CHM1045 Lecture 9 Outline of Last Lecture I Percent Composition of Compounds 1 Example 1 2 Example 2 Percent composition Empirical Formulas 1 Example 3 2 Example 4 3 Example 5 4 Example 6 II Outline of Current Lecture III IV V Chemical i Chemical reaction definition ii Chemical equation definition How to read Chemical Equations Balancing Chemical Equations i Step 1 ii Step 2 iii Step 3 iv Step 4 v Step 5 vi Example 1 Current Lecture Chemical Reaction A process in which one or more substance is changed into one or more new substances Chemical Equation Uses chemical symbols to show what happens during a chemical reaction Reactants products 3 ways of representing the reaction of H2 with O2 to form H2O Two hydrogen molecules One oxygen molecule Two water molecules 2H2 How to Read Chemical Equations O2 2H2O 2Mg O2 2 MgO 2 atoms Mg 1 molecules O2 makes 2 formula units MgO 2 moles Mg mole O2 makes 2 moles MgO 48 6 grams Mg 32 0 grams O2 makes 80 6g MgO Balancing Chemical Equations 1 Write the correct formula s for the reactants on the left side and the correct formula s for the product s on the right side of the equation Ethane reacts with oxygen to form carbon dioxide and water C2H6 CO2 H2O 2 Change the numbers in front of the formulas coefficients to make the number of atoms of each element the same on both sides of the equation Do not change the subscripts 2C2H6 3 Start by balancing those elements that appear in only one reactant and one product C2H6 O2 CO2 H2O 2 carbon 1 carbon on left on right C2 H 6 O 2 6 Hydrogen on left Start with C or H but not O Multiply CO2 by 2 2CO2 H2O 2 hydrogen on right Multiple H2O by 3 C2H6 O2 2CO2 3H2O 4 Balance those elements that appear in two or more reactants or products C2H6 O2 2CO2 3H2O Multiply O2 by 7 2 2 oxygen 4 oxygen 3 oxygen 7 oxygen on right on left 2 2 3 1 C2H6 7 2 O2 2CO2 3H2O remove fraction multiply both sides by 2 2C2H6 7 O2 4CO2 6H2O 5 Check to make sure that you have the same number of each type of atom on both sides of the equation 2C2H6 7O2 4CO2 6H2O Reactants Product 4 C 2 x 2 4C 4C 4C 12 H 2 x 6 12 H 6 x 2 12H 12 H 14 O 7 x 2 14 O 4 x 2 6 14 O 14O Example 1 When aluminum metal is exposed to air a protective layer of aluminum oxide Al2O3 forms on its surface This layer prevents further reaction between aluminum and oxygen and it is the reason that aluminum beverage cans do not corrode In the case of iron the rust or iron III oxide that forms is too porous to protect the iron metal underneath so rusting continues Write a balanced equation for the formation of Al2O3 Al2O3 Al O Al2O3 2 Al 3O2 2 Al2O3 2 Al 3O2 2 Al2O3 4al 3O2 2Al2O3 Al O 4 4 6 6
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