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Percent composition and empirical formulas

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

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Percent composition and empirical formulas

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  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 CaBr2?

  11. Try it. • What is the percent composition of CaBr2? (FM=199.88g/mol)

  12. Try it. • What is the percent composition of CaBr2? (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 CaBr2? (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: • CaO • Na3N • Al2(SO4)3 • NaNO3 • NaNO2

  15. Practice • ? %comp of: • CaO 71.47%Ca 28.53%O • Na3N 83.12%Na16.88%N • Al2(SO4)3 15.77%Al 28.11%S 56.12%O • NaNO3 27.05%Na 16.48%N 56.47%O • NaNO2 33.32%Na 20.30%N 46.38%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.175O.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.175O.350Ti.175/.175O.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.175O.350Ti.175/.175O.350/.175TiO2

  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? CH2O

  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: • What if the smallest number is not 1? • 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.05S.05O.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.05S.05O.075 NaSO1.5

  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.05S.05O.075 NaSO1.5 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.05S.05O.075 NaSO1.5  Na2S2O3

  36. Al.17O.255 • C.89H1.18 • C1.58H 4.22 O1.58 • C1.90H2.38Cl2.38

  37. Al.17O.255AlO1.5 • C.89H1.18 CH1.33 • C1.58H 4.22 O1.58 CH2.67O • C.190H2.38Cl2.38  CH1.25Cl1.25

  38. Al.17O.255AlO1.5  Al2O3 • C.89H1.18 CH1.33  C3H4 • C1.58H 4.22 O1.58 CH2.67O  C3H 8 O3 • C.190H2.38Cl2.38  CH1.25Cl1.25 C4H5Cl5

  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? CH2O • If we know that the FM of the compound is about 60g/mol, what is the molecular formula? • FM(CH2O)=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? CH2O • If we know that the FM of the compound is about 60g/mol, what is the molecular formula? • FM(CH2O)=30g/mol x 2=60 g/mol • CH2O x 2= C2H4O2

  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? CH2O • If we know that the FM of the compound is about 60g/mol, what is the molecular formula? C2H4O2

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