Welcome to chemistry 101HEngr-PhysbldgChem. Engr.ZacheryEngr.Chemistry bldg.CyclotronInst.McD’sRoss StreetUniversity Ave.Spence StreetenterHow can I get a good grade in this course?1. Come to class!2. Study the lecture notes.3. Read the corresponding material in the textbook.4. Work the homework problems.5. Seek help with things you don’t understand.6. Keep up to date.The Scientific MethodExperimental Observations LawTheoryModelFundamental units of measureQuantity Definition Common UnitsMass Quantity of matter kg, g, oz, lb, ton Length Distance between two points m, km, cm, in, yd, ft, miTime Interval between two events s, min, hr, yr Temperature Heat intensity (kinetic energy) oK, oC, oF, oRSI unitsDerived unitsVolume (length)3m3, L, (1L = 1000 mL = 1000 cm3)Force F = ma = m(dv/dt) kg-m/s2= N(ewton)= m(d2x/dt2) g-cm/s2= dyneEnergy EK= ½mv2kg-m2/s2= J(oule)= ½m(dx/dt)2g-cm2/s2= ergPressure P = F/A N/m2= Pa (Pascal)1 bar = 100 kPa1 atm = 101.3 kPaImportant conversion factors 1 in = 2.54 cm 1 lb = 454 g 1 qt = 0.946 L Example conversion factor problems2. Convert 1 kilometer to miles 3. Density (ρ = mass/volume) may be regarded as a conversion factor that relates the mass of a substance to its volume. What volume in liters is occupied by 2 lb of iron if its density is 7.86 g/cm3? 1. Convert 36x1015in. to miles. (36x1015in)(1 ft/12 in)(1 mi/5280 ft) = (36x1015)/[12(5280)] = 5.7x1012mi(1 km)(1000 m/1 km)(100 cm/1 m)(1 in/2.54 cm)(1 ft/12 in)(1 mi/5280 ft)= 0.621 mi(2 lb Fe)(454 g Fe/1 lb Fe)(1 cm3Fe/7.86 g Fe)(1 L Fe/1000 cm3Fe)= 0.116 L Fe1. Calculate the kinetic energy of a baseball having a mass of 4.0 ozand a speed of 60 mi/hr in SI units.Exercises2. The density of silicon is 2.33 g/mL and the mass of a silicon atomis 4.65x10−23g. Calculate the number of silicon atoms in a 1 in. cube of silicon.3. The density of ethanol is 0.794 g/mL at 25oC. What volume ofethanol should be measured out to obtain a sample weighing0.33 lb?Fahrenheit scale Celcius scaleTF-32 TC100180water boilswater freezes212 oF32 oF100 oC0 oC100T18032TCF=−Rules of significant figures1. The digit 0 is significant only when it is preceded by another number(other than 0).106.0 has 4 significant figures0.0016 has 2 significant figures2. Addition and subtraction: round off at the digit closest to the decimalpoint.63.43 − 1.245 + 0.4652 = 62.65 (not 62.6502)3. Multiplication and division: keep only the number of digits held bythe number having the fewest significant figures.(0.156)(84.25)(1.7854)/1.6 = 15 (not 14.6659) 4. Terminal zeros should be expressed in exponential notation unlessthey are significant.93,000,000 expressed to 3 significant figures is 9.30x107ExercisesCarry out the following conversions following the rules of significantfigures.a). 12 oz to mg.b). 5 nickels to Swiss Francs where 1 SF = 0.36 $.c). 