1) The fragment responsible for the peak at m/z 43 is attributed to M(+ losing a neutral fragment of MW = a, 57; b, 43

-which has: a, 57/12 = 4 C's and 57 -4 x 12 = 9 H's; b, 57/12 = 5 C's

-and 57/1.01 = 57 H's and corresponds to the: a, Propyl; b, butyl or isomeric; c, butyl only group.

2) The composition of the fragment at m/z 43 is: a, 43/12 = 4 C's and 43/1.01 = 43 H's; b, 7-4 = 3 C's and 16 - 9 = 7 H's; c, 57 - 43 = 14 C's and 57-14 = 43 H's.

3)The formula of the fragment at m/z 43 is: a, 37 (+); b, C3H7 (.); c, C4H9(+)