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We’ll be working in groups of 2.

OB: Percent composition by mass and mole problem practice day All your stuff, except pen and reference tables on black tables. We’ll be working in groups of 2. Put your stuff off to the side. You need paper, pen, and reference tables and calculators.

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We’ll be working in groups of 2.

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  1. OB: Percent composition by mass and mole problem practice dayAll your stuff, except pen and reference tables on black tables We’ll be working in groups of 2. Put your stuff off to the side. You need paper, pen, and reference tables and calculators.

  2. Everyone in the team works, some of you will talk more and some better ask more questions. The plan is that everyone here will leave in 40 minutes much smarter and much more confident. Question one: What is the percent composition by mass of aluminum in aluminum hydroxide (hint: do the % comp for Al, O, and H, make sure it’s 100% total) Let’s use 3 SF

  3. Question one: What is the percent composition by mass of aluminum in aluminum hydroxide (hint: do the % comp for Al, O, and H, make sure it’s 100% total) Al(OH)3 Al 1 x 27 = 27 Al O 3 x 16 = 48 O H 3 x 1 = 3 H 27g 78g X 100% = 34.6% 48g 78g X 100% = 61.5% 3g 78g X 100% = 3.85% 78 g/mole 99.95% = 100.%

  4. You find a box with a bar of metal that has stamped into it PURE GOLD. The bar weighs 500. grams. How many atoms of gold do you have?

  5. You find a box with a bar of metal that has stamped into it PURE GOLD. The bar weighs 500. grams. How many atoms of gold do you have? 500. g Au1 x 1 mole Au197 g Au = 2.54 moles Au 2.54 moles Au1 6.02 x 1023 atoms Au1 mole Au x = 2.54 x 6.02 x 1023 atoms Au = 15.3 x 1023 = 1.53 x 1024 atoms Au

  6. If you have 50.0 grams of sodium hydroxide, how many grams of those are oxygen?

  7. If you have 50.0 grams of sodium hydroxide, how many grams of those are oxygen? NaOH% Comp by Mass Na 1x23= 23 O 1x16= 16 O H 1x1 = 1 40g/mole 16g 40g X 100% = 40% 50.0 g x .40 = 20.0 grams oxygen

  8. Assume STP Mass of neon? 346 L neon gas

  9. Assume STP Mass of neon? 346 L neon gas 346 L Ne1 1 mole Ne22.4 L Ne x = 15.4 moles Ne 15.4 moles Ne1 20 g Ne1 mole Ne = 308 grams Ne x

  10. Some bigger compounds are not in the simple whole number ratio because the molecules are so big and often are made of repeating components. These larger formulas can be “reduced” mathematically into simpler ratios, but this is just math, these simplified formulas usually don’t exist. For example Glucose formula is C6H12O6 which has a 6:12:6 atomic ratio John Dalton said “simple, whole number ratios” for compounds, but this one exists commonly. It’s reduced, 1:2:1 ratio can exist on paper, in our mind, but not for real. These “reduced” to simple ratio formulas are called EMPIRICAL FORMULAS It’s math, not chemistry.

  11. C6H12O6 C8H18 (octane) C24H48 (candle wax) C2H2 (acetylene gas) H2O2 (hydrogen peroxide) C6H6 (cyclohexene) C10H22 (decane) C6H14 (hexane) C5H10O5 (pentose) CH2O (which doesn’t exist as a real compound, but it’s a real empirical formula) Empirical formulas are mathematical ratios of atoms to atoms in compounds, written as chemical formulas Let’s figure out some empirical formulas now

  12. C6H12O6 C8H18 (octane) C24H48 (candle wax) C2H2 (acetylene gas) H2O2 (hydrogen peroxide) C6H6 (cyclohexene) C10H22 (decane) C6H14 (hexane) C5H10O5 (pentose) CH2O C4H9 CH2 CH HO CH C5H11 C3H7 CH2O Let’s figure out some empirical formulas now

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  14. If you find 1.03 moles of gold and gold is selling for about $56.00/gram, are you rich or just happy, or does it not matter all that much? (hint, how many grams are in that many moles?

  15. If you find 1.03 moles of gold and gold is selling for about $56.00/gram, are you rich or just happy, or does it not matter all that much? (hint, how many grams are in that many moles? 1.03 moles Au1 197 g Au1 mole Au x = 203 g Au 203 g Au1 _$56.00_1 g Au x = $11,368 (wow)

  16. Due today: Mole Lab Due Tuesday: Mole HW #1 Due Wednesday: Mole HW #2 Due Thursday: Bubble Gum Lab Friday: No school Now: bubble gum lab (have fun) We’ll do this in silenceNo kidding

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