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Chapter 3 Chemical Calculations

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Chapter 3 Chemical Calculations Part B Chemical ReactionsUsing % composition by massSlide 3Determine the empirical formula of the compound with the following composition.Calculating an Empirical FormulaCalculating Empirical Formula from % composition…a mnemonic:Slide 7Slide 8Slide 9Slide 10Slide 11Adrenaline Is a Very Important Compound in the Body - IAdrenaline - IISlide 14Ascorbic Acid ( Vitamin C ) - I Contains C , H , and OVitamin C Combustion - IISlide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Chemical EquationsChemical Equation Calculation - IChemical Equation Calculation - IISlide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Limiting ReactantSlide 35Using StoichiometryPercent yieldTheoretical Yield: Which Reactant is Limiting?Slide 39Slide 40Slide 41Slide 42Slide 43Slide 44Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Figure 3.11: Steps in a stoichiometric calculation.Slide 52Figure 3.14: Limiting reactant analogy using cheese sandwiches.Slide 54Conceptual Problem 3.13Slide 56Practice Problem 3.86Slide 58Slide 59Slide 60Slide 61Slide 62Slide 63Slide 64Slide 65Slide 66Slide 67Slide 68Slide 69Slide 70Slide 71Slide 72Other ResourcesChapter 3 Chemical CalculationsPart BPart BChemical ReactionsChemical ReactionsmassA (g) molesA da molesD massD (g) mole ratioUsing % composition by mass1) a sample of clear liquid (3.0000 g) is found to contain 0.1778 g H and 2.8222 g O2) a sample of clear liquid (5.0000 g) is found to contain 0.5595 g H and 4.4405 g O3) a sample of clear liquid (7.0000 g) is found to contain 0.2946 g H and 4.6769 g OAre these the same or different compounds?Atomic mass H: 1.008 C: 12.01 O: 16.00Determine the empirical formula of the compound with the following composition.35.00 % N (14.007 g/mol) 5.04 % H (1.008 g/mol)59.96 % O (15.999 g/mol)Calculating an Empirical Formula%compositionmass of each elementin the samplemoles of each elementin the sampleuse # of moles assubscript in formuladivide all subscriptsby smallest subscriptmultiply by integerso that all subscriptsare whole numbersCalculating Empirical Formula from % composition…a mnemonic: 1) Turn % to mass2) Then turn mass to mole3) divide by the small one4) multiply ‘til wholeStep 1: does % composition change as amount of material changes?Row Row Row Your BoatDetermining the Empirical Formula Beginning with percent composition:•Assume exactly 100 g so percentages convert directly to grams.•Convert grams to moles for each element.•Manipulate the resulting mole ratios to obtain whole numbers.•Manipulating the ratios:•Divide each mole amount by the smallest mole amount.•If the result is not a whole number:•Multiply each mole amount by a factor.For example:If the decimal portion is 0.5, multiply by 2.If the decimal portion is 0.33 or 0.67, multiply by 3. If the decimal portion is 0.25 or 0.75, multiply by 4.•Benzene is composed of 92.3% carbon and 7.7% hydrogen. What is the empirical formula of benzene?Hmol7.64Hg1.008Hmol1Hg7.7Cmol7.685Cg12.01Cmol1Cg92.317.647.6417.647.685Empirical formula: CH•Molecular Formula •A formula for a molecule in which the subscripts are whole-number multiples of the subscripts in the empirical formulaTo determine the molecular formula:•Compute the empirical formula weight.•Find the ration of the molecular weight to the empirical formula weight.•Multiply each subscript of the empirical formula by n.weightformulaempiricalweightmolecularnAdrenaline Is a Very Important Compound in the Body - I•Analysis gives : • C = 56.8 %• H = 6.50 %• O = 28.4 %• N = 8.28 %•Calculate the Empirical FormulaAdrenaline - II•Assume 100g!•C = 56.8 g C/(12.01 g C/ mol C) = 4.73 mol C•H = 6.50 g H/( 1.008 g H / mol H) = 6.45 mol H•O = 28.4 g O/(16.00 g O/ mol O) = 1.78 mol O•N = 8.28 g N/(14.01 g N/ mol N) = 0.591 mol N• Divide by 0.591 = •C = 8.00 mol C = 8.0 mol C or•H = 10.9 mol H = 11.0 mol H •O = 3.01 mol O = 3.0 mol O C8H11O3N•N = 1.00 mol N = 1.0 mol NCombustion Train for the Determination of the Chemical Composition of Organic Compounds.CnHm + (n+ ) O2 = n CO(g) + H2O(g)m 2m 2Fig. 3.4Ascorbic Acid ( Vitamin C ) - I Contains C , H , and O•Upon combustion in excess oxygen, a 6.49 mg sample yielded 9.74 mg CO2 and 2.64 mg H2O•Calculate it’s Empirical formula!•C: 9.74 x10-3g CO2 x(12.01 g C/44.01 g CO2) = 2.65 x 10-3 g C•H: 2.64 x10-3g H2O x (2.016 g H2/18.02 gH2O) = 2.92 x 10-4 g H•Mass Oxygen = 6.49 mg - 2.65 mg - 0.30 mg = 3.54 mg OVitamin C Combustion - II•C = 2.65 x 10-3 g C / ( 12.01 g C / mol C ) = = 2.21 x 10-4 mol C•H = 0.295 x 10-3 g H / ( 1.008 g H / mol H ) = = 2.92 x 10-4 mol H•O = 3.54 x 10-3 g O / ( 16.00 g O / mol O ) = = 2.21 x 10-4 mol O•Divide each by 2.21 x 10-4•C = 1.00 Multiply each by 3 = 3.00 = 3.0•H = 1.32 = 3.96 = 4.0•O = 1.00 = 3.00 = 3.0C3H4O3Vitamin CDetermining a Chemical Formula from Combustion Analysis - IProblem: Erthrose (M = 120 g/mol) is an important chemical compound as a starting material in chemical synthesis, and contains Carbon Hydrogen, and Oxygen. Combustion analysis of a 700.0 mg sample yielded 1.027 g CO2 and 0.4194 g H2O.Plan: We find the masses of Hydrogen and Carbon using the mass fractions of H in H2O, and C in CO2. The mass of Carbon and Hydrogen are subtracted from the sample mass to get the mass of Oxygen. We then calculate moles, and construct the empirical formula, and from the given molar mass we can calculate the molecular formula.Determining a Chemical Formula fromCombustion Analysis - IICalculating the mass fractions of the elements: Mass fraction of C in CO2 = = = = 0.2729 g C / 1 g CO2Mass fraction of H in H2O = = = = 0.1119 g H / 1 g H2OCalculating masses of C and HMass of Element = mass of compound x mass fraction of element mol C x M of Cmass of 1 mol CO21 mol C x 12.01 g C/ 1 mol C 44.01 g CO2 mol H x M of Hmass of 1 mol H2O2 mol H x 1.008 g H / 1 mol H 18.02 g H2ODetermining a Chemical Formula from


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