Using Molar Masses A reaction requires 0.475 mol of Magnesium turnings. How much magnesium, in g, is that? 24.31 g 0.475 mol x = 11.5 g Mg 1 mol “1” always goes with moles, and the number off of the periodic table goes with grams How many atoms of sodium are in 13.0 g of sodium metal? 1 mole 6.022 x 10 23 atoms 13.0 g Na x x = 3.41 x 10 23 atoms Na 22.998 g 1 mole For an experiment that I am performing, I need to add 12.5 mmol of iron (II) sulfate [FeSO4]. How many grams must I weigh out? 1 mol 151.92 g 12.5 mmol x x = 1.90 g 1000 mmol 1 mole Practice Problems What is the mass of 4.91 x 1021 platinum atoms? 4.91 x 10 21 atoms Pt x 1 mole 195.08 g x = 1.59 g Pt 23 1 mole 6.022 x 10 atoms How many atoms of copper are in a 133 kg pure copper statue? 1000 g 1 mol 6.022 x 10 23 atoms Cu 133 kg Cu x x x = 1.26 x 10 27 atoms Cu 1 kg 63.546 g 1 mole Determine the number of sulfur atoms in 1.59 mmol of carbon disulfide. 1 mol 6.022 x 10 23 molecules 2 atoms S 1.59 mmol CS2 x x x = 1.91 x 10 21 1000 mmol 1 mol 1 molec CS2 Determine the number of H atoms are in 9.88 mol of NH3. 6.022 x 10 23 molecules 3 atoms H 9.88 mol NH3 x x = 1.78 x 10 25 atoms H 1 mole 1 molec NH3 Mole-to-Mole Comparisons A reaction requires 10.0 g of iron atoms. How many grams of iron (II) hydroxide are required to complete the reaction? 1 mol Fe(OH)2 1 mol 89.866 g 10.0 g Fe atom x x x = 16.1 g Fe(OH)2 55.85 g 1 mol Fe 1 mol Mass Percents: Determine the mass percent of oxygen in FeSO4. Use that value to determine the grams of oxygen in 98.75 g of FeSO4. Mass percent = 4 x mass Oxygen 4 x 15.9996 amu x 100 = x 100 = 42.126% O mass FeSO 4 151.92 amu 98.75 g FeSO 4 x 0.42126 = 41.60 g O Empirical Formula The pre-hormone androstenedione (commonly called andro) has been in the sports news in recent years owing to its arguable contribution to the breaking of the homerun record. The product has a composition that is 79.68% carbon (C), 9.15% hydrogen (H), and 11.17% oxygen by weight. What is the empirical formula for this compound? assume 100 g: all % ! g 1 mol 79.68 g C x = 6.63 mol = 9.5 x 2 = 19 0.698 mol 12.011 g 1 mol 9.15 g H x = 9.08 mol = 13 x 2 = 26 0.698 mol 1.0079 g 1 mol 11.17 g O x = 0.698 mol =1x2=2 0.698 mol 15.9996 g The empirical formula is C19 H 26 O 2 Empirical Formula Determine the empirical formula for acetominophen, the active ingredient in Tylenol®, from the elemental analysis. C 63.56%, H 6.00%, N 9.27%, O 21.17% assume 100 g: all % ! g 1 mol 63.56 g C x = 5.29 mol =8 0.662 mol 12.011 g 1 mol 6.00 g H x = 5.95 mol =9 0.662 mol 1.0079 g 1 mol 9.27 g N x = 0.662 mol =1 0.662 mol 14.0067 g 1 mol = 1.32 mol =2 0.662 mol 15.9996 g The empirical formula is C8 H 9 NO 2 21.17 g O x Empirical Formula Determine the empirical formula for glucose. C 40.00%, H 6.72%, O 53.29% assume 100 g: all % ! g 1 mol 40.00 g C x = 3.33 mol =1 3.33 mol 12.011 g 1 mol 6.72 g H x = 6.67 mol =2 3.33 mol 1.0079 g 1 mol 53.29 g O x = 3.33 mol =1 3.33 mol 15.9996 g The empirical formula is CH 2 O Empirical Formula → Molecular Formula Use the empirical formula and the molar mass (given) to find the molecular formula for glucose. The molecular formula for glucose is some multiple of the empirical formula: (CH2O)n molar mass n= formula weight (molar mass of the emp. formula) 180.2 amu =6 30.0 amu glucose = (CH 2 O)6 = C6 H12 O6 n= Determining Molecular Formula Adipic acid is used in the commercial manufacture of Nylon. The composition of the acid is 49.3% C, 6.9% H, and 43.8% O by mass. The molecular weight is 146 amu. What is the molecular formula for Adipic Acid? 1 mol = 4.10 mol = 1.50 x 2 = 3 2.74 mol 12.011 g 1 mol 6.9 g H x = 6.85 mol = 2.50 x 2 = 5 2.74 mol 1.0079 g 1 mol 43.8 g O x = 2.74 mol =1x2=2 2.74 mol 15.9996 g The empirical formula is C3 H 5O 2 ! The formula weight = 73.07 amu 49.3 g C x 146 amu =2 73.07 amu The molecular formula is (C3 H 5O 2 )2 = C6 H10 O 4 n=