C14H17N3O4
249 compounds have this formula on CompoundList.
- Molar mass
- 291.31g/mol
- Monoisotopic mass
- 295.1470Da
- Atoms
- 38
- Elements
- 4
- Rings plus double bonds
- 8
The formula C14H17N3O4 describes a molecule made of 14 carbon atoms, 17 hydrogen atoms, 3 nitrogen atoms and 4 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 × 1457.72%12.011 × 14 = 168.154 g/mol
- H Hydrogen × 175.88%1.008 × 17 = 17.136 g/mol
- N Nitrogen × 314.42%14.007 × 3 = 42.021 g/mol
- O Oxygen × 421.97%15.999 × 4 = 63.996 g/mol
Compounds with the formula C14H17N3O4
Isomers we list, best known first. Open one to see its structure, properties, safety data and suppliers.
Showing 201 to 249 of 249 compounds
- 2-[4-(3,4-Dimethoxyphenyl)-1H-1,2,3-triazol-1-yl]-2-methylpropanoic acid2138090-18-1C14H17N3O4
- 2-({[(tert-Butoxy)carbonyl]amino}methyl)-1H-pyrrolo[2,3-b]pyridine-5-carboxylic acid2384450-80-8C14H17N3O4
- 2-{[(tert-Butoxy)carbonyl]amino}-1-methyl-1H-1,3-benzodiazole-6-carboxylic acid2580211-77-2C14H17N3O4
- 3-[3-(2-{[(Benzyloxy)carbonyl]amino}ethyl)-3H-diazirin-3-yl]propanoic acid2580220-06-8C14H17N3O4
- 2-{[(tert-Butoxy)carbonyl]amino}-1-methyl-1H-1,3-benzodiazole-5-carboxylic acid2580201-40-5C14H17N3O4
- 2-(6-{[(tert-Butoxy)carbonyl]amino}imidazo[1,5-a]pyridin-3-yl)acetic acid2580199-80-8C14H17N3O4
- tert-Butyl 2-amino-6-(2,5-dihydroxy-1H-pyrrol-1-yl)pyridine-3-carboxylate2639417-03-9C14H17N3O4
- 2-{[(tert-Butoxy)carbonyl]amino}-1-methyl-1H-1,3-benzodiazole-7-carboxylic acid2639439-91-9C14H17N3O4
- 2-(2-{[(tert-Butoxy)carbonyl]amino}-1H-1,3-benzodiazol-1-yl)acetic acid2680889-01-2C14H17N3O4
- tert-Butyl N-(2-cyanoethyl)-N-(2-nitrophenyl)carbamate2680720-98-1C14H17N3O4
- tert-Butyl N-[2-(4-amino-1,2-oxazol-5-yl)-4-hydroxyphenyl]carbamate2229096-48-2C14H17N3O4
- 2-{1-[(Prop-2-en-1-yloxy)carbonyl]piperidin-4-yl}pyrimidine-5-carboxylic acid2763914-80-1C14H17N3O4
- 6-{4-[(Prop-2-en-1-yloxy)carbonyl]piperazin-1-yl}pyridine-3-carboxylic acid2763912-51-0C14H17N3O4
- (2S)-2-{[(Benzyloxy)carbonyl]amino}-4-(3-methyl-3H-diazirin-3-yl)butanoic acid2763891-45-6C14H17N3O4
- 3-{[(tert-Butoxy)carbonyl]amino}-8-methylimidazo[1,2-a]pyridine-2-carboxylic acid2763925-68-2C14H17N3O4
- 2-{[(Benzyloxy)carbonyl]amino}-4-(3-methyl-3H-diazirin-3-yl)butanoic acid2870645-23-9C14H17N3O4
- 2-{4-[(Prop-2-en-1-yloxy)carbonyl]piperazin-1-yl}pyridine-4-carboxylic acid2913423-04-6C14H17N3O4
- 6-{[(tert-Butoxy)carbonyl]amino}-7-cyano-2,3-dihydro-1H-pyrrolizine-5-carboxylic acid2920134-33-2C14H17N3O4
- (S)-2-((tert-Butoxycarbonyl)amino)-3-(5-cyanopyridin-3-yl)propanoic acid2781825-34-9C14H17N3O4
- Boc[3-(4-methoxyphenyl)-1,2,3-oxadiazol-3-ium-5-yl]amide2173583-97-4C14H17N3O4
- 5-Pyrimidinecarboxylic acid,1,2,3,4-tetrahydro-6-methyl-2-oxo-4-(4-pyridinyl)-,2-methoxyethyl ester512840-69-6C14H17N3O4
- 2-(5,8-Dimethoxy-2-methyl-4-oxoquinolin-1(4H)-yl)acetohydrazide1215674-14-8C14H17N3O4
- tert-Butyl 3-cyano-4-hydroxy-2-oxo-1,5,6,8-tetrahydro-1,7-naphthyridine-7-carboxylate2736598-36-8C14H17N3O4
- rac-(1R,6S)-6-[(4-Methoxy-6-methylpyrimidin-2-yl)carbamoyl]cyclohex-3-ene-1-carboxylic acid1931548-81-0C14H17N3O4
- 2-({[(tert-Butoxy)carbonyl]amino}methyl)-1H-1,3-benzodiazole-7-carboxylic acid2524265-70-9C14H17N3O4
Frequently asked questions
What is the molar mass of C14H17N3O4?
The molar mass of C14H17N3O4 is 291.31 g/mol, calculated from standard atomic weights for 14 carbon atoms, 17 hydrogen atoms, 3 nitrogen atoms and 4 oxygen atoms. The monoisotopic mass is 295.1470 Da.
How many compounds have the formula C14H17N3O4?
CompoundList lists 249 compounds with the molecular formula C14H17N3O4, including 5-(4-Butoxy-3-methoxyphenyl)-4H-1,2,4-triazole-3-carboxylic acid, Ethyl 2-[4-(3,4-dimethoxyphenyl)-1H-1,2,3-triazol-1-yl]acetate, 2-[Benzyl(2-carbamoylazetidine-1-carbonyl)amino]acetic acid. 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 C14H17N3O4?
C14H17N3O4 has a degree of unsaturation of 8. 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.