Download
chemistry sm 1131 week 11 lesson 1 n.
Skip this Video
Loading SlideShow in 5 Seconds..
Chemistry SM-1131 Week 11 Lesson 1 PowerPoint Presentation
Download Presentation
Chemistry SM-1131 Week 11 Lesson 1

Chemistry SM-1131 Week 11 Lesson 1

151 Views Download Presentation
Download Presentation

Chemistry SM-1131 Week 11 Lesson 1

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Chemistry SM-1131Week 11 Lesson 1 Dr. Jesse Reich Assistant Professor of Chemistry Massachusetts Maritime Academy Fall 2008

  2. Class Today • Poem • Test Wednesday on chapter 7&8 • Review Chapter 7 • Review/cover chapter 8

  3. Poem • Fay school motto (English translation): “Where there’s a will there’s a way”

  4. Pre-Chapter 7 • Review- 4 steps to make an ionic compound: symbols, charges switcheroo, reduce • Ionic, molecular, and acid nomenclature • Polyatomics guaranteed to be on the exam: sulfate, phosphate, nitrate, ammonium, carbonate, hydroxide. Know names, charges, and formulas.

  5. Types of Chemical Reactions • There are 5 main types of reactions • Combination aka synthesis • Decomposition • Combustion • Single displacement • Double displacement • (Acid Base, gas evolution, precipitation, oxidation and reduction aka redox)

  6. Combination Reactions • 2 things come together to make 1 thing. • N2 + 3H2 2NH3 • 2Al + 3F2  2AlF3 • P4 + 5O2  P4O10 • SO3 + H2O  H2SO4

  7. Decomposition • When things decompose they break down. This reaction is where 1 molecule breaks down into several molecules. • CaCO3 CaO + CO2 • 2 HgO  2 Hg + O2 • 2 KClO3  2KCl + 3O2

  8. Combustion Reactions • Combustion means burning and fire. What two things does fire require? O2 and something to burn. We normally burn hydrocarbons (Hydro=H, Carbon = C therefore stuff made up of H and C). • The products are always CO2 and H2O. • Methane and Oxygen burn write the equation. • ___CH4+ ___O2 ___ CO2 + ___H2O

  9. Double all the coefficients • 2C2H6 + 7 O2 4 CO2 + 6 H2O 4 C 4_ 12 H 12_ 14 O 14_ If your fuel source has an even number of carbons in it then you will have to double the coefficients of all reactants and products. Otherwise, you’ll have X.5 moles of O2.

  10. Displacements • These are the 2 hardest to tell apart when starting. • Single displacements typically have 1 lone element on both sides of the reaction • Double displacements look like the biggest reactions out there, and you’ll see that the two metals switch places

  11. Single Displacement Examples • 3AgCl + Al  AlCl3 +3 Ag • 2Na + H2O  H2 + 2NaOH • Zn3N2 + 3Mg  Mg3N2 + 3Zn

  12. Double Displacementel double • Silver nitrate and sodium chloride react to form silver chloride and sodium nitrate. Write and balance the reaction. • KEY POINT: FIGURE OUT THE FORMULA FOR EACH INORGANIC PIECE AND DON’T MESS WITH THE FORMULA FOR THE REST OF THE TIME!

  13. El Double • AgNO3 + NaCl  AgCl + NaNO3 • It’s already balanced. • The Ag and Na switched places. • That’s why it’s a double displacement.

  14. Acid Base • These are like double displacement reactions, except one of the compounds is going to be an acid and the other will be an ionorganic salt • Salt just means combination of a cation and anion in a solid form • Hydrochloric acid and sodium hydroxide react together. Write the reaction.

  15. ACID BASE • HCl + NaOH  ____ + ____ • The “metals” swithc places. So H and Na will switch. Na will be with Cl, and H will be with OH • HCl + NaOH  NaCl + H2O

  16. Solubility • Often times we perform a double displacement reaction to actually collect one of the products. We can make certain compounds crash out of (precipitate) an aqueous solution because of how soluble some compounds are. • We’ll take two soluble compounds, they will react, and they will typically make one soluble product and one insoluble product.

  17. Solubility Rules • Any compound with Li, Na, K, or NH4 will always be soluble • Any compound with NO3 or C2H3O2 will always be soluble • Compounds with Cl, Br, I will be soluble except with Ag, Hg or Pb • Compounds with SO4 will be soluble except with Sr, Ba, Pb, or Ca • Hydroxides (OH) are mostly insoluble • Compounds with CO3 and PO4 are insoluble unless with Li, Na, K or NH4

  18. Dissolving • So, what happens when an inorganic compounds dissolves (this is totally different than a molecular compound dissolving)? • Water molecules act as crowbars that split molecules into pieces. The two pieces formed are the cations and the anions. • When you see table salt it’s the compound NaCl. When you dissolve it in water it’s actually Na+ and Cl-.

