Lead(II) Nitrate at a glance
| Property | Value |
|---|
| Molar mass | 331.008 g/mol |
|---|
| Molecular weight | 331.008 u |
|---|
| Chemical formula | Pb(NO3)2 |
|---|
| CAS number | 10099-74-8 |
|---|
| PubChem CID | 24924 |
|---|
How to calculate the molar mass of lead(II) nitrate
1 × 207 + 2 × 14.007 + 6 × 15.999 = 331.008 g/mol
- Count the atoms in Pb(NO3)2: Pb 1, N 2, O 6.
- Multiply each atom count by the atomic weight of the element.
- Add the results to get 331.008 g/mol.
Elemental composition of lead(II) nitrate
| Element | Atoms | Atomic weight | Mass (g/mol) | Mass % |
|---|
| Pb Lead | 1 | 207 | 207 | 62.536% |
|---|
| N Nitrogen | 2 | 14.007 | 28.014 | 8.4632% |
|---|
| O Oxygen | 6 | 15.999 | 95.994 | 29.001% |
|---|
Grams and moles of lead(II) nitrate
| Amount | Equals |
|---|
| 1 g | 0.0030211 mol |
|---|
| 10 g | 0.030211 mol |
|---|
| 100 g | 0.30211 mol |
|---|
| 0.1 mol | 33.101 g |
|---|
| 1 mol | 331.01 g |
|---|
| 1 g, in particles | 1.8193 × 10^21 |
|---|
Calculate other compounds