Question

B. Procedure 1. Measure the [a]obs for both a 10.0 % D-glucose solution and a 10.0 % D-fructose solution. 2. Convert each of the measured [a]os to specific optical activity, [a] o 3. Compare your specific activity to published values and calculate your % error. D-glucose [a] 25 = 52.7° D-fructose[a] 2 = -92.4° Measured [a]obs Convert observed to specifc [a]obs → [a] o 25 % error x 100 experimental - theoretical theoretical C. Additional Questions 1. Calculate the [a]obs one would measure for a 20.0% D-glucose solution. 2. Calculate the [a]obs one would observe for a 20.0% L-glucose solution. 3. A 15.0 gram sample of a monosaccharide was dissolved in enough water to give a 100.0 mL solution. A portion of the solution was transferred to a polarimeter tube that measures 100.cm in length. The [a Jobs was measured to be + 83.1. Determine the identity of the monosaccharide.

Answer 1

25 can - as loo y 52.7 B) Q To answer this question, one has to get the value of [x Jobs by performing the experiment. specific optical activity [x] be calulated wing following formula c = g/mL [x] 25 [<Jobs l= in dm с tor 100% D-glucose 10 on glma Icom c for 10% D - fructose log = 0.1 glme 100mL path length ( specific to instrument) [<Jobs you get from experiment in part You can calculate the % error glucose - fructose /Cajo - (-92.4) % essor 52.7 of error 92.4 & tom [&os from ans [Jobs for 20% D-glucose [x] 25 - [&Jobs (1) - Calculated in cxe c = 20% soln=o2 glul (Cedobs [es cal e = from instrumat similang above of monosaccharide = 15g 0.15 g/m2 path leagth= [<Jobs = +83.1 («Jobs +83.1 = +55.4 cxe 0.15x10 to this specific activity corresponds Dtlactose ans 2 conc. loomL Too ma 10 dm