CHEM 1211 Lecture 6Outline of Last Lecture I. ProblemII. Metal/Non-Metal PropertiesA. Natural StatesB. Two Non-Metal RXNSIII. Molecular NomenclatureA. Greek PrefixesB. ExamplesIV. Ionic CompoundsA. CationB. AnionC. Chemical FormulaD. ExamplesV. The periodic tableA. Main Group Metalsa. Group 1Ab. Group 2Ac. Groups #AB. Transition MetalsC. Non-Metalsa. Group 7Ab. Group 6Ac. Group 5AD. ExamplesVI. Chemistry and Numbers of ThingsA. M&MsB. Carbon AtomsVII. The Mole (mol)Outline of Current Lecture I. MolesII. Formula WeightIII. Calcium CarbonateIV. Percent CompositionV. Elemental CompositionCurrent LectureI. MolesV = l x w x h or a x hYou 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 104m8.4 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 10231mol C = 12.01g C = 6.022 x 1023atoms of C1mol 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 1023Au atoms = 196.96gAu1.00 x 10-6gAu x (1 mol Au / 196.96 gAu) x(6.022 x 1023/1 mol Au)=2.48 x 1016 Au Atoms- How many moles of Mg atoms are in 73.4g of Mg?73.4/24.305 = 3.02 mol MgII. 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 C4H10H = 10.10 amu = 10 x 1.01C = 48.04 amu = 4 x 12.011 mol of C4H10 = 54.14g1. Ca(NO3)2 = formula weightCa= 40.08 = 40.08O= 16.00 x 6 = 96.00N= 14.01 x 2 = 28.02FW = 164.10 amuCompound Compound Mass Molecule / Atom MolesCl210.90g NACl22(NA)Cl21 mol Cl22 mol ClC4H1058.14g 4(NA)C Atoms10(NA)H Atoms4 mol C10 mol HZn(ClO4)2264.28g NAZn(ClO4)2 f.u.NAZn2+ ions2(NA)Cl8(NA)O2(NA)CLO4- ions1 mol Zn(ClO4)22 mol Zn2+ ions2 mol Cl8 mol O2 mol CLO4- ions1 mol Al(OH)3 = 1 mol Al3+ = 3-OHIII. Calcium Carbonate = chalkCa2+ + CO32- = 2.958g 29.58 – 2.740 = 0.218g CaCO3 usedAfter writing on the board: 2.740gCa = 40.08 C = 12.01O = 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 chalkboard2.17 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 board1 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 lefton board2.17 x 10-3mol CaCO3 ( 40.08 gCa / 1 mol Ca) = 0.0869gCa: Mass of Ca left on board1. Calculate the number of O atoms in 26.5g of LiCO31 mol LiCO3 = 6.022 x 10231 mol LiCO3 = 3 mol OUnits LiCO3 = 73.80g LiCO3IV. Percent CompositiongElement / molar mass x 100% = weight percentage1. What is the % composition of H in C4H10 (butane)?Molecular Weight = molar mass = 58.14 g /mol10.10gH / 58.14g C4H10 x 100% = 17.37% H100% - 17.37% = 82.63% CV. Elemental Composition**Lowest whole number ration: empirical formula (determined by finding % comp.)1. Ethane C2H6 = CH32. C2H4 = CH23. C2O4H2= CO2H***Aikenes all have Emp. Form. of
View Full Document
Unlocking...