1. Happy Groundhog Day!! Read for NEXT Wednesday: Read for NEXT Wednesday: Chapter 4: Sections 4-6 Chapter 4: Sections 4-6 HOMEWORK – DUE Monday 2/9/15 HOMEWORK – DUE Monday 2/9/15 HW-BW 3.2 (Bookwork) CH 3 #’s 41, 45, 48, 56-61 all, 68, 70, 72, 73, 78, 79, 85, 88, 95, 96, 104, 119, 124 HW-BW 3.2 (Bookwork) CH 3 #’s 41, 45, 48, 56-61 all, 68, 70, 72, 73, 78, 79, 85, 88, 95, 96, 104, 119, 124 HOMEWORK – DUE Wednesday 2/11/15 HOMEWORK – DUE Wednesday 2/11/15 HW-WS 5 (Worksheet, from course website) HW-WS 5 (Worksheet, from course website) HOMEWORK – DUE Wednesday 2/11/15 HOMEWORK – DUE Wednesday 2/11/15 HW-BW 4.1 (Bookwork) CH 4 #'s 3, 7, 9, 16, 17-31 odd, 42, 44, 46, 56, 62, 64, 66, 70, 72, 149 HW-BW 4.1 (Bookwork) CH 4 #'s 3, 7, 9, 16, 17-31 odd, 42, 44, 46, 56, 62, 64, 66, 70, 72, 149 Lab next Mon./Tues. Lab next Mon./Tues. Lecture in Lab Lecture in Lab Lab next Wed./Thurs. Lab next Wed./Thurs. EXP 5 EXP 5

2. EXAM ON MONDAY! Wait outside until I come get you! Be ready to come in and sit down Backpacks, bags, etc. go to the side/front/back of room Names will be alphabetical SCIENTIFIC calculators only Pencil and calculator only to desk No scantron needed Exam will be video recorded Exam will cover Chapters 1-3 and labs A, 1, 2 DO NOT BE LATE!

3. Step 1: assume 100 g THEN convert each mass to MOLES Step 2: divide EACH by the smallest!! What is the empirical formula of a compound made up of 74.01% C, 5.23% H, and 20.76% O?

4. So we have 4.75 mol C, 4 mol H, and 1 mol O Step 3: If all of the subscripts are NOT whole numbers, then multiply EACH ELEMENT by the smallest number that will make all of them whole numbers C 4.75 H 4 O What is the empirical formula of a compound made up of 74.01% C, 5.23% H, and 20.76% O? multiply EACH of the subscripts by 4! C 4.75 x 4 = 19 H 4 x 4 = 16 O 1 x 4 = 4 C 19 H 16 O 4

5. You burn 2769. grams of an unknown hydrocarbon and collect 8505 grams of CO 2(g) and 3918 grams of H 2 O (g). a)What is the empirical formula of the hydrocarbon? b) If the molar mass of the hydrocarbon is 114.26 g/mol, what is the molecular formula?

6. You burn 2769. grams of an unknown hydrocarbon and collect 8505 grams of CO 2(g) and 3918 grams of H 2 O (g). empirical formula = C 4 H 9

7. You burn 2769. grams of an unknown hydrocarbon and collect 8505 grams of CO 2(g) and 3918 grams of H 2 O (g). b)If the molar mass of the hydrocarbon is 114.26 g/mol, what is the molecular formula? empirical formula = C 4 H 9 = 57.13 g/mol 2(C 4 H 9 )  C 8 H 18 molecular formula = C 8 H 18

8. You burn 1116 g of an unknown hydrocarbon derivative (C x H y O z ) and collect 2617 g of CO 2(g) and 401 g of H 2 O (g). a)What is the empirical formula of this compound? b) If the molar mass of the compound is 300 g/mol, what is the molecular formula?

9. You burn 1116 g of an unknown hydrocarbon derivative (C x H y O z ) and collect 2617 g of CO 2(g) and 401 g of H 2 O (g). Where does all of the carbon in the compound go? Where does all of the carbon in the CO 2 come from? Allows us to use mol-to-mol C/CO 2 Where does all of the hydrogen in the compound go? Where does all of the hydrogen in the H 2 O come from? Allows us to use mol-to-mol H/H 2 O Where does all of the oxygen in the compound go? Cannot use a mol-to-mol ratio for oxygen CO 2 compound H2OH2O everywhere

10. C8H6O3C8H6O3 You burn 1116 g of an unknown hydrocarbon derivative (C x H y O z ) and collect 2617 g of CO 2(g) and 401 g of H 2 O (g).

11. b)If the molar mass of the compound is 300 g/mol, what is the molecular formula? empirical formula = C 8 H 6 O 3 = 150.14 g/mol 2(C 8 H 6 O 3 )  C 16 H 12 O 6 molecular formula = C 16 H 12 O 6

12. Calculating Empirical Formulae What is the empirical formula of a compound that is made up of 32.79% Na, 13.02% Al, and 54.19% F? Step 1: assume 100 g THEN convert each mass to MOLES Step 2: divide EACH by the smallest!!

13. Calculating Empirical Formulae What is the empirical formula of a compound that is made up of 32.79% Na, 13.02% Al, and 54.19% F? Step 1: assume 100 g THEN convert each mass to MOLES Step 2: divide EACH by the smallest!!

14. Calculating Empirical Formulae So we have 3 mol Na, 1 mol Al, and 6 mol F Step 3: if each is a whole number, write the empirical formula Na 3 AlF 6 What is the empirical formula of a compound that is made up of 32.79% Na, 13.02% Al, and 54.19% F?

15. Calculating Empirical Formulae Step 1: assume 100 g THEN convert each mass to MOLES Step 2: divide EACH by the smallest!! What is the empirical formula of a compound that is made up of 62.1% C, 5.21% H, 12.1% N, and 20.7% O?

