OFB Chapter 1 18/22/2006Chapter 1The Atomic Nature of Matter• 1-1 Chemistry: Science of Change• 1-2 The Composition of Matter• 1-3 The Atomic Theory of Matter• 1-4 Chemical Formulas and Relative Atomic Masses• 1-5 The Building Blocks of the Atom• 1-6 Finding Atomic Masses the Modern Way• 1-7 The Mole Concept: Counting and Weighing Atoms and Molecules• 1-8 Finding Empirical and Molecular Formulas the Modern Way• 1-9 Volume and DensityOFB Chapter 1 28/22/2006Atomic Theory of Matter• Law of conservation of mass:Mass is neither created nor destroyed in a chemical reaction• Dalton’s Atomic Theory of Matter (1808):1. All matter consists of solid and indivisible atoms2. All atoms of a given chemical element are identical in mass and in all other properties3. Different elements have different kinds of atoms; these atoms differ in mass from element to element4. Atoms are indestructible and retain their identity in all chemical reactions5. The formation of a compound from its elements occurs through the combination of atoms of unlike elements in small whole-number ratio.OFB Chapter 1 38/22/2006Chemical Formulas and Relative Atomic Masses• Chemical Formulas display symbols for the elements and the relative number of atoms– E.g., NH3, CO2, CH3CO2H or C2H4O2• Molecules are groupings of two or more atoms bound closely together by strong forces that maintain them in a persistent combinationOFB Chapter 1 48/22/2006Building Blocks of the Atom• Electrons, Protons and Neutrons– Electrons discovered in 1897 by Thomson– Rutherford proposed that the atomic nucleus was composed of neutral particles called Neutrons and positively charged particles called protons– Neutron number = N– Atomic number = Z = number of Protons– Atomic mass number = AA = Z + NOFB Chapter 1 58/22/2006http://www.chemsoc.org/viselements/pages/alchemist/alchemy.htmlhttp://www.chemsoc.org/viselements/pages/pertable_j.htmhttp://www.chemsoc.orgOFB Chapter 1 68/22/20061530.794PPhosphorus1428.086SiSilicon714.007NNitrogen612.011CCarbonNonMetalSemiMetalOFB Chapter 1 78/22/2006612.011CCarbonAtomic NumberAtomic MassA = Z + NAtomic Mass = # Protons + # NeutronsFor Carbon, 12 = 6 + NeutronsNeutrons = 6Every Carbon atom has 6 electrons, 6 protons and 6 neutronsOFB Chapter 1 88/22/2006• Theory and ExperimentM + e–(70 eV) M++ 2e–M+lower mass ions• Mass Spectrometeraccelerates ions (or molecular ions) in an electric field and then separates those ions by relative mass in a magnetic fieldMass Spectrometry and IsotopesOFB Chapter 1 98/22/2006Mass Spectrometry and Isotopes• Mass Spectrometeraccelerates ions (or molecular ions) in an electric field and then separates those ions by relative mass in a magnetic fieldMass Spectrometer Separation of Chlorine02040608010035 37Relative MassRelative Amount1735.453ClChlorineOFB Chapter 1 108/22/2006Isotopes of Cl:MASS abund.HalflifeParticle, EnergyDecay Product(s)Isotopic Mass047200 nsecB-/B-n,14.700Ar-47/Ar-4646.987976 0460.22 secB-/B-n,6.900Ar-46/Ar-4545.984111 045400 msecB-/B-n,10.800Ar-45/Ar-4444.979710 0440.43 secB-/B-n,3.920Ar-44/Ar-4343.978539 0433.3 secB-,7.950 MeVAr-4342.974202 0426.8 secB-,9.430 MeVAr-4241.973172 04138.4 secB-,5.730 MeVAr-4140.970649 0401.35 minB-,7.480 MeVAr-4039.970413 03955.6 minB-,3.442 MeVAr-3938.968008 03837.24 minB-,4.917 MeVAr-3837.968010 03724.23%Stable36.96590363.01E+5 yrB-/EC,10.413Ar-36/S-3635.968303575.77%Stable34.96880341.5264 secEC,5.492 MeVS-3433.97376 0332.511 secEC,5.583 MeVS-3332.97745 032 298 msecEC/ECa/ECp,12.685S-32/Si-28/P-3131.985688 031 150 msecEC/ECp,11.980S-31/P-3030.992435Cl contains protons and neutrons37Cl contains protons and neutronsOFB Chapter 1 118/22/2006Atoms• Avogadro’s Number is the number of 12C atoms in exactly 12 grams of carbonN0= 6.0221420 X 1023• The mass, in grams, of Avogadro's number of atoms of an element is numerically equal to the relative atomic mass of that element C atomofass m =OFB Chapter 1 128/22/2006• Relative Molecular Mass of a molecule equals the sum of the relative atomic masses of all of the atoms making up the moleculeMolecules=2CO of massmolecular relative molecule CO mass2OFB Chapter 1 138/22/2006Moles•A mole measures the chemical amount of a substance• Mole is an abbreviation of gram molecular weight• One mole of a substance equals the amount that contains Avogadro's number of atoms, molecules.• One mole = Molar mass (M) of that element or moleculeOFB Chapter 1 148/22/2006Exercise 1-6• Molecules of isoamyl acetate have the formula C7H14O2. Calculate (a) how many moles and (b) how many molecules are present in 0.250 grams of isoamyl acetate.• Strategy:1. Calculate molar mass of C7H14O22. Calculate the number of moles in 0.250 grams3. Using Avogadro’s number to calculate the number of molecules in the number of moles of C7H14O2OFB Chapter 1 158/22/2006Exercise 1-6• Molecules of isoamyl acetate have the formula C7H14O2. Calculate (a) how many moles and (b) how many molecules are present in 0.250g of isoamyl acetate.• Solution:1. Calculate molar mass of C7H14O22. Calculate the number of moles in 0.250 grams3. Using Avogadro’s number calculate the number of molecules in “n” moles of C7H14O2OFB Chapter 1 168/22/2006Percentage Composition from Empirical or Molecular Formula Exercise 1-8• Tetrodotoxin, a potent poison found in the ovaries and liver of the globefish, has the empirical formula C11H17N3O8. Calculate the mass percentages of the four element in this compound.Strategy:1. Calculate molar mass of C11H17N3O, by finding the mass contributed by each element2. Divide the mass for each element by the total mass of the compound.OFB Chapter 1 178/22/2006Exercise 1-8Tetrodotoxin has the empirical formula C11H17N3O8. Calculate the mass percentages of the four element in this compound.Solution:1. Calculate molar mass of C11H17N3O8, by finding the mass contributed by each element2. Divide the mass for each element by the total mass of the compound.319.16127.99%O319.1642.021%N319.1617.134%H319.16132.01%C====319.16 ==MMOFB Chapter 1 188/22/20061 mmol = 1 millimole=1x10-3 mol1mg = 1 milligram =1x10-3gMillimoles and MilligramsOFB Chapter 1 198/22/2006Exercise 1-10Moderate Heating of 97.44 mg of a compound containing nickel, carbon and oxygen and no other elements drives off all of the carbon and oxygen in the form of carbon monoxide (CO) and leaves 33.50 mg of metallic nickel behind. Determine the