C22H13FN2O5
2 compounds have this formula on CompoundList.
- Molar mass
- 404.35g/mol
- Monoisotopic mass
- 404.0808Da
- Atoms
- 43
- Elements
- 5
- Rings plus double bonds
- 17
The formula C22H13FN2O5 describes a molecule made of 22 carbon atoms, 13 hydrogen atoms, 1 fluorine atom, 2 nitrogen atoms and 5 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 × 2265.35%12.011 × 22 = 264.242 g/mol
- H Hydrogen × 133.24%1.008 × 13 = 13.104 g/mol
- F Fluorine × 14.70%18.998 × 1 = 18.998 g/mol
- N Nitrogen × 26.93%14.007 × 2 = 28.014 g/mol
- O Oxygen × 519.78%15.999 × 5 = 79.995 g/mol
Compounds with the formula C22H13FN2O5
Isomers we list, best known first. Open one to see its structure, properties, safety data and suppliers.
Showing 1 to 2 of 2 compounds
Frequently asked questions
What is the molar mass of C22H13FN2O5?
The molar mass of C22H13FN2O5 is 404.35 g/mol, calculated from standard atomic weights for 22 carbon atoms, 13 hydrogen atoms, 1 fluorine atom, 2 nitrogen atoms and 5 oxygen atoms. The monoisotopic mass is 404.0808 Da.
How many compounds have the formula C22H13FN2O5?
CompoundList lists 2 compounds with the molecular formula C22H13FN2O5, including Elraglusib, 1,3-Dioxo-2,3-dihydro-1H-isoindol-2-yl 3-(4-fluorobenzamido)benzoate. 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 C22H13FN2O5?
C22H13FN2O5 has a degree of unsaturation of 17. 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.