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DETERMINATION OF AN EMPIRICAL FORMULA 1. Empirical Formulas Empirical formula (“formula unit”) -the lowest whole number ratio of atoms in an ionic compound.

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Presentation on theme: "DETERMINATION OF AN EMPIRICAL FORMULA 1. Empirical Formulas Empirical formula (“formula unit”) -the lowest whole number ratio of atoms in an ionic compound."— Presentation transcript:

1 DETERMINATION OF AN EMPIRICAL FORMULA 1

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. 1.Using charges (use your ions and this is easy). C CRISS O S S 2.Mathematically (yippee!!!) 2

3 Method #1 – Charges EX: Write the empirical formula for the compound formed by Na & Pc. K & Ne Sr & Cld. Cu & Cl No compound Na 3 P SrCl 2 CuCl or CuCl 2 3

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 4

5 Reference-Empirical Formulas 7.06 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 7.06 g Ag 1.24 g F =.0654 moles Ag =.0653 moles F Found by subtracting! Ag 1 F 1 ANSWER = AgF 5

6 Reference-Empirical Formulas A 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) g K g Mn g O =.629 mole K =.652 moles Mn = 2.53 moles O ANSWER = KMnO 4 6

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 7

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 ), then the value may be rounded to that whole number –Examples: 1.96 can be rounded to cannot be rounded to could be rounded to 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 8

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

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 10

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

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. 35 Cl 17 b. 17 Cl 37 c. 17 Cl 35 12

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 g –d g 13

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

15 Review D 4. Zinc + Phosphorous  Zn ? P ? 4.35g ?g 5.72g –a g –b g –c g –d g 15

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 16

17 Review A 17

18 Molar Mass KC 4 H 5 O 6 is the empirical formula for cream of tartar How many atoms of oxygen are in 1 formula unit of cream of tartar? 6 atoms O 18

19 Molar Mass KC 4 H 5 O 6 How many atoms are in 1 formula unit of cream of tartar? 16 atoms 1 potassium 4 carbon 5 hydrogen 6 oxygen 19

20 Molar Mass KC 4 H 5 O 6 How many formula units are in 1 mole of cream of tartar? 6.02 x formula units 20

21 Molar Mass KC 4 H 5 O 6 How many atoms of oxygen are in (PER) 1 mole of cream of tartar? ? atoms O/mole KC 4 H 5 O 6 = 6 atoms O x 6.02 x10 23 formula unit KC 4 H 5 O 6 1 formula unit 1 mole KC 4 H 5 O 6 KC 4 H 5 O x atoms O/mole KC 4 H 5 O 6 21

22 Molar Mass KC 4 H 5 O 6 How many moles of oxygen atoms are in 1 mole of cream of tartar? Let’s put this in perspective………. 22

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 23

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 H2OH2O 24

25 Molar Mass So……… KC 4 H 5 O 6 How many moles of oxygen atoms are in 1 mole of cream of tartar? 6 moles O 25

26 Molar Mass KC 4 H 5 O 6 What is the mass of the oxygen atoms in (PER) 1 mole of cream of tartar? ? g O/ mole KC 4 H 5 O 6 6 moles O x 16.0 g O = 96.0 g O 1 mole KC 4 H 5 O 6 1 mole O mole KC 4 H 5 O 6 26

27 Let’s Do It!!! How many atoms in one formula unit? Magnesium Acetate: Mg(C 2 H 3 O 2 ) 2 1-Mg 4-C 6-H 4-O 15 total atoms 27

28 Let’s Do It!!! How many formula units in one mole of Magnesium Acetate: Mg(C 2 H 3 O 2 ) x

29 Let’s Do It!!!! How many oxygen atoms are in one formula unit? Magnesium Acetate: Mg(C 2 H 3 O 2 ) 2 4 oxygen atoms 29

30 Let’s Do It!!! How many oxygen atoms are in one mole of this substance? Magnesium Acetate: Mg(C 2 H 3 O 2 ) 2 (4) (6.02 x ) = 2.41 x atoms O 30

31 Let’s Do It!!! How many moles of oxygen atoms are in one mole of this substance? Magnesium Acetate: Mg(C 2 H 3 O 2 ) 2 4 moles of Oxygen 31

32 Let’s Do It!!! What is the mass of oxygen atoms in one mole of this substance? Magnesium Acetate: Mg(C 2 H 3 O 2 ) 2 ? g O= 1 mole Mg(C 2 H 3 O 2 ) 2 x 4 moles O x 16.0 g O 1 mole Mg(C 2 H 3 O 2 ) 2 1 mole O = 64.0 g O = 64.0g/ 1 mole 32

