Determine the structure from the specral data given C5H10O2: IR peak at 1740 cm-1;
NMR(ppm):
1.15 (triplet, 3 H) 2.30 (quartet, 2 H)
1.25 (triplet, 3 H) 4.72 (quartet, 2 H)

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