3. Thursday, May 2 nd : “A” Day Friday, May 3 rd : “B” Day Agenda  Homework Problems/Questions  Finish Section 7.3 (we’re just about done…)  Homework: Sec. 7.3 review, pg. 248: #1-7  Extra Credit Opportunity Next time: Sec. 7.3 quiz Lab write-up

4. Homework Questions/Problems  Practice pg. 245: #1-3  Worksheet pg. 61: 1a-e

5. Percentage Composition  Back on day 1 of this section, we used percentage composition to find the empirical formula for compounds.  Question: How do you calculate the percentage composition?

6. CO and CO 2 are both made up of C and O, but they have different percentage compositions.

7. Rules to Determine Percentage Composition from Chemical Formulas 1. Use the periodic table to find the molar mass of the compound. 2. Divide the mass contributed by each element by the molar mass of the compound. 3.Multiply by 100 and add a % sign.

8. Using a Chemical Formula to Determine Percentage Composition (Sample Problem I, pg. 247)  Calculate the percentage composition of copper (I) sulfide.  You first have to figure out the formula for copper (I) sulfide: Cu 1+ S 2- because copper is 1 + and the sulfide ion is 2 -, two copper cations are needed to balance the charge: Cu 2 S

9. Sample Problem I, continued… 1.Find the molar mass of Cu 2 S: Cu: 2 (63.55 g/mol) = 127.1 g/mol S: 32.07 g/mol Molar mass of Cu 2 S: 159.17 g/mol

10. Sample Problem I, continued… 2. Divide the mass contributed by each element in the compound by the molar mass of the compound. 3. Then, multiply by 100% to get the % composition. Cu = 2 (63.55) X 100% = 79.85 % 159.17 S = 32.07 X 100% = 20.15% 159.17  Check your work, when added together, the sum should be near 100%.  These are the same calculations we did in the jelly bean lab!

11. Example #1  Determine the percentage composition of hydrogen sulfite, H 2 SO 3. 1.Find molar mass of H 2 SO 3. H: 2 (1.01 g/mol) = 2.02 g/mol S: 32.07 g/mol O: 3 (16.00 g/mol) = 48.00 g/mol Molar mass of H 2 SO 3 = 82.09 g/mol

12. Example #1, cont… 2. Divide the mass of each element in the compound by the molar mass of the compound. 3.Then multiply by 100% to get the % composition H = 2 (1.01) X 100% = 2.46% 82.09 S = 32.07 X 100% = 39.07% 82.09 O = 3 (16.00) X 100% = 58.47% 82.09

13. One More Example…  Determine the percentage composition of C 2 H 5 COOH. 1. Find molar mass of C 2 H 5 COOH: C: 3 (12.01 g/mol) = 36.03 g/mol H: 6 (1.01 g/mol) = 6.06 g/mol O: 2 (16.00 g/mol) = 32.00 g/mol Molar mass of C 2 H 5 COOH = 74.09 g/mol

14. One More Example conti… 2. Divide the mass contributed by each element in the compound by the molar mass of the compound. 3.Then, multiply by 100% to get the % composition. C = 3 (12.01) X 100% = 48.63% 74.09 H: 6 (1.01) X 100% = 8.18% 74.09 O: 2 (16.00) X 100% = 43.19% 74.09

15. Homework  Sec. 7.3 review, pg. 248: #1-7  Concept Review: finished chapter! Next Time: Sec. 7.3 quiz Lab Write-Up

16. Extra Credit Opportunity  Attend Morrison High School’s production of the musical “You’re a Good Man Charlie Brown” on either Friday or Saturday night at 7:30 in the MHS auditorium.  Get your program signed by a cast member or by me.  Tickets are $5.00 at Fitzgerald’s Pharmacy or at the door.  Bring it in next class period for 15 pts x-credit!