Unit 1: Atomic Structures and Properties Review Guide

Neeraj Kondragunta 1st Unit 1: Atomic Structures and Properties Review Guide Topic 1.1: Moles and Molar Mass Calculate quantities of a substance or its relative number of particles using dimensional analysis and the mole concept. If they give you a mass of an element like 48g of oxygen, and ask you to find the number of particles, first, you will find the amount of moles. To do this, you would divide the mass by the molar mass of Oxygen, which is 16 g/mol. This would give you 3 moles of Oxygen. After this, you would multiply by Avagadro’s number, which is 6.022 * 10^23 mol^-1. The number of particles in 48g of oxygen is 1.8066* 10 ^24 particles. Topic 1.2: Mass Spectroscopy of Elements Explain the quantitative relationship between the mass spectrum of an element and the masses of the element’s isotopes. The quantitative relationship between the mass spectrum of an element and its element’s isotopes is that the mass spectrum shows the relative abundance of each isotope. The bars each show the abundance of each isotope and how common they are in the natural world. With this, you can find the average mass of an element. Topic 1.3: Elemental Composition of Pure Substances Explain the quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance. The quantitative relationship between the elemental composition by mass and the empirical formula of a pure substance because you need to know the mass of each element in the compound to determine the amount of each element in the compound. So, if you were given a sample that contained 11.11% H and 88.89% O you would assume that the sample is out of 100 grams so that means you would have 11.11 grams of Hydrogen and 88.89 grams of oxygen. Then you would divide each of these numbers by their respective masses to find the number of moles per each element. This would give you roughly 11.022 moles of Hydrogen and 5.555625 moles of Oxygen. Then you would divide by the lowest mole number. So, 11.022/5.555625 and 5.555625/5.555625. This would give you 1.98, which you can round up because its less than a tenth off, and 1. So the empirical formula would be H2O. Topic 1.4: Composition of Mixtures Explain the quantitative relationship between elemental composition by mass and the composition of substances in a mixture. The relationship between elemental composition by mass and the composition of substances in a mixture involves combining the contributions of each part of the mixture based on its proportion and its elements. For example, if you have a mixture that's 40% substance X and 60% substance Y by mass, and substance X is 30% carbon while substance Y is 10% carbon, you can find the overall carbon content of the mixture by calculating how much carbon each substance contributes. For substance A, which makes up 40% of the mixture, the carbon contribution can be calculated by multiplying 0.30 by 0.40, which results in 12%. For substance B, which is 60% of the mixture, the carbon contribution can be calculated by 0.10 * 0.60, which is 6%. Adding these together, the mixture has a total carbon content of 18%. If the mixture has a mass 180 grams, then you can multiply that by 0.18 to find the amount of carbon, in grams in the mixture, which is 32.4 grams of carbon. This shows how the mixture's elemental composition by mass can be used to calculate how much of a substance is in a mixture. Topic 1.5: Atomic Structure and Electron Configuration Represent the electron configuration of an element or ions of an element using the Aufbau principle Aufbau principle is basically that the electrons are added to the lowest subshell first and then build up. So, the electron configuration for C would be: ______ ______ ______ ______ ______ 1s 2s 2p 2p 2p Topic 1.6: Photoelectron Spectroscopy Explain the relationship between the photoelectron spectrum of an atom or ion and the electron configuration of the species and the interactions between the electron and the nucleus. The photoelectron spectrum of an atom or ion relates to the electron configuration and relationship between electrons and the nucleus because it shows the amount of energy it takes to remove an electron with each bar being a subshell. The ones that take the highest amount of energy are the innermost 1s electron shell, and you can figure out the rest from this. This shows the interactions between the electron and the nucleus because the electrons with the highest binding energy have the greatest coulombic attraction, which means that they are the closest to the nucleus and have a greater effective nuclear charge because of this. Topic 1.7: Periodic Trends Explain the relationship between trends in atomic properties of elements and electronic structure and periodicity. The relationship between trends in atomic properties of elements and electron structure and periodicity usually tend to do with the electrons distance from the nucleus. As you move across a period in the periodic table, the atomic radius decreases due to an increasing nuclear charge, which pulls electrons closer to the nucleus, while ionization energy and electronegativity increase as the coulombic attraction increases as more protons are added while the electrons are being added to the same subshell. Moving down a group, the atomic radius increases due to the addition of electron shells, leading to a decrease in ionization energy and electronegativity as the outer electrons are farther from the nucleus and more easily removed due to more shielding and less coulombic attraction. These trends reflect the periodic nature of the elements, where similar valence electron amounts have similar properties. Topic 1.8: Valence Electrons and Ionic Compounds Explain the relationship between trends in the reactivity of elements and periodicity. The reactivity of elements is related to their position on the periodic table because reactivity depends on how close the atom is to having a full valence shell. In general, for metals, which are found on the left side of the periodic table, reactivity increases as you move down a group. This is because the atomic radius increases, making it easier for these elements to lose their outermost electrons and form positive ions. For non-metals, found on the right side of the periodic table, reactivity decreases as you move down a group. This is because the increased atomic radius makes it more difficult for the nucleus to attract additional electrons. Across a period, reactivity decreases for metals and increases for non-metals, reflecting the shift from elements that tend to lose electrons to those that tend to gain them. This periodic difference in reactivity is a result of the changes in atomic structure, particularly the ease with which atoms can lose or gain electrons to achieve a stable electronic configuration.