Final answer:
B) C₂H₅. The hydrocarbon fragment that corresponds to m/z 43 is likely to be C2H5 (ethyl), with a mass of 29 amu, as this is the closest match to the given m/z value among the provided options.
Step-by-step explanation:
The student's question is asking for the hydrocarbon fragment corresponding to a peak with a mass-to-charge ratio (m/z) of 43, based on the atomic masses rounded to the nearest whole number. The hydrocarbon fragment options are CH3, C2H5, CH4, and C2H4.
To answer this, we need to calculate the mass of each hydrocarbon fragment. The atomic masses of carbon (C) and hydrogen (H) are approximately 12 amu and 1 amu respectively when rounded to the nearest whole number.
- CH3: 12 (for C) + 3 (for H) = 15 amu
- C2H5: (2 × 12) (for C) + (5 × 1) (for H) = 24 + 5 = 29 amu
- CH4: 12 (for C) + 4 (for H) = 16 amu
- C2H4: (2 × 12) (for C) + (4 × 1) (for H) = 24 + 4 = 28 amu
Since none of the calculated masses match the given m/z value of 43, we have to consider that the fragments might have an additional mass due to an extra mass of 14 amu from a nitrogen atom (common in mass spectrometry to indicate a charge) or from branching. Thus, C2H7 (which is not one of the options) would have a mass of 29 (for C2H5) + 14 (for 2 additional H atoms) = 43 amu, matching the m/z value. However, since this specific fragment is not an option, and the question might contain a typo, we likely need to identify the closest matching option.
The closest matching option would be C2H5 since 29 amu is the mass value nearest to 43 amu among the listed choices. Therefore, based on the information provided, the best representation of the hydrocarbon fragment corresponding to a peak labeled m/z 43 is C2H5.