determining empirical formula from mass % data to convert the mass % composition obtained from a...

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