C102H140F6N24O37
1 compound has this formula on CompoundList.
- Molar mass
- 2408.36g/mol
- Monoisotopic mass
- 2406.9715Da
- Atoms
- 309
- Elements
- 5
- Rings plus double bonds
- 42
The formula C102H140F6N24O37 describes a molecule made of 102 carbon atoms, 140 hydrogen atoms, 6 fluorine atoms, 24 nitrogen atoms and 37 oxygen atoms. It is written in Hill notation: carbon first, then hydrogen, then the other elements in alphabetical order. The formula tells you which atoms are present, not how they are connected, which is why several different chemicals can share it.
Elemental composition
- C Carbon × 10250.87%12.011 × 102 = 1225.122 g/mol
- H Hydrogen × 1405.86%1.008 × 140 = 141.120 g/mol
- F Fluorine × 64.73%18.998 × 6 = 113.988 g/mol
- N Nitrogen × 2413.96%14.007 × 24 = 336.168 g/mol
- O Oxygen × 3724.58%15.999 × 37 = 591.963 g/mol
Compounds with the formula C102H140F6N24O37
Isomers we list, best known first. Open one to see its structure, properties, safety data and suppliers.
Showing 1 to 1 of 1 compounds
Frequently asked questions
What is the molar mass of C102H140F6N24O37?
The molar mass of C102H140F6N24O37 is 2408.36 g/mol, calculated from standard atomic weights for 102 carbon atoms, 140 hydrogen atoms, 6 fluorine atoms, 24 nitrogen atoms and 37 oxygen atoms. The monoisotopic mass is 2406.9715 Da.
How many compounds have the formula C102H140F6N24O37?
CompoundList lists 1 compound with the molecular formula C102H140F6N24O37, including Bivalirudin trifluoroacetate. Compounds that share a formula are isomers. They contain the same atoms arranged differently, so their properties, uses and prices can be very different.
What is the degree of unsaturation of C102H140F6N24O37?
C102H140F6N24O37 has a degree of unsaturation of 42. That is the total number of rings and pi bonds (a double bond counts as one, a triple bond as two) in any structure with this formula.