Solutions-Manual-Smith-C04

# Ch 3 ch 2 ch 2 ch 2 ch 3 ch 2 ch 3 ch 2 ch 2 ch 3 hc

• Notes
• 10

This preview shows page 6 - 8 out of 10 pages.

-CH 3 -CH 2 CH 2 CH 2 CH 3 -CH 2 CH 3 -CH 2 CH 2 CH 3 HC CH 3 CH 3 C CH 3 CH 3 CH 3 35. For a 5-carbon chain, use pent- . 36. In ethylcyclohexane, the substituent chain has fewer carbon atoms than the ring, so the base name is cyclohexane. In cyclohexyldodecane, the long chain has more carbon atoms than the ring, so the base name is dodecane. ethylcyclohexane 1-cyclohexyldodecane 37. CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 (CH 3 ) 2 CHCH 2 CH 2 CH 3 CH 3 CH 2 CH 2 CH 2 CH 2 CH=CH 2 (a) (b) (c) (d) CH 3 CH 2 C ! C(CH 2 ) 8 CH 3

Subscribe to view the full document.

38. Calculate the % C and H as well as an empirical formula for each of the following using the combustion analysis provided. (a) A sample weighing 0.52g was burned in the presence of oxygen to give 0.6688 g of water and 1.6344 g of CO 2 . g C = 0.2727 x 1.6344 = 0.4457g and g H = 0.1111 x 0.6688= 0.0743 Assume a hydrocarbon, so %C = 0.4457/(0.4457 + 0.0743) = 0.4457/0.5200 x 100 = 85.71% %H = 0.0743/0.52 x 100 = 14.29% Assume 100 g, so 85.71g C and 14.29 g H. moles C = 85.71/12 = 7.14 and moles H = 14.29/1 = 14.29 so, 14.29/7.14= 2H / C = CH 2 = any cylic alkane formula C n H 2n (b) A sample weighing 0.81g was burned in the presence of oxygen to give 0.8578 g of water and 2.6208 g of CO 2 . g C = 0.2727 x 2.6208 = 0.7147g and g H = 0.1111 x 0.8578 = 0.0953 Assume a hydrocarbon, so %C = 0.7147/(0.7147 + 0.0953) = 0.7147/0.8100 x 100 = 88.23% %H = 0.0953/0.81 x 100 = 11.77% Assume 100 g, so 88.23g C and 11.77 g H. moles C = 88.23/12 = 7.35 and moles H = 11.77/1 = 11.77 so, 11.77/7.35= 1.6H / C = CH 1.6 x 2 = C 2 H 3.2 ; x3 = C 3 H 4.8 ; x4 = C 4 H 6.4 ; x5 = C 5 H 8 , The last one makes sense, and matches the C n H 2n-2 formula. So, C 5 H 8 . (c) A sample weighing 1.04 g was burned in the presence of oxygen to give 1.4779 g of water and 3.2116 g of CO 2 . g C = 0.2727 x 3.2116= 0.8758g and g H = 0.1111 x 1.4779 = 0.1642 Assume a hydrocarbon, so %C = 0.8758/(0.8758 + 0.1642) = 0..8758/1.04 x 100 = 84.21% %H = 0.1642/1.04 x 100 = 15.79% Assume 100 g, so 84.21 g C and 15.79 g H. moles C = 84.21/12 = 7.02 and moles H = 15.79/1 = 15.79 so, 15.79/7.02= 12.25H / C = CH 2.5 x 2 = C 2 H 5 , so C 4 H 10 , etc. This fits the genreal

{[ snackBarMessage ]}

###### "Before using Course Hero my grade was at 78%. By the end of the semester my grade was at 90%. I could not have done it without all the class material I found."
— Christopher R., University of Rhode Island '15, Course Hero Intern

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern