may not know the identity of the element, or you may not have a periodic table handy. In these cases, we can actually calculate the atomic mass for that element. Now, toe the atomic mass of an element can be calculated. If you know the isotopic masses and percent abundances off that element, we're going to say here that you're isotopic masses air just the masses of for all the isotopes of a given element, the percent abundances air. Just the percentage is available for each of the isotopes of a given element. Sometimes these percent abundances are also called natural abundances. Yeah, now associated with our percent or natural abundances, we have our fractional abundances. It represents the percent abundance of an isotope divided by 100. So remember when you have a percentage forms such as 25% to get its fractional form, you were divided by 100. That's what's going on here. We have our percent abundances. We can divide them by 100 that will give us our fractional abundances With this piece of information, we can use the atomic mass formula. Now. The atomic mass formula says that your atomic mass equals the isotope mass one. So the mass of your first isotope times it's fractional abundance plus isotope mass. Two times it's fractional abundances now realize that some elements have mawr than two isotopes, so you just continue on with the formula. So you say, plus isotope mass three times. It's fractional abundance and so on and so forth. So just realize that we're gonna utilize this atomic mass formula in order to calculate the atomic mass of an element when we don't have a periodic table handy.
5
example
Atomic Mass Example 2
Video duration:
2m
Play a video:
Everyone. So in this example, question, it says, calculate the atomic mass of lithium if the isotopic masses at abundances are given for its two naturally occurring isotopes. So here we're given the fact that lithium comes in two different isotopic forms. Lithium six and lithium seven. These numbers here represent their isotopic masses and then these are their percent or natural abundances. Now, here we're gonna say for step one, if you are given percent or natural abundances, then divide them by 100 in order to isolate the fractional abundances. So here we'd have 7.59%. We divide that by 100 that gives me 0.0759 then we have 92.41%. We divide that by 100 and that gives me 0.9241 step two. We're going to plug your given variables into the atomic mass formula in order to isolate the missing variable. In this case, we need to find the atomic mass of lithium. So what we're gonna do here is we're going to say that our atomic mass of lithium equals the isotopic mass of the first isotope which is lithium six. So that's 6.015122 AM U times its fractional abundance, which is 0.0759. So remember we divided the percentage by 100 to get this number plus the second isotopic mass which is 7.016004 AM U times its fractional abundance which is 0.9241. When we punch this into our calculators, we're gonna get approximately 6.940037056 AM you something around there. Now, here, if you're rounding, your number is gonna be slightly different from that, we have to look for the answer that matches closest to our answer. And based on our options, given the answer would have to be option B 6.941 AM U. Now, if you go back and look at the periodic table for the mass of lithium, you would see a number that's basically this number here. Remember your atomic mass here is the average of the masses of all the isotopic forms of that given element. So we have two elements or two isotopes for lithium. The average mass or atomic mass is this 6.941 that you see on the periodic table.
6
concept
Atomic Mass
Video duration:
40s
Play a video:
Now, how do we calculate atomic mass when no fractional abundances are given? Well, remember, adding up the fractional abundances off all isotopes for an element gives a total of one. Because remember, all of them together constitute 100% for that particular element. And dividing it by ah 100 gives us it's fractional abundance in that equals one. Now, this is important for atomic mass questions where there are two isotopes with unknown fractional abundances. So basically, if you're gonna have a question where you have to calculate the fraction abundances of all isotopes, it will only be for two isotopes, not three or more.
7
example
Atomic Mass Example 3
Video duration:
5m
Play a video:
Here, it states that iron possesses two naturally occurring isotopic forms. So we have iron 54 has an isotopic mass of 53.939615 atomic mass units. And iron 56 has an isotopic mass of 55.934942 atomic mass units. What is the percent abundance of iron? 56? All right. So here, the way we approach this question is step one, we make the first isotopes fractional abundance X and remember together they add up to 100% of all the iron that exists in the universe. And that's its natural percent abundance. If we were to divide this by 100 we get one. So we know that the first isotope is X. So the second isotope must be one minus X, which is equivalent to 100% minus X here. And we're just changing our percent abundance into our fractional abundance. That's where the one minus X comes from. Now, here we're going to plug in the information that we know in terms of the atomic mass formula. We're gonna say we know the atomic mass of iron based on the periodic table and that's 55.845 atomic mass units. This is gonna equal the isotopic mass of our first isotope iron 54 which is 53.939615 AM U. We don't know it's fractional, but so it's X and plus the second isotope 55.934942 atomic mass units and it'd be one minus X. So we're doing here step two, we're plugging in your given variables into the atomic mass formula in order to isolate our missing variable. Here, we need to find the percent abundance of iron 56. So what we're gonna do here is we're gonna solve this mathematically. We have uh this first isotopic mass times X and then we're gonna distribute this isotopic mass to the one and to the minus X. So our new equation becomes 55.845 equals 53.939615 X plus 55.9349 42 minus 55.934942. X. Let's keep working it out. This number has an X variable. This number has an X variable. So combine them together. So this would be 53.939615 X minus this X variable. When we subtract it to, we get negative 1.995327 X. And then this part stays the same 55.93492 and this still equals 55.845. Remember this is the atomic mass unit of iron that we find on the periodic table. We got to isolate our X variable. So, so we tracked out 55.93 942 from both sides. So there's a lot of numbers here. So always make sure you're looking to make sure everything matches up. This cancels out. So then what we have here on the left side is negative 0.089942 equals negative 1.995327. X divide both sides to get to isolate the X by a negative 1.99532 27. We'll get our X here. So when we do this, we're gonna get X equals 0.045 08. But what exactly is this X? Well, this X, if we go back up to the formula represents the fractional abundance of our first isotope, it gives us a fractional abundance of iron 54. But here they're not asking us for information on iron 54. They want iron 56. So we're gonna deal with this part to find a fraction abundance of iron 56. We found X. So the fractional abundance of iron 56 you can write this way equals one minus X. So one minus 0.04508. So this is 0.95492. But here I don't want the final fractional abundance. I want the percent abundance. So how do you change a fraction into a percentage? That's right. You multiply it by 100. So this gives me 95.492% which are just round to 95.5%. So this is a rough approximation of what the percent abundance would be of, of iron based on these isotopic masses that I've given to you, right. So working it out, we get this final answer, right. So this would be the last answer we place for this particular question.
8
Problem
Problem
Neon possesses three naturally occurring isotopes. 20Ne has a percent abundance of 90.48% and an isotopic mass of 19.99244 amu, 21Ne has a percent abundance of 0.27% and an isotopic mass of 20.99384 amu, and 22Ne has a percent abundance of 9.25%. What is the isotopic mass of the 22Ne isotope?
A
20.9182 amu
B
15.119 amu
C
23.001 amu
D
21.98768 amu
9
Problem
Problem
Three isotopic forms of potassium exist: 39K, 40K and 41K. Potassium has an atomic mass of 39.0983 amu. Potassium-40 has an isotopic mass of 39.9640 amu and natural abundance of 0.0117%. Potassium-41 has an isotopic mass of 40.9618 amu and natural abundance of 6.7302%. What is the isotopic mass of Potassium-39?