1. Percent composition and empirical formulas No matter how great and destructive your problems may seem now, remember, you've probably only seen the tip of them

2. An example. You could claim that NaCl is half sodium and half chlorine—one of each.

3. An example. You could claim that NaCl is half sodium and half chlorine—one of each. Chlorine atoms are heavier than sodium atoms.

4. An example. You could claim that NaCl is half sodium and half chlorine—one of each. Chlorine atoms are heavier than sodium atoms. By mass: Na=22.99g/mol Cl=35.45g/mol

5. An example. You could claim that NaCl is half sodium and half chlorine—one of each. Chlorine atoms are heavier than sodium atoms. By mass: %Na=22.99g/58.44 g x 100% %Cl=35.45g/58.44 g x 100%

6. An example. You could claim that NaCl is half sodium and half chlorine—one of each. Chlorine atoms are heavier than sodium atoms. By mass: %Na=22.99g/58.44 g x 100% %Cl=35.45g/58.44 g x 100% FM of NaCl !

7. An example. You could claim that NaCl is half sodium and half chlorine—one of each. Chlorine atoms are heavier than sodium atoms. By mass: %Na=22.99g/58.44 g x 100%=39.34% and %Cl=35.45g/58.44 g x 100%=60.66%

8. Definition % composition of a compound: % A=mass A in the compound x 100% mass of the compound

9. PS All of the %’s add up to 100% The %’s are constant, no matter how much of the substance! (AKA: the law of definite proportions)

10. Try it. What is the percent composition of CaBr 2 ?

11. Try it. What is the percent composition of CaBr 2 ? (FM=199.88g/mol)

12. Try it. What is the percent composition of CaBr 2 ? (FM=199.88g/mol) %Ca=40.08g/199.88 g x 100% and %Br=2 x 79.90g/199.88 g x 100%

13. Try it. What is the percent composition of CaBr 2 ? (FM=199.88g/mol) %Ca=40.08g/199.88 g x 100%=20.05% and %Br=2 x 79.90g/199.88 g x 100%=79.95%

14. Practice ? %comp of: 1)CaO 2)Na 3 N 3)Al 2 (SO 4 ) 3 4)NaNO 3 5)NaNO 2

15. Practice ? %comp of: 1)CaO71.47%Ca28.53%O 2)Na 3 N 83.12%Na16.88%N 3)Al 2 (SO 4 ) 3 15.77%Al28.11%S56.12%O 4)NaNO 3 27.05%Na16.48%N56.47%O 5)NaNO 2 33.32%Na20.30%N46.38%O

16. So what? Iron (II) oxide is 77.73% iron. Iron (III) oxide is 69.94% iron. An iron oxide that contains 16.09 g of iron and 6.91 g of oxygen has to be…

17. So what? Iron (II) oxide is 77.73% iron. Iron (III) oxide is 69.94% iron. An iron oxide that contains 16.09 g of iron and 6.91 g of oxygen has to be… Uhh… 16.09/(16.09 +6.91) x 100%=…

18. So what? Iron (II) oxide is 77.73% iron. Iron (III) oxide is 69.94% iron. An iron oxide that contains 16.09 g of iron and 6.91 g of oxygen has to be… Iron (III) oxide

19. On the other hand… A sample contains 8.39 g Ti and 5.61 g O. What is the formula of this compound?

20. On the other hand… A sample contains 8.39 g Ti and 5.61 g O. What is the formula of this compound? 8.39 g Ti 5.61 g O

21. On the other hand… A sample contains 8.39 g Ti and 5.61 g O. What is the formula of this compound? 8.39 g Ti x 1mol Ti/47.90 g 5.61 g O x 1mol O/16.00 g

22. On the other hand… A sample contains 8.39 g Ti and 5.61 g O. What is the formula of this compound? 8.39 g Ti x 1mol Ti/47.90 g=.175 mol Ti 5.61 g O x 1mol O/16.00 g=.350 mol O Ti.175 O.350

23. On the other hand… A sample contains 8.39 g Ti and 5.61 g O. What is the formula of this compound? 8.39 g Ti x 1mol Ti/47.90 g=.175 mol Ti 5.61 g O x 1mol O/16.00 g=.350 mol O Ti.175 O.350  Ti.175/.175 O.350/.175

