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Molecular formula

C21H30Cl3FN4

1 compound has this formula on CompoundList.

Molar mass
463.85g/mol
Monoisotopic mass
462.1520Da
Atoms
59
Elements
5
Rings plus double bonds
7

The formula C21H30Cl3FN4 describes a molecule made of 21 carbon atoms, 30 hydrogen atoms, 3 chlorine atoms, 1 fluorine atom and 4 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 × 2154.38%
    12.011 × 21 = 252.231 g/mol
  • H Hydrogen × 306.52%
    1.008 × 30 = 30.240 g/mol
  • Cl Chlorine × 322.93%
    35.45 × 3 = 106.350 g/mol
  • F Fluorine × 14.10%
    18.998 × 1 = 18.998 g/mol
  • N Nitrogen × 412.08%
    14.007 × 4 = 56.028 g/mol
Look up another formula

Compounds with the formula C21H30Cl3FN4

Isomers we list, best known first. Open one to see its structure, properties, safety data and suppliers.

Showing 1 to 1 of 1 compounds

N-[(2E)-3-(4-Fluorophenyl)prop-2-en-1-yl]-N-[(1-methyl-1H-imidazol-2-yl)methyl]-6-azaspiro[2.5]octan-1-amine trihydrochloride chemical structure, C21H30Cl3FN4, CAS 2624153-49-5
N-[(2E)-3-(4-Fluorophenyl)prop-2-en-1-yl]-N-[(1-methyl-1H-imidazol-2-yl)methyl]-6-azaspiro[2.5]octan-1-amine trihydrochloride
CAS 2624153-49-5C21H30Cl3FN4

Frequently asked questions

What is the molar mass of C21H30Cl3FN4?

The molar mass of C21H30Cl3FN4 is 463.85 g/mol, calculated from standard atomic weights for 21 carbon atoms, 30 hydrogen atoms, 3 chlorine atoms, 1 fluorine atom and 4 nitrogen atoms. The monoisotopic mass is 462.1520 Da.

How many compounds have the formula C21H30Cl3FN4?

CompoundList lists 1 compound with the molecular formula C21H30Cl3FN4, including N-[(2E)-3-(4-Fluorophenyl)prop-2-en-1-yl]-N-[(1-methyl-1H-imidazol-2-yl)methyl]-6-azaspiro[2.5]octan-1-amine trihydrochloride. 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 C21H30Cl3FN4?

C21H30Cl3FN4 has a degree of unsaturation of 7. 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.