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Empirical Formula

Empirical Formula . The empirical formula is the simplest whole number ratio between atoms in a compound. It is determined experimentally by measuring the mass of the elements that combine to form a compound. Molecular Formula. Molecular Formula.

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Empirical Formula

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  1. Empirical Formula

  2. The empirical formula is the simplest whole number ratio between atoms in a compound. It is determined experimentally by measuring the mass of the elements that combine to form a compound.

  3. Molecular Formula

  4. Molecular Formula • Is the formula of the molecular unit

  5. Molecular Formula • Is the formula of the molecular unit • Is a multiple of the empirical formula

  6. Molecular Formula • Is the formula of the molecular unit • Is a multiple of the empirical formula Molecular Formula C2H6O2 62.06 g/mol

  7. Molecular Formula • Is the formula of the molecular unit • Is a multiple of the empirical formula Molecular Formula Empirical Formula C2H6O2CH3O 62.06 g/mol 31.03 g/mol

  8. An Example: #1 A compound was found to be composed of 38.7 g C, 9.68 g H and 51.6 g O. Calculate the empirical formula. Steps To Solve: 1.Change the grams to moles of each element.2. Simplify the mole ratio to a whole # by dividing by the smallest # of moles. 3. Write the empirical formula.

  9. 1. Change the grams to moles. 38.7 g C x 1 mole 12.0 g

  10. 1. Change the grams to moles. 38.7 g C x 1 mole 12.0 g 9.68 g H x 1 mole 1.01 g

  11. 1. Change the grams to moles. 38.7 g C x 1 mole 12.0 g 9.68 g H x 1 mole 1.01 g 51.6 g O x 1 mole 16.0 g

  12. 1. Change the grams to moles. 38.7 g C x 1 mole = 3.225 mol 12.0 g 9.68 g H x 1 mole = 9.584 mol 1.01 g 51.6 g O x 1 mole = 3.225 mol 16.0 g

  13. 2. Simplify the mole ratio to a whole # by dividing by the smallest # of moles. 38.7 g C x 1 mole = 3.225 mol 12.0 g 9.68 g H x 1 mole = 9.584 mol 1.01 g 51.6 g O x 1 mole = 3.225 mol 16.0 g

  14. 2. Simplify the mole ratio to a whole # by dividing by the smallest # of moles. 38.7 g C x 1 mole = 3.225 mol 12.0 g 9.68 g H x 1 mole = 9.584 mol It’s a tie for the 1.01 g smallest # moles 51.6 g O x 1 mole = 3.225 mol 16.0 g

  15. 2. Simplify the mole ratio to a whole # by dividing by the smallest # of moles. 38.7 g C x 1 mole = 3.225 mol 12.0 g 3.225 mol 9.68 g H x 1 mole = 9.584 mol 1.01 g 3.225 mol 51.6 g O x 1 mole = 3.225 mol 16.0 g 3.225 mol

  16. 2. Simplify the mole ratio to a whole # by dividing by the smallest # of moles. 38.7 g C x 1 mole = 3.225 mol= 1 12.0 g 3.225 mol 9.68 g H x 1 mole = 9.584 mol= 3 1.01 g 3.225 mol 51.6 g O x 1 mole = 3.225 mol= 1 16.0 g 3.225 mol

  17. 3. Write the empirical formula. 38.7 g C x 1 mole = 3.225 mol = 1 12.0 g 3.225 mol 9.68 g H x 1 mole = 9.584 mol = 3 CH3O 1.01 g 3.225 mol 51.6 g O x 1 mole = 3.225 mol = 1 16.0 g 3.225 mol

  18. #2. a compound is found to contain the following % by mass: 69.58% Ba, 6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula? Step 1: imagine that you have 100 g of the substance. Thus, % will become mass in grams. E.g. 69.58 % Ba becomes 69.58 g Ba. (Some questions will give grams right off, instead of %) Step 2: calculate the # of moles (mol = g /( g/mol) Step 3: express moles as the simplest ratio by dividing through by the lowest number. Step 4: write the simplest formula from mol ratios

  19. mol mol (reduced) #2- 69.58% Ba, 6.090% C, 24.32% O. Whatistheempirical (a.k.a. simplest) formula? 1: 69.58 g Ba, 6.090 g C, 24.32 g O 2: Ba: 69.58 g ¸ 137.33 g/mol = 0.50666 mol Ba C: 6.090 g ¸ 12.01 g/mol = 0.50708 mol C O: 24.32 g ¸ 16.00 g/mol = 1.520 mol O 3: Ba C O 0.50666 0.50708 1.520 0.50666/ 0.50666 = 1 0.50708/ 0.50666 = 1.001 1.520/ 0.50666 = 3.000 4: the simplest formula is BaCO3

  20. #3A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula.

