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Concentration unit converter

Q: How do I convert mg/mL to molarity?

Divide by the molar mass: mg/mL equals g/L, so M = (mg/mL) / (g/mol). For example, 5 mg/mL of glucose (180.16 g/mol) is 5 / 180.16 = 0.02775 M = 27.75 mM.

Q: How do I convert molarity to % w/v?

Multiply by the molar mass to get g/L, then divide by 10: % w/v = M × molar mass / 10. 150 mM NaCl is 0.150 × 58.44 / 10 = 0.8766 % w/v.

Q: How do I convert % w/w to molarity?

You need the density: M = 10 × % w/w × density (g/mL) / molar mass. 37% HCl with density 1.19 g/mL is 10 × 37 × 1.19 / 36.46 = 12.08 M.

Enter one concentration in any unit, with the molar mass, and the converter shows it in every other unit at once: molarity (M, mM, µM, nM), mass concentration (g/L, mg/mL, µg/mL, ng/mL), % w/v, % w/w, ppm and ppb, molality, mole fraction and normality, each with its formula. For example, 150 mM NaCl (58.44 g/mol) is 8.766 g/L = 8.766 mg/mL = 0.8766 % w/v = 8766 ppm. Units based on the mass of solution or solvent (% w/w, molality, mole fraction) use the solution density, and ppm and ppb are taken as mg/L and µg/L for dilute aqueous solutions.

What it does: Converts a concentration from any unit to molarity, mass concentration, % w/v, % w/w, ppm/ppb, molality, mole fraction and normality at once.
Formula: $\begin{aligned} \rho\ (\mathrm{g/L}) &= c\ (\mathrm{mol/L}) \times M \\ \%\ \mathrm{w/v} &= \rho/10, \quad \%\ \mathrm{w/w} = \rho/(10d) \\ \mathrm{ppm} &= \rho \times 1000\ (\mathrm{mg/L}) \\ \text{molality } b &= \frac{1000c}{1000d - cM} \\ x &= \frac{c}{c + (1000d - cM)/M_\mathrm{s}} \\ N &= c \times \mathrm{eq} \end{aligned}$
Example: 37% w/w HCl (M = 36.46 g/mol, density 1.19 g/mL) is 440.3 g/L = 12.08 M, 16.11 mol/kg, mole fraction 0.2249.

Last updated 19 October 2026

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At a glance

Concentration converter at a glance
Input	One concentration in any unit, the molar mass; density and equivalents where needed
Units	M, mM, µM, nM; g/L, mg/mL, µg/mL, ng/mL; % w/v; % w/w; ppm; ppb; mol/kg; mole fraction; N
Assumption	ppm = mg/L and ppb = µg/L (dilute aqueous); density 1.000 g/mL unless entered
Molar masses	CIAAW 2021 standard atomic weights
Output	Every unit at once with its formula; copy or CSV
Runs in	Your web browser; nothing is uploaded
Price	Free, no account needed

What it does

How to use the concentration converter

Enter the concentrationType the value and pick its unit, for example 37 % w/w or 5 mg/mL.
Give the soluteChoose a compound or type its molar mass; for normality set the equivalents per mole.
Check the densityLeave 1.00 g/mL for dilute aqueous solutions, or enter the density for concentrated solutions; then read every unit from the table.

How it works

Every unit is converted through the molar concentration $c$ (mol/L) and the mass concentration $\rho = c \times M$ (g/L), where $M$ is the molar mass of the solute (g/mol). $d$ is the density of the solution (g/mL, numerically kg/L), $M_\mathrm{s}$ the molar mass of the solvent and $\mathrm{eq}$ the equivalents per mole. One litre of solution weighs $1000d$ g, of which $\rho$ g is solute and $1000d - \rho$ g solvent.

Formulas (per litre of solution)
Unit	From $c$ and $\rho$	Back to $c$
mg/mL, g/L	$\rho = c \cdot M$ (1 mg/mL = 1 g/L)	$c = \rho / M$
% w/v (g/100 mL)	$\rho / 10$	$c = 10 \cdot w / M$
% w/w (g/100 g)	$\rho / (10 \cdot d)$	$c = 10 \cdot w \cdot d / M$
ppm, ppb (aqueous)	$\rho \times 10^3$ mg/L, $\rho \times 10^6$ µg/L	$c = \mathrm{ppm} / (1000 \cdot M)$
Molality $b$ (mol/kg solvent)	$1000c / (1000d - cM)$	$c = 1000 \cdot d \cdot b / (1000 + b \cdot M)$
Mole fraction $x$	$c / (c + (1000d - cM)/M_\mathrm{s})$	$c = 1000 \cdot d \cdot x / (x \cdot M + (1 - x) \cdot M_\mathrm{s})$
Normality $N$ (eq/L)	$c \times \mathrm{eq}$	$c = N / \mathrm{eq}$

Assumptions. ppm and ppb are mass ratios; for water and dilute aqueous solutions 1 L weighs about 1 kg, so 1 ppm ≈ 1 mg/L and 1 ppb ≈ 1 µg/L, which is how they are used here. For concentrated solutions or other solvents use % w/w with the real density instead. % w/w, molality and mole fraction depend on the density; the default 1.000 g/mL is only right for dilute aqueous solutions. Molarity, mass concentration, % w/v and normality are per volume and change slightly with temperature; molality and mole fraction do not. IUPAC recommends amount concentration (mol/L) over normality, whose value depends on the reaction ( $\ce{H2SO4}$ is 2 eq/mol as an acid, $\ce{KMnO4}$ 5 eq/mol as an oxidant in acid).

