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  • Determining Empirical Formula from Mass % DataTo convert the mass % composition obtained from a combustion analysis into an empirical formula, we must convert the mass % of each type of atom into the relative number of atoms.To do this, assume that we have 100g of sampleThe mass % will then be in grams

  • Determining Molecular FormulasOnce we know the empirical formula, we need one more piece of information to determine the molecular formula: The Molar MassSay we know the empirical formula of a compound is C3H4O3.All we know about this compound at this point is the ratio of the 3 elements. We dont know the exact number of each type of atom in the molecule.Is the Molecular Formula C6H8O6, C12H16O12 or C18H24O18?

  • G: MolarityHands down one of the most important concepts you need to master is you are going to stay in the sciences. Period.The Molar Concentration, c, of a solute in solution is the number of moles of solute divided by the volume of the solution (in liters).Also referred to as Molarity

  • MolarityThe symbol M is used to denote the molarity of the solution1M NaCl = 1 mole NaCl per liter of H2O

  • G4: DilutionsFrequently in the laboratory, you will need to make dilutions from a stock solution.This involves taking a volume from the stock and bringing it to a new volume with solvent.In order to perform these dilutions, we can use the following equation:c1V1 = c2V2

    Where: c1 = Stock concentrationV1 = Volume removed from stockc2 = Target conc of new solnV2 = Volume of new solution

  • Law of Conservation of MatterMatter can neither be created nor destroyed Antoine Lavoisier, 1774If a complete chemical reaction has occurred, all of the reactant atoms must be present in the product(s)

  • Law of Conservation of Mattera)b)Stoichiometry coefficients are necessary to balance the equation so that the Law of Conservation of Matter is not violated6 molecules of Cl2 react with 1 molecule of P43 molecules of Cl2 react with 2 molecules of Fe

  • Example of Using Stoichiometric Coefficients

  • Balancing Chemical ReactionsLets look at Oxide FormationMetals/Nonmetals may react with oxygen to form an oxide with the formula MxOy

    Example 1: Iron reacts with oxygen to give Iron (III) OxideFe (s) + O2 (g) Fe2O3 (s)

  • How do we solve it?Step 1: Look at the product. There are 3 atoms of oxygen in the product, but we start with an even number of oxygen atoms.Lets convert the # of oxygens in the product to an even numberFe (s) + O2 (g) Fe2O3 (s)Result: Fe (s) + O2 (g) 2Fe2O3 (s)

  • How do we Solve It?Then, balance the reactant side and make sure the number/type of atoms on each side balance.Fe (s) + O2 (g) 2Fe2O3 (s)Balanced Equation: 4Fe (s) + O2 (g) 2Fe2O3 (s)

  • How do we Solve It?Example 2: Sulfur and oxygen react to form sulfur dioxide.S (s) + O2 (g) SO2 (g)

    Step 1: Look at the reaction. We lucked out!Balanced Equation: S (s) + O2 (g) SO2 (g)

  • How do we Solve It?Example 3: Phosphorus (P4) reacts with oxygen to give tetraphosphorus decaoxide.P4 (s) + O2 (g) P4O10 (s)Step 1: Look at the reaction. The phosphorus atoms are balanced, so lets balance the oxygens.Balanced Equation: P4 (s) + 5O2 (g) P4O10 (g)

  • How do we Solve It?Example 4: Combustion of Octane (C8H18).C8H18 (l) + O2 (g) CO2 (g) + H2O (g)

    Step 1: Look at the reaction. Then: Balance the CarbonsC8H18 (l) + O2 (g) 8CO2 (g) + H2O (g)

  • How do we Solve It?Step 2: Balance the HydrogensC8H18 (l) + O2 (g) 8CO2 (g) + 9H2O (g)C8H18 (l) + O2 (g) 8CO2 (g) + H2O (g)Step 3: Balance the OxygensProblem! Odd number of oxygen atomsSolution: Double EVERY coefficient (even those with a value of 1)

  • How do we Solve It?Step 3 (contd): Balance the Oxygens2C8H18 (l) + 25O2 (g) 16CO2 (g) + 18H2O (g)C8H18 (l) + 12.5O2 (g) 8CO2 (g) + 9H2O (g)Step 4: Make sure everything checks out

  • Review of Balancing Equations

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