Concentration Unit Converter — Theory & Relationships
Many laboratory and industrial problems use different, but related, measures of concentration.
This page summarizes the definitions, the links among them, and when additional data
(solute molar mass \(M_r\) and solution density \(\rho\)) are needed to convert between units.
The calculator implements the same relationships and shows each step it uses.
Core definitions
- Molarity \(M\) (temperature-dependent): \( M=\dfrac{n_s}{V} \) (mol of solute per liter of solution).
- Molality \(m\) (temperature-independent): \( m=\dfrac{n_s}{m_\text{solv}/1000} \) (mol per kg of solvent).
- Mass fraction / mass percent:
\( w=\dfrac{m_s}{m_\text{tot}} \), so \( \%\,(w/w)=100\,w \).
- Mass concentration: \( \dfrac{\mathrm{g}}{\mathrm{L}}=\dfrac{m_s}{V} \).
- \% (w/v) (“g per 100 mL”): \( \%\,(w/v)=100\,\dfrac{m_s[\mathrm{g}]}{V[\mathrm{mL}]}\).
Equivalently, \(1\ \%\,(w/v)=10\ \mathrm{g\,L^{-1}}\).
- ppm (mass) (mg per kg of solution): \( \mathrm{ppm}=10^6 \dfrac{m_s}{m_\text{tot}} \).
Thus, \( \%\,(w/w) = \dfrac{\mathrm{ppm}}{10^4} \) and \( \mathrm{ppm}=10^4\times \%\,(w/w)\).
Links that require extra data
- Molar mass \(M_r\) (g·mol⁻¹) converts moles \(\leftrightarrow\) mass:
\( n_s=\dfrac{m_s}{M_r}\), \( m_s = n_s M_r\). The converter extracts \(M_r\) automatically from the entered formula.
- Density \(\rho\) (g·mL⁻¹) converts volume \(\leftrightarrow\) mass of the solution:
\( m_\text{tot}=\rho\,(1000\,V)\) (for \(V\) in L), and \( V=\dfrac{m_\text{tot}}{1000\,\rho}\).
When do I need extra inputs?
- Any conversion between a volume-based unit (M, g·L⁻¹, % w/v) and a mass-based unit (% w/w, ppm, molality) needs \(\rho\).
- Any conversion that crosses mole \(\leftrightarrow\) mass needs \(M_r\).
- Molality always needs the mass of solvent (hence \(\rho\) if a volume is given).
Convenient “basis” choices used by the converter
To keep arithmetic transparent, the calculator normalizes each input to a simple notional sample,
then derives all other units from that basis:
- Given \(M\): take \(V=1.000\ \mathrm{L}\), so \(n_s=M\cdot V\).
- Given g·L⁻¹: take \(V=1.000\ \mathrm{L}\), so \(m_s=\text{g·L⁻¹}\).
- Given % (w/v): take \(V=100\ \mathrm{mL}=0.100\ \mathrm{L}\), so \(m_s=\%\,(w/v)\) in grams.
- Given % (w/w): take \(m_\text{tot}=100\ \mathrm{g}\), so \(m_s=\%\,(w/w)\) in grams and \(m_\text{solv}=100-m_s\).
- Given ppm: take \(m_\text{tot}=1000\ \mathrm{g}\) (1 kg), so \(m_s=\mathrm{ppm}/1000\ \mathrm{g}\).
- Given molality \(m\): take \(m_\text{solv}=1000\ \mathrm{g}\) (1 kg), so \(n_s=m\cdot 1\ \mathrm{kg}\).
Frequently used conversions (closed forms)
- M ↔ g·L⁻¹:
\( \displaystyle \mathrm{g\,L^{-1}} = M\,M_r \), and \( \displaystyle M = \dfrac{\mathrm{g\,L^{-1}}}{M_r} \).
- % (w/v) ↔ g·L⁻¹:
\( \displaystyle \mathrm{g\,L^{-1}} = 10\times \%\,(w/v) \), and \( \%\,(w/v)=\dfrac{\mathrm{g\,L^{-1}}}{10} \).
- % (w/w) ↔ ppm:
\( \%\,(w/w)=\dfrac{\mathrm{ppm}}{10^4} \), \( \mathrm{ppm}=10^4\times \%\,(w/w) \).
Conversions that need density
To go between mass-based and volume-based quantities, you must relate \(m_\text{tot}\) and \(V\) via \(\rho\).
Examples (per liter of solution unless noted):
- Given \(M\) and \(\rho\) → % (w/w):
\[
m_s = M\,M_r,\quad m_\text{tot}=\rho\times 1000,\quad
\%\,(w/w)=100\,\frac{m_s}{m_\text{tot}}.
\]
- Given % (w/w) and \(\rho\) → \(M\):
\[
m_\text{tot}=\rho\times 1000,\quad m_s=\frac{\%\,(w/w)}{100}\,m_\text{tot},\quad
n_s=\frac{m_s}{M_r},\quad M=\frac{n_s}{1\ \mathrm{L}}.
\]
- Given % (w/v) and \(\rho\) → % (w/w):
\[
\text{For }V=100\ \mathrm{mL}:\ m_s=\%\,(w/v)\ \mathrm{g},\
m_\text{tot}=\rho\,V=\rho\times 100,\
\%\,(w/w)=100\,\frac{m_s}{m_\text{tot}}.
\]
Molality links
Molality involves the mass of solvent, not solution. Starting from any basis that gives \(m_s\) and \(m_\text{tot}\),
compute \(m_\text{solv}=m_\text{tot}-m_s\), then
\[
m=\frac{n_s}{m_\text{solv}/1000}=\frac{(m_s/M_r)}{(m_\text{solv}/1000)}.
\]
Assumptions & cautions
- Temperature: Molarity and density depend on temperature. Make sure \(\rho\) corresponds to your concentration and temperature.
- \% (w/v): Despite the “%” symbol, it is not a fraction—it means g per 100 mL by definition.
- Ideal mixing: The converter uses the algebraic definitions (no volume contraction/expansion corrections beyond \(\rho\)). For high concentrations, consult experimental data.
- Significant figures: Report results consistent with the precision of your input data for \(M_r\), \(\rho\), and the given concentration.
Worked miniature example
Convert \(M=0.500\) M NaCl (\(M_r=58.44\ \mathrm{g\,mol^{-1}}\)) with \(\rho=1.00\ \mathrm{g\,mL^{-1}}\) to g·L⁻¹ and % (w/w):
\[
\begin{aligned}
\text{Basis: }&V=1.000\ \mathrm{L},\ n_s=0.500\ \mathrm{mol},\ m_s=n_sM_r=29.22\ \mathrm{g}.\\
&m_\text{tot}=\rho\,(1000V)=1000\ \mathrm{g}.\\[4pt]
&\mathrm{g\,L^{-1}}=\frac{m_s}{V}=29.22, \quad
\%\,(w/w)=100\,\frac{29.22}{1000}=2.922\ \%.
\end{aligned}
\]
Tip: If \(\rho \approx 1.000\ \mathrm{g\,mL^{-1}}\), then for dilute aqueous solutions
ppm \(\approx\) mg·L⁻¹ and % (w/w) \(\times 10^4 \approx\) ppm.
