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)%

Answer 2

The percent of lead in lead(IV) oxide can be calculated by first determining the molar mass of lead(IV) oxide (PbO2) and then calculating the molar mass of lead in that compound. Finally, divide the molar mass of lead by the molar mass of PbO2 and multiply by 100 to get the percentage. Molar mass of lead = 207.2 g/mol and molar mass of PbO2 = 239.2 g/mol, so the percentage of lead in lead(IV) oxide is (207.2/239.2) x 100 ≈ 86.5%.