3x1010cm/s to mi/hr.d). The density of Hg (13.6 g/cm3) to lb/in3.Anatomy of a carbon atomDiffuse cloud containingsix electrons circulatingaround a nucleusnucleusDiameter approx.0.0000000002 m = 2x10-10m = 0.2 nm= 2 ǺDiameter approx.0.000000000000006 m= 6x10-15m = 6 fm(Rn= 1.4A⅓fm)Nucleus of the isotope carbon-126 protons; Z = 6(Z = atomic number)6 neutrons; N = 612 nucleons; A = 12(A = atomic mass number)AZN6126N = A − ZProton +qe1.673x10-27kg 938.2 MeV(1.0075 amu)Neutron 0 1.675x10-27kg 939.5 MeV(1.0087 amu)Electron -qe9.109x10-31kg 0.511 MeV(0.00055 amu)Particle Charge Mass Rest mass energy (m0c2)qe= 1.602x10−19CStructure within the nucleusThe periodic table52Ni53Ni54Ni55Ni56Ni57Ni58Ni59Ni60Ni61Ni62Ni64Ni65Ni66Ni67Ni68Ni69Ni70Ni71Ni72Ni73Ni63Ni74NiNumber of ProtonsNumber of Neutrons6126C175293118xN = ZN = Zν++→+eC N126127νeCB_126125++→(p+→ n + e++ ν)(n → p++ e−+ ν)beta − (β−)beta + (β+)(p++ e−→ n + ν)electron capture (EC)Exercises1. Predict the type of beta decay the unstable isotope 14C undergoesand give the end product of the decay.2. Predict the type of decay and product isotope for the radioactive isotope 36Ca.Average atomic mass (atomic weight) of MgIsotope Atomic mass (amu) relative abundance (%) 24Mg 23.985042 78.9925Mg 24.985837 10.0026Mg 25.982594 11.01Atomic weight = [(0.7899)(23.985042) + (0.1000)(24.985837) + (0.1101)(25.982594)]= 24.31 amuConversion factors for atom counting1 mole of atoms = 6.0221367x1023atomsmass of 1 mole of atoms = atomic weight1 g = 6.0221367x1023amu(1 atom Fe)(55.847 amu Fe/1 atom Fe)(1 g Fe/6.0221x1023amu Fe) = 9.274x10−23g Fe(1 mole Fe )(6.0221x1023atoms Fe/1 mole Fe)(55.847 amu Fe/1 atom Fe)x (1 g Fe/ 6.0221x1023amu Fe) = 55.847 g FeThe radius of a nucleus is approximately given by the simple formula R = 1.4x10−13A1/3cm. Calculate the density of a nucleus of 27Al.ExerciseMoleculesDalton’s hypothesis:1. Atoms are the fundamental particles of which matter is composed.2. Atoms of a particular element are identical in their chemical properties.3. Atoms of one element are distinguishable from atoms of other elements.4. Chemical change is the result of the union or separation of atoms (i.e., molecular transformations).Law of definite proportions: atoms combine in fixed proportionsExample: H2O RH/O= mH/mO= 2/16 = 1/8Law of multiple proportions:Example: R(Fe2O3)/R(FeO) = 2/3Molecular FormulasPotassium dichromate: K2Cr2O7One molecule contains2 atoms of K [mass = 2x39.1 = 78.2 amu]2 atoms of Cr [mass = 2x52.0 = 104 amu]7 atoms of O [mass = 7x16.0 = 112 amu]mass of one molecule = 294 amuOne mole of molecules contains2 moles of K [mass = 2x39.1 = 78.2 g]2 moles of Cr [mass = 2x52.0 = 104 g7 moles of O [mass = 7x16.0 = 112 g]mass of one mole of molecules = 294 gExercises involving molecular formulas1. How many oxygen atoms are in 5.00 g of potassium dichromate (K2Cr2O7)?2. What is the empirical formula of a compound containing 92.3% carbon,and 7.69% hydrogen by weight?3. If the molecular weight of the compound in the preceding exercise is78 amu, what is its molecular formula?Chemical reactions2 Ag+(aq)+ H2S(aq)→ Ag2S(s)+ 2 H+(aq)Chemical reactions conserve:Δm is usually negligible – in this caseΔm = −2.3x10−9 amuΔE = Δmc2= −2.114 eVAtomsChargeMass + energyNuclear reaction: 13 1H + 14 n →27Al∆m = 26.9815 amu – 13(1.007825) amu – 14(1.008665 amu) = − 0.2415 amuΔE = Δmc2= (−0.2415 amu)(931.5 MeV/amu) = −225.0 MeVChemical
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