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Mole Problems Solution Guide. How many water molecules are found in 18.0g water? 602000000000000000000000 6.02x10 23 water molecules If you had 6.02x10.

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Presentation on theme: "Mole Problems Solution Guide. How many water molecules are found in 18.0g water? 602000000000000000000000 6.02x10 23 water molecules If you had 6.02x10."— Presentation transcript:

1 Mole Problems Solution Guide

2 How many water molecules are found in 18.0g water? x10 23 water molecules If you had 6.02x10 23 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x10 23 = $238,616,184,677, (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

3 How many water molecules are found in 18.0g water? x10 23 water molecules If you had 6.02x10 23 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x10 23 = $238,616,184,677, (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

4 How many water molecules are found in 18.0g water? x10 23 water molecules If you had 6.02x10 23 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x10 23 = $238,616,184,677, (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

5 How many water molecules are found in 18.0g water? x10 23 water molecules If you had 6.02x10 23 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x10 23 = $238,616,184,677, (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

6 How many water molecules are found in 18.0g water? x10 23 water molecules If you had 6.02x10 23 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x10 23 = $238,616,184,677, (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min)

7 How many water molecules are found in 18.0g water? x10 23 water molecules If you had 6.02x10 23 dollar bills you would need to spend $238 trillion dollars per second for every day of an 80 year life span to spend that many dollars, if you shared it with each person on the planet they would have to spend $34,000 dollars per second every day of their 80 year life span to spend it all. $6.02x10 23 = $238,616,184,677, (80yrs ) (365 d) ( 24hr) ( 60min) (60s ) second ( 1yr ) ( 1 d) ( 1hr ) (1min) In any measurable sample of matter there are too many particles to count directly! Atoms are exceedingly small!!!

8 1 12 One H has a relative mass of ______ amu. One C has a relative mass of _______ amu H have a relative mass of ______ amu. 10 C have a relative mass of ______ amu H have a relative mass of _____ amu. 100 C have a relative mass of ______ amu H have a relative mass of ______ amu C have a relative mass of ______ amu 1 6

9 1 12 One H has a relative mass of __1___ amu. One C has a relative mass of _______ amu H have a relative mass of ______ amu. 10 C have a relative mass of ______ amu H have a relative mass of _____ amu. 100 C have a relative mass of ______ amu H have a relative mass of ______ amu C have a relative mass of ______ amu 1 6

10 1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu H have a relative mass of ______ amu. 10 C have a relative mass of ______ amu H have a relative mass of _____ amu. 100 C have a relative mass of ______ amu H have a relative mass of ______ amu C have a relative mass of ______ amu 1 6

11 1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu H have a relative mass of _10___ amu. 10 C have a relative mass of ______ amu H have a relative mass of _____ amu. 100 C have a relative mass of ______ amu H have a relative mass of ______ amu C have a relative mass of ______ amu 1 6

12 1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu H have a relative mass of ____ amu. 100 C have a relative mass of ______ amu H have a relative mass of ______ amu C have a relative mass of ______ amu 1 6

13 1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu H have a relative mass of 100 amu. 100_ C have a relative mass of ______ amu H have a relative mass of ______ amu C have a relative mass of ______ amu 1 6

14 1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu H have a relative mass of 100 amu. 100_ C have a relative mass of 1200_ amu H have a relative mass of ______ amu C have a relative mass of ______ amu 1 6

15 1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu H have a relative mass of 100 amu. 100_ C have a relative mass of 1200_ amu H have a relative mass of 1million amu C have a relative mass of ______ amu 1 6

16 1 12 One H has a relative mass of __1___ amu. One C has a relative mass of ___12__ amu H have a relative mass of _10___ amu. 10 C have a relative mass of 120__ amu H have a relative mass of 100 amu. 100_ C have a relative mass of 1200_ amu H have a relative mass of 1million amu C have a relative mass of 12 million amu 1 6

17 1a) The mass of the sample of carbon is always ______ more than an equivalent number of hydrogens

18 Mole Unit Questions 1a) The mass of the sample of carbon is always _12x__ more than an equivalent number of hydrogens

19 Mole Unit Question x10 23 atoms of hydrogen have a mass of 1.0 g that equals the atomic mass of hydrogen found on the periodic table 6.02x10 23 atoms of carbon have a mass of 12.0 g that equals the atomic mass of carbon found on the periodic table

20 Mole Unit Question x10 23 atoms of hydrogen have a mass of 1.0 g that equals the atomic mass of hydrogen found on the periodic table 6.02x10 23 atoms of carbon have a mass of 12.0 g that equals the atomic mass of carbon found on the periodic table

21 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

22 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

23 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

24 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

25 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

26 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

27 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

28 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

29 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

30 Questions What is the mass of 3.01 x carbon–12 atoms? 3.01 x is one half of 6.02 x therefore the mass will be one half of 12 g = 6g 3. What is the mass of 1.51 x carbon–12 atoms? 1.51 x is one fourth of 6.02 x therefore the mass will be one fourth of 12 g = 3g 4. What is the mass of x carbon–12 atoms? x is twice 6.02 x therefore the mass will be twice 12 g = 24 g

31 Questions What is the mass of x carbon–12 atoms? x is four times 6.02 x therefore the mass will be four time 12 g = 48g

32 Questions What is the mass of x carbon–12 atoms? x is four times 6.02 x therefore the mass will be four time 12 g = 48g

33 Questions What is the mass of x carbon–12 atoms? x is four times 6.02 x therefore the mass will be four time 12 g = 48g