  19. Try some more • Ag(NO3) Ag+ and NO3- • Na2(SO4) 2Na+ and SO4-2 • H3PO4 3H+ and PO43- • Na(OH)  Na + and OH- • Li3(PO4)  3Li+ and PO43-

  20. Molecular Compounds • Molecular compounds don’t do that. • C6H6O6(s)  C6H6O6(aq) no change occurs. • All molecular compounds do not break up into individual ions. Ionic compounds in water break up into individual ions.

  21. Spectator • Spectator ions- reactions have components that aren’t that important to the overall effect. We can tell they aren’t that important because they appear on both sides of the chemical equation. They aren’t really participating, they are just hanging out. We call them spectator ions.

  22. Net Ionic Equations • Net just means overall, so we’re trying to figure out what’s the overall reaction. • Aluminum chloride and sodium phosphate undergo a double displacement reaction. What precipitates and what’s the net ionic equation? • AlCl3 + Na3PO4 AlPO4 + NaCl = skeleton • AlCl3 + Na3PO4 AlPO4 + 3NaCl = balanced • AlCl3(aq) + Na3PO4(aq) AlPO4(s) + 3NaCl(aq) total eq

  23. Total Ionic • AlCl3(aq) + Na3PO4(aq) AlPO4(s) + 3NaCl(aq) total eq Al+3(aq) + 3Cl-(aq) + 3Na+(aq) + PO43-(aq)  AlPO4(s) + 3Na+(aq) + 3Cl-(aq) • That’s the total ionic equation • Note the (s) thing is not in pieces. That’s because only things that are (aq) are going to break up like that. • To get the net ionic we have to cancel out the spectator ions from both sides.

  24. Net Al+3(aq) + 3Cl-(aq) + 3Na+(aq) + PO43-(aq)  AlPO4(s) + 3Na+(aq) + 3Cl-(aq) 3Cl-(aq) + 3Na+(aq) appear on both sides of the equation. We’re going to cancel them out. If we do that it leaves us with the “Net Ionic Equation.” • Al+3(aq) + PO43-(aq)  AlPO4(s) is the net equation.

  25. So steps? • 1-Inorganic formula writing for each inorganic compound • 2-skeleton equation • 3-balanced equation • 4-total equation (add solubilities) • 5-Total ionic equation • 6-Cancel out spectator ions • 7-Net ionic equations

  26. Chapter 8 • Review going somewhat backwards • Mole Map • Molar Ratios • Grams to moles • Moles to grams • Limiting Reagent • % yeild

  27. Tips for stoichiometry problem solving • Use the mole map • Make sure you have a balanced RXN to work with • Copy the given to start solving the problem • Only account for the coefficients once, and that’s during the mole to mole conversion using the molar ratio from the balanced RXN • Make sure you know how to calculate molecular masses

  28. Mole Map 1 Multiply by Molar Ratio Use balanced equation Moles unknown Moles known Moles of your known (given) substance Moles of your unknown substance Multiply by Molecular mass unknown 1 mol unknown Multiply by 1 mol known____ Molecular mass known X Grams of your known (given) substance Grams of your unknown substance

  29. Why Care? • 2Cs +F2 2CsF • 2 atoms of Cs and 2 atoms F: same ratio of atoms • But, Cs weighs 133amu and F is only 19 • If you add equal grams because they are equal numbers of atoms you will have a massive excess of fluorine. You need ~6 times as many grams of Cs to balance the reaction even though the number of atoms are the same. • Just because atoms are balanced doesn’t mean the masses will be close. • That’s why we need stoichiometry! So, we use relevant amounts of mols (atoms) and relevant amounts of grams.

  30. Molar Ratios • A  B+ C or A + B  C • A/B, A/C, B/C and reciprocals • A + B  C + D • A/B, A/C, A/D. B/C, B/D, C/D, & reciprocals

  31. Molar Ratio Example • Magnesium chloride and sodium phosphate under a double displacement reaction. What molar ratios exist? • Steps: Skeleton, Balance, Ratios • MgCl2 + Na3PO4 Mg3(PO4)2 + NaCl • 3MgCl2 + 2Na3PO4 Mg3(PO4)2 + 6NaCl • 3 moles MgCl2 to 2 moles Na3PO4 aka • 3 moles MgCl2/2 moles Na3PO4 aka 3MgCl2_ 3MgCl23MgCl2 2Na3PO4 2Na3PO4Mg3(PO4)2 2Na3PO4 Mg3PO4 6NaCl Mg3(PO4)2 Mg3(PO4)2 6NaCl

