Molarity Calculator
Find how many grams to weigh out, the molarity of a solution, the volume to make, or the molar mass. Mix units freely, and see moles, g/L, % w/v and ppm. Advanced adds molar mass from a formula and a dilution helper.
M = (mass ÷ molar mass) ÷ volume
The answer appears in the highlighted box; change its unit to see it another way. Numbers like 2.5e-3 work too.
Dilution helper: C₁ × V₁ = C₂ × V₂
Make a weaker solution from a stronger stock. Pick what to find and fill in the other three. Units can differ.
How your molarity is worked out
Your numbers: turn grams into moles, litres, then share the moles out per litre.
- 1
Grams to moles
5.844 g ÷58.44 g/mol= 0.1 mol
- 2
Volume in litres
100 mL= 0.1 L
- 3
Moles ÷ litres
0.1 mol ÷0.1 L= 1 M
- 4
Other ways to say it
58.44 g/L= 5.844% w/v= 58,440 ppm
Read more: the formula, worked examples, compounds and atomic weights
A mole is a fixed number of particles (6.022 × 10²³), and the molar mass tells you how many grams one mole weighs. So dividing grams by the molar mass counts the moles, and dividing moles by the litres of solution gives the molarity. Every other answer here is the same sum turned around.
The formula
moles n = mass m ÷ molar mass MM
molarity M = moles n ÷ volume V (in litres)
so molarity M = mass ÷ (molar mass × volume)
mass m = molarity × volume × molar mass
volume V = mass ÷ (molar mass × molarity)
molar mass MM = mass ÷ (molarity × volume)
mass concentration (g/L) = molarity × molar mass
% w/v = g/L ÷ 10 (grams per 100 mL)
ppm (mg/L) = g/L × 1,000
dilution: C₁ × V₁ = C₂ × V₂, water to add = V₂ − V₁
Worked example
Molarity (the default): 5.844 g of sodium chloride (58.44 g/mol) made up to 100 mL. Moles: 5.844 ÷ 58.44 = 0.1 mol. Volume: 100 mL = 0.1 L. Molarity: 0.1 ÷ 0.1 = 1 M. The same strength is 1 × 58.44 = 58.44 g/L, 58.44 ÷ 10 = 5.844% w/v and 58.44 × 1,000 = 58,440 ppm (mg/L).
Mass to weigh out: 500 mL of 0.25 M NaCl. 0.25 × 0.5 = 0.125 mol, and 0.125 × 58.44 = 7.305 g.
Volume to make: 10 g of sodium hydroxide (40.00 g/mol) at 0.5 M. 10 ÷ 40 = 0.25 mol, and 0.25 ÷ 0.5 = 0.5 L (500 mL).
Molar mass: 9.807 g in 1 L gives a 0.1 M solution. 0.1 × 1 = 0.1 mol, and 9.807 ÷ 0.1 = 98.07 g/mol (sulfuric acid, H₂SO₄).
Molar mass from a formula: Ca(OH)₂ has 1 Ca, 2 O and 2 H. 40.078 + 2 × 15.999 + 2 × 1.008 = 40.078 + 31.998 + 2.016 = 74.092 ≈ 74.09 g/mol. For the hydrate CuSO₄·5H₂O: 63.546 + 32.06 + 4 × 15.999 + 5 × 18.015 = 63.546 + 32.06 + 63.996 + 90.075 = 249.677 g/mol.
Dilution: 500 mL of 0.1 M from a 1 M stock. V₁ = 0.1 × 500 ÷ 1 = 50 mL of stock, then add 500 − 50 = 450 mL of water. Dilution factor 1 ÷ 0.1 = 10.
% w/v to molarity: 0.9% saline is 0.9 g per 100 mL = 9 g/L, and 9 ÷ 58.44 = 0.154 M (154 mM).
How to make the solution
- Weigh the mass the calculator gives.
- Dissolve it in less water than the final volume (about three quarters).
- Pour into a volumetric flask (or measuring cylinder), rinse the beaker into it, then top up to the line.
- Mix well. Molarity is per litre of finished solution, so don't add the solid to the full volume of water.
For a hydrate (like CuSO₄·5H₂O) use the molar mass including the water, because the water is part of what you weigh. For acids and other chemicals bought as a liquid, use the dilution helper with the stock's strength instead. Always follow the safety sheet: add acid to water, not water to acid.
