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the Mole !

It’s time to learn about . . . the Mole !. Interesting angle to a picture!!!. Stoichiometry: Percent Composition At the conclusion of our time together, you should be able to:. Determine the percent composition of each of the elements in a compound

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the Mole !

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  1. It’s time to learn about . . . the Mole !

  2. Interesting angle to a picture!!!

  3. Stoichiometry: Percent CompositionAt the conclusion of our time together, you should be able to: Determine the percent composition of each of the elements in a compound Use percent composition to determine the mass of each element in a compound

  4. Chemical Formulas of Compounds • Formulas give the relative numbers of atoms or moles of each element in a formula unit - always a whole number ratio (the law of definite proportions). NO2 molecule: 2 atoms of O for every 1 atom of N 1 mole of NO2 : 2 moles of O atoms to every 1 mole of N atoms • If we know or can determine the relative number of moles of each element in a compound, we can determine a formula for the compound.

  5. Types of Formulas • Empirical Formula The formula of a compound that expresses the smallest whole number ratio of the atoms present. Ionic formulas are always empirical formulas • Molecular Formula The formula that states the actual number of each kind of atom found in one molecule of the compound.

  6. What is the percent composition of hydrogen in water? 2.02 g H2 18.02 g H2O Determine the grams in 2 moles of hydrogen and divide that by the grams in 1 mole of H2O, then multiply by 100. x 100 = 11.21 % H2

  7. What Lies At The Bottom Of The Ocean And Twitches? A Nervous Wreck.

  8. Stoichiometry: Percent CompositionLet’s see if you can: Determine the percent composition of each of the elements in a compound Use percent composition to determine the mass of each element in a compound

  9. #1: What is the percent composition of Cu in copper bromide? 63.55 g Cu 223.35 g CuBr2 Determine the grams in 1 mole of copper and divide that by the grams in 1 mole of copper bromide, then multiply by 100. x 100 = 28.45 % Cu

  10. #7: In 75.5 g of water, how many grams of hydrogen are there? x 2.02 g H2 18.02 g H2O Start with the grams of water and multiply by the hydrogen to water ratio. 75.5 g H2O = 8.46 g H2

  11. Lessons from your elders… An old prospector named Ralph, shuffled into town leading an old tired mule.Old Ralph headed straight for the only saloon to clear his parched throat. He walked up and tied his old mule to the hitch rail.  As he stood there, brushing some of the dust from his face and clothes, a young gunslinger stepped out of the saloon with a gun in one hand and a bottle of whiskey in the other. The young gunslinger looked at the old man and laughed, saying, "Hey old man, have you ever danced?"

  12. Lessons from your elders… Ralph looked up at the gunslinger and said, "No, I never did dance.. never really wanted to.” A crowd had gathered as the gunslinger grinned and said,  "Well, you old fool, you're gonna dance now," and started shooting at the old man's feet. The old prospector - not wanting to get a toe blown off - started hopping around like a flea on a hot skillet.  Everybody was laughing, fit to be tied. When his last bullet had been fired, the young gunslinger, still laughing, holstered his gun and turned around to go back into the saloon.

  13. Lessons from your elders… Old Ralph turned to his pack mule, pulled out a double-barreled shotgun, and cocked both hammers. The loud clicks carried clearly through the desert air. The crowd stopped laughing immediately. The young gunslinger heard the sounds too, and he turned around very slowly.  The silence was almost deafening.

  14. Lessons from your elders…  The crowd watched as the young gunman stared at the old timer and the large gaping holes of those twin barrels. The barrels of the shotgun never wavered in Old Ralph's hands, as he quietly said, "Son, have you ever kissed a mule's butt?“ The gunslinger swallowed hard and said, "No sir..... but... I've always wanted to!!!"

  15. Lessons from your elders… Here are a few of the lessons: Never be a cocky punk. Don't waste ammunition. Alcohol makes you think you're smarter than you are. Don't mess with old men; they didn't get old by being stupid.