  32. Using Molar Ratios • Like the mole map says: • Look at the balanced equation, figure out the molar ratios • Copy the given • Multiply by the molar ratio that will allow you to cancel moles of the known/given (that means moles of the given has to be in the denominator) and that gets you into moles of the unknown (has to be in the numerator)

  33. Mole Map 1 Multiply by Molar Ratio Use balanced equation Moles unknown Moles known Moles of your known (given) substance Moles of your unknown substance Multiply by Molecular mass unknown 1 mol unknown Multiply by 1 mol known____ Molecular mass known X Grams of your known (given) substance Grams of your unknown substance

  34. Moles to moles • 3MgCl2 + 2Na3PO4 Mg3(PO4)2 + 6NaCl • You have 15 moles of MgCl2 how many moles of NaCl will you form? 15 moles MgCl2 x 6 moles NaCl = 30 moles NaCl 3 moles MgCl2 • You have 26 moles of Na3PO4 how many moles of Mg3(PO4)2 will form? 26 moles Na3PO4 x 1Mg3(PO4)2 = 13 moles Mg3(PO4)2 2Na3PO4

  35. Mole Map 1 Multiply by Molar Ratio Use balanced equation Moles unknown Moles known Moles of your known (given) substance Moles of your unknown substance Multiply by Molecular mass unknown 1 mol unknown Multiply by 1 mol known____ Molecular mass known X Grams of your known (given) substance Grams of your unknown substance

  36. Grams known to moles unknown • 3MgCl2 + 2Na3PO4 Mg3(PO4)2 + 6NaCl • 38 grams of MgCl2 will make how many grams of Mg3(PO4)2? 38gMgCl2 x 1 mol MgCl2 x 1 Mg3(PO4)2= (copy given) 96g MgCl2 3MgCl2 0.13 moles Mg3(PO4)2

  37. Mole Map 1 Multiply by Molar Ratio Use balanced equation Moles unknown Moles known Moles of your known (given) substance Moles of your unknown substance Multiply by Molecular mass unknown 1 mol unknown Multiply by 1 mol known____ Molecular mass known X Grams of your known (given) substance Grams of your unknown substance

  38. Moles known to grams unknown • 3MgCl2 + 2Na3PO4 Mg3(PO4)2 + 6NaCl • You want to use 4.5 moles of Na3PO4 how many grams of MgCl2 will you need to add to make the reaction run to completion? 4.5moles Na3PO4 x 3MgCl2 x 96g MgCl2 = (copy given) 2Na3PO4 1 mol MgCl2 648g MgCl2

  39. Mole Map 1 Multiply by Molar Ratio Use balanced equation Moles unknown Moles known Moles of your known (given) substance Moles of your unknown substance Multiply by Molecular mass unknown 1 mol unknown Multiply by 1 mol known____ Molecular mass known X Grams of your known (given) substance Grams of your unknown substance

  40. Big Mama: Grams known to grams unknown • 3MgCl2 + 2Na3PO4 Mg3(PO4)2 + 6NaCl • You made 38g of Mg3(PO4)2 how many grams of Na3PO4 did you start with? 38g Mg3(PO4)2 x 1 mole Mg3(PO4)2 x 2 Na3PO4 x 164g Na3PO4 = (copied given) 262 g Mg3(PO4)2 1 Mg3(PO4)2 1 mol Na3PO4 48g Na3PO4

  41. % Yield and limiting reagents • % yield = 100 * actual yield/theoretical yield • It tells you how much product you produced out of the total potential. Higher than 100% typically means you also have impurities. • A limiting reagent is the chemical that is used up first in a reaction and then without it the reaction stops. • In a combustion reaction fuel will burn until one of two things happens. The fuel is all burnt or oxygen is cut off. Either way when one of the reactants is removed the reaction stops.

  42. Limiting reagents • There are two ways to figure out which reagent is a limiting reagent. • 1- Do two stoichiometry problems. The first using the first reagent, the second using the second reagent. Whichever makes less product is the limiting reagent. (more work easier to understand) • 2-Do a stoichiometry problem converting from one reagent to the other. If the amount you come up with is more than what you have then the first reagent is the limiting reagent if it’s less the second reagent is limiting (less work harder to understand)

  43. Limiting reagent • Real simple: 3H2 + N2 2NH3 • You have 9 moles of H2 and 6 moles of N2 which is limiting. • Method 1: 9H2 x 2NH3 = 6NH3 3H2 6N2 x 2NH3 = 12NH3 1N2 H2 is limiting. Even though we have more of it more is consumed in the reaction. You can calculate in terms of moles or grams, but be consistant. Method 2: 9H2* 1N2 = 3N= 3H2 We have more N2 than 3 moles so H2 is limiting

  44. Test Tomorrow • This was your chapter 7&8 review • Everything in this PPT is fair game • New grades will be posted on the Monday after the exam.