Atomic Mass - Online Tutor, Practice Problems & Exam Prep

Atomic mass is the weighted average mass of an element's isotopes, reflecting the masses of protons, neutrons, and electrons. It's typically found on the periodic table below the element's symbol, like hydrogen's 1.008. These values aren't whole numbers due to averaging different isotopic masses. Atomic mass can be expressed in grams per mole or atomic mass units (amu), with 1 amu equaling:

$1.66\times {10}^{-27}\mathrm{kg}$

To calculate atomic mass without a periodic table, use the atomic mass formula, which multiplies each isotope's mass by its fractional abundance (the percent abundance divided by 100). When only two isotopes are involved and no fractional abundances are given, their sum equals one since they represent 100% of the element's natural abundance. Understanding atomic mass is crucial for grasping the composition and characteristics of elements.

Atomic Mass

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 for a given 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 element symbols are related to names. But for now realize that is hydrogen. This number below it, this 1.008 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 of an element is an average of all of its isotopes. So hydrogen has several different isotopes, and from those isotopes we're taking the average mass. That's why we do not have whole numbers. Now we can say here that 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 Daltons. Now recall that an atomic mass unit itself is equal to 1.66×10-27 kg. So just realize when it comes to the periodic table, we have the element 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 atomic number uses the variable Z. 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.

Atomic Mass Example 1

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 in 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 BA again. Later we'll learn about how the names are attached to the element symbol. BA is not in the first column here, it's in the second column. So this cannot be a choice. Then we're going to say next that we have Al 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 CS. It's just Cessium. Here it is right here. It's in the first column. It's pretty low down there. It's 132.91 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 Fr. Notice that in the bottom rows 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 C looks like it's our best choice. If we look at D, we have Li which is up here, not higher in mass, not greater in atomic mass and then we have NA which is right here. So it looks like C 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.

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)

Atomic Mass

Now we can look up the atomic masses of 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 to the atomic mass of an element can be calculated if you know the isotopic masses and percent abundances of that element. We're going to say here that your isotopic masses are just the masses for all the isotopes of a given element. The percent abundances are just the percentages available for each of the isotopes of a given element. Sometimes these percent abundances are also called natural abundances.

Now, associated with R percent or natural abundances, we have R fractional abundances. It represents the percent abundance of an isotope divided by 100. So remember, when you have a percentage form, such as 25%, to get its fractional form you would divide it by 100. That's what's going on here. We have our percent abundances. We can divide them by 100 and that'll 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 its fractional abundance. Plus I just took mass two times its fractional abundances. Now realize that some elements have more than two isotopes, so you just continue on with the formula. So you'd say plus isotope mass, three times its fractional abundance, and so on and so forth.

So just realize that we're going to utilize this atomic mass formula in order to calculate the atomic mass of an element when we don't have a periodic table handy.

Atomic Mass Example 2

Hey, everyone. So in this example question, it says, calculate the atomic mass of lithium if the isotopic masses at percent abundances are given for its 2 naturally occurring isotopes. So here, we're given the fact that lithium comes in 2 different isotopic forms, lithium 6 and lithium 7. These numbers here represent their isotopic masses, and then these are their percent or natural abundances. Now here, we're going to say for step 1, if you are given percent or natural abundances, then divide them by 100 in order to isolate 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 2, 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 6. So that's 6.015122AMU 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 7.016004AMU times its fractional abundance, which is 0.9241. When we punch this into our calculators, we're gonna get approximately 6.940037056 AMU, 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 amu. 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 2 elements or 2 isotopes for lithium. The average mass or atomic mass is this 6.941 that you see on the periodic table.

Atomic Mass

Now how do we calculate atomic mass when no fractional abundances are given? Well remember adding U, the fractional abundances of all isotopes for an element gives a total of 1 because remember all of them together constitute 100% for that particular element and dividing it by 100 gives us its fractional abundance and 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 going to have a question where you have to calculate the fractional abundances of all isotopes, it'll only be for two isotopes, not three or more.

Atomic Mass Example 3

Here it states that Iron possesses 2 naturally occurring isotopic forms. So we have Iron 54 with an isotopic mass of 53.939615 atomic mass units and Iron 56 with an isotopic mass of 55.934942 atomic mass units. What is the percent abundance of Iron 56? Alright. So here, the way we approach this question is:

Step 1: We make the first isotope's fractional abundance \( x \). And remember, together, they add up to 100% of all the iron that exists in the universe. That's its natural percent abundance. If we were to divide this by 100, we get 1. So we know that the first isotope is \( x \), so the second isotope must be \( 1 - x \), which is equivalent to 100% - \( x \) here. And we're just changing our percent abundance into our fractional abundance. That's where the \( 1 - x \) comes from.

Step 2: Now we're going to plug in the information that we know into the atomic mass formula. We know the atomic mass of iron based on the periodic table, and that's 55.845 atomic mass units. This is going to equal the isotopic mass of our first isotope, iron 54, which is 53.939615 AMU. We don't know its fractional, so it's \( x \). And plus the second isotope, 55.934942 atomic mass units, which would be \( 1 - x \). So what we're doing here is plugging your given variables into the atomic mass formula in order to isolate our missing variable.