  21. #3 A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula. 63.55 g Ag x 1 mole 107.9 g

  22. #3 A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula. 63.55 g Ag x 1 mole 107.9 g 8.23 g N x 1 mole 14.0 g

  23. A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula. 63.55 g Ag x 1 mole 107.9 g 8.23 g N x 1 mole 14.0 g 28.24 g O x 1 mole 16.0 g

  24. A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula. 63.55 g Ag x 1 mole= 0.5890 mol 107.9 g 8.23 g N x 1 mole= 0.5879 mol 14.0 g 28.24 g O x 1 mole= 1.765 mol 16.0 g

  25. A compound is found to contain 63.55 % Ag, 8.23 % N and 28.2 % O. Calculate the empirical formula. 63.55 g Ag x 1 mole= 0.5890 mol 107.9 g 8.23 g N x 1 mole= 0.5879 mol 14.0 g smallest moles 28.24 g O x 1 mole= 1.765mol 16.0 g

  26. A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula. 63.55 g Ag x 1 mole = 0.5890 mol 107.9 g 0.5879 mol 8.23 g N x 1 mole = 0.5879 mol 14.0 g 0.5879 mol 28.24 g O x 1 mole = 1.765mol 16.0 g 0.5879 mol

  27. A compound is found to contain 63.55 % Ag, 8.23 % N and 28.24 % O. Calculate the empirical formula. 63.55 g Ag x 1 mole = 0.5890 mol = 1 107.9 g 0.5879 mol 8.23 g N x 1 mole = 0.5879 mol = 1 14.0 g 0.5879 mol 28.24 g O x 1 mole = 1.765 mol = 3 16.0 g 0.5879 mol

  28. A compound is found to contain 63.55% Ag, 8.23% N and 28.24% O. Calculate the empirical formula. 63.55 g Ag x 1 mole = 0.5890 mol = 1 107.9 g 0.5879 mol 8.23 g N x 1 mole = 0.5879 mol = 1 AgNO3 14.0 g 0.5879 mol 28.24 g O x 1 mole = 1.765 mol = 3 16.0 g 0.5879 mol

  29. #4 A compound is found to contain 50.07 % Cu, 16.29 % P and 33.64% O. Calculate the empirical formula. 50.07 g Cu x 1 mole = 0.7885 mol = 1.5 63.5 g 0.5255 mol 16.29 g P x 1 mole = 0.5255 mol = 1 31.0 g 0.5255 mol 33.64 g O x 1 mole = 2.103 mol = 4 16.0 g 0.5255 mol

  30. #4 A compound is found to contain 50.07 % Cu, 16.29 % P and 33.64% O. Calculate the empirical formula. 50.07 g Cu x 1 mole = 0.7885 mol = 1.5 x 2 63.5 g 0.5255 mol 16.29 g P x 1 mole = 0.5255 mol = 1 x 2 31.0 g 0.5255 mol 33.64 g O x 1 mole = 2.103 mol = 4 x 2 16.0 g 0.5255 mol

  31. #4 A compound is found to contain 50.07 % Cu, 16.29 % P and 33.64% O. Calculate the empirical formula. 50.07 g Cu x 1 mole = 0.7885 mol = 3 63.5 g 0.5255 mol 16.29 g P x 1 mole = 0.5255 mol = 2 31.0 g 0.5255 mol 33.64 g O x 1 mole = 2.103 mol = 8 16.0 g 0.5255 mol

  32. #4A compound is found to contain 50.07 % Cu, 16.29 % P and 33.64% O. Calculate the empirical formula. 50.07 g Cu x 1 mole = 0.7885 mol = 3 63.5 g 0.5255 mol 16.29 g P x 1 mole = 0.5255 mol = 2 31.0 g 0.5255 mol 33.64 g O x 1 mole = 2.103 mol = 8 16.0 g 0.5255 mol Cu3P2O8

  33. #4A compound is found to contain 50.07 % Cu, 16.29 % P and 33.64% O. Calculate the empirical formula. 50.07 g Cu x 1 mole = 0.7885 mol = 3 63.5 g 0.5255 mol 16.29 g P x 1 mole = 0.5255 mol = 2 31.0 g 0.5255 mol 33.64 g O x 1 mole = 2.103 mol = 8 16.0 g 0.5255 mol Cu3P2O8 Cu3(PO4)2Only do this for ionic compounds Not for covalent!

  34. Mole ratios and simplest formula Given the following mole ratios for the hypothetical compound AxBy, what would x and y be if the mol ratio of A and B were: A = 1 mol, B = 2.98 mol A = 1.337 mol, B = 1 mol A = 2.34 mol, B = 1 mol A = 1 mol, B = 1.48 mol AB3 A4B3 A7B3 A2B3

  35. 5. A compound consists of 29.1%Na, 40.5% S, and30.4%O. Determinethesimplestformula. 6. A compound is composed of 7.20 g carbon, 1.20 g hydrogen, and 9.60 g oxygen. Find the empirical formula for this compound HOME WORK 3. - 6. Try questions 3 - 6 on page 189.

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