CHEM 109 Week 3
CHEM 109 Week 3 CHEM 109
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This 3 page Class Notes was uploaded by Hannah Rapp on Friday September 9, 2016. The Class Notes belongs to CHEM 109 at University of Nebraska Lincoln taught by Jason Kautz in Fall 2016. Since its upload, it has received 12 views. For similar materials see General Chemistry 1 in Chemistry at University of Nebraska Lincoln.
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Date Created: 09/09/16
MAKE SURE TO STUDY INDEPENDENTLY BEFORE EXAM • Binary covalent compounds • ionic compounds ionic compounds with transition metals • • ionic compounds with polyatomic ions • hydrates • acids • organic function groups alcohol • • OH group attached (hydroxyl) • CH 3H -2H ethanol • CH3-CH-CH 3 isopropyl alcohol (rubbing alcohol) OH •aldehydes • O C H • CH 3C-H acetaldehyde O • H-C-H formaldehyde O •carboxylic acid COOH • • C=O is carbonyl • O-H is hydroxyl O C OH •amine • NH 2roup, NH group, or N CHAPTER 3 Molecular and Formula Masses • Molecular mass is the mass in amu of an individual molecule • multiply atomic mass for each element by the # of atoms and then the total masses • molar mass = atomic mass found on periodic table • molecular mass of H 2 2 *(atomic mass of H) + 1 *(atomic mass of O)= 2(1.008) + 1(15.999) = 18.02 amu Percent Composition • a list of the percent by mass of each element in a compound percent mass = n x atomic mass x 100 • molecular formula mass • 1 mol (mole) is equal to the number of 12C atoms in exactly 12 g of 12C • 1 mole= Avagadro’s Number (6.022 x 10 ) 23 • How many atoms are in 2 moles of Carbon? 2 mol C 6.022 x 10 23 = 1.2 x 1024 1 mol C • How many moles in 3.6 x 10 C atoms? 24 3.6 x 10 atoms 1 mol C = 6 mol 6.022 x 10 23atoms 12 • Molar mass - 12g of C = 1 mol • Mass Moles # of atoms divide by multiply by Molar Mass Avagadro’s # How many atoms of Oxygen are in 47.6 grams of Al (CO 2 ? 3 3 1. Find the mass of each individual element. 2 Al x 26.982 53.964 g 3 C x 12.011 36.033 g 9 O x 15.999 143.991 g 233.988 g Do stoichiometry to ﬁnd the number of atoms. 2. 47.6 g 1 mol 6.022 x 10 23 9 = 1.10 x 10 atoms 233.988 g 1 mol Types of Formulas • Molecular- gives exact number of atoms in a molecule • Empirical- gives simplest ratio of atoms in a compound Compound Molecular Empirical Hydrogen peroxide H2O 2 HO benzene C6H 6 CH water H O H O 2 2 Combustion Analysis • experimental determination of an empirical formula Review of Combustion • • C x +yO 2 CO +2H O 2 • C x Oy+2O 2 CO +2H O 2 • Remember to balance equations Chemical Equations Uses chemical symbols to denote what occurs in a chemical reaction • NH 3 HCl NH C4 ammonia and hydrogen chloride create ammonium chloride • Each chemical to the left of the arrow is called a reactant. • reactants are the substances consumed in the course of the chemical reaction • Each chemical to the right of the arrow is called a product. • products are the substances created during the course of the chemical reaction • Labels are used to indicate physical state • ( g ) - gas ( l ) - liquid • • ( s ) - solid • ( aq ) - aqueous (dissolved in water) • Chemical equations MUST be balanced so the Law of Conservation of Mass applies • balancing equations is achieved by stoichiometric coefﬁcients to the left of the formula • ratio is always conserved • To balance equations: 1) change coefﬁcients of compounds before an element • • 2) treat polyatomic ions as a unit • 3) count atoms/ polyatomic ions carefully (DOUBLE CHECK YOUR WORK) ORIGINAL EQUATION Ba(OH) + H PO3 4 Ba 3PO )4 2HOH BALANCED EQUATION 3Ba(OH) + 2H PO 3 4 Ba 3PO )4 26HOH
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