34 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

35 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

36 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

37 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

38 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

39 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

40 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

41 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

42 Questions How many carbon atoms are in 36.0 g 36.0g is three times 12.0g therefore there are three times 6.02x10 23 C = 18.06x10 23 C = 1.806x10 24 C 7. How many carbon atoms are in 4.0 g 4.0 g is one third of 12.0g therefore one third of 6.02x10 23 C = 2.01x10 23 C 8. How many carbon atoms are in 1.0 g 1.0 g is one twelfth of 12.0g therefore one twelfth 6.02x10 23 C = 5.01x10 22 C

43 Bonus Question 6.02x10 23 C = 12.0 g C 1.0 x10 6 C = x 6.02x10 23 C 12.0 g C Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C = 1.99x g C

44 Bonus Question 6.02x10 23 C = 12.0 g C 1.0 x10 6 C = x 6.02x10 23 C 12.0 g C Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C = 1.99x g C

45 Bonus Question 6.02x10 23 C = 12.0 g C 1.0 x10 6 C = x 6.02x10 23 C 12.0 g C Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C = 1.99x g C

46 Bonus Question 6.02x10 23 C = 12.0 g C 1.0 x10 6 C = x 6.02x10 23 C 12.0 g C Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C = 1.99x g C

47 Bonus Question 6.02x10 23 C = 12.0 g C 1.0 x10 6 C = x 6.02x10 23 C 12.0 g C Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C = 1.99x g C

48 Bonus Question 6.02x10 23 C = 12.0 g C 1.0 x10 6 C = x 6.02x10 23 C 12.0 g C Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C = 1.99x g C

49 Bonus Question 6.02x10 23 C = 12.0 g C 1.0 x10 6 C = x 6.02x10 23 C 12.0 g C Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C = 1.99x g C

50 Bonus Question Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C 1.0 x10 6 C ( 12.0 g C ) = 1.99x g C (6.02x10 23 C )

51 Bonus Question Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C 1.0 x10 6 C ( 12.0 g C ) = 1.99x g C (6.02x10 23 C )

52 Bonus Question Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C 1.0 x10 6 C ( 12.0 g C ) = 1.99x g C (6.02x10 23 C )

53 Bonus Question Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C 1.0 x10 6 C ( 12.0 g C ) = 1.99x g C (6.02x10 23 C )

54 Bonus Question Cross multiply 1.0 x10 6 C multiplied by 12.0 g then divided by 6.0x10 23 C 1.0 x10 6 C ( 12.0 g C ) = 1.99x g C (6.02x10 23 C )

55 6.02 x = P.T. grams 9. How many silver atoms are in 25.0 grams of silver?

56 6.02 x = P.T. grams 9. How many silver atoms are in 25.0 grams of silver? 25.0g Ag

57 6.02 x = P.T. grams 9. How many silver atoms are in 25.0 grams of silver? 25.0g Ag 6.02 x Ag =

58 6.02 x = P.T. grams 9. How many silver atoms are in 25.0 grams of silver? 25.0g Ag 6.02 x Ag = 107.9g

59 6.02 x = P.T. grams 9. How many silver atoms are in 25.0 grams of silver? 25.0g Ag 6.02 x Ag =1.39x10 23 Ag 107.9Ag

60 6.02 x = P.T. grams 10. What is the mass of 1.00 x 109 silver atoms? 1.00 x 10 9 Ag

61 6.02 x = P.T. grams 10. What is the mass of 1.00 x 109 silver atoms? 1.00 x 10 9 Ag ( g Ag ) =

62 6.02 x = P.T. grams 10. What is the mass of 1.00 x 109 silver atoms? 1.00 x 10 9 Ag ( g Ag ) = (6.02 x Ag )

63 6.02 x = P.T. grams 10. What is the mass of 1.00 x 109 silver atoms? 1.00 x 10 9 Ag ( g Ag ) = 1.79 x g Ag (6.02 x Ag )

64 6.02 x = P.T. grams 11. What is the mass of 2.00 x gold atoms? 2.00 x Au ( g Au ) = 65.4 g Au (6.02 x Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x Au = 3.06 X10 23 Au g Au

65 6.02 x = P.T. grams 11. What is the mass of 2.00 x gold atoms? 2.00 x Au ( g Au ) = 65.4 g Au (6.02 x Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x Au = 3.06 X10 23 Au g Au

66 6.02 x = P.T. grams 11. What is the mass of 2.00 x gold atoms? 2.00 x Au ( g Au ) = 65.4 g Au (6.02 x Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x Au = 3.06 X10 23 Au g Au

67 6.02 x = P.T. grams 11. What is the mass of 2.00 x gold atoms? 2.00 x Au ( g Au ) = 65.4 g Au (6.02 x Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x Au = 3.06 X10 23 Au g Au

68 6.02 x = P.T. grams 11. What is the mass of 2.00 x gold atoms? 2.00 x Au ( g Au ) = 65.4 g Au (6.02 x Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x Au = 3.06 X10 23 Au g Au

69 6.02 x = P.T. grams 11. What is the mass of 2.00 x gold atoms? 2.00 x Au ( g Au ) = 65.4 g Au (6.02 x Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x Au = 3.06 X10 23 Au g Au

70 6.02 x = P.T. grams 11. What is the mass of 2.00 x gold atoms? 2.00 x Au ( g Au ) = 65.4 g Au (6.02 x Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x Au = 3.06 X10 23 Au g Au