Sources: Cohen ER et al. Quantities, Units and Symbols in Physical Chemistry (IUPAC Green Book), 3rd ed. RSC Publishing; 2007, doi:10.1039/9781847557889. Molar masses from Prohaska T et al. Standard atomic weights of the elements 2021. Pure Appl Chem 2022;94:573–600, doi:10.1515/pac-2019-0603.

Worked examples: NaCl, concentrated HCl and H₂SO₄

150 mM NaCl ( $M = 58.44\ \mathrm{g/mol}$ , density taken as 1.000 g/mL):

150 mM NaCl in every unit
Unit	Value	Working
g/L = mg/mL	8.766	$0.150 \times 58.44$
µg/mL	8766	$8.766 \times 1000$
% w/v	0.8766	$8.766 / 10$
% w/w	0.8766	$8.766 / (10 \times 1.000)$
ppm (mg/L)	8766	$8.766\ \mathrm{g/L} \times 1000$
Molality	0.1513 mol/kg	$1000 \times 0.150 / (1000 - 8.766)$
Mole fraction	0.002719	$0.150 / (0.150 + 991.234 / 18.015)$
Normality	0.150 N	$0.150 \times 1$

Concentrated hydrochloric acid, 37% w/w ( $M = 36.46\ \mathrm{g/mol}$ ), with the density from the bottle, 1.19 g/mL: $\rho = 10 \times 37 \times 1.19 = 440.3\ \mathrm{g/L}$ , so $c = 440.3 / 36.46 =$ 12.08 M. Per litre there are $1190 - 440.3 = 749.7\ \mathrm{g}$ of water, so the molality is $1000 \times 12.08 / 749.7 = 16.11\ \mathrm{mol/kg}$ and the mole fraction of HCl 0.2249. To make 1 L of 1 M HCl, take $1 / 12.08\ \mathrm{L} = 82.8\ \mathrm{mL}$ of the acid ( $C_1 V_1 = C_2 V_2$ ), adding acid to water.

Concentrated sulfuric acid, 96% w/w ( $M = 98.07\ \mathrm{g/mol}$ , density 1.84 g/mL): 1766.4 g/L, 18.01 M and, at 2 equivalents per mole, 36.02 N.

Quick conversions
From	To	Multiply by
mg/mL	µg/mL	1000
mg/mL	% w/v	0.1
% w/v	g/L	10
mg/L (dilute aqueous)	ppm	1
µM	µg/mL	$M / 1000$
mg/mL	mM	$1000 / M$

Frequently asked questions

How do I convert mg/mL to molarity?

Divide by the molar mass: mg/mL equals g/L, so $M = (\mathrm{mg/mL}) / (\mathrm{g/mol})$ . For example, 5 mg/mL of glucose (180.16 g/mol) is $5 / 180.16 = 0.02775\ \mathrm{M} = 27.75\ \mathrm{mM}$ .

How do I convert molarity to % w/v?

Multiply by the molar mass to get g/L, then divide by 10: $\%\ \mathrm{w/v} = M \times \text{molar mass} / 10$ . 150 mM NaCl is $0.150 \times 58.44 / 10 = 0.8766\ \%\ \mathrm{w/v}$ .

How do I convert % w/w to molarity?

You need the density: $M = 10 \times \%\ \mathrm{w/w}$ $\times\, \text{density (g/mL)} / \text{molar mass}$ . 37% HCl with density 1.19 g/mL is $10 \times 37 \times 1.19 / 36.46 = 12.08\ \mathrm{M}$ .

Is ppm the same as mg/L?

For water and dilute aqueous solutions, yes: 1 L weighs about 1 kg, so $1\ \mathrm{mg/L} \approx 1\ \mathrm{mg/kg} = 1\ \mathrm{ppm}$ , and $1\ \upmu\mathrm{g/L} \approx 1\ \mathrm{ppb}$ . Strictly ppm is a mass ratio, so for dense or non-aqueous solutions use % w/w with the real density.

What is the difference between molarity and molality?

Molarity is moles of solute per litre of solution; molality is moles per kilogram of solvent. They are close for dilute aqueous solutions (150 mM NaCl is 0.1513 mol/kg) but differ a lot for concentrated ones (37% HCl is 12.08 M but 16.11 mol/kg).

How do I calculate normality?

Normality = molarity × equivalents per mole, where an equivalent is the amount that gives or takes one mole of $\ce{H+}$ (acids and bases) or of electrons (redox). 1 M $\ce{H2SO4}$ is 2 N; 0.02 M $\ce{KMnO4}$ used as an oxidant in acid is 0.1 N.

What is % w/v versus % w/w?

% w/v is grams of solute per 100 mL of solution; % w/w is grams per 100 g of solution. They are equal only when the density is 1 g/mL. Commercial concentrated acids are labelled in % w/w.

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