Lead(II) Iodide at a glance
| Property | Value |
|---|
| Molar mass | 460.8 g/mol |
|---|
| Molecular weight | 460.8 u |
|---|
| Chemical formula | PbI2 |
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| CAS number | 13779-98-1 |
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| PubChem CID | 24931 |
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How to calculate the molar mass of lead(II) iodide
1 × 207 + 2 × 126.9 = 460.8 g/mol
- Count the atoms in PbI2: Pb 1, I 2.
- Multiply each atom count by the atomic weight of the element.
- Add the results to get 460.8 g/mol.
Elemental composition of lead(II) iodide
| Element | Atoms | Atomic weight | Mass (g/mol) | Mass % |
|---|
| Pb Lead | 1 | 207 | 207 | 44.922% |
|---|
| I Iodine | 2 | 126.9 | 253.8 | 55.078% |
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Grams and moles of lead(II) iodide
| Amount | Equals |
|---|
| 1 g | 0.0021701 mol |
|---|
| 10 g | 0.021701 mol |
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| 100 g | 0.21701 mol |
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| 0.1 mol | 46.08 g |
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| 1 mol | 460.8 g |
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| 1 g, in particles | 1.3069 × 10^21 |
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