C10H14FN3
101 compounds have this formula on CompoundList.
- Molar mass
- 195.24g/mol
- Monoisotopic mass
- 195.1172Da
- Atoms
- 28
- Elements
- 4
- Rings plus double bonds
- 5
The formula C10H14FN3 describes a molecule made of 10 carbon atoms, 14 hydrogen atoms, 1 fluorine atom and 3 nitrogen 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 × 1061.52%12.011 × 10 = 120.110 g/mol
- H Hydrogen × 147.23%1.008 × 14 = 14.112 g/mol
- F Fluorine × 19.73%18.998 × 1 = 18.998 g/mol
- N Nitrogen × 321.52%14.007 × 3 = 42.021 g/mol
Compounds with the formula C10H14FN3
Isomers we list, best known first. Open one to see its structure, properties, safety data and suppliers.
Showing 101 to 101 of 101 compounds
![cis-6-Fluoro-3,5-dimethyl-2,3,4,5-tetrahydro-1H-pyrido[3,4-E][1,4]diazepine chemical structure, C10H14FN3, CAS 2089651-54-5](/s/MSHZCMXCIIKFJM-RQJHMYQMSA-N.svg)
Frequently asked questions
What is the molar mass of C10H14FN3?
The molar mass of C10H14FN3 is 195.24 g/mol, calculated from standard atomic weights for 10 carbon atoms, 14 hydrogen atoms, 1 fluorine atom and 3 nitrogen atoms. The monoisotopic mass is 195.1172 Da.
How many compounds have the formula C10H14FN3?
CompoundList lists 101 compounds with the molecular formula C10H14FN3, including cis-6-Fluoro-3,5-dimethyl-2,3,4,5-tetrahydro-1H-pyrido[3,4-E][1,4]diazepine. 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 C10H14FN3?
C10H14FN3 has a degree of unsaturation of 5. 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.