PbO2
Depending on the formula ! Two atoms in PbO2, three atoms in Al2O3, etc.
Lead oxides are: PbO, PbO2, Pb3O4, Pb2O3 and the possible Pb12O19.
86.62(8)%The molar masses of the constituent elements must be known.M(Pb) = 2.072(1) x 102 gmol-1M(O) = 1.59994(3) x 10 gmol-1From these and their respective molar ratios in the compound, the total molar mass of the compound must be calculated.M(PbO2) = M(Pb) + 2M(O)M(PbO2) = (2.072 x 10 gmol-1 + 2(1.59994 gmol-1)) x 10M(PbO2) = 2.392(1) x 102 gmol-1From the molar mass of the element in question and its ratio, and the compound, their total masses present in one mole of compound must be calculated.m = nMm(Pb) = 1 mol x 2.072 x 102 gmol-1m(Pb) = 2.072(1) x 102 gm(PbO2) = 1 mol x 2.392(1) x 102 g (always 1 mole)m(PbO2) = 2.392(1) x 102 gFrom these, the percentage of the element in question present in the compound can be calculated.m(Pb) x 100%/m(PbO2) = 2.072 g x 100%/2.392 gm(Pb) x 100%/m(PbO2) = 86.62(8)%
2Pb + O2 ----->2PbO then 2PbO + O2 ------> 2PbO2 some mixed oxide will be formed also 2PbO + PbO2 ----->Pb3O4
Formula: PbO2
PbO2
The ionic chemical formula of lead(IV) oxide is: (Pb)4+ + 2 O-.
PbO2
Lead(IV) oxide would have the formula PbO2
Formula unit is (molar mass) * (6.0221421*10^23)Molar mass of PbO2 is239.2089 g/molMolar mass of 50.0g of PbO2 is 50/239.2089 g/molFormula unit of PbO2 is ( 50g/239.2089 g/mol) *(6.0221421*10^23)=(0.20902232316606948988938120613405mol) *(6.0221421*10^23units/mol)=1.2587621321781923665883669044087*10^23unitsor approx. 1,26.1023 formula units.
The formula for plumbous oxide is PbO.
Depending on the formula ! Two atoms in PbO2, three atoms in Al2O3, etc.
PbO2 is lead(IV) oxide.
Lead oxides are: PbO, PbO2, Pb3O4, Pb2O3 and the possible Pb12O19.
Lead + Oxygen -> Lead (IV) Oxide Formula: Pb (s) + O2 (g) -> PbO2 (s)
‚Black!