Library / Solution chemistry, equilibria and kinetics / SCH-02
Solution chemistry, equilibria and kinetics
Stock molarity from mass fraction and density
Molar concentration of a commercial stock solution, from its density ρ in g/mL, the solute mass fraction w (0 to 1) and the molar mass M in g/mol.
- Ref
- SCH-02
- Rev
- v1
- Field
- Solution chemistry, equilibria and kinetics
- Result
- c [mol/L]
A stock solution’s molarity follows from its density and composition. One litre of solution has a mass of 1000ρ grams (ρ in g/mL), of which the fraction w is solute, so it holds 1000ρw/M moles. Enter w as a fraction, so a 36 % acid is 0.36. The density must be for the stated temperature and concentration.
Molar concentration of a commercial stock solution, from its density ρ in g/mL, the solute mass fraction w (0 to 1) and the molar mass M in g/mol. For a mass percentage, divide by 100 before entering w.
Calculator
- Solute mass per litre, 1000 ρ w240 g/L
- Stock volume for 1 L of 0.1 mol/L solution16.67 mL
- Mass percentage, 100 w20 %
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A solution of density 1.2 g/mL with 20 % by mass solute is 6 mol/L.
- To make 1 L of 0.1 mol/L from it, take 16.67 mL of stock and dilute to the mark.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (mol/L) | Local step | Change in result |
|---|---|---|---|---|---|---|
| M Molar mass of the solute | 1 to 1,000 g/mol | 0.24 | 240 | 239.8 | +10% | −9.09 % |
| w Solute mass fraction (0 to 1) | 0.001 to 1 | 0.03 | 30 | 29.97 | +10% | +10 % |
| ρ Solution density | 0.8 to 2 g/mL | 4 | 10 | 6 | +10% | +10 % |
Sweep of Solute mass fraction (0 to 1)
| Solute mass fraction (0 to 1) | Stock molarity (mol/L) |
|---|---|
| 0.001 | 0.03 |
| 0.002371 | 0.07114 |
| 0.005623 | 0.1687 |
| 0.01334 | 0.4001 |
| 0.03162 | 0.9487 |
| 0.07499 | 2.25 |
| 0.1778 | 5.335 |
| 0.4217 | 12.65 |
| 1 | 30 |
Sweep one variable
Variables and how each one contributes
Stock molarity
What the equation gives you. Everything else feeds it.
Solution density
The density of the solution, in g/mL. Raises the molarity in direct proportion, since more mass fits in a litre. Raising it by 10 % (to 1.32) moves the result by +10 %, from 6 to 6.6 mol/L.
Solute mass fraction (0 to 1)
The solute mass fraction, between 0 and 1. Raises the molarity in direct proportion. Raising it by 10 % (to 0.22) moves the result by +10 %, from 6 to 6.6 mol/L.
Molar mass of the solute
The molar mass of the solute, in g/mol. A heavier molar mass gives fewer moles per gram, so the molarity falls in inverse proportion. Raising it by 10 % (to 44) moves the result by −9.09 %, from 6 to 5.455 mol/L.
Worked example with the values above
- Mass of solution per litre: 1000 × 1.2 = 1,200 g.
- Solute mass: 1,200 × 0.2 = 240 g.
- c = 240 / 40 = 6 mol/L.
Assumptions and limits
- Ideal dilute solutions, with concentrations in mol/L and masses in grams unless stated.
- Complete mixing and no evaporation or contraction on mixing unless the inputs say otherwise.
Common mistakes
- Mixing molality (per kg of solvent) with molarity (per litre of solution).
- Entering a mass percentage where a fraction between 0 and 1 is needed.
- Leaving out counterions when adding up charges.
Statistics and uncertainty
Masses and volumes are usually known to 0.1 % to 1 %, and concentrations from them carry the same relative error. Volumetric glassware tolerances add up through serial dilutions, so each extra stage widens the uncertainty.
Where it is used
- Preparing standards, stocks and dilution series in the laboratory.
- Checking charge balance and ionic strength in water analysis.
Maths you need
- Amount of substance: n = m/M, in mol
- Concentration: c = n/V, in mol/L
- Electroneutrality: total positive charge equals total negative charge