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 the 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 = 53.939615 x + 55.934942 - 55.934942 x

This number has an \( x \) variable. Combine them together:

55.845 = 53.939615 x - 1.995327 x + 55.934942

We need to isolate our \( x \) variable. Subtract \( 55.934942 \) from both sides. We’ll get:

- 0.089097 = - 1.995327 x

Divide both sides to isolate the \( x \) by \( -1.995327 \). We'll get \( x = 0.04508 \). 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, giving 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 need to find the fractional abundance of iron 56:

1 - 0.04508

This is \( 0.95492 \). To convert this fractional abundance to a percent abundance, multiply by 100:

This gives me 95.492%, which I'll just round to 95.5%. So this is a rough approximation of what the percent abundance will be of iron based on these isotopic masses that I've given to you. Alright, so working it out, we get this final answer.

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

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?

• A. 106.905 amu
• B. 38.9637 amu
• C. 106.903 amu

Here’s what students ask on this topic:

What is atomic mass?

Atomic mass, also known as atomic weight, is the average mass of atoms of an element, calculated using the relative abundance of isotopes in a naturally-occurring element. It is expressed in atomic mass units (amu), where one atomic mass unit is defined as one-twelfth of the mass of a carbon-12 atom, which has six protons and six neutrons.

Since an element can have several isotopes—atoms with the same number of protons but different numbers of neutrons—the atomic mass is not a whole number. It takes into account the masses of each isotope and their relative abundance. For example, chlorine has two common isotopes, chlorine-35 and chlorine-37. The atomic mass of chlorine is a weighted average of these isotopes, reflecting the fact that chlorine-35 is more abundant than chlorine-37. This concept is crucial for understanding the composition of elements and their isotopes, as well as for calculations in chemistry involving the amount of substance, like stoichiometry.

How do you calculate atomic mass?

Atomic mass, often referred to as atomic weight, is calculated by taking into account the masses of each isotope of an element and their relative abundance. Each isotope has a certain natural percent abundance and a specific mass number. To calculate the atomic mass of an element, you multiply the mass of each isotope by its relative abundance (expressed as a decimal), then sum these values.

Here's a step-by-step guide:

1. Convert the percentage of each isotope into decimal form by dividing by 100.
2. Multiply the decimal abundance of each isotope by its isotope mass (the number of protons and neutrons in the nucleus).
3. Add the results from step 2 for all isotopes of the element.

The final sum will give you the atomic mass of the element. This value is usually close to the mass number of the most abundant isotope but takes into account the presence of other isotopes and their respective masses. Atomic mass is typically reported in atomic mass units (amu), where 1 amu is defined as 1/12 the mass of a carbon-12 atom.

How do you calculate average atomic mass?

To calculate the average atomic mass of an element, you'll need to consider all the naturally occurring isotopes of that element and their respective abundances. Here's how you do it step by step:

1. List Isotopes: Start by listing each isotope of the element.
2. Find Mass of Each Isotope: Note the atomic mass of each isotope, usually given in atomic mass units (amu).
3. Determine Natural Abundance: Find the natural abundance of each isotope, often expressed as a percentage.
4. Convert Percentages to Decimals: Convert the percentage of each isotope's natural abundance into a decimal by dividing by 100.
5. Multiply Mass and Abundance: For each isotope, multiply the atomic mass by the decimal abundance.
6. Sum the Products: Add up the products from the previous step for all isotopes.

The result will be the average atomic mass of the element. This value is the weighted average based on the abundance of each isotope, which reflects the mass that would typically be found for an atom of that element on Earth.

How do you calculate the atomic mass of isotopes?

To calculate the atomic mass of isotopes, you need to consider the mass of each isotope and its relative abundance. Here's a step-by-step process:

1. Identify the Isotopes: Determine which isotopes of the element you're dealing with. For example, chlorine has two stable isotopes: chlorine-35 and chlorine-37.
2. Find the Mass of Each Isotope: Look up the atomic mass of each isotope, usually given in atomic mass units (amu). For chlorine-35, it's approximately 34.969 amu, and for chlorine-37, it's about 36.966 amu.
3. Determine the Relative Abundance: This is the percentage of each isotope found in nature. For chlorine, about 75.77% is chlorine-35 and 24.23% is chlorine-37.
4. Calculate the Weighted Average: Multiply the atomic mass of each isotope by its relative abundance (expressed as a decimal), then add these values together.

For chlorine:

$(34.969\text{amu}\times 0.7577)+(36.966\text{amu}\times 0.2423)=26.4957\text{amu}+8.9571\text{amu}=35.4528\text{amu}$

So, the atomic mass of chlorine is approximately 35.4528 amu.

What is the relative atomic mass?

Relative atomic mass, often abbreviated as A_r, is a dimensionless quantity (meaning it has no unit) that represents the average mass of atoms of an element, measured in atomic mass units (amu), compared to one-twelfth of the mass of an atom of carbon-12. This average takes into account the different masses of the isotopes of the element and their relative abundance in nature.

To calculate the relative atomic mass of an element, you multiply the mass of each isotope by its relative natural abundance (expressed as a decimal), and then sum these values. For example, chlorine has two main isotopes, ³⁵Cl and ³⁷Cl, with relative abundances of about 75% and 25% respectively. The relative atomic mass of chlorine would be calculated as follows:

$(35\text{amu}\bullet 0.75)+(37\text{amu}\bullet 0.25)=35.5\text{amu}$

Therefore, the relative atomic mass of chlorine is approximately 35.5. This value is often found on the periodic table and is used in chemical calculations involving moles and molar mass.