C56H94O35
1 compound has this formula on CompoundList.
- Molar mass
- 1327.33g/mol
- Monoisotopic mass
- 1326.5576Da
- Atoms
- 185
- Elements
- 3
- Rings plus double bonds
- 10
The formula C56H94O35 describes a molecule made of 56 carbon atoms, 94 hydrogen atoms and 35 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 × 5650.67%12.011 × 56 = 672.616 g/mol
- H Hydrogen × 947.14%1.008 × 94 = 94.752 g/mol
- O Oxygen × 3542.19%15.999 × 35 = 559.965 g/mol
Compounds with the formula C56H94O35
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 C56H94O35?
The molar mass of C56H94O35 is 1327.33 g/mol, calculated from standard atomic weights for 56 carbon atoms, 94 hydrogen atoms and 35 oxygen atoms. The monoisotopic mass is 1326.5576 Da.
How many compounds have the formula C56H94O35?
CompoundList lists 1 compound with the molecular formula C56H94O35, including Prost-13-en-1-oic acid, 11,15-dihydroxy-9-oxo-, (11alpha,13E,15S)-, compd. with alpha-cyclodextrin (1:1). 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 C56H94O35?
C56H94O35 has a degree of unsaturation of 10. 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.