24. On the other hand… A sample contains 8.39 g Ti and 5.61 g O. What is the formula of this compound? 8.39 g Ti x 1mol Ti/47.90 g=.175 mol Ti 5.61 g O x 1mol O/16.00 g=.350 mol O Ti.175 O.350  Ti.175/.175 O.350/.175  TiO 2

25. Try it. A sample contains 2.6366 g C, 0.4425 g H, and 3.5122 g O. What is the empirical formula of this compound?

26. Try it. A sample contains 2.6366 g C, 0.4425 g H, and 3.5122 g O. What is the empirical formula of this compound? CH 2 O

27. Try it. A sample contains 35.378g C, 5.938 g H, and 31.418 g O. What is the formula of this compound?

28. There are two things to watch out for: 1)What if the smallest number is not 1? 2)What if the simplest whole number ratio is smaller than the molecule?

29. 1) What if the smallest number is not 1? A sample contains 1.15 g Na, 1.60 g S, and 1.20 g O.

30. 1) What if the smallest number is not 1? A sample contains 1.15 g Na, 1.60 g S, and 1.20 g O. 1.15 g Na/22.99g/mol 1.60 g S /32.06 g/mol 1.20 g O/16 g/mol

31. 1) What if the smallest number is not 1? A sample contains 1.15 g Na, 1.60 g S, and 1.20 g O. 1.15 g Na/22.99g/mol=.0500 mol Na 1.60 g S /32.06 g/mol=.0500 mol S 1.20 g O/16 g/mol=.0750 mol O

32. 1) What if the smallest number is not 1? A sample contains 1.15 g Na, 1.60 g S, and 1.20 g O. 1.15 g Na/22.99g/mol=.0500 mol Na 1.60 g S /32.06 g/mol=.0500 mol S 1.20 g O/16 g/mol=.0750 mol O  Na.05 S.05 O.075

33. 1) What if the smallest number is not 1? A sample contains 1.15 g Na, 1.60 g S, and 1.20 g O. 1.15 g Na/22.99g/mol=.0500 mol Na 1.60 g S /32.06 g/mol=.0500 mol S 1.20 g O/16 g/mol=.0750 mol O  Na.05 S.05 O.075  NaSO 1.5

34. 1) What if the smallest number is not 1? A sample contains 1.15 g Na, 1.60 g S, and 1.20 g O. 1.15 g Na/22.99g/mol=.0500 mol Na 1.60 g S /32.06 g/mol=.0500 mol S 1.20 g O/16 g/mol=.0750 mol O  Na.05 S.05 O.075  NaSO 1.5 Don’t try to round the decimal away!

35. 1) What if the smallest number is not 1? A sample contains 1.15 g Na, 1.60 g S, and 1.20 g O. 1.15 g Na/22.99g/mol=.0500 mol Na 1.60 g S /32.06 g/mol=.0500 mol S 1.20 g O/16 g/mol=.0750 mol O  Na.05 S.05 O.075  NaSO 1.5  Na 2 S 2 O 3

36. Al.17 O.255 C.89 H 1.18 C 1.58 H 4.22 O 1.58 C 1.90 H 2.38 Cl 2.38

37. Al.17 O.255  AlO 1.5 C.89 H 1.18  CH 1.33 C 1.58 H 4.22 O 1.58  CH 2.67 O C.190 H 2.38 Cl 2.38  CH 1.25 Cl 1.25

38. Al.17 O.255  AlO 1.5  Al 2 O 3 C.89 H 1.18  CH 1.33  C 3 H 4 C 1.58 H 4.22 O 1.58  CH 2.67 O  C 3 H 8 O 3 C.190 H 2.38 Cl 2.38  CH 1.25 Cl 1.25  C 4 H 5 Cl 5

39. Try it. A sample contains 2.6366 g C, 0.4425 g H, and 3.5122 g O. What is the empirical formula of this compound? CH 2 O If we know that the FM of the compound is about 60g/mol, what is the molecular formula? FM(CH 2 O)=30g/mol

40. Try it. A sample contains 2.6366 g C, 0.4425 g H, and 3.5122 g O. What is the empirical formula of this compound? CH 2 O If we know that the FM of the compound is about 60g/mol, what is the molecular formula? FM(CH 2 O)=30g/mol x 2=60 g/mol CH 2 O x 2= C 2 H 4 O 2

41. Try it. A sample contains 2.6366 g C, 0.4425 g H, and 3.5122 g O. What is the empirical formula of this compound? CH 2 O If we know that the FM of the compound is about 60g/mol, what is the molecular formula? C 2 H 4 O 2