Use the mass spectrum of lead to estimate the atomic mass of lead. Estimate the mass and percent intensity values from the graph to three significant figures.

Video transcript

Hey everyone in this example, we need to find the atomic mass of an atom of germanium based on the mass spectrum in the graph below, We're going to be using this graph to determine approximate mass and percent intensity values. And our final answer for our atomic mass of germanium should be 2 to 6 figs. So we want to go ahead and recall that to find the atomic mass of and neutral atom. We would use the average of the masses of the isotopes of that atom. And what we would do to find the sum of the average of the masses of the isotopes is to take the average peak intensity for each isotope and we're going to multiply that by the atomic mass of each isotope. So we should recall that to find average peak intensity. We're going to take the peak intensity of the given isotope in the graph for example, we have a peak intensity of 56 for Germanium 70. So we would take the peak intensity and divide that by the total peak intensity. And this is what we will start out calculating first is our total peak intensity. So to find total peak intensity, we're going to add up all the peak intensities given in the graph 1st 56 for germanium 70. Then we have 75 for germanium 72. So let's go ahead and start adding that. We have 56 plus 75 plus 21 for germanium 73 plus the peak intensity of 100 for germanium 74. And then for germanium 76 we have the peak intensity of 21 again. So what we're going to do is add these up and this is going to give us a sum equal to a value of 273 as our total peak intensity calculation value. And this is what we will use to find the average atomic mass for each of the isotopes of Germanium. So let's start out with again, Germanium 70. So we should first take the peak intensity given for Germanium 70 in the graph which is given as 56. So we have 56 in our numerator. And then in our denominator we want to plug in that total peak intensity which above we set is to 73 for all of our isotopes of Germanium. Now we're going to multiply by the atomic mass of this isotope Germanium 70. Which is actually if we recall given in the name of our isotope. So we should recall that this would have an atomic mass of 70 am use. So when we multiply out are enumerators and divide by the denominator we're going to get a value equal to 14.35 90 Amos. And now we can continue on with our next isotope Germanium 72. So in our numerator we would plug in its average peak intensity. So given in the graph for Germanium 72 we have a peak intensity of 75. So we have 75 in our numerator to 73 in our denominator multiplying by its atomic mass being 72 A. M. U. S. Given in the name of this isotope. And this is going to give us a value equal to 19.78 AM. Use Moving on to our next isotope. We have Germanium 73 In our numerator. The average peak intensity given in the graph for Germanium 73 is 21. We divide that by total peak intensity to 73. We multiply by its atomic mass given in the name being 73 a. m. us. And we're going to get a value equal to 5.61 54 am use. So moving on to our next isotope we have Germanium 74. Its peak intensity in the graph if you recall was 100%. So that was right here for Germanium 74. Sorry that's a four there. So we have 100%. In our numerator we have 2 73 are total peak intensity and the denominator we multiply by our atomic mass being 74 AM use and we're going to get a value here equal to 27.10 62 A. M. Use. And lastly we have Germanium 76. And so in the graph we're given a peak intensity for Germanium 76 being 21%. So we have 21 in our numerator 2 73. In our denominator we multiply by the atomic mass being 76 AM use. And this is going to give us a value equal to 5. AM use. And now what we want to do is take the average of these atomic masses of all of our isotopes by adding them up together. And this gives us a total equal to 72.7 a. M. U. S. As the average or the total of our average atomic masses of each of our isotopes. Now we want our final answer to be rounded to two sig figs. So we're going to round this to about 73 AM use for and adam of germanium. So this would be the atomic mass to complete this example as our final answer. So I hope that everything I explained was clear. If you have any questions, please leave them down below. Otherwise I will see everyone in the next practice video.