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# NCSU CH 201 - Chapter 1: Stoichiometry

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Slide 1Slide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Chapter 1: StoichiometryStoichiometry is the quantitative relationship between the elements in a compound or between the compounds/elements in a chemical reaction.Please review Appendix A2-A4! Significant Figures!“Reporting Quantitative Measurements and Results”Section 1.01Mass of an Atom and Moles of AtomsSection 1.1 For single atoms, amu = atomic mass unit = 1.66 X 10-24 gFor moles of atoms (6.022 x 1023 atoms)the unit is grams.Look at a periodic table.H atom weighs 1.0079 amu or 1.67 x 10-24 g Mole of H atoms weighs 1.0079 g 2A Ca(NO3)2 sample contains 2.04 grams of nitrogen. How much does the Ca(NO3)2 sample weigh?Strategy to Solve:grams N  moles N  moles Ca(NO3)2  grams Ca(NO3)2(ratio)(ratio)(ratio)Section 1.1Ratios are chosen such that 31 mol Ca1 mol Ca(NO3)22 mol N1 mol Ca(NO3)26 mol O1 mol Ca(NO3)21 mol Ca2 mol N1 mol Ca6 mol O2 mol N6 mol O164.086 g Ca(NO3)21 mol Ca(NO3)2Ratios from Chemical Formula: Ca(NO3)2Ratios from Periodic Table40.078 g Ca1 mol Ca14.007 g N1 mol N15.999 g O1 mol OSection 1.14A Ca(NO3)2 sample contains 2.04 grams of nitrogen. How much does the Ca(NO3)2 sample weigh?grams N  moles N  moles Ca(NO3)2  grams Ca(NO3)2(ratio)(ratio)(ratio)Section 1.15Elemental Composition (Mass Percent)% A = (mass A/total mass)×100 H2O2 (hydrogen peroxide):Molecular Weight:(2×1.0079 g/mol H)+(2×15.999 g/ mol O)=34.0138 g/mol H2O2% H =% O =Section 1.26A 7.50 g sample of iron is heated in oxygen to form an iron oxide. If 10.36 g of the oxide is formed, what is the empirical formula of this oxide of iron?We need the ratio moles Fe:moles OEmpirical Formula – the simplest formula, only tells you the ratio of elements in a compound.Section 1.27Now try multiples that lead to whole numbers.Hint: divide larger #by smaller #This is the simplest formula of this compound.For example, the empirical formula of benzene is CH. CH does not exist; the molecular formula of benzene is C6H6.Section 1.2Molecular Formula – the actual number of atoms of each element in a compound.Empirical Formula8Empirical Formula by CombustionIf 12.30 g of a white substance was burned to give 7.40 grams of water and 18.04 grams of carbon dioxide. What is the empirical formula of the substance (CxHyOz)? Figure out mole C in the original sample (solve for X):CxHyOz + ?O2  X CO2 + Y/2 H2O12.30 g18.04 g7.40 gFigure out mole H in the original sample (solve for Y):Section 1.29Empirical Formula by CombustionFigure out how many grams of O had to be in the originalSample and solve for Z:Section 1.2CxHyOz + ?O2  X CO2 + Y/2 H2O12.30 g18.04 g7.40 g10Mole RatioNeed whole numbers so divide each by the smallest.Empirical Formula – the simplest formula, only tells you the ratio of elements in a compound.Section 1.211Molecular FormulaWe need the molecular weight to know the molecular formula.The molecular weight can be determined by other techniques presented next week.Molecular WeightChemical Formula30 g/mol CH2OChemical Formula or Molecular Formula – the actual number of atoms of each element in a compound.Given an empirical formula of CH2O, Section

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