yego.me
💡 Stop wasting time. Read Youtube instead of watch. Download Chrome Extension

Worked example: Identifying an element from its mass spectrum | AP Chemistry | Khan Academy


2m read
·Nov 10, 2024

So let's say that we have some mystery substance here, and we know that it's a pure element. We need to figure out what it is. Well, scientists have a method, and we go into the details or more details in other videos called mass. Sometimes it's known as mass spectrometry or mass spectroscopy.

It's a technique where you could take a sample of a substance and think about the various atomic masses of the different isotopes in that substance. That's what we have right over here. They tell us the mass spectrum for an average sample of a pure element is shown below.

So let's say it's this pure element. This is telling us that this looks like maybe—I don't know—let's call this 82 percent of our sample has an atomic mass of 88 universal atomic mass units. About this looks like about 7 percent of our sample has an atomic mass of 87 universal atomic mass units. It looks like 10 percent has an atomic mass of 86 universal atomic mass units, and it looks like about 1 percent of our sample has an atomic mass of 84 universal atomic mass units.

So from this information, we can try to estimate what the average atomic mass of this mystery element is. We could calculate it as 0.82 times 88 plus—let's call this 7—so 0.07 times 87 plus 0.1 times 86 plus unless it should add up to 100 percent. This is 89, and then this gets us to 99. So then, another 1 percent, 0.01 times 84.

And so if we were to do this calculation, this is our estimate of the average atomic mass of this element. We could type this into a calculator and get some number and then look that up on a periodic table of elements, or we could just try to estimate it.

We can see that it's going to be close to 88 because that's where the highest percentage is. When we're taking the weighted average, we have the highest weight right over there. But these other isotopes, these other versions of the element that have a different number of neutrons, which changes its atomic mass, they're going to bring the average down. So our average atomic mass is going to be a little bit less than 88.

So let's look up a periodic table of elements. What element here has an atomic mass a little bit less than 88? Well, Yttrium is 88.91, but we know it can't be that because none of the isotopes have an atomic mass above 88, so we can rule out Yttrium.

Strontium is looking pretty good—it's exactly what we predicted, a little bit less than 88. And Rubidium is a lot less than 88. So even if we did the calculation, we could feel confident we're not going to be as low as Rubidium.

So I'm feeling very confident—just eyeballing it, just estimating—this is going to be a little bit, have an average atomic mass a little bit less than 88, which tells me that this is Strontium.

More Articles

View All
Example: Transforming a discrete random variable | Random variables | AP Statistics | Khan Academy
Anush is playing a carnival game that involves shooting two free throws. The table below displays the probability distribution of ( x ), the number of shots that Anush makes in a set of two attempts, along with some summary statistics. So here’s the rand…
Tracking the Gray Wolf in Yellowstone | Explorer
The wolf is the world’s largest dog—a top predator and an iconic animal that roamed freely across North America for tens of thousands of years. But in the early 20th century, a ruthless war was waged against these cunning carnivores in an effort to stop t…
BITCOIN TO $500,000 - What You MUST Know
What’s up, Graham? It’s guys here, and I’m not gonna lie, sometimes it feels like we’re living in the golden era of the finance and investment community. Although I realize that “golden era” might not be the proper term here because we’re not talking abou…
Estimating with decimal multiplication
We are asked to estimate what is 2.7 times 4 roughly equal to. Pause this video and see if you can answer that. All right, so we could think of 2.7 times 4 as being roughly equal to, or some people might say as approximately equal to. Let’s see, 2.7, tha…
The SAT Question Everyone Got Wrong
In 1982, there was one SAT question that every single student got wrong. Here it is. In the figure above, the radius of circle A is 1⁄3 the radius of circle B. Starting from the position shown in the figure, circle A rolls around circle B. At the end of h…
2015 AP Calculus AB/BC 3a | AP Calculus AB solved exams | AP Calculus AB | Khan Academy
Johanna jogs along a straight path for (0 \leq t \leq 40). Johanna’s velocity is given by a differentiable function (v). Selected values of (v(t)), where (t) is measured in minutes and (v(t)) is measured in meters per minute, are given in the table above …