Empirical and Molecular Formula PDF

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notes on empirical and molecular formulas...

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Moles bridge macroscopic and submicroscopic - Molar mass helps with thinking abt singular molecular compounds - Glucose(C6H12O6) - What percentage by mass is oxygen? - What is the most basic ratio of elements (empirical formula)? - What is the actual ratio of elements (molecular formula)? Ex/ Using Percent Comp - Are mass percents of 30.54% N and 69.46% consistent with formula N2O4 - Procedure - Assume one mole of compound - Subscripts tell how many moles of each element are present - Use molar masses of elements to determine mass of each element in 1 mole - Calc % by mass of each element - 1 mol N2O4 - 2 mol Nitrogen mass N 14.01 g/mol - 4 mol Oxygen mass O 16.00 g/mol - Molar mass of each element/ molar mass of N2O4 - N 14.01/92.05 = .1522 = 15.22% x 2 = 30.43% - O 16.00/92.05 = .1739 = 17.39% x 4 = 69.53% - Close enough so they are consistent Ex/ Percent by Mass - What is percent comp by mass of glucose - Molar mass 180.156 g/mol C6H12O6 - 6 mol C 12.01 g/mol x 6 = 72.06/180.156 = .4000 = 40% - 12 mol H 1.00 g/mol x 12 = 12/180.156 = .0670 = 6.7% - 6 mol O 16.00 g/mol x 6 = 96/180.156 = .5329 = 53.29% Ex/ sample P and O, percent comp 56.34% P and 43.66% O, is it P4O10? - P (30.97x4)/283.88 = 43.64% - O (16.00x10)/283.88 = 56.36% - No that is not the compound Determining Empirical and Molecular Formulas - Empirical - Simplest ratio of atoms of each element - Obtained from exp analysis - Molecular - Exact comp of one molecule - Exact whole number ratio of atoms of each element in molecule Empirical - Empirical formula from mass data - Determine moles - Div by smallest mole - Mult to get integers

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ex/ black iron oxide, 2.448 g sample, analyzed and found 1.771 g of Fe, and 0.677 g O, calc emp form - Fe 55.85 g/mol 1.771g/55.85g = 0.03171 mol Fe - O 16.00 g/mol 0.677g/16.00g = 0.0423 mol O - 0.03171/0.03171 = 1 x 3 = 3 Fe - 0.0423/0.03171 = 1.33 = 1 ⅓ x 3 = 4 O - Fe3O4 - Find from % comp - Determine mass of each element assuming 100 g - Determine moles of each - Div by smallest mole to get emp form - Mult to get integers - Calc emp form from compound with 43.63% P and 56.36% O. if molar mass is 283.9 g/mol, what is molecular formula - 43.64% P = 43.64 g P / 30.97 g P= 1.409 mol P - 56.36% 0 = 53.36 g O / 16.00 g O = 3.523 mol O - 1.409/1.409 = 1 x 2 = 2 P - 3.523/1.409 = 2.50 x 2 = 5 O - P2O5 - empirical formula - P2O5 = 141.94 g/mol - 283.9/141.94 = 2 so molecular formula is 2(P2O5) = P4O10 - Empirical formula - Accepted formula unit for ionic compounds - Molecular formula - Preferred for molecular compounds - In some cases molecular and empirical formulas are the same - When they are different, the subscripts of molecular formula are integer multiples of those in empirical formula - If empirical formula is AxBy - Molecular formula will be An ×xBn ×y - Determine molecular from empirical - Divide molecular mass by empirical - Need molecular mass and empirical formula - Calc ratio of molecular mass to mass predicted by empirical formula - Round to nearest integer Balancing Formulas- Reaction Stoichiometry - Balanced eq - Critical link b/t substances - Gives relationship b/t amts of reactants used and amounts of products formed - Use the Numeric Coeffs - The mole ratios for reactions - How many individual particles are needed in reaction on microscopic level - How many moles are necessary on macroscopic level

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N2+3H2 = 2NH3 - 1 mol N2 = 2 mol NH3 - 3 mol H2 = 2 mol NH3 - 1 mol N2 = 3 mol H2 Stoichiometry - Mass balance of all formulas - Calc - Conversions from one set of units to another using dimensional analysis•Need to know - Equalities to make conversion factors - Steps to go from starting units to desired units - N2 + 3H2 = 2NH3, how many mols of nitrogen are used when 2.3 moles NH3 produced - 1 mol N2 = 2 mol NH3 - 2.3 mol NH3 x (1 mol N2 / 2 mol NH3) = 1.15 N2 moles = 1.2 moles N2 Mass-to-Mass Conversions Tools to have - Periodic table - Conversion factors - Moles - Chem formulas - Chem equations...