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Moles and Stoichiometry

Moles and Stoichiometry. AP chemistry CCHS, Mr. Bartelt’s class. Mass spectroscopy. Information found on pg 82. What’s amu. Amu stands for atomic mass unit. 1 amu is defined as 1/12 th of the mass of the carbon—12 nucleus.

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Moles and Stoichiometry

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  1. Moles and Stoichiometry AP chemistry CCHS, Mr. Bartelt’s class

  2. Mass spectroscopy Information found on pg 82

  3. What’s amu • Amu stands for atomic mass unit. • 1 amu is defined as 1/12th of the mass of the carbon—12 nucleus. • If you recall from nuclear chemistry, as the size of the nucleus increases some of the mass is converted into energy to keep the nucleus from being blown apart by the repulsive force between the protons. • This causes the mass number (protons and neutrons) to not correspond perfectly with amus. Don’t worry too much about this. Amus will be given when they are needed.

  4. Calculating atomic masses • Look on pg83-85 • Chlorine has two natural isotopes: • Chlorine—35 (75.77% in nature) • Chlorine—37 (24.23% in nature) Determining the molar mass is simple. It’s a mater of performing a weighted average. You do it ever semester when you calculate your final grade. Look • Semester average 86.4 (80% of final grade) • Final exam 96.0 (20% of final grade) So you plug: 86.4*.8+96.0*.2 into your calculator and find your final average to be 88.32. Sorry, you got a “B”

  5. Calculating atomic masses • So think of chlorine just like your grade • Chlorine has two natural isotopes: • Chlorine—35 (75.77% in nature) • Chlorine—37 (24.23% in nature) So you plug: 35*.7577+.37*.2423 into your calculator and find the atomic weight to be 35.48. Which is close to the value on the periodic table 35.453 NOTE: If you use mass number’s you be a little off because amu and mass number are not the same thing. Amus for the different isotopes will be given if you need them

  6. The mole • Pages 90-93 • The molar mass that you look up on the periodic table is the mass in grams that one mole will weigh. • A mole is defined as 6.022*1023 (it has no units) • A mole is like a dozen (12) in that it’s a counting unit. A mole however is significantly larger. • If you gave away a million dollars every second for the last 4.5 billion years (since the earth formed) you would have only given a way ¼ of a mole of dollars. • A mole is HUGE

  7. Conversions • How many moles are in 1.204*1023 molecules of O2? • How much would that weigh in grams? • First calculate the molar mass of O2 • Next find the mass

  8. Sample problems (work in groups) • Find the mass of 1.5*1019 atoms of argon. • How many moles of sugar (C12H22O11) are in 200.0 grams of sugar? • How many alcohol (C2H5OH) molecules are in 14.5 pounds? • What weighs more 400. atoms of krypton or 900. molecules of N2?

  9. % composition • This is easy (93-95) • I’ll show you. Here’s how you find the % composition of cocaine (C17H21NO4) • First, find the mass contribution of all the elements and the total mass

  10. % composition continued • Second, find the percent by dividing each part by the whole • That’s it

  11. Practice • Find the % composition of octane (C8H18) • Find the % composition of MDMA (C11H15NO2) • Find the % composition of prozac (C17H18F3NO)

  12. Empirical formula from % composition • Pg 96-102 • It is possible to determine the empirical formula from % composition data. • A compound is found to be 20.00% C, 53.30% O, 23.34% N, 3.36% H by mass. What is the empirical formula? • This seems like a difficult problem but it’s easy once you learn the technique for solving it. • First assume that you have a 100.0 gram sample. This causes the % signs to become mass in grams

  13. Empirical formula from % composition 20.00g C, 53.30g O, 23.34g N, 3.36g H Next, convert grams into moles

  14. Empirical formula from % composition • Now you have a mole ratio. The easiest way to simplify this ratio is to divide each mole amount by the lowest value. • This means that the empirical formula is: CO2NH2 • The numbers in the simplified ratios are the subscripts in the empirical formula

  15. Sample problems • 63.50% Ag, 28.25% N, 8.25% N • 12.63% Li, 58.21% O, 29.17% S • 27.93% Fe, 48.01% O, 24.06% S • 29.28% C, 52.01% O, 17.08 N, 1.64% H The last one’s a little tricky. Ask if you need help

  16. Molecular formula • Pg 100-101 • As you saw in the bellwork, many different molecules can have the same empirical formula. In order to determine the molecular formula you need more than just the % composition. You also need the molar mass. • If you found that the empirical formula of a molecule was C2O2NH2. That could be the molecular formula, but so could C4O4N2H4 or C6O6N3H6.

  17. Molecular formula • If you know that the molar mass of the molecule with empirical formula C2O2NH2 is 144.1 g/mol then you can determine the molecular formula. • First find the molar mass of the empirical formula

  18. Molecular formula • Now divide144.1 g/mol by the molar mass of the empirical formula, 72.05 g/mol. • Now distribute that number across the empirical formula and you have the molecular formula • And you’re done

  19. Sample problems • You find a sample to be 44.45% C, 19.74% O, 34.57% N, 1.24% H by mass. You know that the molar mass of the molecule is 243.2 g/mol. • Find the empirical and molecular formulas

  20. Balancing equations • Pg 102-108 • Practice makes perfect (balance these) • Zn + HCl ---> ZnCl2 + H2 • C10H16 + Cl2 ---> C + HCl • C7H6O2 + O2 ---> CO2 + H2O • KClO3 ---> KClO4 + KCl • Al(OH)3 + H2SO4 ---> Al2(SO4)3 + H2O • Al2(SO4)3 + Ca(OH)2 ---> Al(OH)3 + CaSO4 • NaOH + Cl2 ---> NaCl + NaClO + H2O And much, much more

  21. Stoichiometry • Most stoichiometry problems will give you grams of one compound and ask for grams of another compound. • Example Given the equation below (unbalanced), how many grams of sulfur hexaflouride will be produced if 85.23 grams of sulfur are reacted? S8 + F2 SF6 A stoichiometry problem has 4 basic steps

  22. Stoichiometry (4 steps) 1) Balance the equation (sometimes this is already done) S8 + F2 SF6 becomes S8 + 24F2 8SF6 2) Convert grams to moles (given: 85.23 grams of sulfur) 3) Perform mole ratio S8 + 24F2 8SF6 X 0.3322

  23. Stoichiometry (Last step) 4) Convert moles to grams What you just found is the theoretical yield. That’s how much product would be produced in a “perfect world”. We don’t live in perfect world so we often times look at the % yield.

  24. Stoichiometry % yield • Example Given the equation below (unbalanced), 155.23 grams of tetraphosphorous decoxide yields only 102.25 grams of phosporic acid, what’s the % yield? P4O10 + H2O  H3PO4 1) Balance P4O10 + 6H2O  4H3PO4 2) Grams to moles

  25. Next 3) Perform mole ratio P4O10 + 6H2O  4H3PO4 4) Convert moles to grams 5) Determine percent yield

  26. Stoichiometry (LR) • Finding the limiting reagent is not very hard. • Example: Given the equation below, if you have 100.0 grams of aluminum and 75.00 grams of oxygen, what will the limiting reagent be? Al +O2 Al2O3 • Balance 4Al +3O2 2Al2O3 • First convert both from grams to moles

  27. Stoichiometry (LR) 4Al +3O2 2Al2O3 • Next, divide both by their coefficient • Oxygen is the limiting reagent. • Solve for the theoretical yield using oxygen. Forget about Al! The oxygen will run out and leave a little Al behind.

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