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Ch. 11 The Mole 11.1 Measuring Matter Mole- SI base unit used to measure the amount of a substance Equal to the number of representative particles (carbon atoms) in exactly 12 grams of carbon-12 Representative particle : Elements –atom Covalent –molecules Ionic - formula units Representative particle Elements Atoms Molecules Compounds Ionic Compound Formula Unit Covalent Compound Molecules Avogadro’s Number Avogadro’s Number = 6.022 X 1023 Is a very large number because it is used to count extremely small particles. Conversion of a Mole to a Particle 1 mole = 6.02 x 1023 # of moles x 6.02 x 1023 representative particles = # of representative particles Ex: How many molecules are in 3.5 moles of sucrose? 3.5 moles sucrose x 6.02 x 1023 molecules = 1 mole sucrose = 2.11 x 1024 molecules of sucrose Conversion of a Particle to a Mole Reverse conversion factor to solve for # of moles Ex: How many moles are in 4.50 x 1024 atoms of zinc? 4.50 x 1024 atoms Zn x = 7.48 mol Zn 1 mol Zn 6.02 x 1023 atoms Zn STOP! YOUR TURN! Practice Problems 11.1 11.2 Mass and the Mole Just as a dozen bricks and a dozen feathers don’t have the same mass, moles of different substances also have different masses. Molar mass- mass in grams of one mole of any pure substance The molar mass of any element is numerically equal to its atomic mass and has the units g/mol Using Molar Mass Ex: What is the mass in grams of 0.0450 moles of chromium? Moles Cr x grams Cr = grams Cr 1 mole Cr 0.0450 mol Cr x 52.00 g Cr = 2.34 g Cr 1 mol Cr Mols Mass Particles (atoms, molecules, formula unit) Converting Mass to Atoms and Atoms to Mass Ex: How many atoms of gold are in a pure nugget having a mass of 25.0 g? Known: Mass = 25.0 g Au Molar mass Au = 196.97 g/mol Au Unknown: Number of atoms = ? Atoms Au Mass Au x 1 mole Au = moles Au # g Au 25.0 g Au x 1 mol Au = ? mol Au 196.67 g Au = 0.127 mol Au Half way there!! Moles Au x 6.02 x 1023 atoms Au = 1 mole Au 0.127 mol Au x 6.02 x 1023 atoms Au 1 mol Au = 7.65 x 1022 atoms Au Mass must always be converted to moles before being converted to atoms, and atoms must be converted moles before calculating their mass. 11.3 Moles of Compounds Chemical formula indicates types of atoms and number of each in one unit of the compound Ex: CCl2F2 Carbon = one atom Chlorine = 2 atoms Fluorine = 2 atoms Ratio of carbon to chlorine to fluorine is 1 : 2 : 2 Conversions with Chemical Formulas How many moles of fluorine atoms are in 5.50 moles of freon (CCl2F2)? 5.50 mol CCl2F2 x 2 mol F atoms = 11.0 mol F atoms 1 mol CCl2F2 Molar Mass of Compounds Mass of a mole of a compound equals the sum of the masses of every particle that makes up the compound. Suppose you want to determine the molar mass of potassium chromate (K2CrO4) # moles x molar mass = # grams 2.000 mol K x 39.10 g K = 78.20 g K 1 mol K 1.000 mol Cr x 52.00 g Cr = 52.00 g Cr 1 mol Cr 4.000 mol O x 16.00 g O = 64.00 g O 1 mol O Molar mass = 78.20 g K 52.00 g Cr + 64.00 g O 194.20 g K2CrO4 Converting Moles of a Compound to Mass Step 1: Calculate molar mass of the compound. Step 2: Convert moles to grams using the molar mass as a conversion factor. Converting Moles to Mass Convert 2.50 mol CHCl3 to mass in grams. Step one: Calculate the number of grams in one mole of CHCl3 = 119.35 grams Step up problem as before What’s given What you want What you want to get rid of 2.5 mol CHCl3 119.35 grams of CHCl3 1 mol of CHCl3 298.38 grams of CHCl Converting Mass of a Compound to Moles Use inverse of mole to mass conversion factor Converting Mass of a Compound to Number of Particles Step 1: Convert given mass to moles by using the molar mass as a conversion factor. Step 2: Convert moles to number of representative particles by multiplying by Avogadro’s number. 11.4 Empirical and Molecular Formulas Percent Composition- percent by mass of each element in a compound % composition = mass of element x 100 mass of compound Percent Composition from the Chemical Formula Percent composition is always the same, regardless of the size of the sample To determine percent composition, assume a sample size of one mole What is the percent composition of water? Percent Hydrogen: 2.02 g H x 100 = 11.2% H 18.02 g H2O Percent Oxygen: 16.00 g O x 100 = 88.80% O 18.02 g H2O STOP! YOUR TURN! Page 331, #45 Empirical Formula Empirical Formula- formula with the smallest whole number mole ratio of the elements Which of these do you think is the empirical formula? HO or H2O2 ? Empirical formula may be different from molecular formula Ex: hydrogen peroxide Empirical formula: HO Molecular formula: H2O2 Finding the Empirical Formula If percent composition is given, assume a 100 g sample, so change % to grams Ex: percent composition is 40.05% S and 59.95% O 100.0 g of the sample is 40.05 g S and 59.95 g O Convert mass of each element to number of moles 40.05 g S x 1 mol S = 1.249 mol S 32.07 g S 59.95 g O x 1 mol O = 3.747 mol O 16.00 g O Ratio of S atoms to O atoms is 1.249 : 3.747 Must convert to whole numbers Since 1.249 is smallest, divide both numbers by that value Continued… 1.249 mol S = 1 mol S 1.249 3.747 mol O = 3 mol O 1.249 Ratio of S atoms to O atoms is 1 : 3 Empirical Formula = SO3 STOP! YOUR TURN! # 46 and 47 page 333 Molecular Formula Molecular Formula- specifies the actual number of atoms of each element in one molecule or formula unit of the substance Molecular formula = (empirical formula) n n is the factor by which the subscripts in the empirical formula must be multiplied to obtain molecular formula Determining Molecular Formula Empirical formula: C2H3O2 Molar Mass C2H3O2 = 59.04 g C2H3O2 (calculated value) Molar Mass succinic acid = 118.1 g (given value) n = molar mass succinic acid molar mass C2H3O2 (empirical formula) n = 118.1 g = 2.00 59.04 g (C2H3O2) 2 = C4H6O4 STOP! YOUR TURN! # 51 and 52 Empirical Formula = HO Given the molar mass is 34.014g/mol. What is the MOLECULAR FORMULA? Class Discussion Why do one mol of sulfur not the same as one mol of hydrogen? CROSS OFF 11.5 11.5 The Formula for a Hydrate Hydrate- a compound with a specific number of water molecules bound to its atoms Analyzing a Hydrate In order to analyze a hydrate, water must be removed (usually from heating) Substance remaining after heating is “anhydrous” (meaning without water)