Calculating Atomic Abundance: Easy Steps for Students

Let's explore atomic mass and isotopes, starting with the fascinating world of carbon atoms!Elements can have different forms called isotopes, which have the same number of protons but different numbers of neutrons.Let's look at carbon's three naturally occurring isotopes: Carbon-12, Carbon-13, and Carbon-14.Each isotope has the same number of protons - six for carbon - but different numbers of neutrons, giving them different mass numbers.On the periodic table, we see carbon's atomic mass as 12.01 atomic mass units.This value is actually a weighted average based on the natural abundance of each isotope.Carbon-12 makes up about 98.93 percent, Carbon-13 about 1.07 percent, and Carbon-14 exists in trace amounts too small to affect the average.Understanding these concepts will help us calculate isotope abundances, which we'll explore next.The abundance formula allows us to calculate the average atomic mass of an element based on its isotopes and their relative abundances.Let's understand what each variable represents in this formula.Abundance can be expressed either as a decimal between zero and one, or as a percentage between zero and one hundred.A fundamental rule is that the abundances of all isotopes must sum to one hundred percent, or one point zero in decimal form.Now, let's see how to rearrange this equation when we need to solve for an unknown abundance.Here are some important tips to remember when working with the abundance formula.Let's solve a real problem using chlorine's isotopes.We'll let x represent the abundance of Chlorine-35, and one minus x will be the abundance of Chlorine-37.First, let's distribute the second term.Now we can rearrange the terms to group them.Combining like terms, negative thirty-six point ninety-seven x plus thirty-four point ninety-seven x equals negative two x.Subtracting thirty-six point ninety-seven from both sides gives us negative one point fifty-two equals negative two x.Finally, dividing both sides by negative two gives us x equals zero point seven six, or seventy-six percent.Therefore, Chlorine-35 has an abundance of seventy-six percent, and Chlorine-37 makes up the remaining twenty-four percent.Let's verify our answer using two important checks.First, check that the abundances sum to one hundred percent.Second, verify that these abundances give us the correct average atomic mass of thirty-five point forty-five.Now you know how to solve and verify isotope abundance problems!