Chemistry: Analyzing Hydrates - Empirical and Molecular Formulas, Lecture notes of Chemistry

Download Chemistry: Analyzing Hydrates - Empirical and Molecular Formulas and more Lecture notes Chemistry in PDF only on Docsity! Chemistry The Mole (5) Mrs. Day-Blattner 26/27 February 2018 Feb 26/27 Agenda Activities of analytical chemists Slip Quiz Analyzing a hydrate - set up and Part 1 of lab procedure Percent composition - uses of Empirical Formulas and Molecular formulas Analyzing a hydrate - Part 2 of lab procedure 11.5 The formula for a hydrate H2O Formula Name 1 (NH4)C2O4·H2O Ammonium oxalate monohydrate 2 CaCl2·2H2O Calcium chloride dihydrate 3 NaC2H2O2·3H2O Sodium acetate trihydrate 4 FePO4·4H2O Iron(III) phosphate tetrahydrate 5 CuSO4·5H2O Copper(II) sulfate pentahydrate 6 CoCl2·6H2O Cobalt(II) chloride hexahydrate 7 MgSO4·7H2O Magnesium sulfate heptahydrate 8 Ba(OH)2·8H2O Barium hydroxide octahydrate 10 Na2CO3·10H2O Sodium carbonate decahydrate How do you think chemists found the formulas for hydrates? How do you think chemists found the formulas for hydrates? To analyze a hydrate we need to drive off the water of hydration (or water of crystallization) to make the crystals anhydrous (without water). Ex. CoCl2·6H2O (s) → CoCl2 (s) + 6 H2O(g) Pink blue Checking the formula for cobalt(II) chloride hexahydrate Safety: All usual safety rules apply. You must wear goggles at all times and be aware of your clothing and arm movements because of the use of open flames today. Do not attempt to move the equipment after heating. Turn off your bunsen burner and sit down while the equipment cools. Empirical and Molecular Formulas • Empirical formula Formula with the smallest whole number ratio of the elements. Empirical formula may be same as molecular formula like water H 2 O Or it may not be: hydrogen peroxide empirical formula is OH and molecular formula H 2 O 2 Calculating Empirical formula • From masses or percent compositions Compound is 40% S 60% O So if we can say that if we had 100g of this compound we would have 40g S and 60g O Calculating Empirical formula 40g S and 60g O Convert to moles 60g O x 1 mole = 60g O x 1 mole = 3.75 mol O mass 1 mole O 16g Calculating Empirical formula 40g S and 60g O Now convert to simplest ratio: 1.25 mol S = 3.75 mol O = 1.25 1.25 Calculating Empirical formula 40g S and 60g O Now convert to simplest ratio: 1.25 mol S = 1 mol S 3.75 mol O = 3 mol O 1.25 1.25 Simplest whole number mole ratio is of S to O is 1:3 Empirical formula SO 3 Calculating Empirical formula. Ex. 1 • From masses or percent compositions Compound is 36.84% Nitrogen 63.16 % O We can say that if we had 100g of this compound we would have Calculating Empirical formula. Ex. 1 • From masses or percent compositions Compound is 36.84% Nitrogen 63.16 % O We can say that if we had 100g of this compound we would have 36.84g N and 63.16g O Ex. 1 Convert to moles (multiply by 1/molar mass) 36.84g N x 1 mole = 2.630 mol N 14.01gN Convert to moles 63.16g O x Ex. 1 Now convert to simplest ratio: 2.630 mol N = 1 N 3.950 mol O = 1.5 O 2.630 2.630 Simplest whole number ratio (x 2) is 2 : 3 Empirical formula N 2 O 3 Ex. 2 Propane is a hydrocarbon, a compound composed only of carbon and hydrogen. It is 81.82% carbon (12.01g/mol) and 18.18% hydrogen (1.008 g/mol). What is the emprical formula? Ex. 2 Convert to moles (multiply by 1/molar mass) 81.82g C x 1 mole = 6.818 mol N 12.01gN Convert to moles 18.18g H x 1 mole = 18.04 mol H 1.008gH Calculating Molecular formula Start from calculated Empirical formula And Molar mass (found by experimentation). Experimentally determined molar mass = n Mass of empirical formula Then n(empirical formula) = molecular formula See Example Problem 11-12 Ex. 1 Calculating Molecular formula The empirical formula for acetylene and benzene is the same: CH. But acetylene is a gas used in welding and benzene is a liquid used as a solvent. They are obviously very different. The molar mass of the empirical formula is (12.01 + 1.008) = 13.02 g/mol The molar mass of acetylene is experimentally found to be 26.04 g/mol and the molar mass of benzene is 78.12 g/mol. Ex. 1 Calculating Molecular formula Acetylene Exp. determined molar mass = 26.04 g/mol = Mass of empirical formula 13.02 g/mol Ex. 1 Calculating Molecular formula Benzene Exp. determined molar mass = 78.12 g/mol = Mass of empirical formula 13.02 g/mol Ex. 1 Calculating Molecular formula Benzene Exp. determined molar mass = 78.12 g/mol = 6 Mass of empirical formula 13.02 g/mol Molecular formula = 6 (CH) Molecular formula benzene is C 6 H 6 Benzene structural formulas Benzene: C,H. H agen HSN (1) (2) (3) Analysis 2) Mass of hydrated cobalt(II) chloride Mass after heating + crucible = g - Mass of crucible = g = mass of anhydrous sample = g Analysis 3) Mass of water lost Mass of hydrated sample = g - mass of anhydrous sample = g = mass of water lost = g Analysis Convert to moles of each Molar mass CoCl2 58.93 + 2(35.45) = Mass g x 1mol = Molar mass