I don't have a clue where to start with this problem: Carbon has two stable isotopes, 12/6C and 13/6C, and fluorine has only one stable isotope, 19/9F. How many peaks would you observe in the mass spectrum of the positive ion of CF4+? Assume that the ion does not break up into smaller fragments.
All 4 of the F's in the molecule would have a mass of 19, and the total for 4 of them would be 19x4 = 76. The mass of the C could be either 12 or 13. Therefore CF4+ could have a mass of either 88 or 89 amu. The charge of CF4+ would be +e. There are only two possible value of the charge/mass ratio, and two mass spectrometer peaks, assuming you have only singly ionized CF4+ and there is no fragmentiation.
prolly 2 idk man
Well, with all those numbers and possibilities, it sounds like the mass spectrum of CF4+ is playing a game of hide-and-seek! But fear not, I'll unravel this puzzle for you.
Since there are two stable isotopes of carbon, and each can have a different mass, you would expect two peaks in the mass spectrum based on the different combinations of isotopes. So that's two peaks for the carbon.
As for fluorine, it only has one stable isotope, so there's only one possibility there. That's one peak for fluorine.
Now, when you add them all together in CF4+, you'll have to consider the different combinations. If carbon has a mass of 12, then the mass of CF4+ would be 12 + (4 x 19) = 88. That's one peak at 88 amu.
But, if carbon has a mass of 13, then the mass of CF4+ would be 13 + (4 x 19) = 89. That's another peak at 89 amu.
So, to sum it up, you'll observe a total of three peaks in the mass spectrum of CF4+. Two peaks for carbon isotopes (at 88 and 89 amu) and one peak for fluorine (at 76 amu). Happy spectrometry-ing!
To determine the number of peaks in the mass spectrum of the positive ion of CF4+, we need to consider the possible values of the mass-to-charge ratio (m/z) for CF4+.
1. The mass of CF4+ can be either 88 amu (if the carbon isotope 12/6C is present) or 89 amu (if the carbon isotope 13/6C is present).
2. The charge of CF4+ is +e, where e is the elementary charge (+1.602 × 10^-19 C).
3. To calculate the m/z ratio, we divide the mass (in amu) by the charge (in C).
- If CF4+ has a mass of 88 amu, the m/z ratio is 88 amu / +e.
- If CF4+ has a mass of 89 amu, the m/z ratio is 89 amu / +e.
4. Since the charge e is a constant, we have two possible values for the m/z ratio, resulting in two distinct peaks in the mass spectrum.
In summary, the mass spectrum of CF4+ would show two peaks corresponding to the two possible values of the m/z ratio, assuming no fragmentation or ionization beyond singly ionized CF4+.
To determine the number of peaks in the mass spectrum of CF4+, we need to consider the possible combinations of masses for the carbon isotope. Carbon has two stable isotopes: 12/6C and 13/6C.
First, let's calculate the mass of one CF4+ ion:
- The mass of each fluorine atom is 19 atomic mass units (amu).
- CF4+ has 4 fluorine atoms, so the total mass of fluorine in CF4+ is 19 amu × 4 = 76 amu.
- The mass of the carbon atom could be either 12 amu (for 12/6C) or 13 amu (for 13/6C).
- Adding the mass of carbon (12 or 13 amu) to the total fluorine mass (76 amu), we get two possibilities for the mass of CF4+: 76 amu + 12 amu = 88 amu or 76 amu + 13 amu = 89 amu.
Now, let's consider the charge of the CF4+ ion. The problem states that the ion has a single positive charge (+e).
In a mass spectrometer, the ratio of charge (in this case, +e) to mass (in this case, amu) is measured. Since there are only two possible values for the charge/mass ratio (1+e/88 amu or 1+e/89 amu), we can conclude that there will be two distinct peaks in the mass spectrum of CF4+. Each peak represents the different charge-to-mass ratio for the two possible masses (88 amu and 89 amu).
Note that we assumed no fragmentation of the CF4+ ion in the mass spectrometer. If the ion were to break up into smaller fragments, there could be additional peaks for those fragments. However, the problem explicitly states that CF4+ does not break up into smaller fragments.
In summary, when analyzing the mass spectrum of CF4+ (assuming no fragmentation), we would observe two peaks corresponding to the two possible masses for CF4+.