Chemistry in Life  You have a future job working for Consumer Reports  Testing advertising claims  An antacid company claims  Neutralizes ten times.

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Presentation transcript:

Chemistry in Life  You have a future job working for Consumer Reports  Testing advertising claims  An antacid company claims  Neutralizes ten times the acid of competitors  How do you test this claim?

CHEMICAL QUANTITIES Chapter 9

Chemical Reactions  Chemical reaction is the rearrangement of atoms to form new substances  Atoms are not created  Atoms are not destroyed  Atoms are shuffled

Chemical Equations  Chemical equations describe chemical reactions  Reactants  Products  Coefficients  Coefficients give us relative numbers of reactants and products  Allows us to predict amounts needed or able to be produced

Nonchemical Analogy  Making deli sandwiches 2 pieces of bread 3 slices of meat 1 slice of cheese  Boss wants 50 sandwiches  50 (2 pieces bread) + 50 (3 slices meat) + 50 (1 slice cheese) 50 sandwiches

Meaning of Chemical Equations Information Conveyed by Balanced Equation 2 molecules H molecule O 2 2 molecules H 2 O 2 dozen H 2 molecules + 1 dozen O 2 molecule 2 dozen H 2 O molecules 2 mol H 2 molecules + 1 mol O 2 molecule 2 mol H 2 O molecules 2 (6.02 x10 23 ) H 2 molecules x10 23 O 2 molecule 2 (6.02 x10 23 ) H 2 O molecules  Chemical equations describe chemical reactions  Coefficients give relative numbers

Coefficients  Coefficients give us relative numbers of moles  Can compare moles of one substance to moles of another in a reaction  Mole ratios  Made into conversion factors  Used in dimensional analysis

Mole Ratios  Balanced chemical equation  Coefficients  Show relative amounts Moles A Moles B Mole Ratios

 What number of moles of O 2 will be needed to produce 5.8 moles of water?  Mole ratio  1 mol O 2 : 2 mol H 2 O  5.8 mol H 2 O 1 mol O 2 2 mol H 2 O O mol O 2 =

Using Mole Ratios  Page 256  Do Example 9.3  Do Self-Check 9.1  Do Example 9.4

Example 9.3

Self-Check 9.1

Example 9.4

Molar Mass  Mass of one mole of a substance  Chemical formula Mass

Mole Ratios  Balanced chemical equation  Coefficients  Show relative amounts Moles A Moles B Mole Ratios

Combustion of Propane C 3 H 8 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g)  What mass of oxygen will be required to react exactly with 44.1 g of propane? C 3 H g C 3 H 5 C 3 H 5 ? mol C 3 H 5 O 2 ? mol O 2 O 2 ? g O 2

Combustion of Propane C 3 H 5 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g)  What mass of oxygen will be required to react exactly with 44.1 g of propane? C 3 H g C 3 H 5 C 3 H mol C 3 H 5 O mol O 2 O g O 2

Basic Strategy  C 3 H 5 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g) Mass A A ? mol A B ? mol B B ? g B Molar Mass Molar Mass Mole Ratios

Stoichiometry  Stoichiometry  Calculation of the quantities of reactants and products involved in a chemical reaction  Based on balanced chemical equations  Based on mole ratios and molar mass

Stoichiometry  Page 260  Do Self-Check 9.2  Do Self-Check 9.3  Get worksheet

Mole – Mole Problems  Asked to moles  Given moles A Moles A B ? mol B Mole Ratios

Moles – Mass Problems  Asked to find Mass of B  Given a Moles of A A Moles A B ? mol B B ? g B Molar Mass Mole Ratios

Mass–Mass Problems  Asked to find mass  Given a mass Mass A A ? mol A B ? mol B B ? g B Molar Mass Molar Mass Mole Ratios

Where are the limits?  Making deli sandwiches 2 pieces of bread 3 slices of meat 1 slice of cheese  Supplies at work 6 pieces bread 12 slices meat 5 slices cheese How many sandwiches can you make? What supply do you run out of?

Where are the limits?  Run out of bread  Can make 3 sandwiches  Excess meat and cheese

Chemicals Limit Reactions  Reactions  Chemical equations  Looks like there is the exact amount  Stoichiometric amounts  This would be called “going to completion”  Doesn’t usually happen  For particles to react: Must collide Proper orientation Enough energy

Chemicals Limit Reactions  Reactions  Don’t often go to completion  Often run out of one substance Limiting reactant  Excess of another substance Excess reactant

2H 2 + O 2 2H 2 O  You combine 10.0 grams of hydrogen gas and 15.0 g of oxygen gas. How many grams of water vapor are made? Which is the limiting reactant?

2H 2 + O 2 2H 2 O  You combine 10.0 grams of hydrogen gas and 15.0 g of oxygen gas. How many grams of water vapor are made? Which is the limiting reactant?

Theoretical Yield C 3 H 5 (g) + 5O 2 (g) 3CO 2 (g) + 4H 2 O(g)  Theoretical Yield  Based on stoichiometry  Calculated amount of product  The maximum amount that can be produced  What we think we should get C 3 H g C 3 H 5 C 3 H 5 ? mol C 3 H 5 O 2 ? mol CO 2 O 2 ? g CO 2

Actual Yield  Actual Yield  Based on measurements  Obtained amount of product  Produced in lab  What we actually got

Percent Yield  Percent Yield  Comparison  Ratio of actual yield to theoretical yield  Rarely 100%  Back and side reactions  % Yield = Actual Yield Theoretical Yield X 100

Percent Yield  Page 278  Do Example 9.9  Do Self-Check 9.8  Get Lab sheet