Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Adrenaline Is a Very Important Compound in the Body - IAdrenaline - IISlide 17Slide 18Ascorbic Acid ( Vitamin C ) - I Contains C , H , and OVitamin C Combustion - IISlide 21Slide 22Slide 23Slide 24Slide 25Chemical FormulasEmpirical Formula - Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements.Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound.Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule.Empirical and Molecular FormulasEmpirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms.Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.Calculating the Moles and Number of Formula Units in a Given Mass of Cpd.Problem: Sodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample?Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients.Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/molConverting mass to moles:Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.235 mol Na3PO4 Formula units = 0.235 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4= 1.42 x 1023 formula unitsSteps to Determine Empirical FormulasMass (g) of ElementMoles of ElementPreliminary FormulaEmpirical FormulaM (g/mol )use no. of moles as subscriptschange to integer subscriptsSome Examples of Compounds with the Same Elemental Ratio’sEmpirical Formula Molecular Formula CH2(unsaturated Hydrocarbons) C2H4 , C3H6 , C4H8OH or HO H2O2 S S8 P P4 Cl Cl2 CH2O (carbohydrates) C6H12O6Determining Empirical Formulas from Masses of Elements - IProblem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is theempirical formula and name of the compound?Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = Moles of Cr = 6.420 g Cr x = Moles of O = 7.902 g O x = 1 mol Na22.99 g NaDetermining Empirical Formulas from Masses of Elements - IProblem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is theempirical formula and name of the compound?Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = Moles of Cr = 6.420 g Cr x = Moles of O = 7.902 g O x = 1 mol Na22.99 g Na 1 mol Cr52.00 g Cr 1 mol O16.00 g ODetermining Empirical Formulas from Masses of Elements - IProblem: The elemental analysis of a sample compound gave the following results: 5.677g Na, 6.420 g Cr, and 7.902 g O. What is theempirical formula and name of the compound?Plan: First we have to convert mass of the elements to moles of the elements using the molar masses. Then we construct a preliminary formula and name of the compound.Solution: Finding the moles of the elements: Moles of Na = 5.678 g Na x = 0.2469 mol Na Moles of Cr = 6.420 g Cr x = 0.12347 mol Cr Moles of O = 7.902 g O x = 0.4939 mol O 1 mol Na22.99 g Na 1 mol Cr52.00 g Cr 1 mol O16.00 g ODetermining Empirical Formulas from Masses of Elements - IIConstructing the preliminary formula:Na0.2469 Cr0.1235 O0.4939Converting to integer subscripts (dividing all by smallest subscript):Na1.99 Cr1.00 O4.02 Rounding off to whole numbers:Na2CrO4 Sodium ChromateDetermining the Molecular Formula from Elemental Composition and Molar Mass - IProblem: The sugar burned for energy in cells of the body is Glucose(M = 180.16 g/mol), elemental analysis shows that it contains 40.00 mass % C, 6.719 mass % H, and 53.27 mass % O. (a) Determine the empirical formula of glucose. (b) Determine the molecular formula.Plan: We are only given mass %, and no weight of the compound so we will assume 100g of the compound, and % becomes grams, and we can do as done previously with masses of the elements.Solution: Mass Carbon = 40.00% x 100g/100% = 40.00 g C Mass Hydrogen = 6.719% x 100g/100% = 6.719g H Mass Oxygen = 53.27% x 100g/100% = 53.27 g O 99.989 g CpdDetermining the Molecular Formula from Elemental Composition and Molar Mass - IIConverting from Grams of Elements to moles: Moles of C = Mass of C x = 3.3306 moles C Moles of H = Mass of H x = 6.6657 moles H Moles of O = Mass of O x = 3.3294 moles OConstructing the preliminary formula C 3.33 H 6.67 O 3.33Converting to integer subscripts, divide all subscripts by the smallest: C 3.33/3.33 H 6.667 / 3.33 O3.33 / 3.33 = CH2O 1 mole C 12.01 g C 1 mol H1.008 g H 1 mol O16.00 g OTwo Compounds with Molecular Formula C2H6OProperty Ethanol Dimethyl EtherM (g/mol) 46.07 46.07Color Colorless
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