Molarity (M) and moles ↔ grams for lab solutions
This calculator helps you move between mass, moles, volume, and molarity using clean unit normalization (mg↔g, mL↔L).
It is designed for typical bio/chem lab prep where you measure solute mass and prepare a final solution volume.
Core definitions
- Molarity (M) means “moles of solute per liter of solution.”
- Moles (n) are the amount of substance.
- Molar mass (MM) converts grams ↔ moles (units: g/mol).
- Volume (V) is the final solution volume (units: L or mL).
Most common lab workflow
Decide target molarity and final volume → compute moles → compute grams to weigh.
Core relations (MathJax)
\[
\begin{aligned}
M &= \frac{n}{V} \\
n &= \frac{m}{MM}
\end{aligned}
\]
In the calculator, the “standard” internal units are:
m in grams (g) and V in liters (L).
Unit normalization used by the calculator
You can type values in common lab units. The calculator converts them to standard units before applying formulas.
Mass
mg
→
g
multiply by
10^{-3}
\[
m_{\mathrm{g}} = m_{\mathrm{mg}}\cdot 10^{-3}
\]
Volume
mL
→
L
multiply by
10^{-3}
\[
V_{\mathrm{L}} = V_{\mathrm{mL}}\cdot 10^{-3}
\]
Important: Use the final solution volume (after dissolving and bringing to volume), not just the solvent volume you started with.
Common tasks (what you usually want to compute)
The calculator’s “Formula map” visualization highlights exactly which path is being used (mass → moles → molarity, or molarity + volume → moles → mass, etc.).
Worked example (solution preparation)
Prepare 250 mL of 0.20 M NaCl. Molar mass \(MM = 58.44\ \mathrm{g\cdot mol^{-1}}\).
Find grams needed.
\[
\begin{aligned}
V_{\mathrm{L}} &= 250\ \mathrm{mL}\cdot 10^{-3} = 0.250\ \mathrm{L} \\
n &= M\cdot V_{\mathrm{L}}
= 0.20\ \mathrm{mol\cdot L^{-1}}\cdot 0.250\ \mathrm{L}
= 0.0500\ \mathrm{mol} \\
m &= n\cdot MM
= 0.0500\ \mathrm{mol}\cdot 58.44\ \mathrm{g\cdot mol^{-1}}
= 2.922\ \mathrm{g}
\end{aligned}
\]
Copyable summary style
“To make 250 mL of 0.20 M NaCl, need 2.92 g.”
Worked example (molarity from grams + volume)
You dissolve 0.500 g of glucose (\(MM = 180.16\ \mathrm{g\cdot mol^{-1}}\)) and bring the final volume to 50.0 mL.
Find the molarity.
\[
\begin{aligned}
V_{\mathrm{L}} &= 50.0\ \mathrm{mL}\cdot 10^{-3} = 0.0500\ \mathrm{L}\\
n &= \frac{m}{MM} = \frac{0.500\ \mathrm{g}}{180.16\ \mathrm{g\cdot mol^{-1}}}
= 2.77\times 10^{-3}\ \mathrm{mol}\\
M &= \frac{n}{V_{\mathrm{L}}} =
\frac{2.77\times 10^{-3}\ \mathrm{mol}}{0.0500\ \mathrm{L}}
= 5.54\times 10^{-2}\ \mathrm{mol\cdot L^{-1}}
\end{aligned}
\]
In practice, you would report this as \(0.0554\ \mathrm{M}\) (or \(5.54\times 10^{-2}\ \mathrm{M}\)), depending on your significant-figure policy.
Practical lab tips
- Volume matters: if you add solute to a flask and then fill to the mark, the final volume is the marked volume.
- Hydrates: if your compound is hydrated (example: CuSO4·5H2O), use the molar mass of the hydrate you actually weighed.
- Purity: if purity is not 100%, the effective moles are lower. (The calculator assumes pure solute.)
- Very concentrated solutions: volume additivity can fail; for routine lab prep this approach is usually acceptable.
Common mistakes
- Using mL directly in the molarity formula without converting to L.
- Mixing up \(MM\) units (it must be in \(\mathrm{g\cdot mol^{-1}}\) if mass is in g).
- Confusing “mass/volume concentration” (like mg/mL) with molarity (mol/L).
- Using solvent volume instead of final solution volume.
Rule of thumb: if you used mL, convert to L before plugging into \(M = \dfrac{n}{V}\).
FAQ
Does this include density?
No. This calculator uses the standard molarity definition based on final solution volume. Density is only needed when converting between mass and volume of the solution itself.
What if my data are in mg/mL?
Convert mass and volume separately to g and L, then compute moles and molarity:
\(m_{\mathrm{g}} = m_{\mathrm{mg}}\cdot 10^{-3}\) and \(V_{\mathrm{L}} = V_{\mathrm{mL}}\cdot 10^{-3}\).
How does the “solve missing variable” mode work?
You provide all but one of \(\{M, n, m, V, MM\}\). The calculator computes the single missing variable using the shortest valid path (for example, grams → moles → molarity).
