BaNb4O12Sr
1 compound has this formula on CompoundList.
- Molar mass
- 788.56g/mol
- Monoisotopic mass
- 789.3753Da
- Atoms
- 18
- Elements
- 4
The formula BaNb4O12Sr describes a molecule made of 1 barium atom, 4 niobium atoms, 12 oxygen atoms and 1 strontium atom. It is written in Hill notation: with no carbon present, every element is listed 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
- Ba Barium × 117.42%137.33 × 1 = 137.330 g/mol
- Nb Niobium × 447.13%92.906 × 4 = 371.624 g/mol
- O Oxygen × 1224.35%15.999 × 12 = 191.988 g/mol
- Sr Strontium × 111.11%87.62 × 1 = 87.620 g/mol
Compounds with the formula BaNb4O12Sr
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 BaNb4O12Sr?
The molar mass of BaNb4O12Sr is 788.56 g/mol, calculated from standard atomic weights for 1 barium atom, 4 niobium atoms, 12 oxygen atoms and 1 strontium atom. The monoisotopic mass is 789.3753 Da.
How many compounds have the formula BaNb4O12Sr?
CompoundList lists 1 compound with the molecular formula BaNb4O12Sr, including Barium strontium niobate. Compounds that share a formula are isomers. They contain the same atoms arranged differently, so their properties, uses and prices can be very different.