Hello, everyone. And thanks for joining me, Miss Hallstrom, For a brief discussion on atomic mass First, atomic math is the weighted average of all naturally occurring occurring isotopes. Here's the equation that we would use to calculate atomic mass. It's equal to the mass of the first isotope times, the relative abundance of the first isotope. We add to that the mass of the second isotope times, the relative abundance of the second isotope. And we do this for each isotope for the element in question. Relative abundance. Okay, it is just percent. With the decimal moved to left. Let me show you if we had a number 71.42% that's our percent number. Our decimal would be 0.7 142 no units. So relative abundance is just the percent with the decimal moved back to to the left. When we add up, the some of all relative abundance is equal to one just like percent. When we add up, all the parts of a percent of equal to 100 to to the left of 100 is one. Let's do our problem for this problem. Were given the following information we have two isotopes, isotope one and isotope to but for isotope one were given a mass of 120 point 9038 AM use and were given a relative a percent abundant abundance of 57.4% which has road abundance of 0.574 Okay, I will switch colors for this isotope to we have a mass of 1 22 0.9042 atomic mass units. And we're not given our relative abundance here, but for step one, let's recall that the some clips to me that's some of relative abundance equals one. So one minus 0.574 equals zero point. Um, where my 4 to 6. So this wasn't given, but we know it. 0.4 to 6. And for step two, we're going to sub insolvent to our equation from the previous page. Okay, I'm going to abbreviate. Average atomic mass is a M. Average atomic mass is equal to the mass of isotope one times a relative abundance of isotope one plus the massive isotope to times a relative abundance of isotope to, and you're probably very capable figuring this the rest of this out on your own. But since it's the first problem, I'll do this. The product of those two numbers is 69.40 Add to that the product of these two numbers, which is 52 0.36. Add them together and I get 1 21.76 and those Aaron units of AM use. The final answer, reported to the correct number of sick gigs, is 1 21 0.8 I am use. And that's our first problem for atomic mass units. Thanks for watching.