Chemistry: Empirical & Molecular Formulas, Atom Ratios, Electronegativity - Prof. Dana E. , Study notes of Chemistry

Download Chemistry: Empirical & Molecular Formulas, Atom Ratios, Electronegativity - Prof. Dana E. and more Study notes Chemistry in PDF only on Docsity! 8 October Review from Wednesday Information in a formula (e.g. AlCl3) Ratios of atoms to each other (1:3) Number of moles of Cl in one mole of AlCl3 Can determine mass percent of Cl Mass of Cl / mass of AlCl3 * 100 Use to determine Cl mass of any sample of AlCl3 Clicker challenge A 0.371 mole sample of a metal oxide (M2O3) weighs 55.4 g. What element is represented by M? M2O3 = 0.371 moles O = 0.371 (3) = 1.11 mol M = 0.371 (2) = 0.742 mol 1.11 mol of O x 16.00 g / 1 mole = 17.76 g 55.4 g – 17.76 g = 37.64 g 37.64 g / 0.742 mol = 50.72 g / mol M is V Determining empirical formulae Remember, compounds do not vary in composition Formula for glucose → C6H12O6 In 1 molecule of glucose, there are 6 carbon atoms In 1 mole of glucose, there are 12 moles of hydrogen atoms Subscripts tell us number (not weight) Empirical formulae can be determined from elemental analysis Compound is decomposed into elements 1. Convert weights to moles 2. Insert moles into a preliminary formula 3. Divide all moles by the lowest number 4. Multiply all values (if necessary) by an integer to get whole numbers Zn 13.74 g x (1 mol / 65.39 g) = 0.21 mol P 4.34 g x (1 mol / 30.97 g) = 0.14 mol O 8.96 g x (1 mol / 16.00 g) = 0.56 mol Zn0.21 / 0.14P0.14 / 0.14 O0.56 / 0.14 Zn1.5P1O4 x 2 Zn3P2O8 = Zn3(PO4)2 Molecular formulae Start with empirical formula Determine empirical mass Compare to molar mass to determine multiple Example → glucose CH2O But its molar mass 180.156 g / mol (12.01) + (2 x 1.008) + 16 = 30.026 g / mol 180.156 / 30.026 = 6 CH2O (x6) C6H12O6 Combustion analysis Generally confined to carbohydrates (C and H containing) Too many other elements make calculation impossible