C24H15Br3
3 compounds have this formula on CompoundList.
- Molar mass
- 543.10g/mol
- Monoisotopic mass
- 539.8724Da
- Atoms
- 42
- Elements
- 3
- Rings plus double bonds
- 16
The formula C24H15Br3 describes a molecule made of 24 carbon atoms, 15 hydrogen atoms and 3 bromine 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 × 2453.08%12.011 × 24 = 288.264 g/mol
- H Hydrogen × 152.78%1.008 × 15 = 15.120 g/mol
- Br Bromine × 344.14%79.904 × 3 = 239.712 g/mol
Compounds with the formula C24H15Br3
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 C24H15Br3?
The molar mass of C24H15Br3 is 543.10 g/mol, calculated from standard atomic weights for 24 carbon atoms, 15 hydrogen atoms and 3 bromine atoms. The monoisotopic mass is 539.8724 Da.
How many compounds have the formula C24H15Br3?
CompoundList lists 3 compounds with the molecular formula C24H15Br3, including 1,3,5-Tri(4-bromophenyl)benzene, 1,3,5-Tris(2-bromophenyl)benzene, 1,3,5-Tris(3-bromophenyl)benzene. 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 C24H15Br3?
C24H15Br3 has a degree of unsaturation of 16. 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.