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

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

3. Reaction stoichiometry means following…..
THE RECIPE

4. 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

5. 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

6. 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 ?

7. 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

8. Simple mol-mol stoichiometry conversions
C3H8 + 5O  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 O mol CO2
1.2 mol CO2
= ? mol O2 2
3) Use reaction stoichiometry coefficients in right ratio to cancel and connect…which ratio ???

9. Mol-mol stoichiometry conversions (continued)
Method 2: `mole ratios’ way
C3H8 + 5O  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= *5
= 2
3)Solve for m

10. More complex stoichiometry problem done 2 ways
MW 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

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

12. Method 2: mole ratio way
22 grams of C3H8 burned with O2 makes how many grams of H2O ?
g/mol
Given:
Problem
stated
visually
C3H8+ 5O2  3CO H2O
?? 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

13. ?? 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 ?
g/mol
Given:
C3H8+ 5O2  3CO H2O
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

14. ?? 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 ?
g/mol
Given:
C3H8+ 5O2  3CO H2O
?? g
22 g
2) Solve for wanted moles m
0.5
= 4 1
0.5 *
*0.5
= 2 mol H2O

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

16. Practicing…clickers and not clickers

17. c)wtwt: How many grams of O2 are needed to burn 1.1 g C3H8?
C3H8 + 5O  3CO2 + 4H2O (BOOM)
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

18. C3H8 + 5O2-------- 3CO2 + 4H2O (BOOM)
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

19. 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

20. =4*1022 How many molecules of CO2 form if 1.592 g H2O results ?
C3H8 + 5O  3CO2 + 4H2O (BOOM)
g/mol
f) weight to count
How many molecules of CO2 form if g H2O results ?
=4*1022

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

22. C3H8 + 5O  3CO2 + 4H2O
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