The mole road trip itinerary... % composition problems and combustion analysis (pp. 94-103) Reaction balancing (pp. 105-108) necessary prequel to moles level 3 Reaction stoichiometry predictions, limiting yields and % yields: moles level 3 ( pp. 108-123)

Reaction Stoichiometry problems: Moles level 3 how chemists cook Outline solution on board for each of the 3 samples…

Reaction stoichiometry means following….. THE RECIPE

12 How many boxes of cream to make 4 souffles ? THE RECIPE… 1 + 2 + 3 300 g/block 500 g/dozen 200 g/box 1900 g/souffle How many boxes of cream to make 4 souffles ? 12

2 1 + 2 + 3 600 grams cheese makes how many souffles ? Cooking and stoichiometry THE RECIPE 1 + 2 + 3 300 g/block 500 g/dozen 200 g/box 1900 g/souffle 600 grams cheese makes how many souffles ? 2

Cooking and stoichiometry THE RECIPE 1 + 2 + 3 300 g/block 500 g/dozen 200 g/box 1900 g/souffle 3000 g How many grams of eggs combine with 9 boxes of cream ?

2 48 eggs combine with how many blocks of cheese? 1 + 2 + 3 Cooking and stoichiometry THE RECIPE 1 + 2 + 3 300 g/block 500 g/dozen 200 g/box 1900 g/souffle 48 eggs combine with how many blocks of cheese? 2

Simple mol-mol stoichiometry conversions C3H8 + 5O2-------- 3CO2 + 4H2O How many moles of O2 will burn to form 1.2 moles of CO2 ? Method 1: factor label way 1) given 2) want * 5 mol O2 3 mol CO2 1.2 mol CO2 = ? mol O2 2 3) Use reaction stoichiometry coefficients in right ratio to cancel and connect…which ratio ???

Mol-mol stoichiometry conversions (continued) Method 2: `mole ratios’ way C3H8 + 5O2-------- 3CO2 + 4H2O 1.2 mol ?? m Problem stated visually How many moles of O2 will burn to form 1.2 moles of CO2 ? 1) ratio `wanted’ moles in numerator to given in denominator m 1.2 5 3 2) Set equal to matching coefficients for compounds given in reaction…which ??? = 1.2* m= 1.2 *5 1.2 3 = 2 3)Solve for m

More complex stoichiometry problem done 2 ways MW 44 32 44 18 g/mol C3H8+ 5O2  3CO2+ 4H2O 22 grams of C3H8 burned with O2 makes how many grams of H2O ? Method 1: factor label 22 g C3H8 x 1 mole C3H8 44 g C3H8 4 mol H2O 1 mole C3H8 x 18 g H2O 1 mol H2O x = ?? g H2O 36

Method 2: mole ratio way (Board first) 22 grams of C3H8 burned with O2 makes how many grams of H2O ? 44 32 44 18 g/mol Given: Problem stated visually C3H8+ 5O2  3CO2+ 4H2O ?? g 22 g

Method 2: mole ratio way 22 grams of C3H8 burned with O2 makes how many grams of H2O ? 44 32 44 18 g/mol Given: Problem stated visually C3H8+ 5O2  3CO2+ 4H2O ?? g 22 g 0) convert all given masses & molecule counts to moles = 0.5 mol C3H8 22 g C3H8 * 1 mol C3H8 44 g C3H8

?? g 22 g = 4 m = mol H2O 1 0.5 mol C3H8 44 32 44 18 g/mol Given: Method 2: mole ratio way 22 grams of C3H8 burned with O2 makes how many grams of H2O ? 44 32 44 18 g/mol Given: C3H8+ 5O2  3CO2+ 4H2O Problem stated visually ?? g 22 g = 0.5 mol C3H8 1) ratio `wanted’ moles m in numerator to given in denominator: = 4 1 m = mol H2O 0.5 mol C3H8

?? g 22 g 2) Solve for wanted moles m 0.5 1 0.5 * *0.5 = 2 mol H2O Method 2: mole ratio way 22 grams of C3H8 burned with O2 makes how many grams of H2O ? 44 32 44 18 g/mol Given: C3H8+ 5O2  3CO2+ 4H2O ?? g 22 g 2) Solve for wanted moles m 0.5 = 4 1 0.5 * *0.5 = 2 mol H2O

Method 2: mole ratio way 22 grams of C3H8 burned with O2 makes how many grams of H2O ? 44 32 44 18 g/mol Given: C3H8+ 5O2  3CO2+ 4H2O ?? g 22 g = 2 mol H2O 3) Convert wanted moles to wanted final units (grams) 2 mol H2O * 18 g H2O 1 mole H2O = 36 g H2O

Practicing…clickers and not clickers

c)wtwt: How many grams of O2 are needed to burn 1.1 g C3H8? C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM) 44 32 44 18 g/mol c)wtwt: How many grams of O2 are needed to burn 1.1 g C3H8? 0.8 4.0 0.25 0.16

C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM) 44 32 44 18 g/mol e) count wt: How many grams of O2 are needed to form 9*1022 molecules of H2O ? count wt 1.0 g 2.0 g 1.5 g 6.0 g

3 mol H2O C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM) 44 32 44 18 g/mol weight to moles How many moles of H2O form if 33 g of C3H8 are burned ? 3 mol H2O

=4*1022 How many molecules of CO2 form if 1.592 g H2O results ? C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM) 44 32 44 18 g/mol f) weight to count How many molecules of CO2 form if 1.592 g H2O results ? =4*1022

C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM) 44 32 44 18 g/mol Mol count: How many molecules of O2 does it take to form 0.020 moles of CO2 in the reaction above? (1 mol count=6*1023) 1.2*1022 7.2*1021 1.0*1022 2.0*1022

C3H8 + 5O2-------- 3CO2 + 4H2O 44 32 44 18 g/mol wtcount How many molecules of CO2 are formed along with 0.40 g H2O in the reaction above. (1 mol count =6*1023) 1.33*1022 1*1022 7.5*1021 4*1022