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Silicon exists as three isotopes: silicon-28, silicon-29, and silicon-30.
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Their isotopic abundances are 92 percent, five percent, and three percent, respectively.
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What is the relative atomic mass of silicon to one decimal place?
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Before we set up the equation to calculate the relative atomic mass of silicon, let’s go over some key terms in this question.
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This question is talking about isotopes, which are versions of an element whose atoms have the same number of protons but different numbers of neutrons.
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The three isotopes that we’re dealing with in this problem are silicon-28, silicon-29, and silicon-30.
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The number following silicon refers to the mass number of that isotope, which is the number of protons and the number of neutrons in the nucleus of atoms of that isotope added together.
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Since silicon is element number 14 on the periodic table, we know that atoms of each of these three isotopes have 14 protons.
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Where they differ is in their number of neutrons.
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Since a proton and a neutron each weigh one unified atomic mass unit, adding the number of protons and the number of neutrons in an atom of an isotope gives us the mass number of the isotope, which can be found in its name.
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Atoms of silicon-28, silicon-29, and silicon-30 have 14, 15, and 16 neutrons, respectively.
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The next key term, isotopic abundance, refers to the relative number of atoms of each isotope you will find in a sample of the element.
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The numbers in this problem more specifically mean that in a sample of silicon, 92 percent of the atoms will be silicon-28 atoms, five percent of the atoms will be silicon-29 atoms, and three percent of the atoms will be silicon-30 atoms.
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The last key term, the relative atomic mass of an element, refers to the average mass of an atom in a sample of that element.
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As we’ve already discussed, in a sample of an element, the different isotopes will have different abundances and atoms of those isotopes will have different masses.
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In order to calculate the relative atomic mass of an element, we need to set up an equation that incorporates the abundances of its different isotopes as well as their mass numbers.
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Our formula looks like this.
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The relative atomic mass equals the percent isotopic abundance of isotope one times the mass number of isotope one plus the percent isotopic abundance of isotope two times the mass number of isotope two plus the percent isotopic abundance of isotope three times the mass number of isotope three.
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Note that this general formula will work for different numbers of isotopes as well.
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In other problems, if there’s only, say, two isotopes, we can simply omit the percent isotopic abundance and the mass of isotope three.
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If there are four isotopes present, we can simply include an additional isotopic abundance and mass of isotope for the fourth isotope.
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As long as the percent isotopic abundances of the different isotopes add up to 100 percent, we can incorporate any number of them into the formula.
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We can insert the numbers from this problem as such: 92 percent times 28 plus five percent times 29 plus three percent times 30.
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We know that we’re incorporating all of the isotopes of silicon because our percentages add up to 100 percent in total.
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As we carry out the calculations, it’s important to remember that three percent as a decimal is 0.03.
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It is a common error for students at this point in the problem to accidentally write three percent as 0.3.
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However, that is the decimal for 30 percent.
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If we add up 25.76, 1.45, and 0.9, we get 28.11.
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Rounded to one decimal place, as the question indicates, our final answer becomes 28.1.
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Interestingly, relative atomic mass does not take a unit.
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So our final answer is simply 28.1.
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Note that of the three mass numbers of the isotopes, 28, 29, and 30, our final answer, 28.1, is closest to 28, the mass number of the isotope with the highest abundance.
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It makes sense that since we’re finding the average mass of an atom in a sample of many atoms, the more silicon-28 atoms there are, the closer to 28 the average is going to be.
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This information can be valuable to keep in mind when checking your answer.
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If your final answer ends up being closer to the mass of an isotope with a relatively small abundance, you may have made a mistake along the way.
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In addition, it is not possible for the average massive of an atom in the sample to be lighter than the lightest atom or heavier than the heaviest atom.
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For this problem, if our answer were below 28 or above 30, we would know that we had made an error.
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When checking our work for this problem, we should ensure that we have properly paired each isotope’s percent abundance with the proper mass number instead of, say, taking the abundance of one isotope and multiplying it by the mass number of a different isotope.
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If we use a calculator to answer this question, we should also ensure that we’re using the proper decimal representations of the percentages indicated.
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For example, the decimal representation of three percent is 0.03.
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In addition, the relative atomic mass of an element is listed on the periodic table.
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Here we can see that the relative atomic mass of silicon is 28.085.
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That number is slightly different than the 28.1 or, unrounded, 28.11 that we calculated in this problem.
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This difference is due to the fact that we’ve used rounded whole-number percentages for our isotopic abundances.
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In addition, we’ve used the mass number as a simplification of the mass of an atom of each isotope.
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If we used more precise numbers with more decimal places, we would see that the actual numbers were slightly higher or slightly lower than the whole numbers listed here.
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As a result, our final answer would more closely match the number found on the periodic table.
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But using the rounded numbers that we have here, we can say that the relative atomic mass of silicon, to one decimal place, is 28.1.