Question

complete the following table

Name molar mass (g/mol) 2-bromoheptane 2-bromohen cacluate from the formula / structure or look up in the table on page 77 density = ? g/ml from PC table 0.22 mL-?g Alkoxide base cacluate from the formula / structure or look up in the table on page 77 heptene isomers cacluate from the formula / structure or look up in the table on page 77 other you calculate you calculate 7.17 target amount target mmoles (1 mmole = 0.001 mole) maximum mmoles of product based on each reactant limiting reagent theoretical yield you calculate you calculate mark one as the limiting reactant you calculate actual amount used fill this in after measuring out chemicals in lab fill this in after measuring out chemicals in lab fill this in after measuring out chemicals in lab fill this in after measuring out chemicals in lab actual mmols updated theoretical yield (mass & mmols) fill this in after measuring out chemicals in lab

Answer 1

12-bramoheptane : •Molar mass (almal) = 179.1 glmal • Density = 1.142 g/ml • Mass & Target amount) = ? Volume = 0.22 ml Mass = density X Volume =10142 g/mL X 0.22 ml = 0.2519 Jarget immoles al mass x 103 Immal = 10 mal - Molar mass mmoles = 0.251g x 103 =1.40 mmal . 179.1 g/mol • Maximum mmoles of product based on 2-bromoheptane L = 1.40 mmal! Alkoxide base " (KOH) Molar mass = 5601 g/mol • Target mmoles = 7017 mmal - Target semount mass) = moles X Molar mass = 7.17 x 103 mol X 56.1 g/mol s i 0.402 G a maximum mmoles of product based on alkoxide. E 70 17 mmol 1 limiting reagent is 2-bromoheptane as it | kao 000005, Миште од моод . 1 mole of 2- bromoheptane will give Imole heptene. 1.40 mmal will give 1.40 mnal heptene. • Molar mass heptene = 98.18 g/mol o Mass heptene & Moles x molar mass = 1.40 HA X 10-3 mol X 98.189 mol Theoretical yield = 00137 g = 0.1379