C37H38BrF3N4O
1 compound has this formula on CompoundList.
- Molar mass
- 691.64g/mol
- Monoisotopic mass
- 690.2181Da
- Atoms
- 84
- Elements
- 6
- Rings plus double bonds
- 19
The formula C37H38BrF3N4O describes a molecule made of 37 carbon atoms, 38 hydrogen atoms, 1 bromine atom, 3 fluorine atoms, 4 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 × 3764.25%12.011 × 37 = 444.407 g/mol
- H Hydrogen × 385.54%1.008 × 38 = 38.304 g/mol
- Br Bromine × 111.55%79.904 × 1 = 79.904 g/mol
- F Fluorine × 38.24%18.998 × 3 = 56.994 g/mol
- N Nitrogen × 48.10%14.007 × 4 = 56.028 g/mol
- O Oxygen × 12.31%15.999 × 1 = 15.999 g/mol
Compounds with the formula C37H38BrF3N4O
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 C37H38BrF3N4O?
The molar mass of C37H38BrF3N4O is 691.64 g/mol, calculated from standard atomic weights for 37 carbon atoms, 38 hydrogen atoms, 1 bromine atom, 3 fluorine atoms, 4 nitrogen atoms and 1 oxygen atom. The monoisotopic mass is 690.2181 Da.
How many compounds have the formula C37H38BrF3N4O?
CompoundList lists 1 compound with the molecular formula C37H38BrF3N4O, including Gsk2193874. 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 C37H38BrF3N4O?
C37H38BrF3N4O has a degree of unsaturation of 19. 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.