Chemistry · Analytical Chemistry
The molecular ion peak [M]•+ of an analyte as measured by Electron Ionization Mass Spectrometry has an m/z of 149 and a relative abundance of 100%. The [M + 1]•+ peak has a relative abundance of 6.7% and the [M + 2]•+ peak has a relative abundance of 5%. Given the isotopic abundances (1H-100%, 12C-98.9%, 13C-1.1%, 14N-99.6%, 15N-0.4%, 16O-99.8%, 18O-0.2%, 32S-95.0%, 33S-0.75%, and 34S-4.2%), what is the most probable molecular formula of the compound?
Verified Solution4 OptionsUpdated: October 2026👁️1 View
Share:
- 1C6H15NOS
- 2C6H13O2S
- 3C5H11NO2S✓
- 4C7H21N2O
Correct answer
C5H11NO2S
Download Single Question PDF
Save this question and verified answer as a clean study sheet with live clickable link.
👥
Share with Friends & Study Groups
Challenge your peer aspirants on WhatsApp or Telegram to solve this question!
Online Test Series
Start Free Mock Test →Preparing for Chemistry?
Practice full-length mock tests with real exam timers, instant score analysis, and all-India ranking.
Chapter Practice
Practice Analytical Chemistry MCQs →Want more practice questions on Analytical Chemistry?
Solve complete topic-wise tests with detailed solutions on GkGet.
Related Practice Questions
Recommended practice questions in Chemistry · Analytical Chemistry