What is the empirical formula of a compound whose percentage composition is found to be 62.60% lead, 8.40% nitrogen, and 29.0% oxygen?

Question

What is the empirical formula of a compound whose percentage composition is found to be 62.60% lead, 8.40% nitrogen, and 29.0% oxygen?

Answer 1

Find moles of each element and get whole number in order to find the empirical formula.

When given % composition, simply assume you have 100 g of unknown and the % = grams.

Pb: 62 g Pb x 1 mol Pb / 207 g = 0.2995 mols Pb

N: 8.40 g N x 1 mol N / 14 g = 0.600 mols N

29.0 g O x 1 mol O / 16 g = 1.813 mols O

To get whole number, try dividing all by the lowest number of moles (0.2995)

Pb = 0.2995 / 0.2995 = 1 mol Pb

N = 0.600 / 0.2995 = 2 mol N

O = 1.813 / 0.2995 = 6 mol O

Empirical formula = PbN₂O₆ = lead(II) nitrate or Pb(NO₃)₂