Announcements - 9/17/02 •No online Quiz this week ☺ •Exam #1 – Wed 7pm 235 Marsh -bring calculator (no PDAs!) and pen/pencil(s) -see Info Page for details •Demo Today! 1 Another Example A solution of sodium chloride was added to a solution of silver nitrate, forming a precipitate of silver chloride 1. sodium choride + silver nitrate → silver chloride + sodium nitrate 2. NaCl + AgNO3 → AgCl + NaNO3 Formulas 3. NaCl + AgNO3 → AgCl + NaNO3 Balanced 4. NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq) Na+ (aq) + Cl- (aq) + Ag+ (aq) + NO3- (aq) → AgCl (s) + Na+ (aq) + NO3- (aq) Ag+ (aq) + Cl- (aq) → AgCl (s) Net Ionic Equation 2 1 Demo! How can we tell an ionic compound from a molecular compound? Measure conductivity (in sol’n)! 3 Let’s Get Quantitative Recall: 1 mole = 6.02 x 1023 particles Atomic Masses (Weights): mass/atom (amu) Molecular/Formula Weights: mass/compound (amu) Molar Masses: mass of a mole of atoms or compound (g) 4 2 Why Fractional Molar Masses? Need to consider the natural abundances of isotopes Example: Chlorine 75.5% 35Cl + 24.5% 37Cl (0.755)(34.97) + (0.245)(36.97) = 35.45 g/mol -This is a weighted average; 1 mol of Cl will have a mass of 35.45 grams 5 General Strategy Be Careful! moles of: atoms? molecules? 6 3 Mole-Based Calculations How many grams of Phosphorous are there in 0.010 mol P2O5? Strategy: mol P2O5 → mol P → g P 0.010 mol P2O5 x 2 mol P x 30.974 g P = 0.61948 g P 1 mol P2O5 1 mol P Round to: 0.62 g Phosphorous 7 How Many Atoms? How many Phosphorous atoms are there in 0.010 mol P2O5? Strategy: mol P2O5 → mol P → #P atoms 0.010 mol P2O5 x 2 mol P x 6.022 x 1023 P atoms 1 mol P2O5 1 mol P = = 1.20440 x 1022 P atoms = 1.2 x 1022 P atoms 8 4 Strategy: Empirical Formula from Percent Composition 9 Empirical Formula from Percent Composition What is the empirical formula for a binary compound which is found to be: 56.4% Oxygen (by mass) 43.6% Phosphorous (by mass)? Strategy: % → grams → mol (% is a relative measure, so DEFINE a sample size (100 g)) In a 100-g sample: 56.4 g O x 1 mol O 43.6 g P 15.999 g O x 1 mol P 30.974 g P = 3.525 mol O = 1.4076 mol P 10 5 Empirical Formula continued This gives: P1.4076O3.525 Dividing: PO2.50 → P2O5 •What about a MOLECULAR formula? -need the molar mass (MW) of the compound Example: MW of P2O5 cmpd is 284 g/mol Empirical Formula Mass ≈ 2x31 + 5x16 = 142 g MW/Emp Form Mass = 284/142 = 2 So: 2 x P2O5 = P4O10 11 6