Common compounds (from rounded standard atomic weights)
| Compound | Formula | Molar mass (g/mol) | Grams for 1 L of 1 M |
|---|---|---|---|
| Sodium chloride (table salt) | NaCl | 58.44 | 58.44 g |
| Glucose | C6H12O6 | 180.16 | 180.16 g |
| Sucrose (table sugar) | C12H22O11 | 342.30 | 342.30 g |
| Sodium hydroxide | NaOH | 40.00 | 40.00 g |
| Hydrochloric acid (HCl) | HCl | 36.46 | 36.46 g |
| Sulfuric acid | H2SO4 | 98.07 | 98.07 g |
| Nitric acid | HNO3 | 63.01 | 63.01 g |
| Acetic acid | CH3COOH | 60.05 | 60.05 g |
| Potassium chloride | KCl | 74.55 | 74.55 g |
| Calcium chloride | CaCl2 | 110.98 | 110.98 g |
| Calcium hydroxide | Ca(OH)2 | 74.09 | 74.09 g |
| Sodium bicarbonate | NaHCO3 | 84.01 | 84.01 g |
| Sodium carbonate | Na2CO3 | 105.99 | 105.99 g |
| Potassium permanganate | KMnO4 | 158.03 | 158.03 g |
| Copper sulfate pentahydrate | CuSO4·5H2O | 249.68 | 249.68 g |
| Magnesium sulfate heptahydrate (Epsom salt) | MgSO4·7H2O | 246.47 | 246.47 g |
| Ammonium chloride | NH4Cl | 53.49 | 53.49 g |
| Potassium dihydrogen phosphate | KH2PO4 | 136.08 | 136.08 g |
| Silver nitrate | AgNO3 | 169.87 | 169.87 g |
| Tris (buffer base) | C4H11NO3 | 121.14 | 121.14 g |
| EDTA disodium dihydrate | C10H14N2Na2O8·2H2O | 372.24 | 372.24 g |
| Ethanol | C2H5OH | 46.07 | 46.07 g |
Standard atomic weights used (rounded)
The formula tool uses rounded textbook values based on the IUPAC conventional and abridged standard atomic weights, for elements from hydrogen to bismuth (except technetium and promethium). They have not been checked against the official IUPAC table here. For elements whose weight varies in nature (like H, Li, B, C, N, O, S, Cl), real samples can differ slightly in the last digit. Some common ones:
| Element | Atomic weight | Element | Atomic weight |
|---|---|---|---|
| H | 1.008 | K | 39.098 |
| C | 12.011 | Ca | 40.078 |
| N | 14.007 | Fe | 55.845 |
| O | 15.999 | Cu | 63.546 |
| Na | 22.99 | Zn | 65.38 |
| Mg | 24.305 | Br | 79.904 |
| P | 30.974 | Ag | 107.87 |
| S | 32.06 | I | 126.9 |
| Cl | 35.45 | Ba | 137.33 |
Notes
- 1 M = 1,000 mM = 1,000,000 µM = 10⁹ nM. 1 mg/mL is the same as 1 g/L.
- % w/v means grams per 100 mL of solution, so 1% w/v = 10 g/L.
- ppm here is mg per litre. That equals parts per million by mass only for dilute water solutions (density about 1 g/mL). For other solvents or strong solutions, ppm by mass differs.
- g/L, mg/mL, % w/v and ppm need the molar mass to turn into molarity. To find the molar mass itself, give the concentration in M, mM, µM or nM.
- Molarity changes slightly with temperature because the volume does. For very exact work, make the solution at the temperature you will use it.
Common questions
What is molarity?
Molarity (M) is how many moles of a substance are dissolved in each litre of the finished solution. 1 M means 1 mole per litre. For sodium chloride (58.44 g/mol), 1 M is 58.44 g in every litre of solution, so 5.844 g made up to 100 mL is also 1 M.
How do I work out molarity from grams?
Divide the mass by the molar mass to get moles, then divide by the volume in litres. 5.844 g of NaCl ÷ 58.44 g/mol = 0.1 mol, and 0.1 mol ÷ 0.1 L = 1 M.
How many grams do I need to make a solution?
Multiply molarity × volume in litres × molar mass. For 500 mL of 0.25 M NaCl: 0.25 × 0.5 × 58.44 = 7.305 g. Dissolve it in less water than you need, then top up to exactly 500 mL.
What is the difference between molarity and molality?
Molarity is moles per litre of solution. Molality (m) is moles per kilogram of solvent. In dilute water solutions they are close, but molarity changes slightly with temperature (the volume changes) while molality does not. This calculator works out molarity.
How do I dilute a stock solution?
Use C1 × V1 = C2 × V2. To make 500 mL of 0.1 M from a 1 M stock: V1 = 0.1 × 500 ÷ 1 = 50 mL of stock, then add 450 mL of water to reach 500 mL. Turn on Advanced to use the dilution tool.
How do I convert molarity to ppm or % w/v?
Multiply molarity by the molar mass to get g/L. Then % w/v = g/L ÷ 10, and ppm (as mg/L) = g/L × 1,000. 1 M NaCl is 58.44 g/L = 5.844% w/v = 58,440 mg/L. The ppm figure assumes a dilute water solution with a density of about 1 g/mL.
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