1. Announcements To join clicker to class today: – Turn on the Clicker (the red LED comes on). – Push “Join” button followed by “20” followed by the “Send” button (switches to flashing green LED if successful). Monday discussion = review. Monday lecture may have intro to chapter 5—if want preview look at sections 5.1&5.2. ● Exam next Wed. ● Quiz on sections 4.2 – 4.8 Monday.

2. Review Calculating MM of compounds. Stoichiometry used to figure out masses or moles of reagents used or products produced in a chemical reaction. Calculating % yield. – Finished with clicker question

3. Final Clicker Question Heat 200.00 g CaCO 3 (limestone) to get CaO + CO 2, get 80.00 g CO 2. What is % yield of CO 2 (or % of CaCO 3 decomposed)? You were given molar masses to save time. 85% got the right answer 90.97 % The most common wrong answer was 40%. Path to correct answer: – Balance equation: CaCO 3 --> CaO + CO 2 – Calculate g of CO 2 expected from 200.00 g of CaCO 3 – Calculate % yield = 100%*(80.00 g CO 2 /expected g CO 2 )

4. Other possible questions with same info. ● How many g of CaO do I get when 80.00 g of CO 2 are produced? – MM(CaO) =56.077 g/mol – MM(CO 2 ) = 44.010 g/mol – MM(CaCO 3 ) = 100.087 g/mol Concrete is made by slaking CaO with water to make Ca(OH) 2.How much Ca(OH) 2 will I get when the CaO is slaked? – MM(Ca(OH) 2 ) = 74.0926 g/mol CaCO 3 --> CaO + CO 2

5. Example Limiting Reagent Calculation RXN: 3Na 2 CO 3 (aq) + 2H 3 PO 4 (aq) ––> 2Na 3 PO 4 (aq) + 3CO 2 (g) + 3H 2 O(l) Have 5.00 g of each reagent. How much CO 2 do we get? –M ( Na 2 CO 3 )= 105.988 g/mol Na 2 CO 3 –M (H 3 PO 4 )=97.994 g/mol H 3 PO 4 –M (CO 2 )=44.010 g/mol CO 2

6. % Composition: useful data for examples M (H 2 O)= 18.0153 g H 2 O/mole H 2 O. Molecules called ethylene and propylene are both 14.372% by mass H. The rest is carbon (85.628%). –M (ethylene)= 28.054 g/mol –M (propylene)= 42.081 g/mol

7. Review Stoichiometry used to figure out masses of reagents used or products produced in a chemical reaction. Percent yield: Reviewed limiting reagents. After balancing chemical equations all of these calculations require doing the following series of conversion given aA + … ––> bB + …: –g A –(÷ M A )–> mol A –([b mol B/a mol A])–> mol B –(x M B )–> g B. –For limiting reagent we do this process up to the moles of B stage for all reagents and only do the last step for the smallest moles of B. –Note that sometimes A and B will be on the same side of the balanced chemical equation: cC + dD ––> aA + bB.