CHEM 210 1st Edition Lecture 7 Outline of Last Lecture I. Dalton’s Atomic TheoryA. Its relatedness to Mass Conservation, Definite Composition and Multiple Proportions II. JJ Thompson and the Cathode Ray TubeIII. Robert Millikan and the Oil Drop ExperimentIV. Ernest Rutherford and the Gold Foil ExperimentV. The Atom TodayA. Isotopes Outline of Current Lecture VI. Calculating Components of IsotopesVII. Calculating Atomic Mass of an ElementVIII. Ionic CompoundsCurrent LectureTo recap from previous lectures, isotopes are elements that have the same number of protons but a different number of neutrons, which results in a difference in mass. There is a way to calculate the amount of neutrons, protons and electrons in different isotopes. For example, if you have silicon 28, silicon 29, and silicon 30, you would first look to the periodic table to locate silicon (Si) and find its number of protons. Silicon has 14 protons and that does not change for any of the three isotopes. When you change the atomic number of an element, you are changing the identity of that element which therefore makes it a different element. Thenumber of protons is always the same, which in these 3 cases, is 14. To calculate the number of neutrons, you simply take the mass and subtract the number of protons because the mass of anelements is the protons and neutrons added together. Silicon 28 has 14 neutrons (28-14=14), silicon 29 has 15 neutrons (29-14=15), and silicon 30 has 16 neutrons (30-14=16). Finally, the number of electrons will always equal the number of protons in order for the element to remainbalanced. In this case, the number of electrons for all three silicon isotopes is 14.To calculate the Atomic Mass of an element, there are a few steps you have to take. If the mass of the element is given, as well as the abundance, you can take the abundance and divide by 100. (It takes 100 percent for it to be whole). Then you multiply by the mass and add both of those values together.These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.For example, you have two isotopes, silver 107 and silver 108. The problem mentions that the mass (amu) for silver 107 is 106.90509, and the mass of silver 108 is 108.90476. The problem also gives you the abundance (%) of silver 107, which is 51.84 and the abundance for silver 108, which is 48.6. You would first take each abundance percentage and divide by 100. Silver 107 = 51.84 / 100 = 0.5184Silver 108 = 48.16 / 100 = 0.4816Then you multiply each abundance value you found and multiply it by the mass.Silver 107 = 0.5184 X 106.905 = 55.42Silver 108 = 0.4816X 108.90476 = 52.45Add the two values together55.42 + 52.45 = 107.87Ionic compounds are elements that essentially form ions in order to make compounds. This process is done so by transferring electrons from the atoms of one element to those of another. Ionic compounds are formed between metals and nonmetals. To name ionic compounds, you first begin with the cation and end with the anion. For example, NaCl is an ionic compound formed by Na (metal) and Cl (nonmetal) transferring electrons to form sodium chloride. Sodium (Na) donates one of its electrons to Chlorine (Cl). Chlorine wants to gain an electron to achieve the octet rule, while sodium wants to lose an electron to satisfy its stability and the octet rule. Factors that influence the strength of ionic bonding are charge and size. As the charge increases, the attraction increases as well. On the other hand, as the size gets bigger, the attraction between the ionic bonds
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