Molar Concentration (Molarity): Theory & Guide
Molar concentration (molarity) measures how many moles of solute are present per liter of solution:
\[
M \;=\; \frac{n_s}{V}
\quad\text{with}\quad
M\;[\mathrm{mol\,L^{-1}}],\; n_s\;[\mathrm{mol}],\; V\;[\mathrm{L}]
\]
Units & Conversions
Useful Rearrangements
Depending on the target, rearrange the core relation \( M=\dfrac{n_s}{V} \):
- Required solute (moles): \( n_{\text{req}} = M^*\,V \)
- Required mass (given formula): \( m_{\text{req}} = (M^*\,V)\,M_r \)
- Required volume: \( V_{\text{req}} = \dfrac{n_s}{M^*} \)
Dilution (Stock → Target)
For simple dilutions of the same solute where solute amount is conserved:
\[
M_1 V_1 \;=\; M_2 V_2
\]
Caveat: Volumes are treated as additive and temperature/volume changes are neglected. For concentrated/strongly
interacting mixtures, density changes can matter.
- Supported syntax: element symbols (e.g., Na, Cl, Ca), parentheses for groups (e.g., Ca(OH)2), and numeric subscripts.
- Optional state symbols are ignored: (s), (l), (g), (aq).
- Charges/electrons are not supported.
- Examples accepted: H2O, C2H6O, Ca(OH)2, Fe2(SO4)3.
Typical Workflow
- Convert any given mass to moles using \( n_s = \dfrac{m_s}{M_r} \) (if needed).
- Convert volume to liters.
- Apply the appropriate formula:
- \( M = \dfrac{n_s}{V} \)
- \( n_{\text{req}} = M^*V \)
- \( V_{\text{req}} = \dfrac{n_s}{M^*} \)
- Dilution: \( M_1 V_1 = M_2 V_2 \)
Worked Examples
Example 1 — Compute \( M \) from mass & volume.
Given \( m_s = 5.85~\mathrm{g} \) NaCl, \( V = 0.100~\mathrm{L} \). NaCl \( M_r \approx 58.44~\mathrm{g\,mol^{-1}} \).
\[
n_s = \frac{5.85}{58.44} \approx 0.100~\mathrm{mol}, \quad
M = \frac{0.100}{0.100} \approx 1.00~\mathrm{mol\,L^{-1}}.
\]
Example 2 — Required mass for target molarity.
Prepare \( 0.250~\mathrm{M} \) NaCl solution, volume \( 0.500~\mathrm{L} \).
\[
n_{\text{req}} = M^* V = 0.250 \times 0.500 = 0.125~\mathrm{mol},
\quad
m_{\text{req}} = n_{\text{req}} M_r = 0.125 \times 58.44 \approx 7.30~\mathrm{g}.
\]
Example 3 — Required volume for target molarity.
You have \( n_s = 0.125~\mathrm{mol} \) NaCl and want \( 0.500~\mathrm{M} \).
\[
V_{\text{req}} = \frac{n_s}{M^*} = \frac{0.125}{0.500} = 0.250~\mathrm{L} = 250~\mathrm{mL}.
\]
Common Pitfalls & Tips
- Forgetting to convert mL → L: Always use liters in \( M=\dfrac{n}{V} \).
- Using solute volume instead of solution volume: \( V \) is the final solution volume, not solvent-only.
- Mass ↔ moles: If a mass is given, you must use the molar mass \( M_r \) derived from the chemical formula.
- Dilution assumption: \( M_1V_1=M_2V_2 \) neglects non-ideal volume changes; acceptable for many labs but not all systems.