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 KC 4 H 5 O 6 /mole KC 4 H 5 O 6 Also called formula weight, gram formula weight, molecular weight 33

34 Reference- Molar Mass Calculate the molar mass of magnesium iodide, MgI 2 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 MgI 2 (for SD use place value!) 34

35 Reference- Molar Mass Calculate the molar mass (grams in one mole) of ammonium sulfite, (NH 4 ) 2 SO 3 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 value 116.2g units always g/mole 1mole (NH 4 ) 2 SO 3 35

36 Molar mass can be used to convert between moles and grams For an element: 1 mole = 6.02x10 23 atoms = atomic mass For a compound: 1 mole = 6.02x10 23 formula units = molar mass Molar Mass 36

37 What is the mass of 1.35 moles of (NH 4 ) 2 SO 3 ? Earlier we found the molar mass: 116.2g (NH 4 ) 2 SO 3 = 1mole (NH 4 ) 2 SO 3 ? g (NH 4 ) 2 SO 3 = 1.35 moles (NH 4 ) 2 SO 3 x g (NH 4 ) 2 SO 3 1 mole (NH 4 ) 2 SO 3 = 157 g (NH 4 ) 2 SO 3 Reference- Molar Mass as a Conversion Factor 37

38 75.2 g (NH 4 ) 2 SO 3 is how many moles of (NH 4 ) 2 SO 3 ? ? moles (NH 4 ) 2 SO 3 = 1 mole (NH 4 ) 2 SO g (NH 4 ) 2 SO 3 X g (NH 4 ) 2 SO 3 =.647 moles (NH 4 ) 2 SO 3 Reference- Molar Mass as a Conversion Factor 38

39 Let’s Do It!!!! How many moles are in 87.4 g of aluminum monohydrogen phosphate Al 2 (HPO 4 ) 3 ? This is a two part problem: remember you first have to find the molar mass! 39

40 Let’s Do It!!! Find the molar mass of aluminum monohydrogen phosphate, Al 2 (HPO 4 ) 3 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)= 54.0 g Al 3.03 g H 93.0 g P 192 g O_ 342 g mole Al 2 (HPO 4 ) 3 MASS OF ONE MOLE!!! So 342 g Al 2 (HPO 4 ) 3 = 1 mole Al 2 (HPO 4 ) 3 40

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 Al 2 (HPO 4 ) 3 = 87.4 g Al 2 (HPO 4 ) 3 x 1 mole Al 2 (HPO 4 ) g Al 2 (HPO 4 ) 3 =.256 moles Al 2 (HPO 4 ) 3 Obviously, this is less than one mole! 41

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 10 23, 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 42

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 43

44 Review B 3. How many elements are in this compound? KC 4 H 5 O 6 –a. 4 –b. 15 –c. 16 –d

45 Review A 4. How many moles of oxygen would be in 3 moles of KC 4 H 5 O 6 ? –a. 3 –b. 6 –c. 18 –d

46 Review C 5. If H 2 O 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 46

47 Review C 47

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 48

49 Reference-% Composition Find percent composition of Al(C 2 H 3 O 2 ) 3 (1 moles Al) (6 moles C) (9 moles H) (6 moles O) (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 g Al(C 2 H 3 O 2 ) 3 /204.1 g Al(C 2 H 3 O 2 ) 3 x 100 = 13.2 % 35.3 % 4.45 % 47.0 % ~100% 27.0 g Al 72.0 g C 9.09 g H 96.0 g O 49

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 = 1.50 moles Al(C 2 H 3 O 2 ) 3 x = 40.5 g Al 1.50 moles Al(C 2 H 3 O 2 ) 3 x 27.0 g Al 1 mole Al = 40.4 g Al OR 50

51 Let’s Do It!!! Find the percent composition of iron (III) dichromate, Fe 2 (Cr 2 O 7 ) 3 (2 moles Fe) (6 moles Cr) (21 moles O) (55.8 g/mole) = (52.0 g/mole) = (16.0 g/mole) = 112 g Fe 312 g Cr 336 g O 76Ō g Fe 2 (Cr 2 O 7 ) 3 /76Ōg Fe 2 (Cr 2 O 7 ) 3 x 100 = 14.7 % 41.1 % 44.2 % ~100% 112 g Fe 312 g Cr 336 g O 51

52 Let’s Do It!!! How many moles of iron can be obtained from 525 g of iron (III) dichromate, Fe 2 (Cr 2 O 7 ) 3 ? ? moles of Fe = 525 g Fe 2 (Cr 2 O 7 ) 3 X = 1.38 moles Fe = 1.39 moles Fe OR 52

