C14H11F5N2O
3 compounds have this formula on CompoundList.
- Molar mass
- 318.25g/mol
- Monoisotopic mass
- 318.0792Da
- Atoms
- 33
- Elements
- 5
- Rings plus double bonds
- 8
The formula C14H11F5N2O describes a molecule made of 14 carbon atoms, 11 hydrogen atoms, 5 fluorine atoms, 2 nitrogen atoms and 1 oxygen atom. 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 × 1452.84%12.011 × 14 = 168.154 g/mol
- H Hydrogen × 113.48%1.008 × 11 = 11.088 g/mol
- F Fluorine × 529.85%18.998 × 5 = 94.990 g/mol
- N Nitrogen × 28.80%14.007 × 2 = 28.014 g/mol
- O Oxygen × 15.03%15.999 × 1 = 15.999 g/mol
Compounds with the formula C14H11F5N2O
Isomers we list, best known first. Open one to see its structure, properties, safety data and suppliers.
Showing 1 to 3 of 3 compounds
Frequently asked questions
What is the molar mass of C14H11F5N2O?
The molar mass of C14H11F5N2O is 318.25 g/mol, calculated from standard atomic weights for 14 carbon atoms, 11 hydrogen atoms, 5 fluorine atoms, 2 nitrogen atoms and 1 oxygen atom. The monoisotopic mass is 318.0792 Da.
How many compounds have the formula C14H11F5N2O?
CompoundList lists 3 compounds with the molecular formula C14H11F5N2O, including Dpc-082, 2(1H)-Quinazolinone, 5,6-difluoro-3,4-dihydro-4-(3-methyl-1-butynyl)-4-(trifluoromethyl)-, 2(1H)-Quinazolinone, 5,6-difluoro-3,4-dihydro-4-(3-methyl-1-butynyl)-4-(trifluoromethyl)-, (4S)-. 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 C14H11F5N2O?
C14H11F5N2O has a degree of unsaturation of 8. 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.