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

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Presentation on theme: "Chemistry SM-1131 Week 11 Lesson 1 Dr. Jesse Reich Assistant Professor of Chemistry Massachusetts Maritime Academy Fall 2008."— Presentation transcript:

1 Chemistry SM-1131 Week 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. N 2 + 3H 2  2NH 3 2Al + 3F 2  2AlF 3 P 4 + 5O 2  P 4 O 10 SO 3 + H 2 O  H 2 SO 4

7 Decomposition When things decompose they break down. This reaction is where 1 molecule breaks down into several molecules. CaCO 3  CaO + CO 2 2 HgO  2 Hg + O 2 2 KClO 3  2KCl + 3O 2

8 Combustion Reactions Combustion means burning and fire. What two things does fire require? O 2 and something to burn. We normally burn hydrocarbons (Hydro=H, Carbon = C therefore stuff made up of H and C). The products are always CO 2 and H 2 O. Methane and Oxygen burn write the equation. ___CH 4 + ___O 2  ___ CO 2 + ___H 2 O

9 Double all the coefficients 2C 2 H O 2  4 CO H 2 O 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 O 2.

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  AlCl 3 +3 Ag 2Na + H 2 O  H 2 + 2NaOH Zn 3 N 2 + 3Mg  Mg 3 N 2 + 3Zn

12 Double Displacement el 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 AgNO 3 + NaCl  AgCl + NaNO 3 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 + H 2 O

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 NH 4 will always be soluble Any compound with NO 3 or C 2 H 3 O 2 will always be soluble Compounds with Cl, Br, I will be soluble except with Ag, Hg or Pb Compounds with SO 4 will be soluble except with Sr, Ba, Pb, or Ca Hydroxides (OH) are mostly insoluble Compounds with CO 3 and PO 4 are insoluble unless with Li, Na, K or NH 4

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(NO 3 )  Ag + and NO 3 - Na 2 (SO 4 )  2Na + and SO 4 -2 H 3 PO 4  3H + and PO 4 3- Na(OH)  Na + and OH - Li 3 (PO 4 )  3Li + and PO 4 3-

20 Molecular Compounds Molecular compounds don’t do that. C 6 H 6 O 6 (s)  C 6 H 6 O 6(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? AlCl 3 + Na 3 PO 4  AlPO 4 + NaCl = skeleton AlCl 3 + Na 3 PO 4  AlPO 4 + 3NaCl = balanced AlCl 3(aq) + Na 3 PO 4(aq)  AlPO 4(s) + 3NaCl (aq) total eq

23 Total Ionic AlCl 3(aq) + Na 3 PO 4(aq)  AlPO 4(s) + 3NaCl (aq) total eq Al +3 (aq) + 3Cl - (aq) + 3Na + (aq) + PO 4 3- (aq)  AlPO 4(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) + PO 4 3- (aq)  AlPO 4(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) + PO 4 3- (aq)  AlPO 4(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 Grams of your known (given) substance Moles of your known (given) substance Grams of your unknown substance Moles of your unknown substance X Multiply by 1 mol known____ Molecular mass known Multiply by Molecular mass unknown 1 mol unknown Multiply by Molar Ratio Use balanced equation Moles unknown Moles known

29 Why Care? 2Cs +F 2  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 MgCl 2 + Na 3 PO 4  Mg 3 (PO 4 ) 2 + NaCl 3MgCl 2 + 2Na 3 PO 4  Mg 3 (PO 4 ) 2 + 6NaCl 3 moles MgCl 2 to 2 moles Na 3 PO 4 aka 3 moles MgCl 2 /2 moles Na 3 PO 4 aka 3MgCl 2 _ 3MgCl 2 3MgCl 2 2Na 3 PO 4 2Na 3 PO 4 Mg 3 (PO 4 ) 2 2Na 3 PO 4 Mg 3 PO4 6NaCl Mg 3 (PO 4 ) 2 Mg 3 (PO 4 ) 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 Grams of your known (given) substance Moles of your known (given) substance Grams of your unknown substance Moles of your unknown substance X Multiply by 1 mol known____ Molecular mass known Multiply by Molecular mass unknown 1 mol unknown Multiply by Molar Ratio Use balanced equation Moles unknown Moles known

34 Moles to moles 3MgCl 2 + 2Na 3 PO 4  Mg 3 (PO 4 ) 2 + 6NaCl You have 15 moles of MgCl2 how many moles of NaCl will you form? 15 moles MgCl 2 x 6 moles NaCl = 30 moles NaCl 3 moles MgCl 2 You have 26 moles of Na 3 PO 4 how many moles of Mg 3 (PO4) 2 will form? 26 moles Na 3 PO 4 x 1Mg 3 (PO 4 ) 2 = 13 moles Mg 3 (PO 4 )2 2Na 3 PO 4

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

36 Grams known to moles unknown 3MgCl 2 + 2Na 3 PO 4  Mg 3 (PO 4 ) 2 + 6NaCl 38 grams of MgCl 2 will make how many grams of Mg 3 (PO 4 ) 2 ? 38gMgCl 2 x 1 mol MgCl 2 x 1 Mg 3 (PO 4 ) 2 = (copy given) 96g MgCl 2 3MgCl moles Mg 3 (PO 4 ) 2

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

38 Moles known to grams unknown 3MgCl 2 + 2Na 3 PO 4  Mg 3 (PO 4 ) 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 Na 3 PO 4 x 3MgCl 2 x 96g MgCl 2 = (copy given) 2Na 3 PO 4 1 mol MgCl 2 648g MgCl 2

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

40 Big Mama: Grams known to grams unknown 3MgCl 2 + 2Na 3 PO 4  Mg 3 (PO 4 ) 2 + 6NaCl You made 38g of Mg 3 (PO 4 ) 2 how many grams of Na 3 PO 4 did you start with? 38g Mg 3 (PO 4 ) 2 x 1 mole Mg 3 (PO 4 ) 2 x 2 Na 3 PO 4 x 164g Na 3 PO 4 = (copied given) 262 g Mg 3 (PO 4 ) 2 1 Mg 3 (PO 4 ) 2 1 mol Na 3 PO 4 48g Na 3 PO 4

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: 3H 2 + N 2  2NH 3 You have 9 moles of H 2 and 6 moles of N 2 which is limiting. Method 1: 9H 2 x 2NH 3 = 6NH 3 3H 2 6N 2 x 2NH 3 = 12NH 3 1N 2 H 2 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: 9H 2 * 1N 2 = 3N= 3H 2 We have more N 2 than 3 moles so H 2 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.


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