Homework 10 due today Homework 11 posted and

• Homework 10 due today • Homework 11 posted and due Friday Monday…Ick

Limiting yield and % yield calculations are the last stop on the mole bus trip

Guided practice limiting yield problems (cont. ) 2 C 4 H 10 + 13 O 2 ---- 8 CO 2 + 10 H 2 O 58 32 44 18 g/mol Weight-weight-mol b)1 gram of C 4 H 10 and 10 grams of O 2 are reacted. What is the maximum yield in H 2 O in moles ? 1 gram C 4 H 10 = 1/58=0. 017 mol C 4 H 10 10 gram O 2 = 10/32=0. 312 mol O 2 0. 017 mol C 4 H 10 produces 10*0. 017/2=0. 085 mol H 2 O 0. 312 mol O 2 produces 10*0. 312/13 =0. 24 mol H 2 O

5. 2 Calculate the maximum yield problems (cont. ) 2 C 4 H 10 + 13 O 2 ---- 8 CO 2 + 10 H 2 O 58 32 44 18 g/mol Weight-weight-molecules c) 5. 8 grams of C 4 H 10 and 160 grams of O 2 are reacted. What is the maximum yield of CO 2 in molecule count? 5. 8/58=0. 1 mol C 4 H 10 160/32=5 mol O 2 0. 1 mol C 4 H 10 produces 8*0. 1/2=0. 4 mol CO 2 5 mol O 2 produces 8*5/13 =3. 07 mol CO 2 C 4 H 10 limits 0. 4*6*1023 = molecules CO 2= 2. 4*1023

5. 2 Calculate the maximum yield problems (cont. ) 2 C 4 H 10 + 13 O 2 ---- 8 CO 2 + 10 H 2 O 58 32 44 18 g/mol Weight-molecules-weight d)116 grams of C 4 H 10 and 1. 66*1024 molecules of O 2 react. What is the maximum yield of H 2 O in grams? 116/58=2 mol C 4 H 10 1. 66*1024/6*1023=1. 1 mol O 2 2 mol C 4 H 10 produces 1. 1 mol O 2 produces O 2 limits 10*2/2=10 mol H 2 O 10*1. 1/13 =0. 85 mol H 2 O Multiply down to calculate: grams H 2 O= 0. 85*18=15. 3 g

U-Try-It on your own Clicker Examples

2 C 4 H 10 + 13 O 2 ---- 8 CO 2 + 10 H 2 O Given 0. 5 mol of C 4 H 10 and 6. 5 mol of O 2, how many mol of CO 2 will form? A. 8 B. 4 C. 2 D. 1

2 C 4 H 10 + 13 O 2 ---- 8 CO 2 + 10 H 2 O 58 32 44 18 g/mol Given 10 g of C 4 H 10 and 100 g of O 2, how many mol of H 2 O will form? A. 2. 4 B. 0. 17 C. 0. 86 D. 3. 12 E. 0. 034

2 C 4 H 10 + 13 O 2 ---- 8 CO 2 + 10 H 2 O 58 32 44 18 g/mol Given 2*1022 molecules of C 4 H 10 and 50 g of O 2, how many g of CO 2 will form? 1 mol count =6*1023 A. 5. 8 g B. 0. 133 g C. 4. 22 g D. 1. 2 g

Given: MW 114 32 44 18 2 C 8 H 18 + 25 O 2 16 CO 2 + 18 H 2 O How many grams of CO 2 are possible if 4. 26*1022 molecules O 2 and 0. 324 g of C 8 H 18 are burned. (1 Mol count=6 EE 23) A. 3. 1 grams B. 2. 0 grams C. 1. 0 grams D. 0. 125 grams E. 0. 071 grams

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% Yield: a non-chemical analogy Billy comes home with his math test. He got 40 out of 75 questions right. Did he `pass’ the test ? Test Score = Test Yield = 100* 40 =53. 3% 75

Examples of chemical % yield 5 O 2 + C 3 H 8 Given mol excess 3 CO 2 + 4 H 2 O 0. 25 Theory product mol 4 x 0. 25 =1 (theory=max) 1 0. 7 (obs) Exp. Product mol 0. 7 % yield =exp. mol x 100 = 0. 7 x 100 theory mol 1 = 70%

Examples of chemical % yield (continued) 5 O 2 + C 3 H 8 3 CO 2 + 4 H 2 O MW 32 44 44 18 Given excess 1. 1 g 0. 5 g Exp. Mol CO 2 Mol C 3 H 8 = 1. 1=0. 025 = 0. 01136 44 44 Product CO 2 = 3 x 0. 025 =0. 075 moles theory 1 % yield = 100 * Exp Mol =0. 01136 x 100% =15. 1% Theory Mol 0. 075

Examples of chemical % yield (continued) Bob reports making 4. 0 g H 2 O from 2. 2 g C 3 H 8. Is Bob crazy or okay ? 5 O 2 + C 3 H 8 3 CO 2 + 4 H 2 O MW 32 44 Given excess 2. 2 g 44 18 Bob Reports 4. 0 g H 2 O Mol C 3 H 8 =2. 2 =0. 050 44 Product H 2 O = 4 x 0. 05 =0. 2 moles theory 1 Maximum g H 2 O=18 g/mol*0. 2 mol=3. 6 g Bob’s an idiot

MW 16 32 18 CH 4 + 2 O 2 CO 2 + 2 H 2 O A 1 g sample of methane (CH 4) is burned in excess O 2 according to the equation above. You collect 1. 125 g of H 2 O. What is your % yield ? A. 50% B. 89 % C. 25 % D. 100% E. I have no idea

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