WWU CHEM 121 - Mass - Mole Relationships of a Compound

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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Slide 18Slide 19Adrenaline Is a Very Important Compound in the Body - IAdrenaline - IISlide 22Slide 23Ascorbic Acid ( Vitamin C ) - I Contains C , H , and OVitamin C Combustion - IISlide 26Slide 27Slide 28Slide 29Slide 30Mass - Mole Relationships of a CompoundMass (g) of Element Molesof Element Atomsof Element Mass (g)of compoundAmount (mol)of compound Molecules(or formula units of compound)Amount (mol)of compoundFor an Element For a CompoundFlow Chart of Mass Percentage CalculationMoles of X in one mole of CompoundMass % of XMass fraction of XMass (g) of X in onemole of compoundM (g / mol) of XDivide by mass (g) of one mole of compoundMultiply by 100Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - IProblem: Sucrose (C12H22O11) is common table sugar. ( a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? Mass Fraction of C = =(a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose & % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd = 0.421046 To find mass % of C = 0.421046 x 100% = 42.105%Calculating Mass Percents and Masses of Elements in a Sample of Compound - II(a) continued Mass % of H = x 100% = x 100% = 6.479% HMass % of O = x 100% = x 100% = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose)Mass (g) of C = 24.35 g sucrose X = 10.25 g C mol H x M of H 22 x 1.008 g Hmass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g Omass of 1 mol sucrose 342.30 g0.421046 g C 1 g sucroseChemical 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


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WWU CHEM 121 - Mass - Mole Relationships of a Compound

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