Chapter 3 Notes
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This 5 page Class Notes was uploaded by Shannan Dillen on Saturday October 1, 2016. The Class Notes belongs to CHEM 1510 at Ohio University taught by Shadrick Paris in Fall 2016. Since its upload, it has received 2 views. For similar materials see Fundamentals of Chemistry I in Chemistry at Ohio University.
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Date Created: 10/01/16
Chapter 3: Stoichiometry of Formulas and Equations The Mole In a chemical reaction, MANY particles participate. Mole (mol) – amount of substance that contains the same number of “entities” as there are atoms in exactly 12 grams of carbon-12. 1 mole = 6.022 * 10 entities (Avogadro’s Number(N )) A “Mole” is “entities” as “adjective” is to “noun” 12 g carbon-12 = 6.022 * 10 carbon-12 atoms 12 g carbon-12 = 1 mole carbon-12 atoms Note: 12 amu carbon-12 = 1 carbon-12 atom Molar Mass (MM) – the mass of a substance that contains 1 mole of its “entities”, in units of g/mol (grams per mole) Molar mass is numerically equivalent to a substance’s atomic, molecular, or formula mass (in amu). Atomic mass (C) = 12.0 amu Molar mass (C) = 12.01 g/mol (12.01 g C = 1 mole of C atoms) Mass Percent – ratio of elements in a compound Mass % X = (mass of X / mass of sample containing X) * 100 Given a pure sample; mass % for each element for a particular compound will be a constant value *Mass can be any unit Determining Formulas Empirical Formula – represents smallest, whole numbered ratio of elements in the compound Ex: CH Molecular Formula – represents the actual formula for a single molecule of a covalent compound Ex: C H 6 6 Empirical Formulas Can be determined with: - Mass of each element in sample (or % mass) - Atomic mass of an element Ex: 2.82 g Na 4.35 g Cl 7.83 g O Covert to moles… (2.82 g Na)(1 mole Na / 22.99 g Na) = 0.122662 mole Na (4.35 g Cl)(1 mole Cl / 35.45 g Cl) = 0.122708 mol Cl (7.83 g O)(1 mol O / 16.00 g O) = 0.489375 mole O Divide by smallest molar value… 1 atom Na 1 atom Cl 4 atoms O NaClO 4 Molecular Formulas Molecular formula of a covalent compound is related to its empirical formula via a whole- number multiple. (empirical)n= molecular N = MM molecular / MM empirical Combustion Analysis -method of determining empirical formula of a hydrocarbon Cm n o+ O -2--combustion----> CO + H2O + 2other” Given mass of CO ,2we can determine mass of C in original sample. Given mass of H O, we can determine mass of H in original sample. 2 Third element – mass makes up remaining amount -add mass of C and mass of H and subtract from mass of original sample Ex: MM(vitamin C) = 176.12 g/mol 1.50 g CO 2 0.41 g H 2 1.000 g sample (1.50 g CO 2(1 mole C / 44.01 g CO )212.01 g C / 1 mole C) = 0.409339 g C (0.41 H O)(2 mole H / 18.02 g H O)(1.008 g H / 1 mole H) = .045869 g H 2 2 .409339 g C + .045869 g H = .455208 g 1.00g sample - .455208 g = .544792 g O (.049339 g C)(1 mol C / 12.01 g C) = .034083 mol C (.045869 g H)(1 mol H / 1.008 g H) = .045505 mol H (.544792 g O)(1 mol O / 16.00 g O) = .034050 mol O Divide all by smallest mole ratio, .034050 1 C 1.33 H 1 O Multiply all by 3 3 C 4 H 3 O Empirical: C3 4 3 176.12 g/mol / 88.06 g/mol = 2 Molecular: C 6 8 6 Chemical Equations -represent chemical reactions H 2g) + F 2g) → 2 HF (g) Reactants Products (s) = solid (l) = liquid (g) = gas (aq) = aqueous solution Equation MUST be balanced – same number of atoms of each element must be present on BOTH sides of the equation The Law of Conservation of Mass must be obeyed Translate a Statement to a Chemical Equation “Magnesium metal was burned in the presence of gaseous oxygen, forming magnesium oxide, a white powder.” 2 Mg (s) + O (g2 → 2 MgO (s) - Atoms of each element must be balanced - Atoms may only be balanced by changing to coefficient of each chemical species in the equation - The equation must be balanced to the smallest, whole number ratio of coefficients 2 N 2 (5) → 4 NO (g) 2 O (g) 2 2 C 8 18) + 25 O (g2 → 16 CO (g) +218 H O (g) 2 Stoichiometry -refers to quantitative relationships between species in a chemical reaction given by its complete and balanced chemical equation C 3 8g) + 5 O (2) → 3 CO (g) 2 4 H O (g)2 1 mol C H3=85 mol O = 3 2ol CO = 4 mol2H O 2 Limiting Reactants Reactant we have the least amount of (based on balanced equation) will: Be fully consumed by the reaction first Limit the amounts of all products that can form Dictate all stoichiometry performed in predicting amounts of products/reactants formed/needed Ex: Cl2(g) + 3 F 2g) → 2ClF (g3 .750 mol Cl 2 3.00 mol F 2 (.750 mol Cl )22 mol ClF / 3 mol Cl ) =21.50 mol ClF 3 (3.00 mol F )22 mol ClF / 3 mol F ) = 3.00 mol ClF 3 Cl2is limiting reactant Form 1.50 mol ClF 3 Reaction Yields Several factors can decrease the amount of product isolated: Poor experimental techniques Human error Side reactions (products unaccounted for) Actual Yield – actual amount of product (moles or mass) obtained experimentally from the reaction Theoretical Yield – expected amount of product (moles or mass) predicted using stoichiometry Less than 100% % yield = (actual / theoretical) * 100 Ex: SiO (s) + 3 C (s) → SiCl (s) + 2 CO (g) 2 100.0 kg SiO 2 51.4 kg SiC 3 3 (100.0 kg S2O )(10 g 2iO / 1 k2 SiO )(1 m2l SiO / 60.28 g SiO )(1 mol SiC2/ 1 mol SiO )(40.09 g SiC / 1 mol SiC)(1 kg SiC / 10 g SiC) = 66.73 kg SiC (51.4 kg / 66.73 kg) * 100 = 77.0%
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