Making a solution to a target concentration

Every “make 250 mL of 0.100 mol/L” problem is the same three-way multiplication, and most of the mistakes come from a unit slipping in one of the three factors rather than from the chemistry.

The three factors

Concentration (mol/L) × volume (L) × molar mass (g/mol) = mass (g). Litres cancel, moles cancel, grams remain. Write the units next to the numbers and the equation checks itself.

Worked example: glucose

c   0.100 mol/L
V   0.250 L          (250 mL)
M   180.156 g/mol    (C6H12O6, anhydrous)

m = 0.100 × 0.250 × 180.156 = 4.504 g

Weigh 4.504 g of anhydrous glucose, dissolve it in less than 250 mL of water, then make up to the 250 mL mark. “Make up to volume” matters: adding 250 mL of water to the solid gives slightly more than 250 mL of solution.

What you are actually weighing

• Hydrates. Glucose is also sold as the monohydrate, C6H12O6·H2O, molar mass 198.171 g/mol: each formula unit brings a water with it. Using 180.156 for the monohydrate under-doses by 9%. The molar mass on the container is the one to use.
• Purity. A solid stated as 98% pure needs more, and the correction is a division, not a 2% addition: 4.504 g ÷ 0.98 = 4.596 g for the example.
• Salts. A hydrochloride weighs more per mole of the parent compound than the free base does. Again, use the molar mass of the form in the bottle.

Other ways concentration is written

• Mass concentration, g/L or mg/mL (1 mg/mL = 1 g/L). Divide by the molar mass to get molarity: 10 mg/mL of glucose is 10 g/L ÷ 180.156 g/mol = 0.0555 mol/L = 55.5 mmol/L.
• Percent w/v. 1% w/v means 1 g per 100 mL, which is 10 g/L or 10 mg/mL.
• Dilutions. c₁V₁ = c₂V₂. For 50 mL of 0.020 mol/L from a 1.00 mol/L stock: V₁ = 0.020 × 50 ÷ 1.00 = 1.0 mL of stock, made up to 50 mL.

The solution calculator does this multiplication for any formula or molar mass, with the purity correction.