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Chem-Chpater 3.docx

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Department
Chemistry
Course Code
CHEM 1320
Professor
Ganley

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Chapter 3: Stoichiometry of Formulas and Equations  Mass Relationships in Chemical Reactions 8/28 Microscopic: atoms and molecules Macroscopic: grams Atomic mass: mass of an atom in amu Carbon=12.01amu Hydrogen=1.008amu Oxygen=16amu Average atomic mass: on other page Mole (mol): Contains as many particles as there are atoms in 12  grams of 12C 1mol=N A=6.022x10^23  How big is a mole? Large burger chain: How long to serve a mole of hamburgers? 6.022x10^23 hamburgers  ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­  = 1.67x10^14 3.6x10^9 hamburgers/day                                                 =458 billion  years! Counting objects of fixed relative mass: 12 apples @ 7g each=84g 12 oranges @4g each=48g 55.85gFe=6.022x10^23 atoms of Fe 32.07gS=6.022x10^23 atoms of S 1 12C atom                     12.00amu                 1.66x10^­24g ­­­­­­­­­­­­­­­    x   ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­   =   ­­­­­­­­­­­­­­­­­­­­­­­   12.00amu          6.022x10^23 12C atoms                  1amu 1amu=1.66x10^­24 or 1 g=6.022x10^23amu conversion table=molar mass in g/mol Molecular mass (or molecular weight) is the sum of the atomic  masses (in amu) in a molecule SO2: 1S=32.07amu 1O=16.00amux2 ­­­­­­­­­­­­­­­­­­­­­­­­ SO2=64.07amu For any molecule: molecular mass (amu)=Molar mass (grams) Table 3.1 How many atoms are in 0.551 g of K? 1mol K=39.10g K 1mol K=6.022x10^23 atoms K 0.551 g K         1 mol K              6.022x10^23 atoms K                      x  ­­­­­­­­­­­­­­­  x  ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­  =                             39.10 g K                    1 mol K =8.49x10^21 atoms K   How many H atoms are in 72.5 g of C3H8O? 1 mol C3H8O=(3x12) + (8x1) +16=60 g C3H8O 1 mol C3H8O molecules=8 mol H atoms 1 mol H=6.002x10^23 atoms H 72.5C3H8O  1mol C3H8O  8 mol H atoms  6.022x10^23 Hatoms                       x ­­­­­­­­­­­­­­  x  ­­­­­­­­­­­­­­­   x   ­­­­­­­­­­­­­­­­­­­­­­­­ =                          60g C3H8O     1mol C3H8O         1mol H atoms 5.82x10^24H atoms 8/30 Formula mass is the sum of the atomic masses (in amu) in a  formula unit of an ionic compound. 1Na=22.99amu + 1Cl=35.45amu = Nacl=55.44amu For any ionic compound, formula mass (amu)=molar mass (grams) 1 formula unit NaCl=58.44 amu 1 mole NaCl=58.44 g mol Formula mass of Ca3(PO4)2 ? 3Ca=3x40.08amu + 2P=2x30.97amu + 8O=8x16.00amu = Ca3(PH4)2=310.18amu Molar mass: 310.18 g mol Percent composition (by mass) 1 mole of CO weighs 28.01 g 12.01 g from C, 16.00 g from O 12.01/28.01x100%=42.88%Carbon 16.00/28.01x100%=57.12%Oxygen Percent composition of an element= n X molar mass of element  ­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­     X 100% molar mass of compound n is the number of moles of the element in 1 mole of the compound C2H6O % of Carbon=  2x12.01g/46.07gx100%=52.14% of Carbon % of Hydrogen= 6x1.008g/46.07gx100%=13.13% of Hydrogen % of Oxygen= 1x16.00g/46.07gx100%=34.73% of Oxygen All equal 100% A compound has a molar mass of 182.17g/mol. By mass, the  compound is 39.56% carbon, 7.74% hydrogen, and 52.70%  oxygen. What is the molecular formula? C6H14O6 Assume 1 mol 182.17g Q   39.56g Carbon           1 mol C                   x  ­­­­­­­­­­­­­­­­­­­  x   ­­­­­­­­­­­­­­­  =   6                           100g Q                    12.01 g C 182.17g Q     7.74g Hydrogen         1 mol H                   x    ­­­­­­­­­­­­­­­­­­­    x   ­­­­­­­­­­­­­­­  =   14                               100g Q                     1.008 g H 182.17g Q     52.70 g Oxygen          1 mol O                  x      ­­­­­­­­­­­­­­­­­­­  x   ­­­­­­­­­­­­­­­  =   6                                100g Q                    16.00 g O Book: assume 100gQ m 39.56gC 1mol/12.01g=3.294mol C/3.294=1 mol Cx3=3 7.74gH 1 mol/1.008g=7.68mol H=2.33 mol Hx3=7 52.70gO 1 mol/16.00g=3.294mol O=1 mol Ox3=3 Ratio: divide by the smallest number, then multiple by fraction C3H7O3 C­3x120.01=36.05 H­7x1.008=7.056 O­3x16.00=48.00 ­­­­­­­­­­­­­­­­­­­­­­­ 91.09 182.17/91.09=2 Now multiple the formula by 2 Ethanol contains only C, H, and O Combust 11.5 g ethanol Collect 22.0g CO2 and 13.5g H20 What is the empirical formula? Need to know grams and moles of C, H, O g CO2­­­m ▯ ol CO2­­­m▯ ol C­­­g▯  C 22.0 g CO2 x     1 mol CO2              1 mol C      ­­­­­­­­­­­­­­­­­­­  x  ­­­­­­­­­­­­­­=0.500 mol C       44.01 g CO2          1 mol CO2 0.500 mol C         12.01g C                         x  ­­­­­­­­­­­­­­­­  =6.01g C                                      mol C g H2O­­­m ▯ ol H2O­­­m ▯ ol H­­­▯  H 13.5 g H2O x     1 mol H2O              2 mol H      ­­­­­­­­­­­­­­­­­­­  x  ­­­­­­­­­­­­­­=1.50 mol H       18.02 g H2O          1 mol H2O 1.50 mol H         1.008 g H                         x  ­­­­­­­­­­­­­­­­  =1.51 g H                                      mol H g of O=g of sample­(g of C + g of H) 11.5 ethanol ­ (6.01g C + 1.51 g H) = 4.0 g O 4.0 g O          1 mol O                x  ­­­­­­­­­­­­­­­­  =0.25 mol O                        16.00 g O Mol C:0.500 Mol H:1.50 Mol O:0.25 Empirical=C0.5H1.5O0.25 Divide by smallest subscript (0.25) Empirical formula C2H6O 9/4 Constitutional Isomers of C2H6O Property                   Ethanol              Dimethyl Ether M  46.07 46.07 Boiling point 78.5 C ­25C Density @ 20C 0.789 g/mL liquid 0.00195 g/mL gas Structural          H     H  
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