CHEM 1211 Lecture 6 Outline of Last Lecture I Problem II Metal Non Metal Properties A Natural States B Two Non Metal RXNS III Molecular Nomenclature A Greek Prefixes B Examples IV Ionic Compounds A Cation B Anion C Chemical Formula D Examples V The periodic table A Main Group Metals a Group 1A b Group 2A c Groups A B Transition Metals C Non Metals a Group 7A b Group 6A c Group 5A D Examples VI Chemistry and Numbers of Things A M Ms B Carbon Atoms VII The Mole mol Outline of Current Lecture I Moles II Formula Weight III Calcium Carbonate IV Percent Composition V Elemental Composition Current Lecture I Moles V l x w x h or a x h You have the V of M Ms the area of U S What is h 6 6 x 1017m3 1 mol x 1 7 82 x 1012m2 8 4 x 104m 8 4 x 104m x 1km 103m x 0 621mi 1km 12 00 g of C 12 contains 6 022 x 1023 atoms of C 12 Atomic Mass of any element g 1 mol of atoms of element 6 022 x 10 23 1mol C 12 01g C 6 022 x 1023atoms of C 1mol O 16 00g O 6 022 x 1023 atoms of O What is the number of gold atoms in 1 00ug of Au 1molAu 6 022 x 1023 Au atoms 196 96gAu 6 1 00 x 10 gAu x 1 mol Au 196 96 gAu x 6 022 x 1023 1 mol Au 2 48 x 1016 Au Atoms II How many moles of Mg atoms are in 73 4g of Mg 73 4 24 305 3 02 mol Mg How do we calculate formula weight of a compound Ionic compounds Each element Determine how many atoms are present Multiply by atomic weight g The molecular weight of C4H10 H 10 10 amu 10 x 1 01 C 48 04 amu 4 x 12 01 1 mol of C4H10 54 14g 1 Ca NO3 2 formula weight Ca 40 08 40 08 O 16 00 x 6 96 00 N 14 01 x 2 28 02 FW 164 10 amu III Compound Compound Mass Molecule Atom Moles Cl2 10 90g NACl2 2 NA Cl2 1 mol Cl2 2 mol Cl C4H10 58 14g 4 NA C Atoms 10 NA H Atoms 4 mol C 10 mol H Zn ClO4 2 264 28g NAZn ClO4 2 f u NAZn2 ions 2 NA Cl 8 NA O 2 NA CLO4 ions 1 mol Zn ClO4 2 2 mol Zn2 ions 2 mol Cl 8 mol O 2 mol CLO4ions 1 mol Al OH 3 1 mol Al3 3 OH Calcium Carbonate chalk Ca2 CO32 2 958g 29 58 2 740 0 218g CaCO3 used After writing on the board 2 740g Ca 40 08 C 12 01 O 16 00 x 3 48 00 100 09 g CaCo3 mol CaCo3 0 218 g CaCO3 x 1 mol CaCO3 100 09g CaCO3 2 17 x 10 3 mol CaCO3 Left on chalkboard 2 17 x 10 3 mol CaCO3 x 6 022 x 1023 f u 1 mol CaCO3 1 31 x 1021 f u formula units of chalk on board 1 mol CaCO3 1 mol Ca2 2 17 x 10 3 mol CaCO3 x 1 mol Ca2 1 mol CaCO3 2 17 x 10 3mol Ca2 atoms of Ca left on board 2 17 x 10 3mol CaCO3 40 08 gCa 1 mol Ca 0 0869gCa Mass of Ca left on board IV V 1 Calculate the number of O atoms in 26 5g of LiCO3 1 mol LiCO3 6 022 x 1023 1 mol LiCO3 3 mol O Units LiCO3 73 80g LiCO3 Percent Composition gElement molar mass x 100 weight percentage 1 What is the composition of H in C4H10 butane Molecular Weight molar mass 58 14 g mol 10 10gH 58 14g C4H10 x 100 17 37 H 100 17 37 82 63 C Elemental Composition Lowest whole number ration empirical formula determined by finding comp 1 Ethane C2H6 CH3 2 C2H4 CH2 3 C2O4H2 CO2H Aikenes all have Emp Form of CH2
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