C27H33F3N4O3
1 compound has this formula on CompoundList.
- Molar mass
- 518.58g/mol
- Monoisotopic mass
- 518.2505Da
- Atoms
- 70
- Elements
- 5
- Rings plus double bonds
- 12
The formula C27H33F3N4O3 describes a molecule made of 27 carbon atoms, 33 hydrogen atoms, 3 fluorine atoms, 4 nitrogen atoms and 3 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 × 2762.54%12.011 × 27 = 324.297 g/mol
- H Hydrogen × 336.41%1.008 × 33 = 33.264 g/mol
- F Fluorine × 310.99%18.998 × 3 = 56.994 g/mol
- N Nitrogen × 410.80%14.007 × 4 = 56.028 g/mol
- O Oxygen × 39.26%15.999 × 3 = 47.997 g/mol
Compounds with the formula C27H33F3N4O3
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 C27H33F3N4O3?
The molar mass of C27H33F3N4O3 is 518.58 g/mol, calculated from standard atomic weights for 27 carbon atoms, 33 hydrogen atoms, 3 fluorine atoms, 4 nitrogen atoms and 3 oxygen atoms. The monoisotopic mass is 518.2505 Da.
How many compounds have the formula C27H33F3N4O3?
CompoundList lists 1 compound with the molecular formula C27H33F3N4O3, including Lirametostat. 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 C27H33F3N4O3?
C27H33F3N4O3 has a degree of unsaturation of 12. 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.