Calculate the empirical and molecular formula of unknown compound A which gave an m/z of 181 in the mass spectrum for the molecular ion peak. Elemental analysis by combustion shows the compound is C 53%, O 35.4%, H 3.9% and N 7.7%.

C atomic mass-12 C percentage by composition: 53% Atomic ratio of C: 53/12= 4.417 H atomic mass-1 H percentage by composition: 3.9% Atomic ratio of H: 3.9/1=3.9 N atomic mass- 14 N percentage by composition:7.7% Atomic ratio of N: 7.7/14 =0.550 O atomic mass-16 O percentage by composition: 35.4% Atomic ratio of O: 35.4/16=2.213 Divide all ratios by the smallest ratio (0.550 in this example)C-4.417/0.550=8.030 N-0.550/0.550=1 H-3.9/0.550=7.091 O-2.213/0.550=4.023Empirical formula: C8H7NO4 molecular weight=181Molecular formula: C8H7NO4 molecular weight=181Molecular formula will be a multiple of the empirical formula molecular weight. Molecular weight=181n. n=1 in this case as the m/z is also 181 therefore the molecular formula is the same as the empirical formula