Molar mass of AgNO3 (silver nitrate)
The molar mass of a compound is the mass of 1 mole of that substance. For an ionic compound like AgNO3, the molar mass is numerically equal to its formula mass expressed in \( \mathrm{g \cdot mol^{-1}} \).
Step 1: Count atoms in the formula
In AgNO3, there are: 1 Ag atom, 1 N atom, and 3 O atoms.
Step 2: Use atomic masses and compute contributions
The general rule is: \[ M(\text{compound})=\sum_i n_i \cdot A_i \] where \(n_i\) is the number of atoms of element \(i\) in the formula and \(A_i\) is its atomic mass.
| Element | Count in AgNO3 | Atomic mass \(A_i\) (\(\mathrm{g \cdot mol^{-1}}\)) | Contribution \(n_i \cdot A_i\) (\(\mathrm{g \cdot mol^{-1}}\)) |
|---|---|---|---|
| Ag | 1 | 107.8682 | \(1 \cdot 107.8682 = 107.8682\) |
| N | 1 | 14.0067 | \(1 \cdot 14.0067 = 14.0067\) |
| O | 3 | 15.999 | \(3 \cdot 15.999 = 47.997\) |
Step 3: Add the contributions
\[ M(\mathrm{AgNO_3}) = 107.8682 + 14.0067 + 47.997 = 169.8719\ \mathrm{g \cdot mol^{-1}} \] Rounded appropriately: \[ M(\mathrm{AgNO_3}) \approx 169.87\ \mathrm{g \cdot mol^{-1}} \]
Atomic masses depend slightly on isotopic composition and reference tables, so small differences in the last digits are normal. Using consistent atomic-mass values and reporting a sensible number of significant digits is the key.
Quick application (mass ↔ moles)
Once the molar mass is known, conversion between mass and amount of substance uses: \[ n=\frac{m}{M} \] For example, if \(m=5.00\ \mathrm{g}\) of \(\mathrm{AgNO_3}\), \[ n=\frac{5.00}{169.87}=0.0294\ \mathrm{mol}\ \ (\text{to 3 s.f.}) \]