71 6.02 x = P.T. grams 11. What is the mass of 2.00 x gold atoms? 2.00 x Au ( g Au ) = 65.4 g Au (6.02 x Au ) 12. How many gold atoms are in 100 grams of gold? 100g Au 6.02 x Au = 3.06 X10 23 Au g Au

72 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

73 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

74 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

75 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

76 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

77 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

78 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

79 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

80 6.02 x = P.T. grams 13. What is the mass 2.00 x platinum atoms? 2.00 x Pt ( g Pt ) = 648 g Pt (6.02 x Pt ) 14. What is the mass of 2.00 x silicon atoms? 2.00 x Si ( 28.1g Si ) =.93 g Si (6.02 x Si )

81 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

82 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

83 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

84 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

85 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

86 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

87 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

88 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

89 6.02 x = P.T. grams 15. How many silicon atoms are in 12.0 grams of silicon atoms? 12g Si 6.02 x Si = 2.57X10 23 Si 28.1g Si 16. How many lead atoms are in 12.0 grams of lead? 12.0g Pb 6.02 x Pb = 3.49 X10 22 Pb g Pb

90 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

91 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

92 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

93 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

94 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

95 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

96 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 ) )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

97 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

98 6.02 x = P.T. grams 17. What is the mass of 2.0 x10 9 Pb atoms? 2.0 x10 9 Pb 207.2g Pb =6.88x g Pb 6.02 x Pb 18. What is the mass of 1.51 x10 23 C 6 H 12 O 6 ? 1.51 x10 23 C 6 H 12 O 6 (180 g C 6 H 12 O 6 )= 45.0g C 6 H 12 O 6 (6.02 x C 6 H 12 O 6 )

99 6.02 x = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO 3 ) 2 454g Cu(NO 3 ) x Cu(NO 3 ) 2 =1.46x10 24 Cu(NO 3 ) g Cu(NO 3 ) 2

100 6.02 x = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO 3 ) 2 454g Cu(NO 3 ) x Cu(NO 3 ) 2 =1.46x10 24 Cu(NO 3 ) g Cu(NO 3 ) 2

101 6.02 x = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO 3 ) 2 454g Cu(NO 3 ) x Cu(NO 3 ) 2 =1.46x10 24 Cu(NO 3 ) g Cu(NO 3 ) 2

102 6.02 x = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO 3 ) 2 454g Cu(NO 3 ) x Cu(NO 3 ) 2 =1.46x10 24 Cu(NO 3 ) g Cu(NO 3 ) 2

103 6.02 x = P.T. grams 19. How many copper (II) nitrate are in 454g Cu(NO 3 ) 2 454g Cu(NO 3 ) x Cu(NO 3 ) 2 =1.46x10 24 Cu(NO 3 ) g Cu(NO 3 ) 2

104 Mole Concept 6.02 x is known as _____________________ number x particles is known as __________of particles just as 12 particles is known as a _______ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x of those particles or ___________ of those particle. __________ g = 1 mole of particles = 6.02 x particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x C atoms 6

105 Mole Concept 6.02 x is known as ___Avogadro’s___ number x particles is known as __________of particles just as 12 particles is known as a _______ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x of those particles or ___________ of those particle. __________ g = 1 mole of particles = 6.02 x particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x C atoms 6

106 Mole Concept 6.02 x is known as ___Avogadro’s___ number x particles is known as _1mole__of particles just as 12 particles is known as a _______ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x of those particles or ___________ of those particle. __________ g = 1 mole of particles = 6.02 x particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x C atoms 6

107 Mole Concept 6.02 x is known as ___Avogadro’s___ number x particles is known as _1mole__of particles just as 12 particles is known as a _dozen__ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x of those particles or ___________ of those particle. __________ g = 1 mole of particles = 6.02 x particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x C atoms 6

108 Mole Concept 6.02 x is known as ___Avogadro’s___ number x particles is known as _1mole__of particles just as 12 particles is known as a _dozen__ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x of those particles or __1mole__ of those particle. __________ g = 1 mole of particles = 6.02 x particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x C atoms 6

109 Mole Concept 6.02 x is known as ___Avogadro’s___ number x particles is known as _1mole__of particles just as 12 particles is known as a _dozen__ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x of those particles or __1mole__ of those particle. ___P.T.____ g = 1 mole of particles = 6.02 x particles atomic mass / molecular mass / formula mass 12 ___________ g C = 1.00 moles of C = 6.02 x C atoms 6

110 Mole Concept 6.02 x is known as ___Avogadro’s___ number x particles is known as _1mole__of particles just as 12 particles is known as a _dozen__ The mass of 1 particle in atomic mass units (amu) is equivalent to the mass of 6.02 x of those particles or __1mole__ of those particle. ___P.T.____ g = 1 mole of particles = 6.02 x particles atomic mass / molecular mass / formula mass 12 __12.0____ g C = 1.00 moles of C = 6.02 x C atoms 6

111 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x10 20 C? 7.5x10 20 C ( 12.0g C ) =.015 g C ( 6.02x10 23 C)

112 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x10 20 C? 7.5x10 20 C ( 12.0g C ) =.015 g C ( 6.02x10 23 C)

113 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x10 20 C? 7.5x10 20 C ( 12.0g C ) =.015 g C ( 6.02x10 23 C)

114 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x10 20 C? 7.5x10 20 C ( 12.0g C ) =.015 g C ( 6.02x10 23 C)

