CompoundList
Molecular formula

C22H17Cl3N2O5S

1 compound has this formula on CompoundList.

Molar mass
527.80g/mol
Monoisotopic mass
525.9924Da
Atoms
50
Elements
6
Rings plus double bonds
14

The formula C22H17Cl3N2O5S describes a molecule made of 22 carbon atoms, 17 hydrogen atoms, 3 chlorine atoms, 2 nitrogen atoms, 5 oxygen atoms and 1 sulfur 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 × 2250.07%
    12.011 × 22 = 264.242 g/mol
  • H Hydrogen × 173.25%
    1.008 × 17 = 17.136 g/mol
  • Cl Chlorine × 320.15%
    35.45 × 3 = 106.350 g/mol
  • N Nitrogen × 25.31%
    14.007 × 2 = 28.014 g/mol
  • O Oxygen × 515.16%
    15.999 × 5 = 79.995 g/mol
  • S Sulfur × 16.07%
    32.06 × 1 = 32.060 g/mol
Look up another formula

Compounds with the formula C22H17Cl3N2O5S

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 C22H17Cl3N2O5S?

The molar mass of C22H17Cl3N2O5S is 527.80 g/mol, calculated from standard atomic weights for 22 carbon atoms, 17 hydrogen atoms, 3 chlorine atoms, 2 nitrogen atoms, 5 oxygen atoms and 1 sulfur atom. The monoisotopic mass is 525.9924 Da.

How many compounds have the formula C22H17Cl3N2O5S?

CompoundList lists 1 compound with the molecular formula C22H17Cl3N2O5S, including 4-(2-((4-CL-PH)Sulfonyl)carbohydrazonoyl)-2-ethoxyphenyl 2,4-dichlorobenzoate. 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 C22H17Cl3N2O5S?

C22H17Cl3N2O5S has a degree of unsaturation of 14. 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.