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DETERMINATION OF AN EMPIRICAL FORMULA

DETERMINATION OF AN EMPIRICAL FORMULA. Empirical Formulas. E mpirical formula (“formula unit”) -the lowest whole number ratio of atoms in an ionic compound. There are two ways to determine the empirical formula for an ionic compound.

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DETERMINATION OF AN EMPIRICAL FORMULA

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  1. DETERMINATION OF AN EMPIRICAL FORMULA

  2. Empirical Formulas • Empirical formula (“formula unit”) -the lowest whole number ratio of atoms in an ionic compound. • There are two ways to determine the empirical formula for an ionic compound. • Using charges (use your ions and this is easy). C CRISS O SS • Mathematically (yippee!!!)

  3. Method #1 – Charges EX: Write the empirical formula for the compound formed by Na & P c. K & Ne Sr & Cl d. Cu & Cl No compound Na3P CuCl or CuCl2 SrCl2

  4. Method #2 – Mathematically • Step 1:Use the information given in the problem and dimensional analysis with the atomic mass of the element (from periodic table- round to 3 SDs) to find the number of moles you have • Step 2:Take all the mole values and divide them by the SMALLEST one to figure out a ratio • Step 3: Use the answers as subscripts in the empirical formula

  5. Reference-Empirical Formulas7.06 g of silver combine with an excess of fluorine to produce 8.30 g of a compound Silver + Fluorine  Ag?F? 7.06g 1.24g 8.30g Found by subtracting! = 1.00 or 1 X 7.06 g Ag 1.24 g F = .0654 moles Ag Ag1F1 = 1.00 or 1 X = .0653 moles F ANSWER = AgF

  6. Reference-Empirical FormulasA compound contains 24.58% K, 35.81% Mn, and 40.50% O. Find the empirical formula (assume working with 100 grams of the compound and change percentages to grams) X = 1.00 or 1 = .629 mole K 24.58 g K 35.81 g Mn 40.50 g O X = 1.04 or 1 = .652 moles Mn = 4.02 or 4 X = 2.53 moles O ANSWER = KMnO4

  7. Uneven Empirical Formulas • When figuring empirical formulas mathematically, sometimes the resulting numbers don’t come out so clean • You can’t just assume and round how you choose

  8. Reference- .05 Rule • Values used in these problems are obtained by experimentation. The 0.05 rule allows for experimental error: • If the value is within .05 of a whole number (+0.05 or - 0.05), then the value may be rounded to that whole number • Examples: 1.96 can be rounded to 2 • 1.07 cannot be rounded to 1 • 3.02 could be rounded to 3 • 1.93 cannot be rounded to 2 • IF one of the values is not within .05 of a whole number, all the values must be multiplied by an integer so that all values fall within .05 of whole numbers

  9. Reference-Uneven Empirical Formulas 4.35 g sample of zinc is combined with an excess of the element phosphorus. 5.72 g of compound are formed. Calculate the empirical formula. Zinc + Phosphorous  Zn?P? 4.35g 1.37g 5.72g Found by subtracting! = 1.50 X = .0665 moles Zn X 2 = 3.00 or 3 4.35 g Zn 1.37 g P Not within .05 of a whole number X = 1.00 = .0442 moles P X 2 = 2.00 or 2 ANSWER = Zn3P2

  10. Let’s Do It!!! A compound is found to contain 72.3% Fe and 27.7% O by weight. Calculate the empirical formula. Assume in 100 g of compound there would be 72.3 g Fe and 27.7 g O

  11. 1 mole Fe X —————— 55.8 g Fe 72.3g Fe = 1.30 mole Fe 1.30 mole 1.30 mole = 1.00 X 3 = 3.00 = 3 Fe3O4 1 mole O X —————— 16.0 g O 27.7g O = 1.73 mole O 1.73 mole 1.30 mole =1.33 X 3 = 3.99 = 4

  12. Review • Number your paper from 1-5 and answer the following questions. Two will be cumulative review! • 1. Which of these is the correct symbol for Chlorine-35? • a. 35Cl 17 • b. 17Cl 37 • c. 17Cl 35

  13. Review • A • 2. Which is the correct answer with the right number of SD’s if we add 75g and 25.0g? • a. 105 g • b. 10Ō g • c. 100.0 g • d. 105.0 g

  14. Review • B • 3. Which is correct if we consider our rounding rule? • a. 4.02 couldn’t be rounded to 4 but 1.93 can be rounded to 2 • b. 4.02 couldn’t be rounded to 4 and 1.93 can’t be rounded to 2 • c. 4.02 can be rounded to 4 and 1.93 can be rounded to 2 • d. 4.02 can be rounded to 4 and 1.93 can’t be rounded to 2

