1. M OLE / S TOICHIOMETRY By: Tyler Lewis, Michael Stylc and Erich Johnson

2. D EFINITION OF S TOICHIOMETRY The relationship between the relative quantities of substances taking part in a reaction or forming a compound, typically a ratio of whole integers.

3. A VOGADRO ’ S N UMBER 6.02 x 10^23

4. W HAT IS A MOLE ? The mole is a unit of measurement used in chemistry to express amounts of a chemical substance.

5. M OLAR M ASS Definition: The mass, in grams, of one mole of a substance

6. M ASS TO M OLE CONVERSION 1 mol = g-formula-mass (periodic table) Problem #1: How many moles in 28 grams of CO 2 ? Gram-formula-mass of CO 2 1 C = 1 x 12.01 g = 12.01 g 2 O = 2 x 16.00 g =32.00 g 44.01 g/mol 28 g CO 2 x 1 mole CO 2 = 0.64 mole CO 2 44.01 g CO 2

7. Problem #2: How many moles in 480. grams of Fe 2 O 3 ? 2 Fe = 2 x 55.85 g = 111.7 g 3 O = 3 x 16.00 g = 48.0 g 159.7 g/mol 480. gram Fe 2 O 3 x 1 moleFe 2 O 3 = 3.01 mole Fe 2 O 3 159.7 gram Fe 2 O 3

8. Problem #3: Find the number of moles of argon in 452 g of argon. Ar = 1 x 39.95 g = 39.95 g 39.95 g/mol 452 gram Ar x 1 mole Ar = 11.3 mole Ar 39.95 gram Ar

9. P ARTICLE TO M OLE CONVERSION 1 mol = 6.02 x 10 23 particles Problem #1: How many moles in 39.0 x 10 28 particles of CO 2 ? 39.0 x 10 28 particles CO 2 x 1 mole CO 2 = 6.48 x 10 28 mole CO 2 6.02 x 10 23 particles CO 2

10. Problem #2: How many moles are 1.20 x 10 25 particles of phosphorous? 1.20 x 10 25 particles P x 1 mole P = 19.9 mole P 6.02 x 10 23 particles P

11. Problem #3: How many moles in 48.0 x 10 23 particles of Cs ? 48.0 x 10 23 particles Cs x 1 mole Cs = 7.97 mole CS 6.02 x 10 23 particles Cs

12. V OLUME TO M OLE CONVERSION 1 mol = 22.4 L for a gas at STP Problem #1: How many moles in 22.4 Liters of CO 3 ? 22.4 L CO 3 x 1 mole CO 3 = 1 mole CO 3 22.4 L CO 3

13. Problem #2: How many moles of argon atoms are present in 11.2 L of argon gas at STP? 11.2 L Ar x 1 mole Ar =.500 mole CO 3 22.4 L Ar

14. Problem #3: How many moles of argon atoms are present in 403.2 L of Xenon gas at STP? 403.2 L Xe x 1 mole Xe = 18.00 mole Xe 22.4 L Xe

15. M OL -M OL CALCULATIONS N 2 + 3 H 2 ---> 2 NH 3 Problem #1: if we have 2 mol of N 2 reacting with sufficient H 2, how many moles of NH 3 will be produced? ratio to set up the proportion: That means the ratio from the equation is: NH 3 N 2 2121 2 mole N 2 x 2 mole NH 3 = 4mole NH 3 1 mole N 2 Initial amountFinal Amount

16. N 2 + 3 H 2 ---> 2 NH 3 Problem #2: Suppose 6 mol of H 2 reacted with sufficient nitrogen. How many moles of ammonia would be produced? 6 mole H 2 x 2 mole NH 3 = 4mole NH 3 3 mole H 2

17. N 2 + 3 H 2 ---> 2 NH 3 Problem #2: We want to produce 2.75 mol of NH 3. How many moles of nitrogen would be required? 2.75 mole NH 3 x 1 mole N 2 = 1.38 mole NH 3 2 mole NH 3

18. M ASS -M ASS CALCULATION 2 AuCl 3 ---> 2 Au + 3 Cl 2 Problem #1: How many grams of chlorine can be liberated from the decomposition of 64.0 g. of AuCl 3 ? 6 4.0 gram AuCl 3 x 1 mole AuCl 3 =.211 mole AuCl 3 303.35 grams AuCl 3 The molar ratio of chemical The amount of moles of initial chemical.211 mole AuCl 3 x 3 mole Cl 2 =.3165 mole Cl 2 2 mole AuCl 3 The mole ratio of chemicals.3165 mole Cl x 70.9 gram Cl 2 = 22.4 gram Cl 2 1 mole Cl 2

19. 3 AgNO 3 + AlCl 3 --> 3 AgCl + Al(NO 3 ) 3 Problem #2: Calculate the mass of AgCl that can be prepared from 200. g of AlCl 3 and sufficient AgNO 3 ? 200. gram AlCl 3 x 1 mole AlCl 3 = 1.50 mole AlCl 3 133.33 grams AlCl 3 1.50 mole AlCl 3 x 3 mole AgCl = 4.5 mole AgCl 1 mole AlCl 3 4.5 mole AgCl x 143.35 gram AgCl = 645. gram Cl 2 1 mole AgCl