53 Reference- Advanced Molar Mass How many grams of Mg are needed to combine with 2.00 g of N to make magnesium nitride? Magnesium + Nitrogen  Magnesium nitride (Mg 3 N 2 ) ?g 2.00 g (3 moles Mg) (2 mole N) (24.3 g Mg/mole) = (14.0 g N/mole) = 72.9 g Mg 28.0 g N g Mg 3 N g N x 72.9 g Mg 28.0 g N = 5.21g Mg 72.9 g Mg g N = Mg 3 N 2 53

54 Reference- Advanced Molar Mass OR THE EASY WAY!!! Magnesium + Nitrogen  Magnesium nitride (Mg 3 N 2 ) ?g 2.00 g ? g Mg = 2.00 g N x 1 mole N x 3 mole Mg x 24.3 g = 14.0 g N 2 mole N 1 mole Mg = 5.21g Mg 72.9 g Mg g N = Mg 3 N 2 54

55 How many grams of nitrogen are needed to combine with 5.00 g Mg to make magnesium nitride? Magnesium + Nitrogen  Magnesium nitride Mg 3 N g ?g ? g N = 5.00 g Mg x 28.0 g N 72.9 g Mg = 1.92 g N Reference- Advanced Molar Mass 72.9 g Mg g N = Mg 3 N 2 55

56 OR THE EASY WAY!!! Magnesium + Nitrogen  Magnesium nitride Mg 3 N g ?g ? g N = 5.00 g Mg x 1 mole Mg x 2 mole N x 24.3 g = 24.3 g Mg 3 mole Mg 1 mole Mg = 1.92 g N Reference- Advanced Molar Mass 56

57 Reference- Advanced Molar Mass How many grams of Mg are needed to make 25 g of magnesium nitride? Magnesium + Nitrogen  Magnesium nitride – Mg 3 N 2 ?g 25g 72.9 g Mg = 28.0 g N = Mg 3 N 2 ? g Mg= 25 g Mg 3 N 2 x 72.9 Mg = 18 g Mg g Mg 3 N 2 57

58 Reference- Advanced Molar Mass OR THE EASY WAY!!! How many grams of Mg are needed to make 25 g of magnesium nitride? Magnesium + Nitrogen  Magnesium nitride Mg 3 N 2 ?g 25g ? g Mg = 25 g Mg 3 N 2 x 1 mole Mg 3 N 2 x 3 mole Mg x 24.3 g = g Mg 3 N 2 1 mole Mg 3 N 2 1 mole Mg molar mass = 18 g Mg 58

59 Let’s Do It!!! How many grams of magnesium nitride can be made from 12.5 g of magnesium? Magnesium + Nitrogen  Magnesium nitride – Mg 3 N g ?g 72.9 g Mg = 28.0 g N = Mg 3 N 2 59

60 Let’s Do It!!! How many grams of magnesium nitride can be made from 12.5 g of magnesium? Magnesium + Nitrogen  Magnesium nitride – Mg 3 N g ?g 72.9 g Mg = 28.0 g N = Mg 3 N g Mg x g Mg 3 N g Mg = 17.3 g Mg 3 N 2 60

61 Let’s Do It!!! OR THE EASY WAY!!! How many grams of magnesium nitride can be made from 12.5 g of magnesium? Magnesium + Nitrogen  Magnesium nitride Mg 3 N g ?g ? g Mg 3 N 2 = molar mass 12.5 g Mg x 1 mole Mg x 1 mole Mg 3 N 2 x 100.9g Mg 3 N g Mg 3 mole Mg 1 mole Mg 3 N 2 = = 17.5 g Mg 3 N 2 61

62 Number your paper from 1-5 and answer the following questions. Two will be cumulative review! 1. Atoms with the same atomic number but different mass numbers are a. different elements b. ions c. isotopes of the same element d. isotopes of different elements Review 62

63 Review D 2.An alpha particle a. has a negative charge b. can penetrate a sheet of aluminum c. is identical to a helium nucleus d. all of the above 63

64 Review C 3. How many moles of Cr are in a mole of Fe 2 (Cr 2 O 7 ) 3 –a. 6 –b. 3 –c. 2 –d. 1 64

65 Review A 4. What is the percent composition of hydrogen and oxygen in water? –a. 80% O and 20% H –b. 88.9% O and 11.2% H –c. 75% O and 35% H –d. 66.6% H and 33.3% O 65

66 Review B 1 mole O x 16.0 g O = 16.0 g 16.0/18.0 x mole O 2 moles H x 1.01 g H = /18.0 x mole H 5. How many grams of water can be made from 10 grams of hydrogen and excess oxygen? –a. 90 g –b. 10 g –c. 20 g –d. 0 g 66