115 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x10 20 C? 7.5x10 20 C ( 12.0g C ) =.015 g C ( 6.02x10 23 C)

116 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x10 20 C? 7.5x10 20 C ( 12.0g C ) =.015 g C ( 6.02x10 23 C)

117 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 20. What is the mass of 2 moles C? 2 mol C ( 12.0g C ) = 24.0g C ( 1mol C ) 21. What is the mass of 7.5x10 20 C? 7.5x10 20 C ( 12.0g C ) =.015 g C ( 6.02x10 23 C)

118 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x10 25 C atom sample? 1.806x10 25 C(1mole C ) = 30 moles C (6.02x10 23 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x10 23 C) = 5.04x10 24 C atoms ( 12.0g C )

119 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x10 25 C atom sample? 1.806x10 25 C(1mole C ) = 30 moles C (6.02x10 23 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x10 23 C) = 5.04x10 24 C atoms ( 12.0g C )

120 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x10 25 C atom sample? 1.806x10 25 C(1mole C ) = 30 moles C (6.02x10 23 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x10 23 C) = 5.04x10 24 C atoms ( 12.0g C )

121 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x10 25 C atom sample? 1.806x10 25 C(1mole C ) = 30 moles C (6.02x10 23 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x10 23 C) = 5.04x10 24 C atoms ( 12.0g C )

122 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x10 25 C atom sample? 1.806x10 25 C(1mole C ) = 30 moles C (6.02x10 23 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x10 23 C) = 5.04x10 24 C atoms ( 12.0g C )

123 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x10 25 C atom sample? 1.806x10 25 C(1mole C ) = 30 moles C (6.02x10 23 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x10 23 C) = 5.04x10 24 C atoms ( 12.0g C )

124 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x10 25 C atom sample? 1.806x10 25 C(1mole C ) = 30 moles C (6.02x10 23 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x10 23 C) = 5.04x10 24 C atoms ( 12.0g C )

125 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 22. How many moles of carbon are in 1.806x10 25 C atom sample? 1.806x10 25 C(1mole C ) = 30 moles C (6.02x10 23 C) 23. How many atoms of carbon are in 100g C? 100gC ( 6.02x10 23 C) = 5.04x10 24 C atoms ( 12.0g C )

126 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 25. How many atoms of carbon are in.750 mole sample?.750 mole C ( 6.02 x10 23 C) = 4.52 x10 23 C 1mole C

127 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 25. How many atoms of carbon are in.750 mole sample?.750 mole C ( 6.02 x10 23 C) = 4.52 x10 23 C 1mole C

128 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 25. How many atoms of carbon are in.750 mole sample?.750 mole C ( 6.02 x10 23 C) = 4.52 x10 23 C 1mole C

129 Mole Concept II 6.02 x = P.T. grams = 1.0 mole 25. How many atoms of carbon are in.750 mole sample?.750 mole C ( 6.02 x10 23 C) = 4.52 x10 23 C 1mole C

130 6.02 x = P.T. grams 26. What is the mass of 3.50 moles of C 2 H 5 OH? 3.50 mole C 2 H 5 OH 46 g C 2 H 5 OH=161 g C 2 H 5 OH 1mole C 2 H 5 OH 27. How many molecules are in.250 mole of C 2 H 5 OH?.250 mole C 2 H 5 OH 6.02 x C 2 H 5 OH= 1.51 X mole C 2 H 5 OH

131 6.02 x = P.T. grams 26. What is the mass of 3.50 moles of C 2 H 5 OH? 3.50 mole C 2 H 5 OH 46 g C 2 H 5 OH=161 g C 2 H 5 OH 1mole C 2 H 5 OH 27. How many molecules are in.250 mole of C 2 H 5 OH?.250 mole C 2 H 5 OH 6.02 x C 2 H 5 OH= 1.51 X mole C 2 H 5 OH

132 6.02 x = P.T. grams 26. What is the mass of 3.50 moles of C 2 H 5 OH? 3.50 mole C 2 H 5 OH 46 g C 2 H 5 OH=161 g C 2 H 5 OH 1mole C 2 H 5 OH 27. How many molecules are in.250 mole of C 2 H 5 OH?.250 mole C 2 H 5 OH 6.02 x C 2 H 5 OH= 1.51 X mole C 2 H 5 OH

133 6.02 x = P.T. grams 26. What is the mass of 3.50 moles of C 2 H 5 OH? 3.50 mole C 2 H 5 OH 46 g C 2 H 5 OH=161 g C 2 H 5 OH 1mole C 2 H 5 OH 27. How many molecules are in.250 mole of C 2 H 5 OH?.250 mole C 2 H 5 OH 6.02 x C 2 H 5 OH= 1.51 X mole C 2 H 5 OH

134 6.02 x = P.T. grams 26. What is the mass of 3.50 moles of C 2 H 5 OH? 3.50 mole C 2 H 5 OH 46 g C 2 H 5 OH=161 g C 2 H 5 OH 1mole C 2 H 5 OH 27. How many molecules are in.250 mole of C 2 H 5 OH?.250 mole C 2 H 5 OH 6.02 x C 2 H 5 OH= 1.51 X mole C 2 H 5 OH

135 6.02 x = P.T. grams 26. What is the mass of 3.50 moles of C 2 H 5 OH? 3.50 mole C 2 H 5 OH 46 g C 2 H 5 OH=161 g C 2 H 5 OH 1mole C 2 H 5 OH 27. How many molecules are in.250 mole of C 2 H 5 OH?.250 mole C 2 H 5 OH 6.02 x C 2 H 5 OH= 1.51 X mole C 2 H 5 OH

