General Chemistry 1110-001 Mohseni
General Chemistry 1110-001 Mohseni CHEM 1110 - 01
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This 7 page Class Notes was uploaded by Leah Washburn on Sunday September 18, 2016. The Class Notes belongs to CHEM 1110 - 01 at East Tennessee State University taught by Mohseni in Fall 2016. Since its upload, it has received 63 views. For similar materials see Gen Chem in Chemistry at East Tennessee State University.
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Date Created: 09/18/16
1 Chapter 0 A Very Brief History of Chemistry Objectives: Atomic Theory Subatomic Particles Isotopes -Atoms: Smallest representative sample of an element and a compound as a substance that contains two or more elements. -Scientific support for the concept of atoms awaited the discovery of the law of definite proportions and the law of conservation of mass. Law of Definite proportion: Atoms are combined in a definite proportion. When a compound is formed, ……atoms……. always combine in the same proportions by mass. Example: Hydrogen and oxygen react together to form…… H O……… 2 Law of Conservation of Mass: Mass is neither lost nor created during a chemical rxn. Example: Place hydrogen and oxygen into a sealed container and initiate the reaction to form water. The mass of water and whatever hydrogen and oxygen is left over is the same as the mass of the hydrogen and oxygen we started up with. -The laws mentioned above served as the experimental foundation for the atomic theory. Dalton’s Atomic Theory 2 1. Matter consists of tiny particles called atoms. 2. In any sample of a pure element all the atoms are identical in mass and other properties. 3. The atoms of different elements differ in mass and other properties. 4. When atoms of different elements combine to form compounds, new and more complex particles form. However, in a given compound the constituent atoms are always present in the same fixed numerical ratio. 5. Atoms are indestructible. Atoms are re-arranged in a chemical reaction. Internal Structure of the Atom Subatomic particles Atomic Symbols: Backbone of chemical formula and chemical rxns. Law of Definite Proportions Different samples of the same compound always retain their constituent elements in the same proportion by mass. Ex. Carbon monoxide has chemical formula of CO regardless how it is synthesized. Law of Multiple Proportions: If two elements can combine to form more than one compound, the masses of one element that combine w/ a fixed mass of the other element are in ratios of small whole numbers. Ex. CO and CO 2 3 If 1.000 g C is mixed and reacted w/ oxygen, two compounds are formed: one contains 1.332 g O, and one with 2.664 g O. What is the ratio of O between the two compounds? CO CO 2 Law of Conservation of Mass: Matter can be neither created nor destroyed. Atoms: Dalton postulated that atoms were indivisible particles. But, we now know that an atom consists of a nucleus (positively charged), and electron. An electron is a very light with negative charge spinning around the nucleus. Electrons Negatively28arged particles; mass of 9.10 x 10 g. Protons Positively charged particles; mass of 1.67262 x 10 g. Protons are much heavier than electrons. Neutrons Do not have any charge24 Mass of 1.67493 x 10 g Where are these particles located? Most of the mass of the atom (99.95%) is concentrated in a positively charged center. Let’s call this center “nucleus”. 4 Negatively charged particles, electrons, move around the nucleus. Nucleus has a diameter of about 0.0010 pm. Atomic diameter is about 100 pm. Name the two heaviest particles in an atom. Protons and neutrons Protons and Neutrons are the most massive particles in an atom. Yet the nucleus has a very small volume. So the nucleus is very dense. If it were possible to have a nucleus w/ a volume of an average grape, that nucleus would have a mass greater than 9 million metric tons. Atomic Number, Mass Number, and Isotopes: A X Z Atomic Number (Z): The atoms of a particular element have the same number of protons in the nucleus. This number is called atomic number and shown as Z. A is the atomic mass or mass number. In a neutral atom, the number of protons is the same as the number of electrons outside of the nucleus. Problem: How many protons and electrons does an element with Z=11 have? 11 electrons and protons Problem: Calculate the number of protons if an atom consists of 20 electrons. 5 20 protons Problem: Answer the following questions regarding 23Na 11 Name the element Sodium How many electrons and protons does Na have? 11 electrons and protons Calculate the number of neutrons in the atom. Problems: 1. The charge on one electron is 1 2. What subatomic particle has a charge opposite to that of the electron? proton 3. If a hydrogen atom contains one electron, it must also contain one proton if it is to be electrically neutral. Atomic Mass Units (amu): To establish a uniform mass scale for atoms, C12 is used as a reference. Atomic weight is based on C12 mass. The mass of an atom is compared to the C12 isotope, which is assigned a mass of exactly 12 atomic mass units. One amu is a mass unit equal to exactly 1/12 the mass of a C12 atom. 1 atom C has a mass of 12 u. Isotopes: Atoms with the same number of electrons and protons but a different number of neutrons are called isotopes. 6 Table 0.4 (Page 17) A naturally occurring element may be a mixture of isotopes, each isotope having its own characteristic mass. Ex: 1H 2H 3H 1p 1e 1p 1e 1p 1e 0n 1n 2n Problem: Determine the number of electrons, protons, and neutrons in 23 24 Na Na 11e 11p 12n 11p 11e 13n 23 23 Problem: 11a and 12Mg (are / are not) isotopes of the same element. 235 238 Problem: 92U and 92U (are / are not) isotopes of the same element. How is the average atomic mass obtained? Avg atomic mass = (Isotope 1)(Isotope 1 abundance)+ (Isotope 2)(Isotope 2 abundance)+….. Ex: Calculate the avg atomic mass of B if there are two isotopes of B have found, i.e., B10.0129 at 19.8% and B11.0093 at 80.2%. (10.0129) (19.8) + (11.0093) (80.2) = 1.9825542 + 8.8294586 = 10.8120128 Example: Calculate the average mass of chlorine from the data given below. Assume there are two isotopes for Cl. 7 Isotopes Mass % abundance Cl-35 34.9689 75.77 Cl-37 36.9659 ------ (34.9689) (75.77) + (36.9659) (x)= X= 26.4959
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