  15. Review • D • 4. Zinc + Phosphorous  Zn?P? 4.35g ?g 5.72g • a. 4.35 g • b. 2.37 g • c. 1.37 g • d. 1.47 g

  16. Review • C • 5. If I have 32.0g of O, then how many moles is this? • a. 2.0 moles • b. 2.5 moles • c. 2 moles • d. 20 moles

  17. Review • A

  18. Molar Mass • KC4H5O6is the empirical formula for cream of tartar • How many atoms of oxygen are in 1 formula unit of cream of tartar? 6 atoms O

  19. Molar Mass • KC4H5O6 • How many atoms are in 1 formula unit of cream of tartar? 16 atoms 1 potassium 4 carbon 5 hydrogen 6 oxygen

  20. Molar Mass • KC4H5O6 • How many formula units are in 1 mole of cream of tartar? 6.02 x 1023 formula units

  21. Molar Mass • KC4H5O6 • How many atoms of oxygen are in (PER) 1 mole of cream of tartar? • ? atoms O/mole KC4H5O6 = 6 atoms O x 6.02 x1023 formula unit KC4H5O6 1 formula unit 1 mole KC4H5O6 KC4H5O6 3.61 x 1024 atoms O/mole KC4H5O6

  22. Molar Mass • KC4H5O6 • How many moles of oxygen atoms are in 1 mole of cream of tartar? • Let’s put this in perspective……….

  23. How many wheels are in a dozen bicycles? 1 bicycle = 2 wheels 1 dozen bicycles = 2 dozen wheels 1 mole bicycles = 2 mole wheels

  24. How many dozen hydrogen atoms are in 1 dozen water molecules? 1 water = 2 hydrogen 1 dozen water = 2 dozen hydrogen 1 mole water = 2 mole hydrogen H2O

  25. Molar Mass • So……… • KC4H5O6 • How many moles of oxygen atoms are in 1 mole of cream of tartar? 6 moles O

  26. Molar Mass • KC4H5O6 • What is the mass of the oxygen atoms in (PER) 1 mole of cream of tartar? ? g O/ mole KC4H5O6 6 moles O x 16.0 g O = 96.0 g O 1 mole KC4H5O61 mole O mole KC4H5O6

  27. Let’s Do It!!! • How many atoms in one formula unit? Magnesium Acetate: Mg(C2H3O2)2 1-Mg 4-C 6-H 4-O 15 total atoms

  28. Let’s Do It!!! • How many formula units in one mole of Magnesium Acetate: Mg(C2H3O2)2 6.02 x 1023

  29. Let’s Do It!!!! • How many oxygen atoms are in one formula unit? Magnesium Acetate: Mg(C2H3O2)2 4 oxygen atoms

  30. Let’s Do It!!! • How many oxygen atoms are in one mole of this substance? Magnesium Acetate: Mg(C2H3O2)2 (4) (6.02 x 1023) = 2.41 x 1024 atoms O

  31. Let’s Do It!!! • How many moles of oxygen atoms are in one mole of this substance? Magnesium Acetate: Mg(C2H3O2)2 4 moles of Oxygen

  32. Let’s Do It!!! • What is the mass of oxygen atoms in one mole of this substance? Magnesium Acetate: Mg(C2H3O2)2 • ? g O= 1 mole Mg(C2H3O2)2 x 4 moles O x 16.0 g O 1 mole Mg(C2H3O2)2 1 mole O = 64.0 g O = 64.0g/ 1 mole

  33. Molar Mass • We can use this information about what makes up a compound to figure out a compound’s total molar mass • Molar mass- the mass in grams of 1 mole of a compound • It’s the mass of one mole: grams KC4H5O6/mole KC4H5O6 • Also called formula weight, gram formula weight, molecular weight

  34. Reference- Molar Mass • Calculate the molar mass of magnesium iodide, MgI2 from its parts. How many grams are in 1 mole? Mg 1 mole (24.3g Mg/mole )= 24.3 g Mg I 2 mole (127 g I/mole) = 254 g I 278 g/mole MgI2 (for SD use place value!)

