Prof Gilbert LECTURE 5 CHEM 1211 Fall 10 Announcements Reminders Results of clicker quiz Last time Dalton s law of multiple proportions Ch 2 Names and formulas of acids Discovery of Sub atomic Particles Homework Assignment 1 is due this Sunday 9 19 Smartwork seems to be running OK now Relative mass ratios of the elements in small molecule compounds made of the same elements are simple who numbers Ex ratio of O to C in CO2 is twice what it is in CO J J Thomson s experiments with cathode ray tubes led to discovery and characterization of the mass to charge ratio of electrons Robert Millikan s oil drop experiments defined the mass and the charge of an electron Rutherford s gold foil experiments discredited Thomson s plum pudding model of the atom and produced the modern version This time Isotopes One of J J Thomson s students Fred Aston built a modified cathode ray tube and detected a beam of rays much heavier particles that turned out to be atomic nuclei When the tube contained Ne gas he got 2 beams one with a mass of 20 amu particles and a weaker beam of 22 amu particles the two principal isotopes of Ne 20Ne and 22Ne 2 stable isotopes of H 1H and 2H The neutron gray sphere and protons red spheres are called nucleons The number of protons in the nucleus of an atom defines its atomic number Z The combined number of protons and neutrons equals its mass number A The symbol of the isotope is or simply AX The weighted average of the masses of the isotopes defines the average atomic mass of an element Naturally occurring C is 98 892 1 108 Ave atomic mass 0 98892 12 0000 0 0108 13 00335 12 011 CLICKER QUIZ 2 Periodic Tables of the Elements The Rosetta Stone of Chemistry Earliest successful attempt 1860s Dmitri Mendeleev elements arranged in order of increasing atomic mass and by the formulas of the compounds they form with O and H Compare this early table against a modern one What is missing from Mendeleev s table Why How was he able to use only 8 columns instead of 18 CHAPTER 3 Molar Mass M mass in grams of a mole mol of a substance that is Avogadro s number NA 6 022 1023 particles atoms ions or molecules of the substance 1 For an element M atomic mass in grams e g 12 011g mol of C which is the mass of 6 022 1023 atoms of carbon 2 For a molecular compound M is the weighted sum of molar masses of the elements in the formula Ex Molar mass of glucose C6H12O6 6 mol C 12 011 g mol C 72 066 g C 12 mol H 1 008 g mol H 12 096 g H 16 000 g mol O 96 000 g O 6 mol O which is the mass of 6 022 1023 molecules of glucose 180 162 g mol of glucose Ex Molar mass of aluminum oxide Al2O3 3 For an ionic M is also the weighted sum of molar masses of the elements in the formula 2 mol Al 26 98 g mol Al 53 96 g Al 3 mol O 101 96 g mol of aluminum oxide 16 00 g mol O 48 00 g O Keep in mind there are 2 moles of aluminum ions Al3 and 3 moles of oxide ions O2 in one mole of Al2O3 Percent Composition Consider two of the minerals found in iron ore w stite FeO and hematite Fe2O3 Question What is the iron content composition of each Applying the COAST approach C O We know the formulas of the compounds and we can get their molar masses from the periodic table A The formulas tell us how many moles of Fe are in a mole of each of the compounds We can use the molar masses of Fe and O to calculate how many grams of Fe are in a molar mass of each compound and those ratios can tell us which has a higher percentage of Fe by mass S In a mole of FeO we have 1 mol Fe 1 mol O or 55 85 g of Fe 16 00 g of O 71 85 g mol of FeO The iron content of a mole of FeO is In a mole of Fe2O3 we have 2 mol Fe 3 mol O or and the iron content is T FeO is richer in Fe than Fe2O3 which makes sense because the mole ratios of O to Fe are 1 1 in FeO and 1 5 1 Fe2O3 Chemical Formulas from Percent Composition Consider another iron oxide mineral magnetite An assay of a sample of it for iron discloses that is 72 36 Fe Question What is the formula of this compound C O We know Fe of the compound and need to determine its chemical formula The only other element in the compound is O A If the only elements in the compound are Fe and O then the O content of the compound must be 100 00 72 36 27 64 O One way to translate values into moles is to assume that we have 100 grams of the compound then the values are the same as grams and can be converted into moles S The values of Fe and O turn into the same values in grams 72 36 g Fe and 27 64 g O The the number of moles of Fe and O in these masses is Calculate the ratio of these values by dividing both by the smaller one This mole ratio 1 333 1 000 is the same as 4 3 Therefore the chemical formula of the mineral is Fe3O4 This is an empirical formula