Question

Fill out and attach this sheet to your lab report. Table 1: Data for the Equimolar Mixtures. GC Data for the Equimolar Mixtures Mol% Retention Molar Adjusted Mass % Response Area Compound Of time (RT) in mixture in mixture Area of Peak Factor mass Peak 142,5귀 14.72 34,02 123.54 17 1.00 14.72 50.00% n-butyl chloride 4.39 n-butyl bromide 2.55 14.72 50.00% 92-97 30.46 197.02 74.5 2.19 sec-butyl chloride 24.48 2.15 50.000 36. 48 2.89 sec-butyl bromide 92-5448 75 131.02 79-83lo4 50.00 2.30 1. 00 48. 75 t-butyl chloride 50, 0010 48.75 t-butyl bromide

Answer 1

Mass = moles × molar mass

Given mol % ratio in the mixture = 50:50

Hence , moles of the substance are equal .

Now,

Let moles of each substance in the mixture = 1 .

Then, mass of each sample = molar mass

Then,

Compound	mass	total mass (g)	mass percentage (mass of each×100/total mass)
n-butyl chloride	(1×92.57) = 92.57	92.57+137.02 = 229.59	(92.57×100)/229.59 = 40.32 %
n-butyl bromide	(1×137.02) = 137.02	229.59	(137.92×100/229.59) = 59.68 %
Sec butyl chloride	92.57	(92.57+137.02) = 229.59	(92.57× 100/229.59) = 40.32 %
Sec butyl bromide	137.02	229.59	(197.02×100/229.59) = 59.68 %
t-butyl chloride	92.57	229.59	40.32 %
t-butyl bromide	137.02	229.59	59.68 %