136 6.02 x = P.T. grams 26. What is the mass of 3.50 moles of C 2 H 5 OH? 3.50 mole C 2 H 5 OH 46 g C 2 H 5 OH=161 g C 2 H 5 OH 1mole C 2 H 5 OH 27. How many molecules are in.250 mole of C 2 H 5 OH?.250 mole C 2 H 5 OH 6.02 x C 2 H 5 OH= 1.51 X mole C 2 H 5 OH

137 6.02 x = P.T. grams = 1.0 mole 28. How many moles of C 2 H 5 OH are in a 1.0x10 3 molecule sample of C 2 H 5 OH? 1.00 x 10 3 C 2 H 5 OH 1mole C 2 H 5 OH= 1.67x moles 6.02 x C 2 H 5 OH 29. How many moles of C 2 H 5 OH are in 12.0 grams of C 2 H 5 OH? 12.0 gC 2 H 5 OH 1mole C 2 H 5 OH=.26 moles C 2 H 5 OH 46.0 g C 2 H 5 OH

138 6.02 x = P.T. grams = 1.0 mole 28. How many moles of C 2 H 5 OH are in a 1.0x10 3 molecule sample of C 2 H 5 OH? 1.00 x 10 3 C 2 H 5 OH 1mole C 2 H 5 OH= 1.67x moles 6.02 x C 2 H 5 OH 29. How many moles of C 2 H 5 OH are in 12.0 grams of C 2 H 5 OH? 12.0 gC 2 H 5 OH 1mole C 2 H 5 OH=.26 moles C 2 H 5 OH 46.0 g C 2 H 5 OH

139 6.02 x = P.T. grams = 1.0 mole 28. How many moles of C 2 H 5 OH are in a 1.0x10 3 molecule sample of C 2 H 5 OH? 1.00 x 10 3 C 2 H 5 OH 1mole C 2 H 5 OH= 1.67x moles 6.02 x C 2 H 5 OH 29. How many moles of C 2 H 5 OH are in 12.0 grams of C 2 H 5 OH? 12.0 gC 2 H 5 OH 1mole C 2 H 5 OH=.26 moles C 2 H 5 OH 46.0 g C 2 H 5 OH

140 6.02 x = P.T. grams = 1.0 mole 28. How many moles of C 2 H 5 OH are in a 1.0x10 3 molecule sample of C 2 H 5 OH? 1.00 x 10 3 C 2 H 5 OH 1mole C 2 H 5 OH= 1.67x moles 6.02 x C 2 H 5 OH 29. How many moles of C 2 H 5 OH are in 12.0 grams of C 2 H 5 OH? 12.0 gC 2 H 5 OH 1mole C 2 H 5 OH=.26 moles C 2 H 5 OH 46.0 g C 2 H 5 OH

141 6.02 x = P.T. grams = 1.0 mole 28. How many moles of C 2 H 5 OH are in a 1.0x10 3 molecule sample of C 2 H 5 OH? 1.00 x 10 3 C 2 H 5 OH 1mole C 2 H 5 OH= 1.67x moles 6.02 x C 2 H 5 OH 29. How many moles of C 2 H 5 OH are in 12.0 grams of C 2 H 5 OH? 12.0 gC 2 H 5 OH 1mole C 2 H 5 OH=.26 moles C 2 H 5 OH 46.0 g C 2 H 5 OH

142 6.02 x = P.T. grams = 1.0 mole 28. How many moles of C 2 H 5 OH are in a 1.0x10 3 molecule sample of C 2 H 5 OH? 1.00 x 10 3 C 2 H 5 OH 1mole C 2 H 5 OH= 1.67x moles 6.02 x C 2 H 5 OH 29. How many moles of C 2 H 5 OH are in 12.0 grams of C 2 H 5 OH? 12.0 gC 2 H 5 OH 1mole C 2 H 5 OH=.26 moles C 2 H 5 OH 46.0 g C 2 H 5 OH

143 6.02 x = P.T. grams = 1.0 mole 28. How many moles of C 2 H 5 OH are in a 1.0x10 3 molecule sample of C 2 H 5 OH? 1.00 x 10 3 C 2 H 5 OH 1mole C 2 H 5 OH= 1.67x moles 6.02 x C 2 H 5 OH 29. How many moles of C 2 H 5 OH are in 12.0 grams of C 2 H 5 OH? 12.0 gC 2 H 5 OH 1mole C 2 H 5 OH=.26 moles C 2 H 5 OH 46.0 g C 2 H 5 OH

144 6.02 x = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C 2 H 5 OH? 2.50 mole C 2 H 5 OH 46.0g C 2 H 5 OH = 115 g C 2 H 5 OH 1mole C 2 H 5 OH 31. How many molecules are found in 454 g C 2 H 5 OH 454 gC 2 H 5 OH 6.02 X C 2 H 5 OH= 5.94X g C 2 H 5 OH

145 6.02 x = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C 2 H 5 OH? 2.50 mole C 2 H 5 OH 46.0g C 2 H 5 OH = 115 g C 2 H 5 OH 1mole C 2 H 5 OH 31. How many molecules are found in 454 g C 2 H 5 OH 454 gC 2 H 5 OH 6.02 X C 2 H 5 OH= 5.94X g C 2 H 5 OH

