PowerPoint PresentationSlide 2Slide 3Slide 4Slide 5Slide 6Slide 7Slide 8Slide 9Slide 10Slide 11Slide 12Slide 13Slide 14Slide 15Slide 16Slide 17Chapter 2 StoichiometryStoichiometry: the science dealing with quantitative relationships involving the mass of substances and the number of particles.(1) Convert mass to moles and moles to mass(2) Compute % elements in compounds(3) Compute empirical formulas (4) Compute molecular formulas (5) Balance chemical equations(6) Mass (mole) relationships for chemical reactions(7) Gas volume relationships for chemical reactions(8) Limiting reagent in chemical reactionsGeneral strategy for computing empirical formulas:(1) Given or compute the % elements in a substance.(2) From the % of each element, compute the number of mol of each element in the compound. The number of mol is directly related to the number of atoms in the substance.(3) Express the number of mol of each element in a chemical formula using the smallest possible whole numbers.(1) Assume a sample of 100 g for the computation (any mass will work, but selecting 100 makes the computation straightforward)(2) Translate the % mass into g (Example: element X is 10% of the total mass of a substance. For a 100 g sample of the substance, the sample contains 10 g of X).(3) Compute the number of mol of each element in the 100 g sample by dividing the mass of the element in the sample by the atomic weight of the element.(4) The ratio of the molar masses of the elements in the substance is directly proportional (within round off error) to the ratio of the atoms in the substance.(5) Express the number of mol of each element in a chemical formula using the smallest possible whole numbers.CommonName%H%OMoles H in 100 gof substanceMoles O in100gMolarRatio H/OEmpiricalFormulaMolecularFormulaWater11%89%11 mol(11g/1gmol-1)5.6 mol(89g/16 gmol-1)1.95 ~ 2/1H2OH2OHydrogenPeroxide6.0%94%6.0 mol(6g/1 gmol-1)5.9 mol(94g/16 gmol-1)1.01 ~ 1/1HOH2O2HydrogenTrioxide4.0%96%4.0 mol(4g/1 gmol-1)6.0 mol(96g/16 gmol-1).0.67 ~ 2/3H2O3H2O3Exemplar: Computation of the empirical formulas for three hydrogen oxides.(1) Assume a sample of 100 g for the computation (any mass will work, but selecting 100 makes the computation straightforward)(2) Translate the % mass into g (Example: Suppose O is 89% of the total mass of a substance. For a 100 g sample of the substance, the sample contains 89 g of O).(3) Compute the number of mol of each element in the 100 g sample by dividing the mass of the element in the sample by the atomic weight of the element.(4) The ratio of the molar masses of the elements in the substance is directly proportional (within round off error) to the ratio of the atoms in the substance.(5) Express the number of mol of each element in a chemical formula using the smallest possible whole numbers.CommonName%N%OMoles N in 100 gof substanceMoles O in100gMolar RatioN/OEmpiricalFormulaMolecularFormulaNitricOxide47%53%3.4(47g/14gmol-1)3.3(53g/16 gmol-1)1.03 ~ 1/1NONONitrousOxide64%36%4.6(64g/14 gmol-1)2.3(36g/16 gmol-1)2.0 ~ 2/1N2ON2ONitrogenDioxide30%70%2.1(30g/14 gmol-1)4.4(70g/16 gmol-1)0.48 ~ 1/2NO2NO2DinitrogenDioxide47%53%3.4(47g/14 gmol-1)3.3(53g/16 gmol-1)1.03 ~ 1/1NON2O2Exemplar: Computation of the empirical formulas for four nitrogen oxides.(1) Assume a sample of 100 g for the computation (any mass will work, but selecting 100 makes the computation straightforward)(2) Translate the % mass into g (Example: Suppose N is 47% of the total mass of a substance. For a 100 g sample of the substance, the sample contains 47 g of N).(3) Compute the number of mol of each element in the 100 g sample by dividing the mass of the element in the sample by the atomic weight of the element.(4) The ratio of the molar masses of the elements in the substance is directly proportional (within round off error) to the ratio of the atoms in the substance.(5) Express the number of mol of each element in a chemical formula using the smallest possible whole numbers.From empirical formulas to molecular formulas through Avogadro’s hypothesisEqual volumes of different gases contain the same number of particles (atoms or molecules).Logic: If equal volumes contain equal numbers of particles, the ratio of the masses of equal volumes is the same as the ratio of the masses of the particles.Thus, with the selection of a standard “particle”, the masses of equal volumes of gases provides a simple basis for establishing atomic and molecular weights.The substance hydrogen (molecular weight = 2) was selected as the standard.Hydrogen as a standard for molecular weightsWith the H2 (MW = 2 g) standard, the molecular weight is given by the density of the gas times the volume of a mole of the gas (22.4 L).Molecular weight = density (gL-1) x 22.4 LExample:Density of hydrogen gas = 0.090 gL-1MW of hydrogen defined as 2 (H2), i.e., MW (H2) = 0.090 gL-1 x 22.4 L = 2.0 gComputing molecular weight of gases from densitiesExemplars: oxygen and ozoneProblem: density of oxygen gas = 1.43 gL-1. What is the MW of oxygen “particles”?Answer: MW of oxygen particles is 1.43 gL-1 x 22.4 L = 32 gProblem: density of ozone gas is 2.14 gL-1. What is the molecular weight of ozone?Answer: MW of ozone particles is 2.14 gL-1 x 22.4 L = 48 gThese data are all consistent with the AW of hydrogen atoms = 1 g, the AW of oxygen atoms = 16 g and the MW of hydrogen (H2) gas = 2 g, the MW of oxygen (O2) gas = 32 g and the MW of ozone (O3) gas = 48 g.From empirical formula to molecular weightAnother exemplarProblem: A hydrocarbon gas has an empirical formula of CH. The gas has a density of 1.16 gL-1. What is the molecular weight of the gas?Answer:(1) We symbolize the molecular formula as (CH)n. We need to solve for n.(2) The MW of the hydrocarbon gas is given by the density of the gas time the molar volume: MW = 1.16 gL-1 x 22.4 L = 26 g.(3) The empirical formula CH corresponds to an atomic mass of 13. Dividing this empirical weight into the molecular weight gives the multiplier that takes the empirical formula into the molecular formula: 26/13 = 2.(4) Thus, n = 2 so that (CH)n becomes (CH)2 or written in the accepted way for a molecular formula or molecular composition: C2H2. (5) There is only one substance with the composition C2H2. That substance is acetylene whose molecular structure is HC CHLimiting reagent problem:Balanced Equation: 2 C2H6 + 7 O2  4 CO2 + 6 H2OProblem: Yield of CO2 if O2 = 0 mol, when C2H6 = 2 mol?Answer: 0 mol of