  16. Stoichiometry: Percent CompositionAt the conclusion of our time together, you should be able to: Determine the percent composition of each of the elements in a compound Use percent composition to determine the mass of each element in a compound Use percent compositions of a compound to determine the empirical and/or molecular formula

  17. Why Do Gorillas Have Big Nostrils? • Because They Have Big Fingers.

  18. To Obtain an Empirical Formula 1. Determine the mass in grams of each element present, if necessary. 2. Calculate the number of moles of each element. 3. Divide each by the smallest number of moles to obtain the simplest whole number ratio. 4. If whole numbers are not obtainedin step 3, multiply through by the smallest number that will give all whole numbers

  19. A sample of a brown gas, a major air pollutant, is found to contain 2.34 g N and 5.34g O. Determine the formula for this substance. This requires mole ratios, so convert grams to moles moles of N = 2.34 g of N 14.01 g/mole = 0.167 moles of N moles of O = 5.34 g 16.00 g/mole = 0.334 moles of O Formula:

  20. Empirical Formula from % Composition A substance has the following composition by mass: 60.80 % Na ; 28.60 % B ; 10.60 % H What is the empirical formula of the substance? Consider a sample size of 100 grams This would contain 60.80 g of Na, 28.60 grams of B and 10.60 grams H Determine the number of moles of each Determine the simplest whole number ratio

  21. Empirical Formula from % Composition Determine the number of moles of each 60.80 g Na x 1 mol/22.99 g = 2.64 mol Na 28.60 g B x 1 mol/10.81 g = 2.65 mol B 10.60 g H x 1 mol/1.01 g = 10.50 mol H Determine the simplest whole number ratio NaBH4

  22. Calculation of the Molecular Formula A compound has an empirical formula of NO2. The colorless liquid, used in rocket engines has a molar mass of 92.0 g/mole. What is the molecular formula of this substance? Formula mass = 14.01 + 32.00 = 46.01 g/mol 92.0 g/mol 46.01 g/mol = ~2 N2O4

  23. Bill Gates' RulesHere is a list of 11 things that many high school and college graduates did not learn in school. In his book, Bill Gates talks about how feel-good, politically-correct teachings created a full generation of kids with no concept of reality and how this concept has set them up for failure in the real world. RULE 7 Before you were born, your parents weren't as boring as they are now. They got that way from paying your bills, cleaning your clothes and listening to you talk about how cool you are. So before you save the rain forest from the parasites of your parents' generation, try "delousing" the closet in your own room.

  24. Stoichiometry: Percent CompositionLet’s see if you can: Determine the percent composition of each of the elements in a compound Use percent composition to determine the mass of each element in a compound Use percent compositions of a compound to determine the empirical and/or molecular formula

  25. Check out my latest invention…

  26. Percent Composition What is the percent carbon in C5H8NO4 (the glutamic acid used to make MSG monosodium glutamate), a compound used to flavor foods and tenderize meats? a) 8.22 % C b) 24.3 % C c) 41.1 % C d) not listed

  27. Empirical Formulas #10: 0.556 g C and 0.0933 g H. Determine the formula for this substance. moles of C = 0.556 g C 12.01 g/mole = 0.0463 moles of C moles of H = 0.0933 g H 1.01 g/mole = 0.0924 moles of H Formula:

  28. Stoichiometry: Percent CompositionLet’s see if you can: Determine the percent composition of each of the elements in a compound Use percent composition to determine the mass of each element in a compound Use percent compositions of a compound to determine the empirical and/or molecular formula

  29. Law of Mechanical Repair - After your hands become coated with  grease, your nose will begin to itch and you'll have to pee.

  30. Empirical Formulas #9: 72% iron and 28% oxygen Determine the formula for this substance. moles of Fe 72 g Fe 55.85 g/mole = 1.29 moles Fe moles of O = 28 g H 16.00 g/mole = 1.75 moles O Formula:

  31. Empirical Formulas #9: 72% iron and 28% oxygen Determine the formula for this substance. Multiply subscripts by 3 to get

  32. Example #4: Empirical Formula from % Composition • A substance has the following composition by mass: 40.0% C, 6.7% H, and 53.3% O(by mass). Calculate the empirical formula and the molecular formula of this compound given that the molar mass is 60.00 g/mol. • Consider a sample size of 100 grams Determine the number of moles of each Determine the simplest whole number ratio

  33. Empirical Formula from % Composition Determine the number of moles of each 40.0 g x 1 mol/12.01 g = 3.33 mol C 6.7 g x 1 mol/1.01 g = 6.63 mol H 53.3 g x 1 mol/16.00 g = 3.33 mol O CH2O

  34. Molecular Formula from Empirical Formula CH2O Empirical Molar Mass = 30.03 g/mol Molecular Molar Mass = 60.00 g/mol 30.03 g/mol ~2 C2H4O2

  35. 15 Helpful Hints On The Lab Report from Mr. T’s Vast Lab Experience!!! Hint #10. If an experiment works, you must be using the wrong equipment.

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