146 6.02 x = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C 2 H 5 OH? 2.50 mole C 2 H 5 OH 46.0g C 2 H 5 OH = 115 g C 2 H 5 OH 1mole C 2 H 5 OH 31. How many molecules are found in 454 g C 2 H 5 OH 454 gC 2 H 5 OH 6.02 X C 2 H 5 OH= 5.94X g C 2 H 5 OH

147 6.02 x = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C 2 H 5 OH? 2.50 mole C 2 H 5 OH 46.0g C 2 H 5 OH = 115 g C 2 H 5 OH 1mole C 2 H 5 OH 31. How many molecules are found in 454 g C 2 H 5 OH 454 gC 2 H 5 OH 6.02 X C 2 H 5 OH= 5.94X g C 2 H 5 OH

148 6.02 x = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C 2 H 5 OH? 2.50 mole C 2 H 5 OH 46.0g C 2 H 5 OH = 115 g C 2 H 5 OH 1mole C 2 H 5 OH 31. How many molecules are found in 454 g C 2 H 5 OH 454 gC 2 H 5 OH 6.02 X C 2 H 5 OH= 5.94X g C 2 H 5 OH

149 6.02 x = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C 2 H 5 OH? 2.50 mole C 2 H 5 OH 46.0g C 2 H 5 OH = 115 g C 2 H 5 OH 1mole C 2 H 5 OH 31. How many molecules are found in 454 g C 2 H 5 OH 454 gC 2 H 5 OH 6.02 X C 2 H 5 OH= 5.94X g C 2 H 5 OH

150 6.02 x = P.T. grams = 1.0 mole 30. What is the mass of 2.50 moles of C 2 H 5 OH? 2.50 mole C 2 H 5 OH 46.0g C 2 H 5 OH = 115 g C 2 H 5 OH 1mole C 2 H 5 OH 31. How many molecules are found in 454 g C 2 H 5 OH 454 gC 2 H 5 OH 6.02 X C 2 H 5 OH= 5.94X g C 2 H 5 OH

151 6.02 x = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO 3 ) 2 ? 0.10 g Ca(NO 3 ) 2 1mole Ca(NO 3 ) 2 = mol Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.125 moles Ca(NO 3 ) 2 ?.125 moles Ca(NO 3 ) g Ca(NO 3 ) 2 =20.5 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2

152 6.02 x = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO 3 ) 2 ? 0.10 g Ca(NO 3 ) 2 1mole Ca(NO 3 ) 2 = mol Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.125 moles Ca(NO 3 ) 2 ?.125 moles Ca(NO 3 ) g Ca(NO 3 ) 2 =20.5 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2

153 6.02 x = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO 3 ) 2 ? 0.10 g Ca(NO 3 ) 2 1mole Ca(NO 3 ) 2 = mol Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.125 moles Ca(NO 3 ) 2 ?.125 moles Ca(NO 3 ) g Ca(NO 3 ) 2 =20.5 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2

154 6.02 x = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO 3 ) 2 ? 0.10 g Ca(NO 3 ) 2 1mole Ca(NO 3 ) 2 = mol Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.125 moles Ca(NO 3 ) 2 ?.125 moles Ca(NO 3 ) g Ca(NO 3 ) 2 =20.5 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2

155 6.02 x = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO 3 ) 2 ? 0.10 g Ca(NO 3 ) 2 1mole Ca(NO 3 ) 2 = mol Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.125 moles Ca(NO 3 ) 2 ?.125 moles Ca(NO 3 ) g Ca(NO 3 ) 2 =20.5 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2

156 6.02 x = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO 3 ) 2 ? 0.10 g Ca(NO 3 ) 2 1mole Ca(NO 3 ) 2 = mol Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.125 moles Ca(NO 3 ) 2 ?.125 moles Ca(NO 3 ) g Ca(NO 3 ) 2 =20.5 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2

157 6.02 x = P.T. grams = 1.0 mole 32. How many moles are in 0.10g Ca(NO 3 ) 2 ? 0.10 g Ca(NO 3 ) 2 1mole Ca(NO 3 ) 2 = mol Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.125 moles Ca(NO 3 ) 2 ?.125 moles Ca(NO 3 ) g Ca(NO 3 ) 2 =20.5 g Ca(NO 3 ) 2 1 mole Ca(NO 3 ) 2

158 6.02 x = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO 3 ) 2 ? 454g Ca(NO 3 ) X Ca(NO 3 ) 2 = 1.67x10 24 Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.50 moles Ca(NO 3 ) 2 ?.50 moles Ca(NO 3 ) 2 (164.1g Ca(NO 3 ) 2 =82.1g Ca(NO 3 ) 2 (1mol Ca(NO 3 ) 2

159 6.02 x = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO 3 ) 2 ? 454g Ca(NO 3 ) X Ca(NO 3 ) 2 = 1.67x10 24 Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.50 moles Ca(NO 3 ) 2 ?.50 moles Ca(NO 3 ) 2 (164.1g Ca(NO 3 ) 2 =82.1g Ca(NO 3 ) 2 (1mol Ca(NO 3 ) 2

160 6.02 x = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO 3 ) 2 ? 454g Ca(NO 3 ) X Ca(NO 3 ) 2 = 1.67x10 24 Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.50 moles Ca(NO 3 ) 2 ?.50 moles Ca(NO 3 ) 2 (164.1g Ca(NO 3 ) 2 =82.1g Ca(NO 3 ) 2 (1mol Ca(NO 3 ) 2