  35. Reference- Molar Mass Calculate the molar mass (grams in one mole) of ammonium sulfite, (NH4)2SO3 In 1 mole of the compound there are: 2 moles of N X 14.0 g N/mole= 28.0g N 8 moles of H X 1.01 g H/mole = 8.08gH 1 mole of S X 32.1 g S/mole =32.1 g S 3 moles of O X 16.0 g O/mole=48.0g O SD by place value116.2g units always g/mole 1mole (NH4)2SO3

  36. Molar Mass • Molar mass can be used to convert between moles and grams • For an element: • 1 mole = 6.02x1023 atoms = atomic mass • For a compound: • 1 mole = 6.02x1023 formula units = molar mass

  37. Reference- Molar Mass as a Conversion Factor • What is the mass of 1.35 moles of (NH4)2SO3? • Earlier we found the molar mass: 116.2g (NH4)2SO3= 1mole (NH4)2SO3 ? g (NH4)2SO3 = 1.35 moles (NH4)2SO3 x 116.2 g (NH4)2SO3 1 mole (NH4)2SO3 = 157 g (NH4)2SO3

  38. Reference- Molar Mass as a Conversion Factor • 75.2 g (NH4)2SO3 is how many moles of (NH4)2SO3? ? moles (NH4)2SO3 = 1 mole (NH4)2SO3 75.2 g (NH4)2SO3 X ---------------------- 116.2 g (NH4)2SO3 = .647 moles (NH4)2SO3

  39. Let’s Do It!!!! • How many moles are in 87.4 g of aluminum monohydrogen phosphate Al2(HPO4)3 ? • This is a two part problem: remember you first have to find the molar mass!

  40. Let’s Do It!!! • Find the molar mass of aluminum monohydrogen phosphate, Al2(HPO4)3 54.0 g Al 3.03 g H 93.0 g P 192 g O_ 2 moles Al 3 moles H 3 moles P 12 moles O (27.0g Al/mole) = (1.01g H/mole) = (31.0 g P/mole) = (16.0 g O/mole)= • 342 g • mole Al2(HPO4)3 MASS OF ONE MOLE!!! So 342 g Al2(HPO4)3 =1 mole Al2(HPO4)3

  41. Let’s Do It!!!! • Then find the number of moles in 87.4 g of aluminum monohydrogen phosphate • Molar mass = 342 g/mole • ? moles Al2(HPO4)3 = 87.4 g Al2(HPO4)3 x 1 mole Al2(HPO4)3 342 g Al2(HPO4)3 Obviously, this is less than one mole! = .256 moles Al2(HPO4)3

  42. Review • Number your paper from 1-5 and answer the following questions. Two will be cumulative review! • 1. What does Avogadro’s number, 6.02 x 1023, mean? • a. There are that many grams in a mole • b. There are that many moles in a gram • c. There are that many moles in an atom • d. There are that many atoms in a mole

  43. Review • D • 2. Which of these is false? • a. protons repel protons • b. protons attract neutrons • c. protons attract electrons • d. neutrons don’t repel anything because they have no charge

  44. Review • B • 3. How many elements are in this compound? KC4H5O6 • a. 4 • b. 15 • c. 16 • d. 10

  45. Review A • 4. How many moles of oxygen would be in 3 moles of KC4H5O6? • a. 3 • b. 6 • c. 18 • d. 48

  46. Review • C • 5. If H2O has a molar mass of 18, what does this mean? • a. There are 18 moles of water in a gram • b. There are 18 moles of oxygen in a mole of water • c. There are 18 grams in a mole of water • d. There are 10 moles of oxygen and 8 moles of hydrogen in each water molecule

  47. Review • C

  48. % Composition • As we learned before, when we have multiple elements that make up a compound, each one has a certain ratio • We call this the compound’s empirical formula • From an empirical formula we can figure out each element’s percent composition

  49. Reference-% Composition • Find percent composition of Al(C2H3O2)3 (27.0 g/mole) = (12.0 g/mole) = (1.01 g/mole) = (16.0 g/mole) = 27.0 g Al 72.0 g C 9.09 g H 96.0 g O (1 moles Al) (6 moles C) (9 moles H) (6 moles O) 204.1 g Al(C2H3O2)3 • 27.0 g Al • 72.0 g C • 9.09 g H • 96.0 g O 13.2 % 35.3 % 4.45 % 47.0 % ~100% /204.1g Al(C2H3O2)3 x 100 = /204.1 g Al(C2H3O2)3 x 100 = /204.1 g Al(C2H3O2)3 x 100 = /204.1g Al(C2H3O2)3 x 100 =

  50. % Composition • We can use this information further • How many grams of aluminum can be obtained from 1.50 moles of aluminum acetate? • ? grams Al = X 1.50 moles Al(C2H3O2)3 x OR = 40.5 g Al X 1.50 moles Al(C2H3O2)3 x 27.0 g Al 1 mole Al = 40.4 g Al

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