Glycerol Solution Preparation

How many grams of glycerol to weigh for any molarity and volume, with a full recipe table.

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To prepare a solution of glycerol you need the grams, and grams is molarity × volume in litres × 92.094 g/mol. A litre of 1 M takes 92.094 g; 500 mL of 0.1 M takes 4.6047 g; 250 mL of 0.01 M takes 230.23 mg. Enter the strength you want and the volume you want it in, and the calculator returns the mass to weigh together with the same figure in milligrams and as a percentage strength.

The figure comes from the formula. Glycerol is 3 × 12.011 (C) + 8 × 1.008 (H) + 3 × 15.999 (O), which is 92.094 g/mol, taking standard atomic weights, and every page here computes that sum rather than quoting it. Oxygen makes up the largest share of the mass: 3 of the 14 atoms in a formula unit are oxygen, and they account for 47.997 g of the 92.094 g, or 52.12% by mass. The derivation table below breaks the whole molecule down element by element.

Order of operations matters more than precision here. Dissolve the solid in rather less than the final volume, then make up to the mark once it has all gone into solution and come back to room temperature — warm solutions read low when they cool. If the calculated mass is under about 20 mg, prepare ten times the strength and dilute it tenfold by pipette instead, because C₁V₁ = C₂V₂ carries less error than a marginal weighing.

Glycerine: a humectant in food and cosmetics, a cryoprotectant, and the backbone of every triglyceride. Sold as a 99.5% syrup. At 92.094 g/mol it is heavier than 8 of the 21 organic compounds covered here, and that ranking matters more than it looks: benzene has a molar mass of 78.114 g/mol, so a gram of it contains 17.9% more formula units than a gram of glycerol. Weigh by mass and you are not weighing equal amounts of substance.

Purity is the usual gap between the calculation and the balance. A reagent sold at 98% means 1.8419 g of a nominal 92.094 g is something else, so for exact work you divide the weighed mass by the assay figure on the certificate. For most purposes the difference is smaller than the error in reading the meniscus, but it is systematic rather than random, so it does not average out.

The formula

m = c × (V ÷ 1000) × 92.094
c
The molarity you want, in mol/L
V
The volume you want to make, in millilitres
92.094
The molar mass of glycerol in g/mol, derived from its formula
m
The mass of glycerol to weigh out, in grams

How it works, step by step

  1. Enter the molarity you want, in mol/L.
  2. Enter the volume you want to make, in millilitres.
  3. The two are multiplied together and by 92.094 g/mol.
  4. The result is the mass to weigh; dissolve it in part of the volume and then make up to the mark.

Worked examples

250 mL of 0.1 M glycerol

Weigh 2.30235 g. That is 0.1 M × 0.25 L × 92.094 g/mol, and it comes to 2302.35 mg if your balance is set to milligrams. Dissolve in about 175 mL first, then make up to 250 mL.

500 mL of 0.5 M glycerol

Weigh 23.0235 g. That is 0.5 M × 0.5 L × 92.094 g/mol, and it comes to 23,023.5 mg if your balance is set to milligrams. Dissolve in about 350 mL first, then make up to 500 mL.

1000 mL of 1 M glycerol

Weigh 92.094 g. That is 1 M × 1 L × 92.094 g/mol, and it comes to 92,094 mg if your balance is set to milligrams. Dissolve in about 700 mL first, then make up to 1000 mL.

How to read your score

0–0NothingNothing to weigh.
0–0.05Below a sensible weighingUnder 50 mg. Weigh ten or a hundred times this and dilute; the balance error does not shrink with the sample.
0.05–10Weighable directlyFrom 50 mg to 10 g, which is where a two- or three-decimal balance gives better than 1% on the mass.
10–—Large batchMore than 10 g of glycerol. Worth checking solubility and, for a solid that generates heat on dissolving, adding it in portions.

Frequently asked questions

What is the molar mass of glycerol (C3H8O3)?

It is 92.094 g/mol. That is the sum of the standard atomic weights of every atom in the formula: 3 × 12.011 (C) + 8 × 1.008 (H) + 3 × 15.999 (O). One mole of glycerol therefore weighs 92.094 g, and a gram of it is 10.858 mmol.

What percentage of glycerol is oxygen?

52.12% by mass. Each formula unit contains 3 oxygen atoms contributing 47.997 g of the 92.094 g total, so 100 g of glycerol contains 52.12 g of oxygen and a kilogram contains 521.17 g of it.

How much glycerol do I need for 1 litre of 1 M solution?

92.094 g — the molar mass in grams, which is what a 1 molar solution means. For 1 L of 0.1 M it is 9.2094 g, and for 1 L of 0.01 M it is 0.92094 g.

How much for 100 mL of 0.1 M?

0.92094 g. The volume is a tenth of a litre and the strength a tenth of molar, so the mass is a hundredth of 92.094 g. In milligrams that is 920.94 mg.

Do I dissolve the solid first or make up the volume first?

Dissolve first, in perhaps 70% of the final volume, then make up to the mark. Adding solid to a full flask overshoots the volume and leaves the solution weak, and undissolved solid at the mark means the strength keeps changing as it goes in.

Can I dilute a stock solution instead of weighing?

Yes, and it is more accurate for small amounts. C₁V₁ = C₂V₂: to get 500 mL of 0.01 M from a 0.5 M stock, take 10 mL of stock and make up to 500 mL. Weighing the 460.47 mg that 500 mL of 0.01 M needs directly is the harder of the two.

Solution recipes for glycerol

How the molar mass of glycerol is arrived at
ElementAtomsAtomic weightContribution (g/mol)By mass
Oxygen (O)315.99947.99752.12%
Carbon (C)312.01136.03339.13%
Hydrogen (H)81.0088.0648.756%
Total — one mole of glycerol92.094100%

Standard atomic weights, IUPAC 2021. The contribution column is atoms × atomic weight, and the total is the molar mass this page uses: 92.094 g/mol.

Grams of glycerol needed for a standard solution
Targetfor 100 mLfor 250 mLfor 500 mLfor 1 L
0.001 M9.2094 mg23.0235 mg46.047 mg92.094 mg
0.005 M46.047 mg115.117 mg230.235 mg460.47 mg
0.01 M92.094 mg230.235 mg460.47 mg920.94 mg
0.05 M460.47 mg1.15118 g2.30235 g4.6047 g
0.1 M920.94 mg2.30235 g4.6047 g9.2094 g
0.15 M1.38141 g3.45352 g6.90705 g13.8141 g
0.2 M1.84188 g4.6047 g9.2094 g18.4188 g
0.25 M2.30235 g5.75587 g11.5117 g23.0235 g
0.5 M4.6047 g11.5117 g23.0235 g46.047 g
1 M9.2094 g23.0235 g46.047 g92.094 g
2 M18.4188 g46.047 g92.094 g184.188 g

Each cell is molarity × volume in litres × 92.094 g/mol. Dissolve the solid first and then make up to the marked volume — adding solid to a full volume overshoots it.

Glycerol among other organic compounds, by molar mass
CompoundFormulaMolar mass (g/mol)Millimoles in 1 g
Ethylene glycolC2H6O262.06816.111
GlycineC2H5NO275.06713.321
BenzeneC6H678.11412.802
Glycerol (this page)C3H8O392.09410.858
TolueneC7H892.14110.853
CreatinineC4H7N3O113.128.8402
OctaneC8H18114.2328.7541

Ordered by molar mass. The last column is 1000/M, which is the number a weighed gram actually gives you.

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