20. 2 Au + 3 Cl 2 ---> 2 AuCl 3 Problem #4: How many grams of AuCl 3 can be made from 100.0 grams of chlorine? 100. gram Cl 2 x 1 mole Cl 2 = 1.41mole Cl 2 70.9 grams Cl 2 1.41 mole Cl 2 x 2 mole AuCl 3 =.94mole AuCl 3 3 mole Cl 2.94 mole AuCl 3 x 303.35 grams AuCl 3 = 285 gram AuCl 3 1 mole AuCl 3

21. P ARTICLE -P ARTICLE CALCULATION 2 AuCl 3 ---> 2 Au + 3 Cl 2 Problem #1: How many particles of chlorine can be liberated from the decomposition of 6.02x10 23 particles of AuCl 3 ? 6.02x10 23 particles AuCl 3 x 1 mole AuCl 3 = 1 mole AuCl 3 6.02x10 23 particles AuCl 3 The partial ratio of chemical The amount of moles of initial chemical 1 mole AuCl 3 x 3 mole Cl 2 = 1.5mole Cl 2 2 mole AuCl 3 The mole ratio of chemicals 1.5 mole Cl 2 x 6.02x10 23 particles Cl 2 = 9.03x10 23 particles Cl 2 1 mole Cl 2

22. 2 AuCl 3 ---> 2 Au + 3 Cl 2 Problem #2: How many particles of chlorine can be liberated from the decomposition of 150. particles of AuCl 3 ? 150. particles AuCl 3 x 1 mole AuCl 3 = 2.492x10 -22 mole AuCl 3 6.02x10 23 particles AuCl 3 2.492x10 -22 mole AuCl 3 x 3 mole Cl 2 = 3.738x10 -22 mole Cl 2 2 mole AuCl 3 3.738x10 -22 mole Cl 2 x 6.02x10 23 particles Cl 2 = 225 particles Cl 2 1 mole Cl 2

23. 3 AgNO 3 + AlCl 3 --> 3 AgCl + Al(NO 3 ) 3 Problem #3: Calculate the particles of AgCl that can be prepared from 132particals of AlCl 3 and sufficient AgNO 3 ? 132. article AlCl 3 x 1 mole AlCl 3 = 2.19x10 24 mole AlCl 3 6.02x10 23 particle AlCl 3 2.19x10 24 mole AlCl 3 x 3 mole AgCl = 6.57x10 24 mole AgCl 1 mole AlCl 3 6.57x10 24 mole AgCl x 6.02x10 23 particle AgCl =396. particle Cl 2 1 mole AgCl

24. V OLUME -V OLUME CALCULATION 2 AuCl 3 ---> 2 Au + 3 Cl 2 Problem #1: How many particles of chlorine can be liberated from the decomposition of 22.4 liters of AuCl 3 ? 22.4 liters AuCl 3 x 1 mole AuCl 3 = 1 mole AuCl 3 22.4 liters AuCl 3 The volume ratio The amount of moles of initial chemical 1 mole AuCl 3 x 3 mole Cl 2 = 1.5mole Cl 2 2 mole AuCl 3 The mole ratio of chemicals 1.5 mole Cl 2 x 22.4 liters Cl 2 = 33.6 liters Cl 2 1 mole Cl 2

25. N 2 + 3 H 2 ---> 2 NH 3 Problem #2: Suppose 5.6 liters of H 2 reacted with sufficient nitrogen. How many liters of ammonia would be produced? 5.6 liters H 2 x 1 mole H 2 =.25 moles H 2 22.4 liters H 2.25 mole H 2 x 2 mole NH 3 =.167mole NH 3 3 mole H 2.167 mole NH 3 x 22.4 liters NH 3 = 3.7 liters NH 3 1 mole NH 3

26. 2 Na + Cl 2 ---> 2 NaCl Problem #3: How many liters of Na are required to react completely with 75.0 liters of chlorine? 75. liters Cl 2 x 1 mole Cl 2 = 3.35moles Cl 2 22.4 liters Cl 2 3.35 mole Cl 2 x 2 mole Na = 6.7 mole Na 1 mole Cl 2 6.7 mole Na x 22.4 liters Na = 150. liters Na 1 mole Na

27. P ERCENT C OMPOSITION F ORMULAS Mass of element in sample of compound X 100= % Element in Compound Mass of sample of compound OR Mass of element in 1 mol of compound X 100= % Element in Compound Molar mass of compound

28. P ERCENT C OMPOSITION C ALCULATION Given: Formula Cu 2 S Problem #1 : Percent Composition of Sulfur 2 Mol Cu X 63.55 Cu = 127.1 g Cu mol Cu 1 mol S X 32.07g S = 32.07 g S mol S Molar mass of Cu 2 S = 159.2 g 32.07 g S x 100 = 20.15% S 159.2 g Cu 2 S

29. Problem #2 : Calculate the percent composition of carbon in the following: C 6 H 12 O 6 6 Mol C x 12.01 g/mole = 72.06 g C 12 Mol H x 1.008 g/mole=12.096 g H 6 Mol O x 16.00 g/mole =96.00 g O 180.2 g C 6 H 12 O 6 72.06 g C x 100 = 39.99% C 180.2 g C 6 H 12 O 6

30. Problem #3 : Calculate the percent composition of carbon in the following: CO 2 1 Mol C x 12.01 g/mole = 12.01 g C 2 Mol O x 16.00 g/mole =32.00 g O 44.01 g CO 2 12.01 g C x 100 = 27.29% C 44.01 g CO 26