1. Review

2. Stoichiometry u Greek for “measuring elements” u The calculations of quantities in chemical reactions based on a balanced equation. u We can interpret balanced chemical equations several ways. 2

3. In terms of Particles  Element- atoms  Molecular compound (non- metals)- molecule  Ionic Compounds (Metal and non-metal) - formula unit (ion ratio) 3

4. 2H 2 + O 2   2H 2 O u Two molecules of hydrogen and one molecule of oxygen form two molecules of water.  2 Al 2 O 3  Al + 3O 2 2 formula units Al 2 O 3 form 4 atoms Al And 3molecules O 2 2Na + 2H 2 O  2NaOH + H 2 2 sodium reacts with 2 water to form 2 sodium hydroxide and 1 hydrogen 4

5. Counting in Chemistry u We use certain words to indicate number. Examples: Dozen means a collection of 12 Gross means a collection of 144 Ream a package of 500 sheets of paper In Chemistry the particles that we deal with are very small. To collect a usable sample we need a very large number of particles – 6.02 x 10 23. This number is given the name mole (mol). 5

6. Look at it differently  2H 2 + O 2   2H 2 O u 2 dozen molecules of hydrogen and 1 dozen molecules of oxygen form 2 dozen molecules of water. u 2 moles of hydrogen and 1 mole of oxygen form 2 moles of water. 6

7. In terms of Moles  2 Al 2 O 3  Al + 3O 2  2Na + 2H 2 O  2NaOH + H 2 u The coefficients tell us how many moles of each kind 7

8. We can’t count to measure moles!! u The particles are too small. What can we do? u If you job was to roll pennies, there are three ways to complete this task: 1. We can count the pennies directly 2. Count by using mass (use average mass of a penny to count 50). 3. Use volume (Plastic rolls are sized to hold 50). Number 2 & 3 count indirectly by using mass and volume. This is how a chemist counts – uses mass and volume. 8

9. Counting Atoms by mass u On the periodic table the average atomic mass for carbon is 12.01 u. u 1 mole of carbon atoms has a mass of 12.01g u The mass of I mole is called the molar mass. u The molar mass of 1 mole of atoms for any element is the same numeric value as the atomic mass expressed in grams. Atomic mass H = 1.01u Molar mass H = 1.01g Atomic mass O = 16.00u Molar mass O = 16.00g Atomic mass N = 14.01u Molar mass N = 14.01g 9

10. Conversion factors u “A ratio of equivalent measurements” u Start with two things that are the same u one meter is one hundred centimeters u As an equation u 1 m = 100 cm u can divide by each side by 100cm to equal the number 1 u Called conversion factors because they allow us to convert (change) units. u really just multiplying by one, in a creative way. 100 cm1 m= 100 cm = 1 10

11. Conversion Factors u 75 cm x 1 m = 0.75 m 100cm 36.03 g C x 1 mole C = 3.000 mole C 12.01g C 2.50 mole C x 12.01 g C = 30.0g C 1 mole C 11

12. In terms of mass u The law of conservation of mass applies u We can check using moles  2H 2 + O 2   2H 2 O 2 moles H 2 2.02 g H 2 1 mole H 2 = 4.04 g H 2 1 moles O 2 32.00 g O 2 1 mole O 2 = 32.00 g O 2 36.04 g H 2 O 2 moles H 2 O18.02 g H 2 O 1 mole H 2 O = 12

13. Mole to mole conversions  2 Al 2 O 3  Al + 3O 2 u every time we use 2 moles of Al 2 O 3 we make 3 moles of O 2 2 moles Al 2 O 3 3 mole O 2 or 2 moles Al 2 O 3 3 mole O 2 13

14. Mole to Mole conversions u How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose?  2 Al 2 O 3  Al + 3O 2 3.34 mol Al 2 O 3 2 mol Al 2 O 3 3 mole O 2 5.01 moles O 2 x mole O 2 = = x 3 mole O 2 2 mol Al 2 O 3 3.34 mol Al 2 O 3 = x mole O 2 O2O2 Al 2 O 3 = 14

15. Your Turn  2C 2 H 2 + 5 O 2  4CO 2 + 2 H 2 O u If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? u O 2 = 5 = x x = 9.60 mol O 2 u C 2 H 2 2 3.84 mol u How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? u C 2 H 2 = 2 = x x = 8.95 mol C 2 H 2 u H 2 O 2 8.95 mol u If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? u CO 2 = 4 = x x = 4.94 mol CO 2 u C 2 H 2 2 2.47 mol 15

16. How do you get good at this? 16

17. Mass in Chemical Reactions How much do you make? How much do you need?

18. For example... u If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how much solid copper would form?  Fe + CuSO 4  Fe 2 (SO 4 ) 3 + Cu  2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu 10.1 g Fe 55.85 g Fe 1 mol Fe = 0.181 mol Fe 18

19. 2Fe + 3CuSO 4  Fe 2 (SO 4 ) 3 + 3Cu Fe 0.181 mol Fe 2 mol Fe 3 mol Cu = 0.272 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu Cu = x mol Cu = x mol Cu 0.181 mol Fe x 3 mol Cu 2 mol Fe x mol Cu = 19

20. Could have done it 10.1 g Fe 55.85 g Fe 1 mol Fe 2 mol Fe 3 mol Cu 1 mol Cu 63.55 g Cu =17.3 g Cu 20

21. We can also change u Liters of a gas to moles u At STP (0ºC and 101.3 kPa) u At STP 22.4 L of a gas = 1 mole u If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP? 21

22. For Example u If 6.45 grams of water are decomposed, how many liters of oxygen will be produced at STP?  2H 2 O  2H 2 + 1O 2 6.45 g H 2 O 18.02 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol O 2 22.4 L O 2 = 0.3579 mol H 2 O x mol O 2 0.3579 mol H 2 O = x mol O 2 = 0.1790 mol O 2 = 4.01 L O 2 at STP 22

23. Your Turn u How many liters of CO 2 at STP will be produced from the complete combustion of 23.2 g C 4 H 10 ? (58.14g/mol) u What volume of oxygen will be required? 2C 4 H 10 + 13O 2 → 8CO 2 + 10H 2 O 0.399mole C 4 H 10 = 1.60 mol CO 2 = 35.8L @STP = 2.59 mol O 2 = 58.1L @STP 23

24. Gases and Reactions A few more details

25. Example u How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ?  1CH 4 + 2O 2  1CO 2 + 2H 2 O 17.5 L O 2 22.4 L O 2 1 mol O 2 2 mol O 2 1 mol CH 4 22.4 L CH 4 = 8.75 L CH 4 at STP =0.7812 mol O 2 x mol CH 4 0.7812 mol O 2 = = 0.3906 mol CH 4 0.3906 mol CH 4 25

26. Avogadro told us u Equal volumes of gas, at the same temperature and pressure contain the same number of particles. u Moles are numbers of particles u You can treat reactions as if they happen liters at a time, as long as you keep the temperature and pressure the same. 26

27. Example u How many liters of CO 2 at STP are produced by completely burning 17.5 L of CH 4 ?  1CH 4 + 2O 2  1CO 2 + 2H 2 O 17.5 L CH 4 1 L CH 4 1 L CO 2 = 17.5 L CO 2 at STP x L CO 2 = 27