What is the percent of lead in lead IV oxide Or how do i calculate it?

Answer 1

86.62(8)%

• The molar masses of the constituent elements must be known.
  

M(Pb) = 2.072(1) x 102 gmol-1

M(O) = 1.59994(3) x 10 gmol-1

• From 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-1

• From 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 = nM

m(Pb) = 1 mol x 2.072 x 102 gmol-1

m(Pb) = 2.072(1) x 102 g

m(PbO2) = 1 mol x 2.392(1) x 102 g (always 1 mole)

m(PbO2) = 2.392(1) x 102 g

• From 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 g

m(Pb) x 100%/m(PbO2) = 86.62(8)%