The detector used in this experiment is called a Flame Ionization Detector (FID). As the different sample components come out of the end of the column and enter the detector, they pass into a flame and generate free electrons. The electrons are generated as a result of carbon atoms being converted to positively charged carbon ions during the combustion process. The response measured (i. e.# of electrons generated per unit time; the electrical current) is directly proportional to the carbon content of the component entering the detector. Predict which of the following compounds will generate the largest detector response per unit concentration. Assume the concentration of each compound is the same in the sample mixture analyzed by GC-FID. a. 2-methyl-1-propanol.
b. 1-butanol.
c. 2-methyl-1-butanol.
d. ethyl acetate.

c. 2-methyl-1-butanol.

Explanation:

As depicted in the question; the number of electrons produced is straightforwardly relative and directly proportional to the content and substance of carbon in the atom. So 2-methyl-1-butanol has the most highest and noteworthy carbon content (5carbons) among the accompanying species. So it will generate and create the biggest detector reaction response.

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally