CHEMISTRY UNIT 7 (Chemical Formulas and Compounds) Monatomic ions are ions formed from a single atom. Ex: Mg+2, FBinary compounds are compounds composed of two different elements. Ex: Mg F2; H2O Naming: To name compounds you simply put the two names together. The cation is always named first. This naming system requires that you have knowledge of ions and balancing. Ex: Fe+2 and ClIron II & Chloride becomes FeCl2 - Iron II chloride Chemist also use an older form of nomenclature which takes advantage of prefixes to indicate the number of each ion present in the compound Ex: CO2 -Carbon dioxide CO -carbon monoxide 7-2 Prefixes: mono-1 hexa-6 di-2 tri-3 tetra-4 penta-5 hepta-7 octa-8 nona-9 deca-10 Ex: carbon tetrachloride CCl4 dinitrogen monoxide N2O dinitrogen pentoxide N2O5 tetraphosphorus decasulfide P4O10 The Combining of Ions to form Compounds (newest style): Sodium chloride Na+ + ClNa+ClSodium chlorate Na+ + ClO3Na+ClO3Aluminum oxide Al+3 + O-2 Al2+3 O3-2 Calcium hydroxide Ca+2 + OHCa+2(OH)2Ammonium phosphate NH4+1 + PO4-3 (NH4+1)3PO4-3 Mercury (I) phosphate Hg2+2 + PO4-3 (Hg2+2)3(PO4-3)2 7-3 The Oxidation state of an element is represented by a signed number called the oxidation number Oxidation number rules: 1. The oxidation number of a free element is zero. 2. The oxidation number of a mono-atomic ion is equal to its charge. 3. The algebraic sum of the oxidation numbers of all the atoms in a compound is zero. A chemical reaction in which an element attains a more positive oxidation state (loss of electrons) is called oxidation A chemical reaction in which an element attains a more negative oxidation state (gain of electrons) is called reduction The substance that is reduced is called the oxidizing agent The substance that is oxidized is called the reducing agent 7-4 Ex: In the reaction between sodium and sulfur (Na0 + S0 Na2+1S-2) Describes the following: S is reduced by Na Na is oxidized by S S is the oxidizing agent Na is the reducing agent The formula mass of any molecule is the sum of the average atomic masses of all the atoms represented in its formula. (that is, add up all the numbers off the chart. The molar mass of a substance is the mass of one mole of atoms (or molecules) of that substance Ex: The molar mass of carbon is 12.011 g/mole The gram-molecular weight (G.M.W.) of a molecular substance is the mass of one mole of molecules of the substance Ex: The G.M.W. of water (H2O) is 18.0148 g/mole H-2 (1.0079) = 2.0158 + O-1 (15.999) =15.999 18.0158 g/mole Note: For all practical situations atomic mass, grammolecular weight (G.M.W.), and molar mass are equivalent. 7-5 The percent composition of a compound is the percentage by mass of each element in the compound. Calculating Percent Composition: Mass of Element % Comp. of E = x 100% G.M.W of Compound Ex: Water is made up of Hydrogen and Oxygen. So is many other compounds. What percent of water is oxygen? What percentage of water is hydrogen? We must first calculate the G.M.W. of water (H2O): H-2 (1.0079) = 2.0158 + O-1 (15.999) =15.999 18.0158 g/mole 2.0158 %H = x 100% = 12.5% 18.0158 15.999 %O = x 100% = 88.8% 18.0158 7-6 Ex: Find the percent composition for each constituent of Iron (III) hydroxide. Iron (III) hydroxide Fe+3(OH-1)3 Calculating G.M.W.: Fe-1 (55.847) = 55.847 O-3 (15.999) = 47.997 + H-3 (1.0079) = 3.0237 106.8677 g/mole 55.847 % Fe = x 100% = 52.3% 106.8677 47.997 %O = x 100% = 44.9% 106.8677 3.0237 %H = x 100% = 2.8% 106.8677 These are the percent compositions for Iron (III) hydroxide. The percent compositions for Iron (II) hydroxide are different. 7-7 An Empirical formula consists of the symbols for the elements combined in a compound, with subscripts showing the smallest whole-number ratio of the different elements in the compound Ex: Na17Cl17 is a Molecular formula NaCl is an Empirical formula Frequently when evaluating unknown chemicals it is possible to experimentally determine the percent composition of constituent elements. This data yields only the percent composition (or masses) of each constituent substance in the unknown. How can we then use this information to predict the identity (formula) of the unknown compound? How can we "Calculate the Empirical Formula? Ex: The decomposition of 11.47g of a compound yields 9.16g of copper and 2.31g of oxygen. What is the Empirical formula of the compound? Ex: A compound contains 81.7% carbon and 18.3% Hydrogen. Find the empirical formula.