16. Calculating Empirical Formulae So we have 6 mol C, 6 mol H, 1 mol N, and 1.5 mol O Step 3: if each is a whole number, write the empirical formula C 12 H 12 N 2 O 3 What is the empirical formula of a compound that is made up of 62.1% C, 5.21% H, 12.1% N, and 20.7% O? O 1.5 x 2 = 3 C 6 x 2 = 12 H 6 x 2 = 12 N 1 x 2 = 2 C 6 H 6 NO 1.5

17. Molecular Formulae: The Next Step What is the empirical formula of: C6H9N3C6H9N3 C2H3NC2H3N ratio = 3 to 1 What are the molar masses of the molecular and empirical formulae? C 6 H 9 N 3 = 123.18 g/mol C 2 H 3 N = 41.06 g/mol ratio = 3 to 1

18. Molecular Formulae: The Next Step Molecular formula and empirical formula will always have the same ratio as the molecular mass and the empirical mass! If a compound has a molar mass of 78.1150 g/mol and an empirical formula of CH, what is the molecular formula? empirical mass = 13.02 g/mol mass ratio = 6 to 1 formula ratio = 6 to 1 6(CH)  C 6 H 6 molecular formula = C 6 H 6

19. Caffeine is my life blood! Give the empirical and molecular formulae based on: 40.09g C, 4.172g H, 23.41g N, and 13.37g O. The molar mass of caffeine is 194.19 g/mol Molecular Formulae: The Next Step C4H5N2OC4H5N2O

20. C4H5N2OC4H5N2O Caffeine is my life blood! Give the empirical and molecular formulae based on: 40.09g C, 4.172g H, 23.41g N, and 13.37g O. The molar mass of caffeine is 194.19 g/mol Molecular Formulae: The Next Step C 4 H 5 N 2 O = 97.11 g/mol mass ratio = 2 to 1 formula ratio = 2 to 1 2(C 4 H 5 N 2 O)  C 8 H 10 N 4 O 2 molecular formula = C 8 H 10 N 4 O 2

21. Give the empirical and molecular formulae for a compound based on: 6.85g C, 0.432g H, 3.41g O, and 11.4g Br. The molar mass of the compound is 929.88 g/mol Molecular Formulae: The Next Step C 8 H 6 O 3 Br 2 ~ 4 mol C ~ 3 mol H ~ 1.5 mol O 1 mol Br x2

22. C 8 H 6 O 3 Br 2 = 309.96 g/molC 8 H 6 O 3 Br 2 Molecular Formulae: The Next Step mass ratio = 3 to 1 formula ratio = 3 to 1 3(C 8 H 6 O 3 Br 2 )  C 24 H 18 O 9 Br 6 molecular formula = C 24 H 18 O 9 Br 6 Give the empirical and molecular formulae for a compound based on: 6.85g C, 0.432g H, 3.41g O, and 11.4g Br. The molar mass of the compound is 929.88 g/mol

23. 31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction? K 2 SO 4(aq) + Pb(C 2 H 3 O 2 ) 2(aq)  2 KC 2 H 3 O 2(aq) + PbSO 4(s) 174.27 g/mol325.3 g/mol98.15 g/mol303.3 g/mol limiting reactanttheoretical yield of KC 2 H 3 O 2 theoretical yield = 15.09g KC 2 H 3 O 2 Pb(C 2 H 3 O 2 ) 2 is the L.R. (0 left) K 2 SO 4(aq) + Pb(C 2 H 3 O 2 ) 2(aq)  2 KC 2 H 3 O 2(aq) + PbSO 4(s)

24. (Hint: start with the given amount of the limiting reactant.) Now calculate how much of the other product(s) will be formed. K 2 SO 4(aq) + Pb(C 2 H 3 O 2 ) 2(aq)  2 KC 2 H 3 O 2(aq) + PbSO 4(s) 174.27 g/mol325.3 g/mol98.15 g/mol303.3 g/mol 31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction?

25. 23.31 g 15.09 g 31.89 g of potassium sulfate is reacted with 25.00 g of lead (II) acetate. What are the masses of all reactants and products after the reaction? reactantsproducts K 2 SO 4 = 18.49 g Pb(C 2 H 3 O 2 ) 2 = 0 g (L.R.) KC 2 H 3 O 2 = PbSO 4 = Mass must be the same before and after the reaction!! mass before reaction  31.89 + 25.00 = 56.89 g mass after reaction  15.09 + 23.31 = 38.40 g difference will be the mass of excess reactant left over = 18.49 g K 2 SO 4(aq) + Pb(C 2 H 3 O 2 ) 2(aq)  2 KC 2 H 3 O 2(aq) + PbSO 4(s)

26. theoretical yield = 19.66g PbCl 2(s) Pb(NO 3 ) 2 is the L.R. (0 left) CrCl 3 =Pb(NO 3 ) 2 =PbCl 2 = 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) Cr(NO 3 ) 3 = 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s)

27. 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) (Hint: start with the given amount of the limiting reactant.) Now calculate how much of the other product(s) will be formed. CrCl 3 =Pb(NO 3 ) 2 =PbCl 2 =Cr(NO 3 ) 3 =

28. 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 19.66 g 11.22 g reactantsproducts CrCl 3 = 7.24 g Pb(NO 3 ) 2 = 0 g (L.R.) Cr(NO 3 ) 3 = 11.22 g PbCl 2 = 19.66 g Mass must be the same before and after the reaction!! mass before reaction  14.71 + 23.41 = 38.12 g mass after reaction  11.22 + 19.66 = 30.88 g difference will be the mass of excess reactant left over = 7.24 g 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s)