161 6.02 x = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO 3 ) 2 ? 454g Ca(NO 3 ) X Ca(NO 3 ) 2 = 1.67x10 24 Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.50 moles Ca(NO 3 ) 2 ?.50 moles Ca(NO 3 ) 2 (164.1g Ca(NO 3 ) 2 =82.1g Ca(NO 3 ) 2 (1mol Ca(NO 3 ) 2

162 6.02 x = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO 3 ) 2 ? 454g Ca(NO 3 ) X Ca(NO 3 ) 2 = 1.67x10 24 Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.50 moles Ca(NO 3 ) 2 ?.50 moles Ca(NO 3 ) 2 (164.1g Ca(NO 3 ) 2 =82.1g Ca(NO 3 ) 2 (1mol Ca(NO 3 ) 2

163 6.02 x = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO 3 ) 2 ? 454g Ca(NO 3 ) X Ca(NO 3 ) 2 = 1.67x10 24 Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.50 moles Ca(NO 3 ) 2 ?.50 moles Ca(NO 3 ) 2 (164.1g Ca(NO 3 ) 2 =82.1g Ca(NO 3 ) 2 (1mol Ca(NO 3 ) 2

164 6.02 x = P.T. grams = 1.0 mole 34. How many units are in 454g Ca(NO 3 ) 2 ? 454g Ca(NO 3 ) X Ca(NO 3 ) 2 = 1.67x10 24 Ca(NO 3 ) g Ca(NO 3 ) What is the mass of.50 moles Ca(NO 3 ) 2 ?.50 moles Ca(NO 3 ) 2 (164.1g Ca(NO 3 ) 2 =82.1g Ca(NO 3 ) 2 (1mol Ca(NO 3 ) 2

165 6.02 x = P.T. grams = 1.0 mole 36. How many units are in.250 mol Ca(NO 3 ) 2 ?.250 mole Ca(NO 3 ) x10 23 Ca(NO 3 ) 2 =1.51x10 23 Ca(NO 3 ) 2 1mole Ca(NO 3 ) How many moles are in 1.0x10 9 Ca(NO 3 ) 2 ? 1.0x10 9 Ca(NO 3 ) 2 ( 1mole Ca(NO 3 ) 2 = 1.66x mol 6.02x10 23 Ca(NO 3 ) 2

166 6.02 x = P.T. grams = 1.0 mole 36. How many units are in.250 mol Ca(NO 3 ) 2 ?.250 mole Ca(NO 3 ) x10 23 Ca(NO 3 ) 2 =1.51x10 23 Ca(NO 3 ) 2 1mole Ca(NO 3 ) How many moles are in 1.0x10 9 Ca(NO 3 ) 2 ? 1.0x10 9 Ca(NO 3 ) 2 ( 1mole Ca(NO 3 ) 2 = 1.66x mol 6.02x10 23 Ca(NO 3 ) 2

167 6.02 x = P.T. grams = 1.0 mole 36. How many units are in.250 mol Ca(NO 3 ) 2 ?.250 mole Ca(NO 3 ) x10 23 Ca(NO 3 ) 2 =1.51x10 23 Ca(NO 3 ) 2 1mole Ca(NO 3 ) How many moles are in 1.0x10 9 Ca(NO 3 ) 2 ? 1.0x10 9 Ca(NO 3 ) 2 ( 1mole Ca(NO 3 ) 2 = 1.66x mol 6.02x10 23 Ca(NO 3 ) 2

168 6.02 x = P.T. grams = 1.0 mole 36. How many units are in.250 mol Ca(NO 3 ) 2 ?.250 mole Ca(NO 3 ) x10 23 Ca(NO 3 ) 2 =1.51x10 23 Ca(NO 3 ) 2 1mole Ca(NO 3 ) How many moles are in 1.0x10 9 Ca(NO 3 ) 2 ? 1.0x10 9 Ca(NO 3 ) 2 ( 1mole Ca(NO 3 ) 2 = 1.66x mol 6.02x10 23 Ca(NO 3 ) 2

169 6.02 x = P.T. grams = 1.0 mole 36. How many units are in.250 mol Ca(NO 3 ) 2 ?.250 mole Ca(NO 3 ) x10 23 Ca(NO 3 ) 2 =1.51x10 23 Ca(NO 3 ) 2 1mole Ca(NO 3 ) How many moles are in 1.0x10 9 Ca(NO 3 ) 2 ? 1.0x10 9 Ca(NO 3 ) 2 ( 1mole Ca(NO 3 ) 2 = 1.66x mol 6.02x10 23 Ca(NO 3 ) 2

170 6.02 x = P.T. grams = 1.0 mole 36. How many units are in.250 mol Ca(NO 3 ) 2 ?.250 mole Ca(NO 3 ) x10 23 Ca(NO 3 ) 2 =1.51x10 23 Ca(NO 3 ) 2 1mole Ca(NO 3 ) How many moles are in 1.0x10 9 Ca(NO 3 ) 2 ? 1.0x10 9 Ca(NO 3 ) 2 ( 1mole Ca(NO 3 ) 2 = 1.66x mol 6.02x10 23 Ca(NO 3 ) 2

171 6.02 x = P.T. grams = 1.0 mole 36. How many units are in.250 mol Ca(NO 3 ) 2 ?.250 mole Ca(NO 3 ) x10 23 Ca(NO 3 ) 2 =1.51x10 23 Ca(NO 3 ) 2 1mole Ca(NO 3 ) How many moles are in 1.0x10 9 Ca(NO 3 ) 2 ? 1.0x10 9 Ca(NO 3 ) 2 ( 1mole Ca(NO 3 ) 2 = 1.66x mol 6.02x10 23 Ca(NO 3 ) 2