67 Review A 5. How many grams of water can be made from 10 grams of hydrogen? EASY WAY!! –a. 90 g –b. 10 g –c. 20 g –d. 0 g ? g H2O = molar mass 10 g H x 1 mole H x 1 mole H 2 O x 18.0 g H 2 O 1.01 g H 2 mole H 1 mole H 2 O 67

68 Hydrates Hydrated compounds- compounds with molecules of water held in their crystal structure These compounds contain an anhydrous (non- water) part and a hydrous (water based) part 68

69 Hydrates Ex.) CaSO 4. 7H 2 O Compounds with molecules of water held in their crystal structure Very common in nature Water can be removed by heating, leaving behind what is called the anhydrous compound. Anhydrous compound Water 69

70 Reference-Naming Hydrated Compounds The following is tacked on the name obtained from the ions H 2 O monohydrate 2 H 2 O dihydrate 3 H 2 O trihydrate 4 H 2 O tetrahydrate 5 H 2 O pentahydrate 6 H 2 O hexahydrate 7 H 2 O heptahydrate 8 H 2 O octahydrate 9 H 2 O nonahydrate 10 H 2 O dekahydrate CaSO 4  7 H 2 O -- named as calcium sulfate heptahydrate 70

71 Reference-Empirical Formula of Hydrates Find the empirical formula of a hydrate of CaSO 4 hydrate that is 28.5% H 2 O To solve this problem, find the simplest mole ratio between the anhydrous part of the compound (CaSO 4 ) and the water (H 2 O) H 2 O =28.5 % CaSO 4 =71.5 % (100% %) 71

72 Reference-Empirical Formula of Hydrates 71.5 g CaSO 4 x 1 mole CaSO 4 =.525 mole CaSO g CaSO 4 molar mass of CaSO g H 2 O x 1 mole H 2 O = 1.58 mole H 2 O 18.0 g H 2 O = 1.00 = = 3.01 = CaSO 4  3 H 2 O When you are finding formulas of hydrates they ALWAYS come out even! 72

73 Reference- Molar Mass of a Hydrate Find the molar mass of CaSO 4  7 H 2 O First we find the molar mass for H 2 O and treat the water like it’s an element! 2 mole H X 1.01 g/mole = 2.02 g 1 mole O X 16.0 g/mole = 16.0 g 18.0 g/mole (MEMORIZE THIS) 73

74 Reference- Molar Mass of a Hydrate CaSO 4  7 H 2 O 1 mole Ca x 40.1 g/mole = 40.1 g 1 mole S x 32.1 g/mole = 32.1 g 4 mole O x 16.0 g/mole = 64.0 g 7 mole H 2 O x 18.0 g/mole =126 g 262 g/mole (SD by place value) 74

75 Reference- % Composition of a Hydrate Find the % composition of CaSO 4  7 H 2 O The water is treated like an element! % Ca = 40.1 g Ca X 100% = 15.3 % Ca 262 g compound % S = 32.1 g S X 100% = 12.3 % S 262 g compound % O = 64.0 g O X 100% = 24.4 % O 262 g compound % H 2 O = 126 g H 2 O X 100% = 48.1 % H 2 O 262 g compound 75

76 Using Molar Mass or % Composition of a Hydrate How much water do we get when we heat 2.00 g of CaSO 4  7 H 2 O? 2.00 g x.481 =.962 g H 2 O OR 2.00 g CaSO 4  7 H 2 O x 126 g H 2 O = g H 2 O 262 g CaSO 4  7 H 2 O 76

77 Review Number your paper from 1-5 and answer the following questions. Two will be cumulative review! –1. What is the actual mass of a hydrogen-3 atom? a x g b x g c x g d x g 77

78 Review D 2. Which of these is true about the discovery of Millikan’s oil drop experiment –a. He discovered the electron –b. He discovered the mass of the neutron –c. He discovered the mass and the charge of the electron –d. He discovered the proton 78

79 Review C 3. What is CaSO 4  5 H 2 O called? –a. calcium sulfate pentahydrate –b. calcium sulfate heptahydrate –c. calcium sulfate hydrate –d. calcium sulfate trihydrate 79

80 Review A 4. What is the mass of water in CaSO 4  5 H 2 O? –a. 5 g –b g –c. 5.0 g –d. 18 g 80

81 Review B 5 mole H 2 O x 18.0 g/mole = 90.0 g 5. Which of these would have the most water given off when heated? –a. calcium sulfate pentahydrate –b. calcium sulfate heptahydrate –c. calcium sulfate hydrate –d. calcium sulfate trihydrate 81

82 Review B 82


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