Input interpretation
![bis(trifluoromethyl)peroxide](../image_source/70a6fa2ec60f3aa0464051c1b62648ea.png)
bis(trifluoromethyl)peroxide
Chemical names and formulas
![formula | C_2F_6O_2 name | bis(trifluoromethyl)peroxide IUPAC name | trifluoro-(trifluoromethylperoxy)methane alternate names | di(trifluoromethyl)peroxide | hexafluorodimethyl peroxide | trifluoromethyl peroxide mass fractions | C (carbon) 14.1% | F (fluorine) 67% | O (oxygen) 18.8%](../image_source/8a7243601f4c14f4e54af9e75997fd9c.png)
formula | C_2F_6O_2 name | bis(trifluoromethyl)peroxide IUPAC name | trifluoro-(trifluoromethylperoxy)methane alternate names | di(trifluoromethyl)peroxide | hexafluorodimethyl peroxide | trifluoromethyl peroxide mass fractions | C (carbon) 14.1% | F (fluorine) 67% | O (oxygen) 18.8%
Lewis structure
![Draw the Lewis structure of bis(trifluoromethyl)peroxide. Start by drawing the overall structure of the molecule: Count the total valence electrons of the carbon (n_C, val = 4), fluorine (n_F, val = 7), and oxygen (n_O, val = 6) atoms: 2 n_C, val + 6 n_F, val + 2 n_O, val = 62 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), fluorine (n_F, full = 8), and oxygen (n_O, full = 8): 2 n_C, full + 6 n_F, full + 2 n_O, full = 80 Subtracting these two numbers shows that 80 - 62 = 18 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 9 bonds and hence 18 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 62 - 18 = 44 electrons left to draw: Answer: | |](../image_source/f987b67e8bd7d9606938b15b045fb5db.png)
Draw the Lewis structure of bis(trifluoromethyl)peroxide. Start by drawing the overall structure of the molecule: Count the total valence electrons of the carbon (n_C, val = 4), fluorine (n_F, val = 7), and oxygen (n_O, val = 6) atoms: 2 n_C, val + 6 n_F, val + 2 n_O, val = 62 Calculate the number of electrons needed to completely fill the valence shells for carbon (n_C, full = 8), fluorine (n_F, full = 8), and oxygen (n_O, full = 8): 2 n_C, full + 6 n_F, full + 2 n_O, full = 80 Subtracting these two numbers shows that 80 - 62 = 18 bonding electrons are needed. Each bond has two electrons, so the above diagram has all the necessary bonds. There are 9 bonds and hence 18 bonding electrons in the diagram. Lastly, fill in the remaining unbonded electrons on each atom. In total, there remain 62 - 18 = 44 electrons left to draw: Answer: | |
3D structure
![3D structure](../image_source/1525db35c5af8a36d14c2126ab6559d6.png)
3D structure
Basic properties
![molar mass | 170.01 g/mol phase | gas (at STP) boiling point | -2 °C solubility in water | insoluble](../image_source/35688b6da35d5be0b431142f9df85a74.png)
molar mass | 170.01 g/mol phase | gas (at STP) boiling point | -2 °C solubility in water | insoluble
Units
Chemical identifiers
![CAS number | 927-84-4 Beilstein number | 1753686 PubChem CID number | 70228 SMILES identifier | C(OOC(F)(F)F)(F)(F)F InChI identifier | InChI=1/C2F6O2/c3-1(4, 5)9-10-2(6, 7)8 EU number | 213-165-0](../image_source/e2db43d4b66026b97bd5a2f162870f74.png)
CAS number | 927-84-4 Beilstein number | 1753686 PubChem CID number | 70228 SMILES identifier | C(OOC(F)(F)F)(F)(F)F InChI identifier | InChI=1/C2F6O2/c3-1(4, 5)9-10-2(6, 7)8 EU number | 213-165-0
NFPA label
![NFPA label](../image_source/1cdae2dba89d1e2a5f9a47ac2b4d4fa8.png)
NFPA label
![NFPA hazards | oxidizing agent](../image_source/0be67f424561020c0e2989fe0816ace2.png)
NFPA hazards | oxidizing agent