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Mass–Molar Flow Converter

Convert mass flow to molar flow, or molar flow to mass flow, using a supplied molar mass. Keep the substance or mixture basis explicit, distinguish mol from kmol and lbmol, and follow the normalized calculation steps.

MATERIAL BALANCESExplicit molar mass · mass flow ↔ molar flow

01 Define the stream basis

Changing direction carries the last valid result into the new input. Otherwise, enter a new flow.

Flow magnitude for the component or total mixture described by your molar mass.

Use a positive dimensional molar mass. For total mixture flow, supply that mixture's mean molar mass.

1 kmol = 1000 mol · 1 lbmol = 453.59237 mol.
g/mol, kg/kmol, and lbm/lbmol share the same numerical molar-mass value.

Changing units converts entered values. No substance or mixture properties are assumed.

Calculations run in your browser. Input values are not sent to the server.

02 Read the converted flow

Molar flow rate · ṅ

—kmol/h

Enter flow and molar mass to begin.

Engine 1.0.0 · Units 1.0.0Read the method, stream basis, and sources below.

Formula and method

Molar mass is mass per amount of substance, M = m/n. For the same stream and chemical-entity basis, molar flow ṅ = ṁ/M and mass flow ṁ = ṅM.

The engine normalizes mass flow to kg/s, molar flow to mol/s, and molar mass to kg/mol. Dividing kg/s by kg/mol gives mol/s; multiplying mol/s by kg/mol gives kg/s. The chosen output unit then sets the amount or mass scale and the time basis.

Molar-mass and flow units

QuantityUnit basis
Mass flowkg/s, kg/min, kg/h, t/h, lbm/s, lbm/min, lbm/h
Molar flowmol, kmol, or lbmol per s, min, or h
Molar massg/mol, kg/kmol, kg/mol, lbm/lbmol

1 kmol = 1000 mol and 1 lbmol = 453.59237 mol. Here lbm is the international avoirdupois pound, 0.45359237 kg, and t is a metric tonne, 1000 kg. One minute is 60 s and one hour is 3600 s.

Molar-mass values in g/mol, kg/kmol, and lbm/lbmol have the same numerical value: 18 g/mol = 18 kg/kmol = 18 lbm/lbmol = 0.018 kg/mol. A mol-to-kmol change contributes a factor of 1000; do not drop it when mixing kg/mol with kg/kmol. Enter a dimensional molar mass, not an unlabelled relative molecular mass.

Component and mixture basis

  • One species or component: use the flow of that species with its molar mass. Specify the same molecules, atoms, ions, or formula units in the amount basis and molar mass. A total mixture flow cannot be converted using just one component’s molar mass.
  • Total mixture flow: supply the mean molar mass of that same mixture and composition basis. From mole fractions, M̄ = ΣxᵢMᵢ; from mass fractions, M̄ = 1 / Σ(wᵢ/Mᵢ), with fractions summing to 1 and consistent molar-mass units. The converter does not calculate composition or mean molar mass.
  • Use matching wet/dry, phase, and stream definitions. Reactions or changing composition can change the appropriate mean molar mass. The conversion is an algebraic relation for the supplied basis, not a material-balance or reaction calculation.

Assumptions and limits

  • Molar mass must be positive. Flow is a nonnegative magnitude; zero produces zero mass and molar flow. Assign signs separately when placing a stream in a material balance.
  • No compound, atomic-weight, isotope, or composition database is used. The molar mass is your explicit input. The example value is illustrative and does not declare an exact property of water or another substance.
  • No density, gas temperature, pressure, or compressibility is inferred. Converting volumetric or standard gas flow requires a separate property/state calculation.
  • Arithmetic uses double precision. Results display up to 8 significant digits, steps up to 10. Normalized molar mass, mass flow, molar flow, and output must be finite and representable; overflow and nonzero underflow are rejected.

Worked examples

Mass to molar flow: take 1800 kg/h and a supplied M = 18 g/mol = 0.018 kg/mol. Normalize ṁ = 1800/3600 = 0.5 kg/s. Then ṅ = 0.5/0.018 = 27.77777778 mol/s, or 100 kmol/h. Reversing 100 kmol/h with the same M returns 1800 kg/h.

Imperial basis: 100 lbm/h divided by 20 lbm/lbmol gives 5 lbmol/h, or 2267.96185 mol/h. The molar flow is not 5 mol/h. Changing units preserves the physical flow and molar mass; changing direction carries the last valid result into the new input.

Mixture basis: a hypothetical 50 mol% mixture of species with M₁ = 20 g/mol and M₂ = 40 g/mol has M̄ = 30 g/mol. A total flow of 300 kg/h is therefore 10 kmol/h. A 50 mass% mixture of the same species instead has M̄ = 26.66666667 g/mol and 11.25 kmol/h; the fraction basis matters.

Sources and calculation version

Changelog: engine 1.0.0 introduces bidirectional mass/molar flow, explicit dimensional molar mass, SI normalization, component/mixture basis notes, unit conversion, zero flow, examples, and range checks. Shared mass/time factors 1.0.0 and amount/molar-mass definitions 1.0.0 are versioned locally. No chemical-property dataset is used. Engineering review status: independent review pending.