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Library / Chemical / CHE-01

Chemical

Amount and concentration

Moles and molar concentration from mass, molar mass and volume.

Ref
CHE-01
Rev
v1
Field
Chemical
Result
C [mol/L]

nA amount in mol, mA mass, MA molar mass, CA concentration and V the solution volume. Mass in grams and molar mass in g/mol give moles; volume in litres gives mol/L.

Chemists count molecules in moles. Dividing a mass by the molar mass gives the number of moles, and dividing that by the volume gives the concentration.

Calculator

Solve for
mol/L
g
0.11,000
g/mol
1500
L
0.01100
  • Amount n0.2002 mol
  • Concentration in mol/m³400.4 mol/m³

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.

Concentration (mol/L) current value 0.4004 mol/L
mMass of soluteKey driver
0.0034234.2
MMolar mass
0.046823.4
VVolume
0.00220
  • 1 mol/L equals 1000 mol/m³, a conversion that matters in SI reactor calculations.
  • Mixing mass and molar units (kg with g/mol) is the commonest error.
  • Volume changes on dissolving; for precise work use the final solution volume.
Chart data as tables

Ranked variables

VariableTypical rangeLowest resultHighest resultSwing (mol/L)Local stepChange in result
m Mass of solute0.1 to 1,000 g0.00342234.2234.22+10%+10 %
M Molar mass1 to 500 g/mol0.046823.423.35+10%−9.09 %
V Volume0.01 to 100 L0.00200220.0220.02+10%−9.09 %

Sweep of Volume

Volume (L)Concentration (mol/L)
0.0120.02
0.031626.331
0.12.002
0.31620.6331
10.2002
3.1620.06331
100.02002
31.620.006331
1000.002002

Sweep one variable

Concentration (mol/L)Hover, touch, or use the arrow keys on the chart
05101520 0.010.1110100 Volume (L), log scale now 0.4004

Variables and how each one contributes

C

Concentration

Result · mol/L

What the equation gives you. Everything else feeds it.

m

Mass of solute

g · typical 0.1 to 1,000

Mass dissolved. Directly proportional.

M

Molar mass

g/mol · typical 1 to 500

Mass of one mole (NaCl 58.44). Inversely proportional.

V

Volume

L · typical 0.01 to 100

Volume of the solution. Inversely proportional.

Worked example with the values above

  1. Amount: n = m/M = 11.7 g / 58.44 g/mol = 0.2002 mol.
  2. Divide by volume: 0.2002 mol / 0.5 L.
  3. C = 0.4004 mol/L.

Assumptions and limits

  • Pure solute and a known molar mass.
  • Volume is the final solution volume.
  • Consistent units throughout.

Common mistakes

  • Using the volume of the solvent instead of the solution.
  • Mixing kg and g.
  • Using the wrong molar mass (hydrates).

Statistics and uncertainty

A balance accurate to ±0.01 g on 11.7 g gives 0.1% error; the volumetric flask is usually the larger error, around ±0.2%. They combine as δC/C ≈ √[(δm/m)² + (δV/V)²].

Where it is used

  • Preparing electrolytes and reagents.
  • Reactor feed concentrations.
  • Corrosion and decontamination solutions.

Maths you need

  • Moles: a counting unit for molecules
  • Units: g, L and mol

Related formulas

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