172 Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S 1.0 ( 32 g S ) g O ( 1 mol O ) = 3.13 mol O 2.0 ( 16 g O ) 1.56 SO 2

173 Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S 1.0 ( 32 g S ) g O ( 1 mol O ) = 3.13 mol O 2.0 ( 16 g O ) 1.56 SO 2

174 Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S 1.0 ( 32 g S ) g O ( 1 mol O ) = 3.13 mol O 2.0 ( 16 g O ) 1.56 SO 2

175 Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S 1.0 ( 32 g S ) g O ( 1 mol O ) = 3.13 mol O 2.0 ( 16 g O ) 1.56 SO 2

176 Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S 1.0 ( 32 g S ) g O ( 1 mol O ) = 3.13 mol O 2.0 ( 16 g O ) 1.56 SO 2

177 Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S 1.0 ( 32 g S ) g O ( 1 mol O ) = 3.13 mol O 2.0 ( 16 g O ) 1.56 SO 2

178 Empirical Formula Extra Credit 50 g S 50 g O 50g S ( 1mole S) = 1.56 mol S 1.0 ( 32 g S ) g O ( 1 mol O ) = 3.13 mol O 2.0 ( 16 g O ) 1.56 SO 2

179 Empirical formula Count the number of particles of each element by determining the number of moles of each element Divide by the smallest number of moles in order to determine the simplest whole number ratio that the elements combine in to form the compound If the ratios are not whole numbers multiply each ratio by a common factor until they are whole number ratios

180 Empirical formula Count the number of particles of each element by determining the number of moles of each element Divide by the smallest number of moles in order to determine the simplest whole number ratio that the elements combine in to form the compound If the ratios are not whole numbers multiply each ratio by a common factor until they are whole number ratios

181 Empirical formula Count the number of particles of each element by determining the number of moles of each element Divide by the smallest number of moles in order to determine the simplest whole number ratio that the elements combine in to form the compound If the ratios are not whole numbers multiply each ratio by a common factor until they are whole number ratios

182 Empirical formula Count the number of particles of each element by determining the number of moles of each element Divide by the smallest number of moles in order to determine the simplest whole number ratio that the elements combine in to form the compound If the ratios are not whole numbers multiply each ratio by a common factor until they are whole number ratios

183 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

184 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

185 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

186 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

187 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

188 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

189 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

190 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

191 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

192 Empirical Formula A) 35.6g Co and 64.4 g Cl 35.6gCo(1mole Co)=.603 mol Co 1.0 (59.0gCo ) gCl ( 1mole Cl)=1.81 mol Cl 3.0 ( 35.5 g Cl).603 CoCl 3

193 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

194 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

195 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

196 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

197 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

198 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

199 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

200 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

201 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

202 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

203 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

204 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

205 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

206 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

207 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

208 Empirical Formula B) CoCl 2 C) 52.9gAl and 47.1gO 52.9gAl(1mole Al)= 1.96 mol Al 1.0 (27.0g Al ) gO ( 1mole O)=2.94 mol O 1.5 ( 16.0 g O) 1.96 Al 1.0 O 1.5 x 2 = Al 2 O 3

209 Empirical Formula D) MgO E) 2.83gK,.435g C, g O 2.83 g K ( 1 mole K ) =.0724 mole K 2 ( 39.1 g K ) g C ( 1 mole C ) =.0363 mole C ( 12.0 g C ) g O ( 1 mole O ) =.108 mole O 3 ( 16.0 g O ).0363

210 Empirical Formula E) 2.83gK,.435g C, g O 2.83 g K ( 1 mole K ) =.0724 mole K 2 ( 39.1 g K ) g C ( 1 mole C ) =.0363 mole C ( 12.0 g C ) g O ( 1 mole O ) =.108 mole O 3 ( 16.0 g O ).0363

211 Empirical Formula E) 2.83 g K ( 1 mole K ) =.0724 mole K 2 ( 39.1 g K ) g C ( 1 mole C ) =.0363 mole C ( 12.0 g C ) g C ( 1 mole O ) =.108 mole O 3 ( 16.0 g O ).0363

212 Empirical Formula E) 2.83 g K ( 1 mole K ) =.0724 mole K 2 ( 39.1 g K ) g C ( 1 mole C ) =.0363 mole C ( 12.0 g C ) g C ( 1 mole O ) =.108 mole O 3 ( 16.0 g O ).0363

213 Empirical Formula E) 2.83 g K ( 1 mole K ) =.0724 mole K 2 ( 39.1 g K ) g C ( 1 mole C ) =.0363 mole C ( 12.0 g C ) g C ( 1 mole O ) =.108 mole O 3 ( 16.0 g O ).0363

214 Empirical Formula E) 2.83 g K ( 1 mole K ) =.0724 mole K 2 ( 39.1 g K ) g C ( 1 mole C ) =.0363 mole C ( 12.0 g C ) g C ( 1 mole O ) =.108 mole O 3 ( 16.0 g O ).0363

215 Empirical Formula F) ZnSO 3 G) CaC 2 O 4 H) NaCN I) Al 2 S 3 O 12  Al 2 (SO 4 ) 3 J) Fe 2 S 3 O 12  Fe 2 (SO 4 ) 3 K) MgCl 2 O 8  Mg(ClO 4 ) 2


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