By knowing the natural abundance of minor isotopes, it's possible to calculate the relative heights of M+ and M+1 peaks. If natural abundances are 12C - 98.9% and 13C - 1.10%, what are the relative heights, to the nearest 0.1%, of the M+ and M+1 peaks in the mass spectrum of ribose, C5H10O5?
Ignore the contributions of isotopes like 2H (deuterium; 0.015% natural abundance) and 17O (0.04% natural abundance) that are small.
I initially just substituted 13 for 12 when calculating the molar mass but that didn't work. How else should I go about this problem.
5.5 is the answer
Hey Saj how did you get 5.5 as the answer?
To calculate the relative heights of the M+ and M+1 peaks in the mass spectrum of ribose, you need to consider the natural abundance of the isotopes involved.
First, let's calculate the molar mass of ribose (C5H10O5). The atomic mass of carbon-12 (12C) is about 12 amu, and the atomic mass of carbon-13 (13C) is about 13 amu. Similarly, the atomic mass of hydrogen-1 (1H) is about 1 amu, and the atomic mass of oxygen-16 (16O) is about 16 amu.
Molar mass of ribose:
(5 * atomic mass of carbon) + (10 * atomic mass of hydrogen) + (5 * atomic mass of oxygen)
= (5 * 12 amu) + (10 * 1 amu) + (5 * 16 amu)
= 60 amu + 10 amu + 80 amu
= 150 amu
Next, let's calculate the contributions of each isotope to the M+ and M+1 peaks.
For the M+ peak, we consider the most abundant isotope, which is carbon-12 (98.9% natural abundance). The contributions from each atom are as follows:
- Carbon (C): 5 atoms * (98.9% * 100%) = 494.5% contribution
- Hydrogen (H): 10 atoms * (100% * 100%) = 100% contribution
- Oxygen (O): 5 atoms * (100% * 100%) = 100% contribution
Therefore, the relative height of the M+ peak would be 494.5% + 100% + 100% = 694.5% (to the nearest 0.1%).
For the M+1 peak, we consider the less abundant isotope, which is carbon-13 (1.10% natural abundance). The contributions from each atom are as follows:
- Carbon (C): 5 atoms * (1.10% * 100%) = 5.5% contribution
- Hydrogen (H): 10 atoms * (100% * 100%) = 100% contribution
- Oxygen (O): 5 atoms * (100% * 100%) = 100% contribution
Therefore, the relative height of the M+1 peak would be 5.5% + 100% + 100% = 205.5% (to the nearest 0.1%).
To confirm your calculations, the ratio of the heights of the M+1 to M+ peaks should be equal to the ratio of their contributions. In this case, the ratio of contributions is 5.5% / 494.5%, which is approximately 0.0111. Similarly, the ratio of the heights of M+1 to M+ peaks is 205.5% / 694.5%, which is also approximately 0.0111.