Molar MassMolar mass of lead(II) nitrate (Pb(NO3)2)
The molar mass of lead(II) nitrate, Pb(NO3)2, is 331.21 g/mol. One mole of it, which is 6.022 × 1023 formula units, has a mass of 331.21 grams.
331.21 g/mol
207.2 + 2 × 14.007 + 6 × 15.999 = 331.208 · 9 atoms in the formula · also called lead nitrate
About lead(II) nitrate
Lead(II) nitrate is one of the few lead salts that dissolve well in water. Heated in a test tube it crackles and gives brown fumes of nitrogen dioxide, a standard school demonstration. The bracket means two nitrate groups: two nitrogen and six oxygen atoms.
Other names: lead nitrate.
The working
The formula Pb(NO3)2 has 1 lead atom, 2 nitrogen atoms and 6 oxygen atoms, 9 atoms in all. Multiply each count by the atomic weight of the element and add:
| Element | Atoms | Atomic weight | Total | Share by mass |
|---|---|---|---|---|
| Lead (Pb) | 1 | 207.2 | 207.200 | 62.56% |
| Nitrogen (N) | 2 | 14.007 | 28.014 | 8.46% |
| Oxygen (O) | 6 | 15.999 | 95.994 | 28.98% |
207.2 + 2 × 14.007 + 6 × 15.999 = 331.208 g/mol
With the rounded atomic masses used in many school books (Pb = 207, N = 14, O = 16) the answer is 331 g/mol. The difference from 331.21 is too small to matter in most school sums.
Percentage composition
By mass, Pb(NO3)2 is 62.56% lead, 8.46% nitrogen and 28.98% oxygen. The largest share comes from lead. In a 100 g sample there are 62.6 g of lead, 8.5 g of nitrogen and 29.0 g of oxygen.
Notice that nitrogen has more atoms in the formula (2) than lead (1), yet gives a smaller share of the mass. Percentage by mass depends on how heavy each atom is, not only on how many there are.
Worked examples
Grams to moles. How many moles are in 10 g of Pb(NO3)2? Divide the mass by the molar mass: 10 ÷ 331.21 = 0.03019 mol.
Moles to grams. What is the mass of 1.5 mol of Pb(NO3)2? Multiply: 1.5 × 331.21 = 496.8 g.
Counting particles. 10 g contains 0.03019 mol × 6.022 × 1023 = 1.818 × 1022 formula units.
Making a solution. A 0.1 mol/L solution contains 0.1 × 331.21 = 33.12 g of Pb(NO3)2 in each litre. For 250 mL, weigh 8.28 g and make it up to the mark with water.
Quick conversion tables
| Amount | Mass |
|---|---|
| 0.1 mol | 33.12 g |
| 0.5 mol | 165.6 g |
| 1 mol | 331.2 g |
| 2 mol | 662.4 g |
| 5 mol | 1656 g |
| Mass | Moles |
|---|---|
| 1 g | 0.003019 mol |
| 10 g | 0.03019 mol |
| 100 g | 0.3019 mol |
| 1 kg | 3.019 mol |
For other amounts use the grams and moles converter.
Questions people ask
What is the molar mass of Pb(NO3)2?
The molar mass of lead(II) nitrate (Pb(NO3)2) is 331.21 g/mol, or 331.208 g/mol before rounding, using IUPAC abridged atomic weights.
How many moles are in 100 g of Pb(NO3)2?
100 ÷ 331.21 = 0.3019 mol.
What is the mass of one formula unit of Pb(NO3)2?
Divide the molar mass by the Avogadro constant: 331.21 ÷ 6.022 × 1023 = 5.500 × 10-22 g.
What is the percentage of lead in Pb(NO3)2?
Lead makes up 62.56% of the mass of Pb(NO3)2: 1 × 207.2 ÷ 331.208 × 100.
Which values? This site uses the abridged standard atomic weights published by IUPAC (CIAAW), read on 12 November 2026. Your textbook or question paper may round them differently, for example H = 1 and O = 16, so its answer can differ slightly. In an exam, use the values the paper gives.