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

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-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
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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
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