Sodium citrate Solution Preparation

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

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To prepare a solution of sodium citrate you need the grams, and grams is molarity × volume in litres × 258.06831 g/mol. A litre of 1 M takes 258.0683 g; 500 mL of 0.1 M takes 12.903 g; 250 mL of 0.01 M takes 645.17 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. Sodium citrate is 3 × 22.9898 (Na) + 6 × 12.011 (C) + 5 × 1.008 (H) + 7 × 15.999 (O), which is 258.06831 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: 7 of the 21 atoms in a formula unit are oxygen, and they account for 111.993 g of the 258.0683 g, or 43.4% 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.

The anticoagulant in blood bags, the emulsifying salt in smooth cheese sauce, and the buffer in effervescent tablets. The dihydrate is 294.10 g per mole. At 258.0683 g/mol it is heavier than 29 of the 34 salts covered here, and that ranking matters more than it looks: copper(II) sulfate pentahydrate has a molar mass of 249.677 g/mol, so a gram of it contains 3.36% more formula units than a gram of sodium citrate. 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 5.1614 g of a nominal 258.0683 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) × 258.06831
c
The molarity you want, in mol/L
V
The volume you want to make, in millilitres
258.06831
The molar mass of sodium citrate in g/mol, derived from its formula
m
The mass of sodium citrate 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 258.06831 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 sodium citrate

Weigh 6.45171 g. That is 0.1 M × 0.25 L × 258.0683 g/mol, and it comes to 6451.71 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 sodium citrate

Weigh 64.5171 g. That is 0.5 M × 0.5 L × 258.0683 g/mol, and it comes to 64,517.1 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 sodium citrate

Weigh 258.068 g. That is 1 M × 1 L × 258.0683 g/mol, and it comes to 258,068 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 sodium citrate. 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 sodium citrate (Na3C6H5O7)?

It is 258.06831 g/mol. That is the sum of the standard atomic weights of every atom in the formula: 3 × 22.9898 (Na) + 6 × 12.011 (C) + 5 × 1.008 (H) + 7 × 15.999 (O). One mole of sodium citrate therefore weighs 258.0683 g, and a gram of it is 3.8749 mmol.

What percentage of sodium citrate is oxygen?

43.4% by mass. Each formula unit contains 7 oxygen atoms contributing 111.993 g of the 258.0683 g total, so 100 g of sodium citrate contains 43.4 g of oxygen and a kilogram contains 433.97 g of it.

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

258.06831 g — the molar mass in grams, which is what a 1 molar solution means. For 1 L of 0.1 M it is 25.8068 g, and for 1 L of 0.01 M it is 2.58068 g.

How much for 100 mL of 0.1 M?

2.58068 g. The volume is a tenth of a litre and the strength a tenth of molar, so the mass is a hundredth of 258.0683 g. In milligrams that is 2580.68 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 1290.3 mg that 500 mL of 0.01 M needs directly is the harder of the two.

Solution recipes for sodium citrate

How the molar mass of sodium citrate is arrived at
ElementAtomsAtomic weightContribution (g/mol)By mass
Oxygen (O)715.999111.99343.4%
Carbon (C)612.01172.06627.93%
Sodium (Na)322.9897768.9693126.73%
Hydrogen (H)51.0085.041.953%
Total — one mole of sodium citrate258.06831100%

Standard atomic weights, IUPAC 2021. The contribution column is atoms × atomic weight, and the total is the molar mass this page uses: 258.06831 g/mol.

Grams of sodium citrate needed for a standard solution
Targetfor 100 mLfor 250 mLfor 500 mLfor 1 L
0.001 M25.8068 mg64.5171 mg129.034 mg258.068 mg
0.005 M129.034 mg322.585 mg645.171 mg1.29034 g
0.01 M258.068 mg645.171 mg1.29034 g2.58068 g
0.05 M1.29034 g3.22585 g6.45171 g12.9034 g
0.1 M2.58068 g6.45171 g12.9034 g25.8068 g
0.15 M3.87102 g9.67756 g19.3551 g38.7102 g
0.2 M5.16137 g12.9034 g25.8068 g51.6137 g
0.25 M6.45171 g16.1293 g32.2585 g64.5171 g
0.5 M12.9034 g32.2585 g64.5171 g129.034 g
1 M25.8068 g64.5171 g129.034 g258.068 g
2 M51.6137 g129.034 g258.068 g516.137 g

Each cell is molarity × volume in litres × 258.06831 g/mol. Dissolve the solid first and then make up to the marked volume — adding solid to a full volume overshoots it.

Sodium citrate among other salts, by molar mass
CompoundFormulaMolar mass (g/mol)Millimoles in 1 g
Epsom saltMgSO4·7H2O246.4664.0574
Sodium thiosulfate pentahydrateNa2S2O3·5H2O248.17154.0295
Copper(II) sulfate pentahydrateCuSO4·5H2O249.6774.0052
Sodium citrate (this page)Na3C6H5O7258.06833.8749
Iron(II) sulfate heptahydrateFeSO4·7H2O278.0063.597
Zinc sulfate heptahydrateZnSO4·7H2O287.5413.4778
Potassium dichromateK2Cr2O7294.18183.3993

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

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